{
  "schema": "cie-topical-past-papers/answers-v1",
  "subject": "9231",
  "subject_name": "Further Mathematics",
  "generated": "2026-10-21",
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    "unavailable_mark_scheme_entries": 0
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  "records": [
    {
      "id": "9231-2021-mj-11-q01",
      "question_id": "9231-2021-mj-11-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 24 +31−2=45 is divisible by 15 B1 Checks base case.\nAssume that 24k +31k −2 is divisible by 15 for some positive integer k. B1 States inductive hypothesis.\nThen 24k+4 +31k+1−2=(15+1)24k +(30+1)31k −2 M1 A1 Separates 24k +31k −2 or considers difference.\nis divisible by 15 because 15×24k +30×31k is divisible by 15. A1\nHence, by induction, true for every positive integer n. A1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-11-q02",
      "question_id": "9231-2021-mj-11-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) n−1n(n+1)−1n(n+1)(2n+1) M1A1 Substitutes correct formulae from MF19.\n2 6\nExpanding brackets correctly.\n1n−n2 −1n3 A1 Simplifies by collecting terms.\n3 3\n3\n© UCLES 2021 Page 6 of 14\n\n2(b) r+1 r (r+1) r2 +1 −r r2 +2r+2 1−r−r2 M1 A1 Puts over a common denominator and expands, AG.\n− =\n( )( )\n=\n( )( )\n(r+1)2 +1 r2 +1 r2 +2r+2 r2 +1 r2 +2r+2 r2 +1\nn 1−r−r2 n  r+1 r  M1 A1 Shows at least three complete terms including first and\n = −\n( )( )   last. Cancellation may be implicit.\nr=1\nr2 +2r+2 r2 +1\nr=1\n(r+1)2 +1 r2 +1\n2 1 3 2 4 3 n+1 n\n= − + − + − ++ −\n5 2 10 5 17 10 (n+1)2 +1 n2 +1\n1 n+1 A1 ISW\n=− +\n2 ( n+1 )2 +1\n5\n\n2(c) −1 B1 FT FT from their answer to part (b).\n2\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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        "../answer-assets/9231_s21_ms_11-p07.png"
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    },
    {
      "id": "9231-2021-mj-11-q03",
      "question_id": "9231-2021-mj-11-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) y=x3 B1 Correct substitution.\ny 4 3 −2y−1=0 y4 =(2y+1)3 =8y3 +12y2 +6y+1 M1 Obtains an equation not involving radicals.\ny4 −8y3−12y2 −6y−1=0 A1\nα3+β3 +γ3 +δ3 =8 B1 FT\n4\n© UCLES 2021 Page 7 of 14\n\n3(b) 1 + 1 + 1 + 1 = α3β3δ3 +α3β3γ3 +β3γ3δ3 +α3γ3δ3 = 6 M1 Relates 1 + 1 + 1 + 1 to coefficients.\nα3 β3 γ3 δ3 α3β3γ3δ3 −1 A1 FT α3 β3 γ3 δ3\n−6 A1\n3\n\n3(c) α4 +β4 +γ4 +δ4 =2(α3+β3+γ3+δ3)+4 M1 Uses original equation.\n=20 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2021-mj-11-q04",
      "question_id": "9231-2021-mj-11-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "4(a) cos60 −sin60  1 − 3 B1 Rotation 60° anticlockwise about the origin.\n \n= 2 2 \nsin60 cos60   3 1 \n \n2 2\nd 0 B1 One-way stretch in the x-direction, scale factor d.\n \n0 1\nd 0  1 − 3  d −d 3 M1 A1 Correct order.\nM=\n \n 2 2 = 2 2 \n0 1  3 1   3 1 \n   \n2 2 2 2\n4\n\n4(b) d = 1d2 M1 Uses value of detM.\n2\nd ≠0d =2 A1 AG\n2\n© UCLES 2021 Page 8 of 14\n\n4(c) 1  1 3 B1 FT Inverse of their M.\nM −1 =  2 \n2− 3 1 \n \n2\n1 1 1  1 31 1 M1 Multiplies on the left by the inverse of their M.\nN=M −1   =  2  \n 1 2 1 2  2  − 2 3 1   1 2 1 2 \n11+ 3 1+ 3 A1 CAO\n=  \n4 1− 3 1− 3  \n3\n\n4(d) 1 1x  x+ y  B1 x X\n  1 1    y   =   1x+ 1 y   Transforms  y   to  Y   .\n2 2 2 2\n1x+ 1mx=m ( x+mx ) M1 A1 Uses y=mx and Y =mX .\n2 2\n1+m=2m+2m2 2m2 +m−1=0 A1\ny= 1x and y=−x A1 WWW\n2\n5\n© UCLES 2021 Page 9 of 14",
      "source_pages": [
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        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2021-mj-11-q05",
      "question_id": "9231-2021-mj-11-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) θ=0 B1 B1 Initial line with correct position and shape of curve.\nSCB1 if correct shape with wrong starting position.\nPole position must be clear.\nacot1π=2 3a=2 B1 Substitutes θ= 1π. AG.\n6 6\n3\n\n5(b) 1 1 6 π 4cot2 ( 1π−θ ) dθ M1 Uses 1r2dθ with correct limits.\n2\n2 3\n0\n( ) ( )\n2 0 1 6 π cosec2 ( 1 3 π−θ ) −1dθ=2  cot ( 1 3 π−θ ) −θ  0 1 6 π M1 A1 Uses cot2 1 3 π−θ =cosec2 1 3 π−θ −1 and integrates.\n( )\n2 3−1π−1 3 = 4 3−1π A1 OE, must be exact.\n6 3 3 3\n4\n\n5(c) cosπcosθ+sinπsinθ M1 Uses the identities for cos(A−B) and sin(A−B). Or\nr =2 3 3\nsinπcosθ−cosπsinθ identity for tan (A – B).\n3 3\nx2 + y2 =2 rcosθ+ 3rsinθ M1 Applies r = x2 + y2,x=rcosθ and y=rsinθ.\n3rcosθ−rsinθ\n( ) ( )\n2 x+ y 3 = x 3−y x2 + y2 A1 AG.\n3\n© UCLES 2021 Page 10 of 14",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_11-p10.png"
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    },
    {
      "id": "9231-2021-mj-11-q06",
      "question_id": "9231-2021-mj-11-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) 0 t  −1 B1 Finds vector from any point on l 1 to any point on l 2 .\n     \n1 − 1 =t 0\n     \n     \nt 0  1 \ni j k  −1 M1 A1 Finds common perpendicular.\n \n−2 −1 0 = 2\n \n0 −2 1   4  \n\n−1\n\n−1 M1 A1 Uses formula for shortest distance.\nt    \n0  2 = 21t = 21=4.2\n21     5\n   \n 1   4 \n5\n\n6(b) r= 21i+j+λ( 2i+j )+μ(−2j+k ) B1 FT Using their value of t.\n5\n1\n\n6(c)  0   5  M1 A1 Uses dot product of −2j+k and normal.\n    19\n−2  −6 = 5 110cosαleading to cosα=\n   \n    5 110\n 1   7 \nAcute angle between l and Π is 90−α=54.1 ° A1\n2 2\n3\n© UCLES 2021 Page 11 of 14\n\n6(d)  −1  5  M1 Dot product using their normal to Π .\n1\n 2    −6  = 21 110cosαleading to cosα= 11 A1 FT\n   \n    21 110\n 4   7 \n76.8 ° A1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_11-p11.png",
        "../answer-assets/9231_s21_ms_11-p12.png"
      ]
    },
    {
      "id": "9231-2021-mj-11-q07",
      "question_id": "9231-2021-mj-11-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1 B1 States vertical asymptote.\nx(x+1)+9 9 M1 Finds oblique asymptote.\ny= =x+\nx+1 x+1\ny= x A1\n3\n\n7(b) dy M1 A1 Differentiates and sets derivative equal to 0.\n=1−9(x+1) −2 =0(x+1)2 =9\ndx\n( )\n2,5 A1\n(−4,−7 )\nA1\n4\n© UCLES 2021 Page 12 of 14\n\n7(c) B1 Axes labelled and correct asymptotes drawn.\nB1 Upper branch with (0,9) stated or shown on diagram.\n9\nB1 Lower branch correct and good approach to asymptotes\nthroughout, no extra branches.\nx\n3\n© UCLES 2021 Page 13 of 14\n\n7(d) B1 FT FT from sketch in (c) with asymptotes shown.\n9\nx\n13 13 M1 M1 Finds critical points, award M1 for each case.\nx2 +x+9= (x+1) or x2 +x+9=− (x+1)\n2 2 May state that x2 +x+9=−13(x+1) has no real\n2\nsolutions since 7>13.\nx2 −11x+ 5 =0 or x2 +15x+ 31 =0 2\n2 2 2 2\nx= 1,5 A1\n2\nx< 1 and x>5. A1\n2\n5\n© UCLES 2021 Page 14 of 14",
      "source_pages": [
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        13,
        14
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2021-mj-12-q01",
      "question_id": "9231-2021-mj-12-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 24 +31−2=45 is divisible by 15 B1 Checks base case.\nAssume that 24k +31k −2 is divisible by 15 for some positive integer k. B1 States inductive hypothesis.\nThen 24k+4 +31k+1−2=(15+1)24k +(30+1)31k −2 M1 A1 Separates 24k +31k −2 or considers difference.\nis divisible by 15 because 15×24k +30×31k is divisible by 15. A1\nHence, by induction, true for every positive integer n. A1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_12-p06.png"
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    },
    {
      "id": "9231-2021-mj-12-q02",
      "question_id": "9231-2021-mj-12-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) n−1n(n+1)−1n(n+1)(2n+1) M1A1 Substitutes correct formulae from MF19.\n2 6\nExpanding brackets correctly.\n1n−n2 −1n3 A1 Simplifies by collecting terms.\n3 3\n3\n© UCLES 2021 Page 6 of 14\n\n2(b) r+1 r (r+1) r2 +1 −r r2 +2r+2 1−r−r2 M1 A1 Puts over a common denominator and expands, AG.\n− =\n( )( )\n=\n( )( )\n(r+1)2 +1 r2 +1 r2 +2r+2 r2 +1 r2 +2r+2 r2 +1\nn 1−r−r2 n  r+1 r  M1 A1 Shows at least three complete terms including first and\n = −\n( )( )   last. Cancellation may be implicit.\nr=1\nr2 +2r+2 r2 +1\nr=1\n(r+1)2 +1 r2 +1\n2 1 3 2 4 3 n+1 n\n= − + − + − ++ −\n5 2 10 5 17 10 (n+1)2 +1 n2 +1\n1 n+1 A1 ISW\n=− +\n2 ( n+1 )2 +1\n5\n\n2(c) −1 B1 FT FT from their answer to part (b).\n2\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-12-q03",
      "question_id": "9231-2021-mj-12-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) y=x3 B1 Correct substitution.\ny 4 3 −2y−1=0 y4 =(2y+1)3 =8y3 +12y2 +6y+1 M1 Obtains an equation not involving radicals.\ny4 −8y3−12y2 −6y−1=0 A1\nα3+β3 +γ3 +δ3 =8 B1 FT\n4\n© UCLES 2021 Page 7 of 14\n\n3(b) 1 + 1 + 1 + 1 = α3β3δ3 +α3β3γ3 +β3γ3δ3 +α3γ3δ3 = 6 M1 Relates 1 + 1 + 1 + 1 to coefficients.\nα3 β3 γ3 δ3 α3β3γ3δ3 −1 A1 FT α3 β3 γ3 δ3\n−6 A1\n3\n\n3(c) α4 +β4 +γ4 +δ4 =2(α3+β3+γ3+δ3)+4 M1 Uses original equation.\n=20 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s21_ms_12-p08.png"
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    },
    {
      "id": "9231-2021-mj-12-q04",
      "question_id": "9231-2021-mj-12-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "4(a) cos60 −sin60  1 − 3 B1 Rotation 60° anticlockwise about the origin.\n \n= 2 2 \nsin60 cos60   3 1 \n \n2 2\nd 0 B1 One-way stretch in the x-direction, scale factor d.\n \n0 1\nd 0  1 − 3  d −d 3 M1 A1 Correct order.\nM=\n \n 2 2 = 2 2 \n0 1  3 1   3 1 \n   \n2 2 2 2\n4\n\n4(b) d = 1d2 M1 Uses value of detM.\n2\nd ≠0d =2 A1 AG\n2\n© UCLES 2021 Page 8 of 14\n\n4(c) 1  1 3 B1 FT Inverse of their M.\nM −1 =  2 \n2− 3 1 \n \n2\n1 1 1  1 31 1 M1 Multiplies on the left by the inverse of their M.\nN=M −1   =  2  \n 1 2 1 2  2  − 2 3 1   1 2 1 2 \n11+ 3 1+ 3 A1 CAO\n=  \n4 1− 3 1− 3  \n3\n\n4(d) 1 1x  x+ y  B1 x X\n  1 1    y   =   1x+ 1 y   Transforms  y   to  Y   .\n2 2 2 2\n1x+ 1mx=m ( x+mx ) M1 A1 Uses y=mx and Y =mX .\n2 2\n1+m=2m+2m2 2m2 +m−1=0 A1\ny= 1x and y=−x A1 WWW\n2\n5\n© UCLES 2021 Page 9 of 14",
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2021-mj-12-q05",
      "question_id": "9231-2021-mj-12-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) θ=0 B1 B1 Initial line with correct position and shape of curve.\nSCB1 if correct shape with wrong starting position.\nPole position must be clear.\nacot1π=2 3a=2 B1 Substitutes θ= 1π. AG.\n6 6\n3\n\n5(b) 1 1 6 π 4cot2 ( 1π−θ ) dθ M1 Uses 1r2dθ with correct limits.\n2\n2 3\n0\n( ) ( )\n2 0 1 6 π cosec2 ( 1 3 π−θ ) −1dθ=2  cot ( 1 3 π−θ ) −θ  0 1 6 π M1 A1 Uses cot2 1 3 π−θ =cosec2 1 3 π−θ −1 and integrates.\n( )\n2 3−1π−1 3 = 4 3−1π A1 OE, must be exact.\n6 3 3 3\n4\n\n5(c) cosπcosθ+sinπsinθ M1 Uses the identities for cos(A−B) and sin(A−B). Or\nr =2 3 3\nsinπcosθ−cosπsinθ identity for tan (A – B).\n3 3\nx2 + y2 =2 rcosθ+ 3rsinθ M1 Applies r = x2 + y2,x=rcosθ and y=rsinθ.\n3rcosθ−rsinθ\n( ) ( )\n2 x+ y 3 = x 3−y x2 + y2 A1 AG.\n3\n© UCLES 2021 Page 10 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2021-mj-12-q06",
      "question_id": "9231-2021-mj-12-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) 0 t  −1 B1 Finds vector from any point on l 1 to any point on l 2 .\n     \n1 − 1 =t 0\n     \n     \nt 0  1 \ni j k  −1 M1 A1 Finds common perpendicular.\n \n−2 −1 0 = 2\n \n0 −2 1   4  \n\n−1\n\n−1 M1 A1 Uses formula for shortest distance.\nt    \n0  2 = 21t = 21=4.2\n21     5\n   \n 1   4 \n5\n\n6(b) r= 21i+j+λ( 2i+j )+μ(−2j+k ) B1 FT Using their value of t.\n5\n1\n\n6(c)  0   5  M1 A1 Uses dot product of −2j+k and normal.\n    19\n−2  −6 = 5 110cosαleading to cosα=\n   \n    5 110\n 1   7 \nAcute angle between l and Π is 90−α=54.1 ° A1\n2 2\n3\n© UCLES 2021 Page 11 of 14\n\n6(d)  −1  5  M1 Dot product using their normal to Π .\n1\n 2    −6  = 21 110cosαleading to cosα= 11 A1 FT\n   \n    21 110\n 4   7 \n76.8 ° A1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2021-mj-12-q07",
      "question_id": "9231-2021-mj-12-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1 B1 States vertical asymptote.\nx(x+1)+9 9 M1 Finds oblique asymptote.\ny= =x+\nx+1 x+1\ny= x A1\n3\n\n7(b) dy M1 A1 Differentiates and sets derivative equal to 0.\n=1−9(x+1) −2 =0(x+1)2 =9\ndx\n( )\n2,5 A1\n(−4,−7 )\nA1\n4\n© UCLES 2021 Page 12 of 14\n\n7(c) B1 Axes labelled and correct asymptotes drawn.\nB1 Upper branch with (0,9) stated or shown on diagram.\n9\nB1 Lower branch correct and good approach to asymptotes\nthroughout, no extra branches.\nx\n3\n© UCLES 2021 Page 13 of 14\n\n7(d) B1 FT FT from sketch in (c) with asymptotes shown.\n9\nx\n13 13 M1 M1 Finds critical points, award M1 for each case.\nx2 +x+9= (x+1) or x2 +x+9=− (x+1)\n2 2 May state that x2 +x+9=−13(x+1) has no real\n2\nsolutions since 7>13.\nx2 −11x+ 5 =0 or x2 +15x+ 31 =0 2\n2 2 2 2\nx= 1,5 A1\n2\nx< 1 and x>5. A1\n2\n5\n© UCLES 2021 Page 14 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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        "../answer-assets/9231_s21_ms_12-p14.png"
      ]
    },
    {
      "id": "9231-2021-mj-13-q01",
      "question_id": "9231-2021-mj-13-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) sin(r+1) sinr sin(r+1)cosr−cos(r+1)sinr sin(r+1−r) M1 Applies all relevant correct addition formulae\n− = =\ncos(r+1) cosr cos(r+1)cosr cos(r+1)cosr from MF19 for the route chosen and combines\ninto a single fraction.\nsin(r+1) sinr sinrcos1+cosrsin1 sinr\nOR − = −\ncos(r+1) cosr cos(r+1) cosr\nsinrcosrcos1+cos2rsin1−sinrcosrcos1+sin2rsin1\n=\ncos(r+1)cosr\nsin1 A1 SC B2 for this route completely correct to AG.\n=\ncos(r+1)cosr\ntanr+tan1 tanr+tan1−tanr+tan2rtan1\n−tanr =\n1−tanrtan1 1−tanrtan1\ntan1sec2r tan1 sin1\n= = =\n1−tanrtan1 cosr(cosr−sinrtan1) cosr(cosrcos1−sinrsin1)\nsin1\n=\ncosrcos(r+1)\n2\n\n1(b) n M1 A1 Shows enough terms for cancellation to be clear.\n[ ]\nsin1 u =tan2−tan1+tan3−tan2+...+tan(n+1)−tann There must be three complete terms including first\nr\nr=1 and last. Cancelling may be implied.\nn tan(n+1)−tan1 A1 OE and ISW.\nu =\nr sin1\nr=1\n3\n© UCLES 2021 Page 6 of 16\n\n1(c) tan(n+1) oscillates as n→∞ so u +u +u + does not converge. B1 States ‘oscillates’ or refers to diverging values of\n1 2 3\ntan(n+1), or states that tan(n+1)does not tend to\na limit.\n1",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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        "../answer-assets/9231_s21_ms_13-p07.png"
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    },
    {
      "id": "9231-2021-mj-13-q02",
      "question_id": "9231-2021-mj-13-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "2(a) S =2 B1\n1\nS =S2 −2(0) M1 Uses formula for sum of squares.\n2 1\n=4 A1 Correct answer implies M1A1.\n3\n\n2(b)(i) S =2S −3S B1 CAO or as a single fraction.\nn+3 n+2 2 n\n1\n\n2(b)(ii) S =2S −3S =2(2S −3S )−3S M1 Uses their recursive formula from part (i) to find\n4 3 2 1 2 2 0 2 1\nS S = 7.\n4  3 2\n=4 A1\n2\n\n2(c) x=2− y B1 SOI\n2(2− y)3 −4(2− y)2 +3=0 M1 Makes their substitution.\n2y3−8y2 +8y−3=0 A1 OE but must be an equation.\n3\n© UCLES 2021 Page 7 of 16\n\n2(d) 8 M1 1 1 1 α'β'+β'γ'+γ'α'\n2 Uses + + = .\n−3 α' β' γ' α'β'γ'\n2\nOR\nOr use 2S −8S +8S −3S =0 with substitution of their values\n2 1 0 −1\n8 A1 FT FT from 2(c).\n=\n3\n2\n© UCLES 2021 Page 8 of 16",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2021-mj-13-q03",
      "question_id": "9231-2021-mj-13-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 5×14 +12 = 1 (2)2 ( 2+1 ) [ =6 ] so H is true. B1 Checks base case.\n2 1\nAssume that  k   ( 5r4 +r2 )   = 1 2 k2(k+1)2 ( 2k+1 ) B1 S th t e a t a e l s g i e n b d r u ai c c t i f v o e r m hy . p I o f t s h a e y s s i s f o [f r o A r L so L m k e , k th ] e in n c B lu 0 d . ing\nr=1\n k+1  ( 5r4 +r2 ) = 1 k2(k+1)2 ( 2k+1 )+5(k+1)4 +(k+1)2 M1 Considers sum to k +1.\n  2\nr=1\n( )\n1(k+1)2 2k3 +k2 +10(k+1)2 +2 M1 Take out the factor of (cid:4666)𝑘+1(cid:4667)(cid:2870) OR expands the\n2 summation expression and the target expression\nfor 𝑘+1 and collects like terms for both.\n1(k+1)2(2k3+11k2 +20k+12)= 1(k+1)2(k+2)2(2k+3) A1 Factorises or having expanded, checks explicitly.\n2 2 At least one intermediate step seen following the\naward of M1 before reaching the answer.\nSo H is true. By induction, H is true for all positive integers n. A1 States conclusion. Implication must be clearly\nk+1 n\nexpressed.\n6\n\n3(b) n M1 Uses correct formula for r2.\n5r4 +1n(n+1)(2n+1)= 1n2(n+1)2(2n+1)\n6 2\nr=1\n[ ] n M1 Makes r4 the subject and takes out all linear\n5 r4 = 1n(n+1)(2n+1)(3n(n+1)−1)\n6\nfactors and the remaining term is of correct form.\nr=1\nn A1 CAO\nr4 = 1 n(n+1)(2n+1)(3n2 +3n−1)\n30\nr=1\n3\n© UCLES 2021 Page 9 of 16",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-13-q04",
      "question_id": "9231-2021-mj-13-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) 2 k k1 0 2+k k  M1 A1 Multiplies two matrices correctly.\n 0 1 1    0 1 1 \n  −1 2 0    5 −1 3     0 1   =   −1 2 0     8 −1  \n1 0 11 0  2 0 \n2 k k1 0 1 0\n 0 1 1   6 −1 4  \nOr   −1 2 0     5 −1 3     0 1   =  8 −k−2 −k+6     0 1  \n1 0 11 0 1 0\n 10 −1  A1\n \n\n−k+14 −k−2\n3\n\n4(b) −1 3 5 3 5 −1 M1 A1 Sets determinant equal to zero and forms linear\n2 −k +k =0 leading to −2−2k+k =0 equation.\n0 1 1 1 1 0\nk =−2 A1\n3\n© UCLES 2021 Page 10 of 16\n\n4(c) 10 −1x 10x− y M1 x\n   =   Transforms  .\n16 0 y  16x  y\n1 \nAllow q  where q is x, t or a nonzero number.\nm\n10x−mx= X and 16x=mX M1 A1 Uses y=mx and Y =mX .\nExpect 16x=m ( 10x−mx ) .\n16=10m−m2 [m2 −10m+16=0] A1 OE\ny=2x and y=8x A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2021-mj-13-q05",
      "question_id": "9231-2021-mj-13-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Approximately correct curve passing through the\npole, O, with correct domain.\nB1 r strictly increasing over the domain 0 to π .\n2\nB1 Correct form at O.\nApprox tangential to initial line at O.\n[O] [θ = 0]\n3\n© UCLES 2021 Page 11 of 16\n\n5(b) 1π 1 1 2 M1 Uses 1r2 dθ with correct limits.\n12\n\n−\n\ndθ 2\n2 0 π−θ π\n1π 1 2 1 M1 Expands.\n12 − + dθ\n2 0 ( π−θ)2 π ( π−θ) π2\n1 1 2 θ  1 2 π M1 A1 Integrates all terms to obtain correct form.\n+ ln(π−θ)+\n2  π−θ π π2 \n0\n12 2 π 1 1 2  1 3 2 1 M1 Substitute limits into an expression of the correct\n + ln + −  + lnπ  =  + ln  form and simplify the log terms.\n2π π 2 2π π π  22π π 2\n3−4ln2 A1 AG\n=\n4π\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-13-q06",
      "question_id": "9231-2021-mj-13-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a)\n−(−2i−2j+4k )+(−i−2j+k )=i−3k\nB1 OE. Finds direction vector from P to a point of l.\n1\nr=−2i−2j+4k+λ( 2i−3j )+μ( i−3k ) B1 FT OE. FT their i – 3k\nr =−i−2j+k+λ( 2i−3j )+μ( i−3k )\n2\n© UCLES 2021 Page 12 of 16\n\n6(b) i j k  −9 M1 A1 OE. Finds vector perpendicular to Π .\n2\n \n2 −3 0 = −6\n \n3 −1 3  7  \n−9(3)−6(0)+7(−2)=−41 M1 Uses point on Π .\n2\n9x+6y−7z=41 A1\n4\n\n6(c) i j k 9 3 M1 A1 Finds vector perpendicular to Π .\n1\n   \n2 −3 0 = 6 ~ 2\n   \n1 0 −3  3    1  \n3  9  DM1 Dot product using their normal vectors.\n 2    6  = 14 166cosαleading to cosα= 32 A1 FT\n   \n 1     −7   14 166\n48.4° A1 0.845 rad\n5\n© UCLES 2021 Page 13 of 16\n\n6(d)  −2 9   10+9t  M1 A1 FT Scales OP by a factor of ±5 and uses multiple of\n  \nOF =OQ+QF =−5  −2  +t  6  =  10+6t  their normal to Π .\n      2\n  4     −7     −20−7t  \n9 ( 10+9t )+6 ( 10+6t )−7 (−20−7t )=41leading to290+166t=41 DM1 Substitutes into the equation of Π .\n2\n−7  A1\n 2\n \nt=−3 leading toOF = 1\n \n2\n−19\n \n2\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        12,
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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        "../answer-assets/9231_s21_ms_13-p14.png"
      ]
    },
    {
      "id": "9231-2021-mj-13-q07",
      "question_id": "9231-2021-mj-13-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) 1− 5 1+ 5 B1 Vertical asymptotes. Must be exact.\nx= , x=\n2 2\ny=−1 B1 Horizontal asymptote.\n2\n\n7(b) dy (1+x−x2)(2x−1)−(x2 −x−3)(1−2x) M1 dy\n= Finds .\ndx ( 1+x−x2 )2 dx\n(2x−1)(−2)=0 M1 dy\nSets their equal to 0 and forms equation.\ndx\n( )\n1,−13 A1 WWW\n2 5\n3\n© UCLES 2021 Page 14 of 16\n\n7(c) y B1 Both axes labelled and correct asymptotes shown.\nB1 Correct shape and position, with all asymptotic\nbehaviour clear.\nx\n( ) ( )\n1 +1 13,0 , 1 −1 13,0 , (0,−3) B1 States exact coordinates of intersections with axes.\n2 2 2 2\n3\n© UCLES 2021 Page 15 of 16\n\n7(d) B1 FT FT from sketch in (c).\nB1 Correct shape as x tends to infinity and\nintersections with x axis.\nx2 −x−3 x2 −x−3 M2 Finds critical points, award M1 for each case.\n=3 or =−3\n1+x−x2 1+x−x2\n4x2 −4x−6=0 or −2x2 +2x=0\nx= 1 +1 7, x= 1 −1 7, x=0, x=1 A1 Must be exact.\n2 2 2 2\nx< 1 −1 7, 0<x<1, x> 1 +1 7 A1 FT Must be three distinct regions and strict\n2 2 2 2 inequalities.\n6\n© UCLES 2021 Page 16 of 16",
      "source_pages": [
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        15,
        16
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s21_ms_13-p15.png",
        "../answer-assets/9231_s21_ms_13-p16.png"
      ]
    },
    {
      "id": "9231-2021-mj-21-q01",
      "question_id": "9231-2021-mj-21-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) a 3 1 M1 A1 Finds determinant. If solving equations x, y, z in\n2 1 3 =−11a+26 terms in a.\n−1 2 −5\ndetA≠0 leads to unique solution. A1 States that detA≠0. (Since a is an integer.)\n(M1 A0 A0 if one of x, y, z found in terms of a.)\n(M1 A1 A0 if two of x, y, z found in terms of a.)\nThe three planes intersect at a single point. B1\n4\n\n1(b) a=4 B1\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-21-q02",
      "question_id": "9231-2021-mj-21-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) m2 +3m+2=0 leading to m = -1, -2 M1 Auxiliary equation.\ny= Ae −x +Be −2x A1 Complimentary function.\ny= p+qx leading to y'=q leading to y''=0 B1 Particular integral and its derivatives.\n3q+2(p+qx)=2x+1 M1 Substitutes and equates coefficients.\nq=1 p=−1 A1\ny= Ae −x +Be −2x +x−1 A1 Must see ‘y =’.\n6\n© UCLES 2021 Page 6 of 13\n\n2(b) y = x −1 B1 FT Accept y ≈ x −1.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_21-p06.png",
        "../answer-assets/9231_s21_ms_21-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-21-q03",
      "question_id": "9231-2021-mj-21-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a)  1 x3dx< ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3+ ( 1 )( n )3 M1 A1 Forms the sum of the areas of the n rectangles.\nn n n n n n n n Need last term for A1.\n0\n1  n r3 = n2(n+1)2 =   n+1  2 M1 A1 Applies  n r3 = 1n2(n+1)2, AG.\nn4 4n4  2n  4\nr=1 r=1\n4\n\n3(b)  1 x3dx> ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n rectangles.\n0\nNeed the last term for A1.\n1  n−1 r3 = (n−1)2n2 =   n−1  2 M1 A1 Applies  n r3 = 1n2(n+1)2.\nn4 4n4  2n  4\nr=1 r=1\nn+1 2 1\nAccept   − .\n 2n  n\n4\n\n3(c)   n+1  2 −   n−1  2 = 1 <10 −3 leading to n>103 M1 Simplifies their U n − L n to k n .\n 2n   2n  n\nLeast value of n is 1001. A1\n2\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_21-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-21-q04",
      "question_id": "9231-2021-mj-21-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4 dy tan1θ B1 Divides through by sinθ.\n+cosecθy = 2\ndθ sinθ\ne cosecθdθ =elntan1 2 θ= tan 1θ M1 A1 Finds integrating factor.\n2\nd ( ) tan2 1θ  1sin1θ M1 Correct form on LHS and uses an appropriate\nytan1θ = 2 = 1tan1θsec2 1θ  = 2 2  identity.\ndθ 2 sinθ 2 2 2  cos3 1\n2\nθ \n( )\nytan1θ= 1tan2 1θ+C or 1sec2 1θ+C M1 A1 Integrates RHS.\n2 2 2 2 2\n1 M1 Substitutes initial conditions.\n1= +C\n2\ny=1 ( tan1θ+cot1θ )[ =cosecθ ] M1 A1 Divides through by their integrating factor.\n2 2 2\n9\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_21-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-21-q05",
      "question_id": "9231-2021-mj-21-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) z−zn+1 zn+1 − z B1\nor .\n1−z z −1\n1\n\n5(b) zn =1 and z≠1 leading to z+z2 +z3 +...+zn =0 leading to M1 A1 Must see zn =1.\n1+z+z2 +...+zn−1 =0\n2\n\n5(c) ∞ z cosθ+isinθ M1 A1 Applies sum to infinity and substitutes for z.\nzm = =\n1−z 3−cosθ−isinθ\nm=1\n( cosθ+isinθ)( 3−cosθ+isinθ)\nM1 A1 Rationalises denominator.\n( 3−cosθ)2 +sin2θ\n3cosθ−cos2θ−sin2θ+isinθcosθ+isinθ( 3−cosθ) M1 Applies sin2θ+cos2θ=1 or eiθ+e−iθ= 2cosθ.\n9−6cosθ+cos2θ+sin2θ\n ∞  3cosθ−1 M1 A1 Takes the real part, AG.\nRe zm  =\n10−6cosθ\nm=1 \n7\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_21-p09.png"
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    },
    {
      "id": "9231-2021-mj-21-q06",
      "question_id": "9231-2021-mj-21-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Eigenvalues of A are 5, −6and 1. B1 Upper diagonal matrix or characteristic equation.\ni j k 88 1 M1 Uses vector product (or equations) to find\nλ=5: 0 −22 8 =  0    0  corresponding Eigenvectors.\n3    \n0 −11 0  0    0   A1 A1 A1 for each correct Eigenvector.\ni j k\n−154\n2\nA1\n3\n   \nλ=−6: 11 −2 3 2 8 =  −77  3 \n   \n0 0 7 0 0\n   \ni j k  56   2 \n   \nλ=1: 4 −22 8 = 0  0\n3    \n0 −7 0   −28     −1  \n1 2 2  25 0 0 M1 A1 Or correctly matched permutations of columns.\nThus P=  0 3 0  and D=  0 36 0  (Accept scalar multiples of the eigenvectors\n   \nshown here.) P must have at least two non-zero\n 0 0 −1     0 0 1   columns.\n7\n\n6(b) (λ−5 )(λ+6 )(λ−1 )=λ3−31λ+30=0 =0 B1 Finds characteristic equation.\nA3 −31A +30I = 0 A3 =31A −30I M1 Finds A3 in terms of A. Allow missing I.\n125 −682 248 M1A1 Substitutes for A.\n 3 \nA3 = 0 −216 0\n \n 0 0 1 \n \n4\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2021-mj-21-q07",
      "question_id": "9231-2021-mj-21-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a)\nsech y =\n1\n= x+ 1  cosh y =\n(\nx+ 1\n)−1 B1 Relates to cosh y,\ncosh y 2 2\ndy B1 Differentiates LHS.\nsinhy\ndx\n− ( x+ 1 )−2 B1 Differentiates RHS. AG.\n2\n3\n\n7(b) d2y dy 2  1 −3 M1 A1 M1 A1 for LHS.\nsinh +coshy   =2  x+  B1 for RHS.\ndx2 dx  2\nB1\ny(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form.\n2\n4 16 M1 Evaluates derivatives at x=0.\ny'(0)=− y''(0)=\n3 3 3\n( ) 4 8 A1\ny =ln 2+ 3 − x+ x2\n3 3 3\nAlternative method for question 7(b)\ndy 1 1 B1\n= − = −\ndx ( x+ 1 )2 ( x+ 1 )−2 −1 ( x+ 1 ) 3 − x− x2\n2 2 2 4\nd2y = 1 ( x+ 1 )−1 ( 3 − x− x2 )−3 2 (−1−2x )+ ( 3 − x− x2 )−1 2 ( x+ 1 )−2 M1 A1\ndx2 2 2 4 4 2\ny(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form.\n2\n© UCLES 2021 Page 11 of 13\n\n7(b) 4 16 M1 Evaluates derivatives at x=0.\ny'(0)=− y''(0)=\n3 3 3\n( ) 4 8 A1\ny =ln 2+ 3 − x+ x2\n3 3 3\n7\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_21.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-21-q08",
      "question_id": "9231-2021-mj-21-q08",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) dx B1\n=2sinht\ndt\ndy ( ) M1 A1 Applies 2sinh2t = cosh2t −1, AG.\n= 3−1cosh2t =1− 1cosh2t−1 =1−sinh2t\ndt 2 2 2 2\n3\n\n8(b)(i) 2 2 M1 2 2\ndx dy dx dy\n  +   =4s2 +(s2 −1)2 =4s2 +s4 −2s2 +1=(s2 +1)2 Factorises   +   .\ndt  dt  dt  dt \ncosh4t A1\n2π 1( 3t−1sinh2t ) cosh2t dt =π 1( 3t−1sinh2t )( cosh2t+1 ) dt M1 A1 Correct formula 2 for surfa 2 ce area, AG. A0 if limits\n2 4 2 4 dx dy\n+\n0 0 missing.     does not need to be\ndt   dt \nsimplified for M1.\n4\n© UCLES 2021 Page 12 of 13\n\n8(b)(ii)  1 1sinh2t ( cosh2t+1 ) dt=1 (1+cosh2t)2 1 = 1 ( 1+cosh2 )2−1 M1 A1 Integrates.\n0 4 16  0 16 4\n3 1 t ( cosh2t+1 ) dt=3t ( 1sinh2t+t )  1 −3 1 1sinh2t+tdt M1 A1 Integrates by parts.\n2 0 2 2  0 2 0 2\n3t ( 1sinh2t +t ) − 1cosh2t − 1t2 1 = 3 ( 1sinh2− 1cosh2+ 3 ) A1\n2 2 4 2  0 2 2 4 4\n( )\nπ 3sinh2−3cosh2− 1 (1+cosh2)2 +11 A1 OE. Must be exact.\n4 8 16 8 (Decimal answer is 3.980131435…)\n6\n© UCLES 2021 Page 13 of 13",
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        "../answer-assets/9231_s21_ms_21-p13.png"
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    },
    {
      "id": "9231-2021-mj-22-q01",
      "question_id": "9231-2021-mj-22-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) a 3 1 M1 A1 Finds determinant. If solving equations x, y, z in\n2 1 3 =−11a+26 terms in a.\n−1 2 −5\ndetA≠0 leads to unique solution. A1 States that detA≠0. (Since a is an integer.)\n(M1 A0 A0 if one of x, y, z found in terms of a.)\n(M1 A1 A0 if two of x, y, z found in terms of a.)\nThe three planes intersect at a single point. B1\n4\n\n1(b) a=4 B1\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-22-q02",
      "question_id": "9231-2021-mj-22-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) m2 +3m+2=0 leading to m = -1, -2 M1 Auxiliary equation.\ny= Ae −x +Be −2x A1 Complimentary function.\ny= p+qx leading to y'=q leading to y''=0 B1 Particular integral and its derivatives.\n3q+2(p+qx)=2x+1 M1 Substitutes and equates coefficients.\nq=1 p=−1 A1\ny= Ae −x +Be −2x +x−1 A1 Must see ‘y =’.\n6\n© UCLES 2021 Page 6 of 13\n\n2(b) y = x −1 B1 FT Accept y ≈ x −1.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p06.png",
        "../answer-assets/9231_s21_ms_22-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-22-q03",
      "question_id": "9231-2021-mj-22-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a)  1 x3dx< ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3+ ( 1 )( n )3 M1 A1 Forms the sum of the areas of the n rectangles.\nn n n n n n n n Need last term for A1.\n0\n1  n r3 = n2(n+1)2 =   n+1  2 M1 A1 Applies  n r3 = 1n2(n+1)2, AG.\nn4 4n4  2n  4\nr=1 r=1\n4\n\n3(b)  1 x3dx> ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n rectangles.\n0\nNeed the last term for A1.\n1  n−1 r3 = (n−1)2n2 =   n−1  2 M1 A1 Applies  n r3 = 1n2(n+1)2.\nn4 4n4  2n  4\nr=1 r=1\nn+1 2 1\nAccept   − .\n 2n  n\n4\n\n3(c)   n+1  2 −   n−1  2 = 1 <10 −3 leading to n>103 M1 Simplifies their U n − L n to k n .\n 2n   2n  n\nLeast value of n is 1001. A1\n2\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p07.png"
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    },
    {
      "id": "9231-2021-mj-22-q04",
      "question_id": "9231-2021-mj-22-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4 dy tan1θ B1 Divides through by sinθ.\n+cosecθy = 2\ndθ sinθ\ne cosecθdθ =elntan1 2 θ= tan 1θ M1 A1 Finds integrating factor.\n2\nd ( ) tan2 1θ  1sin1θ M1 Correct form on LHS and uses an appropriate\nytan1θ = 2 = 1tan1θsec2 1θ  = 2 2  identity.\ndθ 2 sinθ 2 2 2  cos3 1\n2\nθ \n( )\nytan1θ= 1tan2 1θ+C or 1sec2 1θ+C M1 A1 Integrates RHS.\n2 2 2 2 2\n1 M1 Substitutes initial conditions.\n1= +C\n2\ny=1 ( tan1θ+cot1θ )[ =cosecθ ] M1 A1 Divides through by their integrating factor.\n2 2 2\n9\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-22-q05",
      "question_id": "9231-2021-mj-22-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) z−zn+1 zn+1 − z B1\nor .\n1−z z −1\n1\n\n5(b) zn =1 and z≠1 leading to z+z2 +z3 +...+zn =0 leading to M1 A1 Must see zn =1.\n1+z+z2 +...+zn−1 =0\n2\n\n5(c) ∞ z cosθ+isinθ M1 A1 Applies sum to infinity and substitutes for z.\nzm = =\n1−z 3−cosθ−isinθ\nm=1\n( cosθ+isinθ)( 3−cosθ+isinθ)\nM1 A1 Rationalises denominator.\n( 3−cosθ)2 +sin2θ\n3cosθ−cos2θ−sin2θ+isinθcosθ+isinθ( 3−cosθ) M1 Applies sin2θ+cos2θ=1 or eiθ+e−iθ= 2cosθ.\n9−6cosθ+cos2θ+sin2θ\n ∞  3cosθ−1 M1 A1 Takes the real part, AG.\nRe zm  =\n10−6cosθ\nm=1 \n7\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p09.png"
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    },
    {
      "id": "9231-2021-mj-22-q06",
      "question_id": "9231-2021-mj-22-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Eigenvalues of A are 5, −6and 1. B1 Upper diagonal matrix or characteristic equation.\ni j k 88 1 M1 Uses vector product (or equations) to find\nλ=5: 0 −22 8 =  0    0  corresponding Eigenvectors.\n3    \n0 −11 0  0    0   A1 A1 A1 for each correct Eigenvector.\ni j k\n−154\n2\nA1\n3\n   \nλ=−6: 11 −2 3 2 8 =  −77  3 \n   \n0 0 7 0 0\n   \ni j k  56   2 \n   \nλ=1: 4 −22 8 = 0  0\n3    \n0 −7 0   −28     −1  \n1 2 2  25 0 0 M1 A1 Or correctly matched permutations of columns.\nThus P=  0 3 0  and D=  0 36 0  (Accept scalar multiples of the eigenvectors\n   \nshown here.) P must have at least two non-zero\n 0 0 −1     0 0 1   columns.\n7\n\n6(b) (λ−5 )(λ+6 )(λ−1 )=λ3−31λ+30=0 =0 B1 Finds characteristic equation.\nA3 −31A +30I = 0 A3 =31A −30I M1 Finds A3 in terms of A. Allow missing I.\n125 −682 248 M1A1 Substitutes for A.\n 3 \nA3 = 0 −216 0\n \n 0 0 1 \n \n4\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2021-mj-22-q07",
      "question_id": "9231-2021-mj-22-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a)\nsech y =\n1\n= x+ 1  cosh y =\n(\nx+ 1\n)−1 B1 Relates to cosh y,\ncosh y 2 2\ndy B1 Differentiates LHS.\nsinhy\ndx\n− ( x+ 1 )−2 B1 Differentiates RHS. AG.\n2\n3\n\n7(b) d2y dy 2  1 −3 M1 A1 M1 A1 for LHS.\nsinh +coshy   =2  x+  B1 for RHS.\ndx2 dx  2\nB1\ny(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form.\n2\n4 16 M1 Evaluates derivatives at x=0.\ny'(0)=− y''(0)=\n3 3 3\n( ) 4 8 A1\ny =ln 2+ 3 − x+ x2\n3 3 3\nAlternative method for question 7(b)\ndy 1 1 B1\n= − = −\ndx ( x+ 1 )2 ( x+ 1 )−2 −1 ( x+ 1 ) 3 − x− x2\n2 2 2 4\nd2y = 1 ( x+ 1 )−1 ( 3 − x− x2 )−3 2 (−1−2x )+ ( 3 − x− x2 )−1 2 ( x+ 1 )−2 M1 A1\ndx2 2 2 4 4 2\ny(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form.\n2\n© UCLES 2021 Page 11 of 13\n\n7(b) 4 16 M1 Evaluates derivatives at x=0.\ny'(0)=− y''(0)=\n3 3 3\n( ) 4 8 A1\ny =ln 2+ 3 − x+ x2\n3 3 3\n7\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p11.png",
        "../answer-assets/9231_s21_ms_22-p12.png"
      ]
    },
    {
      "id": "9231-2021-mj-22-q08",
      "question_id": "9231-2021-mj-22-q08",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) dx B1\n=2sinht\ndt\ndy ( ) M1 A1 Applies 2sinh2t = cosh2t −1, AG.\n= 3−1cosh2t =1− 1cosh2t−1 =1−sinh2t\ndt 2 2 2 2\n3\n\n8(b)(i) 2 2 M1 2 2\ndx dy dx dy\n  +   =4s2 +(s2 −1)2 =4s2 +s4 −2s2 +1=(s2 +1)2 Factorises   +   .\ndt  dt  dt  dt \ncosh4t A1\n2π 1( 3t−1sinh2t ) cosh2t dt =π 1( 3t−1sinh2t )( cosh2t+1 ) dt M1 A1 Correct formula 2 for surfa 2 ce area, AG. A0 if limits\n2 4 2 4 dx dy\n+\n0 0 missing.     does not need to be\ndt   dt \nsimplified for M1.\n4\n© UCLES 2021 Page 12 of 13\n\n8(b)(ii)  1 1sinh2t ( cosh2t+1 ) dt=1 (1+cosh2t)2 1 = 1 ( 1+cosh2 )2−1 M1 A1 Integrates.\n0 4 16  0 16 4\n3 1 t ( cosh2t+1 ) dt=3t ( 1sinh2t+t )  1 −3 1 1sinh2t+tdt M1 A1 Integrates by parts.\n2 0 2 2  0 2 0 2\n3t ( 1sinh2t +t ) − 1cosh2t − 1t2 1 = 3 ( 1sinh2− 1cosh2+ 3 ) A1\n2 2 4 2  0 2 2 4 4\n( )\nπ 3sinh2−3cosh2− 1 (1+cosh2)2 +11 A1 OE. Must be exact.\n4 8 16 8 (Decimal answer is 3.980131435…)\n6\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_22-p12.png",
        "../answer-assets/9231_s21_ms_22-p13.png"
      ]
    },
    {
      "id": "9231-2021-mj-23-q01",
      "question_id": "9231-2021-mj-23-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) a=1, b=i B1\n1\n\n1(b) z=ei2\n5\nkπ M1 Finds one fifth root of unity.\nz=ei2 5 kπ, k =0,1,2,3,4 A1 Gives all fifth roots of unity. Accept 1 not in exponential form.\nA0 if r =1 not seen or implied.\nArgument of i is 1π. B1\n2\nz=ei1 6 π M1 A1 Finds one root of z3 =i.\nA0 if first root not given in exponential form.\nz=ei5\n6\nπ, ei3\n2\nπ A1 Finds other two roots of z3 =i. A0 if r =1 not seen or implied.\nWithhold final A1 if a and b reversed in part (b) but all other\nwork correct.\n6 SC If z=e i ( 1 1 0 π+2 5 kπ ) , k =0,1,2,3,4 award M1 A1 A1 only.\n( )\nSC If z=e i 2 3 kπ , k =0,1,2 award M1 A1 only.\n© UCLES 2021 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p06.png"
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    },
    {
      "id": "9231-2021-mj-23-q02",
      "question_id": "9231-2021-mj-23-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 dy B1 Finds first derivative.\n=tanhx\ndx\nd2y B1 Finds second derivative.\n=sech2x\ndx2\nd3y sinhx M1 A1 Finds third derivative.\n=−2tanhxsech2x=−2\ndx3 cosh3x\nd4y cosh2 ( x )−3sinh2 ( x ) B1 Finds fourth derivative. Alternative:\ndx4 =−2 cosh4 ( x ) =−2sech4 x+4tanh2 xsech2 x 6tanh2 xsech2 x−2sech2 x.\ny(0)= y(1)(0)= y(3)(0)=0 y(2)(0)=1 y(4)(0)=−2 M1 Evaluates derivatives at x=0.\ny= 1 x2 − 1 x4 A1 CWO\n2 12\n7\n© UCLES 2021 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-23-q03",
      "question_id": "9231-2021-mj-23-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) N r N r M1 A1 Compares with sum of the areas of the rectangles.\n =1+\n2r2 −1 2r2 −1\nr=1 r=2\nN x M1 Compares with integral.\n< 1+ dx\n1 2x2 −1\nN x ( ) N M1 A1 Finds integral.\n dx=1ln 2x2 −1 \n1 2x2 −1 4  1\nN r ( ) M1 A1 Inserts limits, AG.\n <1+ 1ln 2N2 −1\n2r2 −1 4\nr=1\n7\n\n3(b) N r N−1 r N N x N M1 A1 Compares with integral.\n = + > dx+\n2r2 −1 2r2 −1 2N2 −1 1 2x2 −1 2N2 −1\nr=1 r=1\n( ) N A1 CWO.\n1ln 2N2 −1 +\n4 2N2 −1\nAlternative method for question 3(b)\nN r N+1 x M1 A1 Compares with integral.\n > dx\n2r2 −1 1 2x2 −1\nr=1\n( )\n1ln 2 ( N +1 )2 −1 A1 CWO.\n4\n3\n© UCLES 2021 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p08.png"
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    },
    {
      "id": "9231-2021-mj-23-q04",
      "question_id": "9231-2021-mj-23-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4 z−z −1 =2isinθand z+z −1 =2cosθ B1 Use of z−z −1 =2isinθ and z+z −1 =2cosθ\n( z−z −1 )5 = ( z5 −z −5 ) −5 ( z3−z −3 ) +10 ( z−z −1 ) M1 A1 Expands and groups. A0 if grouping not shown.\n( z+z −1 )5 = ( z5 +z −5 ) +5 ( z3 +z −3 ) +10 ( z+z −1 ) M1 A1 Expands and groups. A0 if grouping not shown.\n25isin5θ 2isin5θ−5(2isin3θ)+10(2isinθ) M1 Substitutes zn +z −n =2cosnθand zn −z −n =2isinnθ.\n=\n25cos5θ 2cos5θ+5 ( 2cos3θ)+10 ( 2cosθ)\nsin5θ−5sin3θ+10sinθ A1 Justifies cancellation of constants.\ntan5θ= .\ncos5θ+5cos3θ+10cosθ\n7\n© UCLES 2021 Page 9 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p09.png"
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    },
    {
      "id": "9231-2021-mj-23-q05",
      "question_id": "9231-2021-mj-23-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) y=kxe −x leading to y'=ke −x ( 1−x ) leading to y''=ke −x ( x−2 ) B1 B1 Differentiates particular integral.\nke −x ( x−2 )−2ke −x ( 1−x )−3kxe −x =4e −x leading to −4k =4 M1 Substitutes.\nk =−1 A1\n4\n\n5(b) m2 −2m−3=0 leading to m=−1,3 M1 Auxiliary equation.\ny= Ae −x +Be3x A1 Complementary function. (Can be awarded if seen in the\ngeneral solution.)\ny= Ae −x +Be3x −xe −x =(A−x)e −x +Be3x A1 FT General solution. FT on their k. Must have “y=”.\ny'=−(A−x)e −x −e −x +3Be3x B1 Derivative of general solution.\nA+B= 1, −A−1+3B= 1 leading to A=0, B= 1 M1 Substitutes initial conditions into general solution.\n2 2 2\ny= 1e3x −xe −x A1 Must have “y=”.\n2\n6\n© UCLES 2021 Page 10 of 15",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p10.png"
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    },
    {
      "id": "9231-2021-mj-23-q06",
      "question_id": "9231-2021-mj-23-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) sinhx= 1 ex −e −x coshx= 1 ex +e −x B1 Writes in exponential from.\n2 2\n1 ( ex −e −x )2 = 1 ( e2x −2+e −2x ) = 1 ( e2x +e −2x ) −1 M1 Expands.\n2 2 2\ncosh2x−1 A1 AG.\n3\n\n6(b) e cothx =elnsinhx =sinhx M1 A1 Finds integrating factor.\nd M1 Correct form on both sides.\n(ysinhx)=4sinh2x\ndx\n( )\nysinhx=2 1sinh2x−x +C=sinh2x−2x+C M1 A1 Integrates sinh2xusing identity.\n2\n4 40 M1 Substitutes initial conditions into their solution.\nsinhln3=sinhln9−2ln3+C leading to = −2ln3+C\n3 9\n28 A1\nysinhx=sinh2x−2x+2ln3−\n9\n7\n© UCLES 2021 Page 11 of 15",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p11.png"
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    },
    {
      "id": "9231-2021-mj-23-q07",
      "question_id": "9231-2021-mj-23-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) 3( )−1 ( ) 3 M1 A1 Recognises integral or uses appropriate substitution.\nI =2 4+x2 2 =sinh −1 1x 2\n1  2 \n0 0\nln2 A1 Insert limits, must simplify.\n3\n\n7(b) d  x ( 4+x2 )−1 2 n  =−nx2 ( 4+x2 )−1 2 n−1+ ( 4+x2 )−1 2 n M1 A1 Uses the product rule to differentiate.\ndx \n−n ( 4+x2 −4 )( 4+x2 )−1 2 n−1+ ( 4+x2 )−1 2 n M1 Applies x2 =4+x2 −4.\n x ( 4+x2 )−1 2 n 3 2 =−nI +4nI +I M1 Integrates both sides using the limits given.\n  n n+2 n\n \n0\n3 ( 2 )n =(1−n)I +4nI 4nI = 3 ( 2 )n +(n−1)I A1 Substitutes limits and rearranges, AG.\n2 5 n n+2 n+2 2 5 n\nAlternative method for question 7(b)\nI = x ( 4+x2 )−1 2 n 3 2 +n 3 2x2 ( 4+x2 )−1 2 n−1 dx M1 A1 Integrates by parts.\nn  \n  0\n0\nI = x ( 4+x2 )−1 2 n2 3 +n2 3( 4+x2 −4 )( 4+x2 )−1 2 n−1 dx M1 Applies x2 =4+x2 −4.\nn  \n  0\n0\nI = 3 ( 2 )n +nI −4I 4nI = 3 ( 2 )n +(n−1)I M1 A1 Substitutes limits and rearranges, AG.\nn 2 5 n n+2 n+2 2 5 n\n5\n© UCLES 2021 Page 12 of 15\n\n7(c) I = 3 B1 Applies reduction formula with n=1.\n3 20\n12I = 3 ( 8 ) +2I I = 33 =0.033 M1 A1 Applies reduction formula with n=3.\n5 2 125 3 5 1000\n3",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_23-p12.png",
        "../answer-assets/9231_s21_ms_23-p13.png"
      ]
    },
    {
      "id": "9231-2021-mj-23-q08",
      "question_id": "9231-2021-mj-23-q08",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) 13 18 −28 M1 Finds determinant.\n−4 −a 8 =9a−24\n2 6 −5\n8 A1 24\na= Accept .\n3 9\n2\n\n8(b) 13 18 −282  −2 B1\n    \n−4 −1 8 0 = 0 λ=−1.\n    \n  2 6 −5    1   −1 \n1\n© UCLES 2021 Page 13 of 15\n\n8(c) 13−λ 18 −28 B1 Sets det ( A−λI )=0.\n−4 −1−λ 8 =0\n2 6 −5−λ\n−λ3 +7λ2 −7λ−15=0 (λ+1 )(λ−3 )(λ−5 )=0 M1 Expands determinant and factorises.\nλ=−1, λ=3, λ=5 A1\ni j k  −16 1 M1 Uses vector product (or equations) to find corresponding\nλ=3: −4 −4 8 =  −16    1  eigenvectors.\n   \n2 6 −8   −16    1   A1 A1 A1 for each correct Eigenvector.\ni j k  12   −1\n   \nλ=5: −4 −6 8 = −24  2\n   \n2 6 −10   −12    1  \n2 1 −1  −1 0 0 M1 A1 Or correctly matched permutations of columns. Accept scalar\nThus P=  0 1 2  and D=  0 3 0  multiples of eigenvectors. At least two (non-zero) Eigenvectors\n    for M1.\n   \n1 1 1   0 0 5\n8\n© UCLES 2021 Page 14 of 15\n\n8(d) −A3 +7A2 −7A−15I=0 M1 Substitutes A into characteristic equation and multiplies\nthrough by A\n−1.\nAllow missing I.\nA −1 =\n1 (\n−A2 +7A−7I\n) A1 13 18 −28\n \n15 Accept with A= −4 −1 8 and\n \n  2 6 −5  \n 41 48 −80\n \nA2 = −32 −23 64 substituted to give\n \n  −8 0 17  \n43 78 −116\nA −1 = 1  4 9 −8  .\n \n15\n 22 42 −59  \n2\n© UCLES 2021 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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        "../answer-assets/9231_s21_ms_23-p15.png"
      ]
    },
    {
      "id": "9231-2021-mj-31-q01",
      "question_id": "9231-2021-mj-31-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 dv 10dt M1 Separate variables.\n=−\nv ( t+1 )2\n10 M1 A1 Attempt to integrate.\n2 v = + A\nt+1\nt =0, v=25, A=0 M1 Use correct initial condition.\n25 A1 CAO\nv=\n( t+1 )2\n5\n© UCLES 2021 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-31-q02",
      "question_id": "9231-2021-mj-31-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 ↑ Rcosθ=mg B1\nmv2 B1\n→ Rsinθ=\nr\nr =asinθ B1\n( )\n8cosθ=3 1−( cosθ)2 M1 Quadratic equation in cosθ .\n1 A1\ncosθ=\n3\n2 A1\nx= a\n3\n6\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-31-q03",
      "question_id": "9231-2021-mj-31-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) kmg ( x−a ) 4a B1 AG. Shown convincingly.\nUse Hooke’s Law: 4mg = leading to k =\na\nx−a\n1\n\n3(b) Gain in KE + gain in EPE = loss in GPE B1 One correct EPE term seen.\n1 ×6m× ga + 1 kmg     x+ a −a  2 −( x−a )2  =6mg× a M A 1 1 A Al l l l t 3 e r ty m p s e c s o o r f r e e c n t e . rgy included in energy equation.\n2 9 2 a   3   3\n \nSimplify and substitute for k from part (a) M1\nObtain linear equation in x and a M1\n5 A1 (k = 6)\nx = a\n3\n6\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-31-q04",
      "question_id": "9231-2021-mj-31-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) Centre of mass M1\nVolume from base of\ncylinder\n1 kh\nCone πr2kh h+\n3 4\nh\nCylinder πr2h\n2\nCombined πr2h   1 k +1   x\n3 \nTake moments about base: A1 2 terms correct.\nπr2h   1 k +1   x = 1 πr2kh  h+ kh  + πr2h h A1 All terms correct.\n3  3  4  2\n( )\nh k2 + 4k +6 A1 AG. Shown convincingly.\nx =\n4 ( 3+ k )\n4\n\n4(b) r M1\ntanθ=\nx\n4 6h ( k +3 ) M1 4\n= Equate to and simplify to quadratic.\n( )\n3 h k2 +4k +6 3\n2k2 −k −15=0\nk=3 A1 CAO. No other solutions.\n3\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p09.png"
      ]
    },
    {
      "id": "9231-2021-mj-31-q05",
      "question_id": "9231-2021-mj-31-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5 mu2 B1\nT −mgcosθ=\nA a\nmag B1\nT +mgcosθ=\nB a\nmu2  mag M1 Use given relationship and combine.\nT A = 7T B so\nmgcosθ+ =7\n\n−mgcosθ+\n\na  a \nu2 =ag ( 7−8cosθ)\nEnergy: 1 mu2 − 1 mag =mg ( acosθ+acosθ) M1 A1 Energy equation.\n2 2\n( )\nSo\nu2=ag 4cosθ+1\nEquate expressions for u2 M1\n1 A1 CAO\ncosθ= , θ =60°\n2\nu= 3ga A1 CAO\n8\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p10.png"
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    },
    {
      "id": "9231-2021-mj-31-q06",
      "question_id": "9231-2021-mj-31-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Along line of centres, speeds v and v M1 Momentum (condone missing masses).\n1 2\nmv +mv = mucosα−mucosβ\n1 2\n( )\nv −v =eu cosβ+ cosα M1 Restitution.\n2 1\nBoth correct, masses seen. A1\nv = 0 so A has no speed along line of centres: moves A1 AG.\n1\nperpendicular to line of centres\n4\n\n6(b) 1 M1 Both components.\n(v = u cosα=ucosβ )\n2 2\n( )\nKE of B after collision is\n1\nm v 2\n+( usinβ)2\n2 2\nKE of A after collision =\n1\nm\n( usinα)2\n2\n3 M1\nAdd both KEs and equate to mu2\n4\nSimplify to equation in sinα M1\nsinα=\n1\n,\nα=45° A1\n2\n4\n© UCLES 2021 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p11.png"
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    },
    {
      "id": "9231-2021-mj-31-q07",
      "question_id": "9231-2021-mj-31-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) At greatest height 0=100sinθ− gt M1\nt = 8 A1\nTherefore times at height H are t=3 (and t =13) B1\n1 M1\nSubstitute into H =100sinθt− gt2\n2\nH =195 A1\nAlternative method to question 7(a)\n1 M1\n↑H =100sinθt− gt2\n2\nAnd H =100sinθ( t+10 )− 1 g ( t+10 )2 A1\n2\nSubtract: 1000sinθ= 1 g ( 20t+100 ) M1\n2\nt=3 B1\nH =195 A1\n© UCLES 2021 Page 12 of 13\n\n7(a) Alternative method to question 7(a)\n1 B1\n↑H =100sinθt− gt2\n2\nM1 A1\n(100sinθ)2−2gH\nDifference between roots =\n1\ng\n2\nEquate to 10 and rearrange to find H M1\nH =195 A1\n5\n\n7(b) Time to required point = 15 s B1\n( )\n↑v=100sinθ−10×15 =−70 B1 Both components.\n→v=100cosθ=60\nMagnitude = 92.2 B1\nAngle below horizontal = tan -1 (70/60) = 49.4° B1\n4\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
        12,
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_31-p12.png",
        "../answer-assets/9231_s21_ms_31-p13.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q01",
      "question_id": "9231-2021-mj-32-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 dv 10dt M1 Separate variables.\n=−\nv ( t+1 )2\n10 M1 A1 Attempt to integrate.\n2 v = + A\nt+1\nt =0, v=25, A=0 M1 Use correct initial condition.\n25 A1 CAO\nv=\n( t+1 )2\n5\n© UCLES 2021 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q02",
      "question_id": "9231-2021-mj-32-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 ↑ Rcosθ=mg B1\nmv2 B1\n→ Rsinθ=\nr\nr =asinθ B1\n( )\n8cosθ=3 1−( cosθ)2 M1 Quadratic equation in cosθ .\n1 A1\ncosθ=\n3\n2 A1\nx= a\n3\n6\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q03",
      "question_id": "9231-2021-mj-32-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) kmg ( x−a ) 4a B1 AG. Shown convincingly.\nUse Hooke’s Law: 4mg = leading to k =\na\nx−a\n1\n\n3(b) Gain in KE + gain in EPE = loss in GPE B1 One correct EPE term seen.\n1 ×6m× ga + 1 kmg     x+ a −a  2 −( x−a )2  =6mg× a M A 1 1 A Al l l l t 3 e r ty m p s e c s o o r f r e e c n t e . rgy included in energy equation.\n2 9 2 a   3   3\n \nSimplify and substitute for k from part (a) M1\nObtain linear equation in x and a M1\n5 A1 (k = 6)\nx = a\n3\n6\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q04",
      "question_id": "9231-2021-mj-32-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) Centre of mass M1\nVolume from base of\ncylinder\n1 kh\nCone πr2kh h+\n3 4\nh\nCylinder πr2h\n2\nCombined πr2h   1 k +1   x\n3 \nTake moments about base: A1 2 terms correct.\nπr2h   1 k +1   x = 1 πr2kh  h+ kh  + πr2h h A1 All terms correct.\n3  3  4  2\n( )\nh k2 + 4k +6 A1 AG. Shown convincingly.\nx =\n4 ( 3+ k )\n4\n\n4(b) r M1\ntanθ=\nx\n4 6h ( k +3 ) M1 4\n= Equate to and simplify to quadratic.\n( )\n3 h k2 +4k +6 3\n2k2 −k −15=0\nk=3 A1 CAO. No other solutions.\n3\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p09.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q05",
      "question_id": "9231-2021-mj-32-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5 mu2 B1\nT −mgcosθ=\nA a\nmag B1\nT +mgcosθ=\nB a\nmu2  mag M1 Use given relationship and combine.\nT A = 7T B so\nmgcosθ+ =7\n\n−mgcosθ+\n\na  a \nu2 =ag ( 7−8cosθ)\nEnergy: 1 mu2 − 1 mag =mg ( acosθ+acosθ) M1 A1 Energy equation.\n2 2\n( )\nSo\nu2=ag 4cosθ+1\nEquate expressions for u2 M1\n1 A1 CAO\ncosθ= , θ =60°\n2\nu= 3ga A1 CAO\n8\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p10.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q06",
      "question_id": "9231-2021-mj-32-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Along line of centres, speeds v and v M1 Momentum (condone missing masses).\n1 2\nmv +mv = mucosα−mucosβ\n1 2\n( )\nv −v =eu cosβ+ cosα M1 Restitution.\n2 1\nBoth correct, masses seen. A1\nv = 0 so A has no speed along line of centres: moves A1 AG.\n1\nperpendicular to line of centres\n4\n\n6(b) 1 M1 Both components.\n(v = u cosα=ucosβ )\n2 2\n( )\nKE of B after collision is\n1\nm v 2\n+( usinβ)2\n2 2\nKE of A after collision =\n1\nm\n( usinα)2\n2\n3 M1\nAdd both KEs and equate to mu2\n4\nSimplify to equation in sinα M1\nsinα=\n1\n,\nα=45° A1\n2\n4\n© UCLES 2021 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p11.png"
      ]
    },
    {
      "id": "9231-2021-mj-32-q07",
      "question_id": "9231-2021-mj-32-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) At greatest height 0=100sinθ− gt M1\nt = 8 A1\nTherefore times at height H are t=3 (and t =13) B1\n1 M1\nSubstitute into H =100sinθt− gt2\n2\nH =195 A1\nAlternative method to question 7(a)\n1 M1\n↑H =100sinθt− gt2\n2\nAnd H =100sinθ( t+10 )− 1 g ( t+10 )2 A1\n2\nSubtract: 1000sinθ= 1 g ( 20t+100 ) M1\n2\nt=3 B1\nH =195 A1\n© UCLES 2021 Page 12 of 13\n\n7(a) Alternative method to question 7(a)\n1 B1\n↑H =100sinθt− gt2\n2\nM1 A1\n(100sinθ)2−2gH\nDifference between roots =\n1\ng\n2\nEquate to 10 and rearrange to find H M1\nH =195 A1\n5\n\n7(b) Time to required point = 15 s B1\n( )\n↑v=100sinθ−10×15 =−70 B1 Both components.\n→v=100cosθ=60\nMagnitude = 92.2 B1\nAngle below horizontal = tan -1 (70/60) = 49.4° B1\n4\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_32-p12.png",
        "../answer-assets/9231_s21_ms_32-p13.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q01",
      "question_id": "9231-2021-mj-33-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 Centre of mass B1 All distances correct.\nArea\nfrom DB\nABD 24a2 – a ABCD can be split in other ways, for example ADC and ABC.\nBCD 48a2 2a\nCombined 72a2 x\nTaking moments about DB: M1 Moments equation with masses in correct ratio.\n72a2x = 24a2 ×−a + 48a2 ×2a\nOR\nTaking moments about A:\n72a2x = 24a2×2a+48a2 ×5a\nOR\nTaking moments about G:\n24a 2( x+a )= 48a2 ×( 2a−x )\nx =a A1 CWO\nAlternative method for question 1\n6a−3a B1 One calculation.\nADC: distance of centre of mass from BD = =a\n3\n6a−3a\nABC: distance of centre of mass from BD = =a\n3\nSecond calculation or statement about symmetry M1\nx =a A1\n3\n© UCLES 2021 Page 6 of 16",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q02",
      "question_id": "9231-2021-mj-33-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1\n×\n36x2\n(=\n22.5x2) B1 EPE correct.\n2 0.8\nLoss in GPE + loss in KE = gain in EPE *M1 Energy equation with only GPE, EPE and KE terms, allow sign\n1 1 36x2 errors, allow missing g for M1 only, weight must be resolved\nx mgsinα+ m×2= ×\n2 2 0.8 (allow sin or cos).\nAll terms correct A1\n\n\n6 x+ 1 = 9 x2 \n\n[leading to45x2 −24x−4 = 0] DM1 Simplify to 3-term quadratic and attempt to solve.\n5 5 4 \n( )( )\n3x−2 15x+2 =0 A1\n2\nx= only\n3\n5\n© UCLES 2021 Page 7 of 16",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q03",
      "question_id": "9231-2021-mj-33-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3(a) For B: T =mg B1 May be embedded.\nFor A: R+Tcosθ=3mg M1 All 3 terms, allow sign errors, allow sin/cos mix.\n3 A1\nUse given R to obtain cosθ=\n5\n3\n\n3(b) 3mv2 M1 May be seen in part (a), allow sin/cos mix.\nTsinθ=\nr\nr= ARsinθ B1 Or equivalent.\n3a 1 A1\n[Combine to give AR= , so] BR= a\n4 4\n3\n© UCLES 2021 Page 8 of 16",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q04",
      "question_id": "9231-2021-mj-33-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 2 *M1 Energy equation, with 2 KE terms and a two-part GPE term,\n1 mu2 − 1 m   u  =mg ( acosθ+asinθ) allow cos/sin mix.\n2 2 2\n3 ( ) A1\nu2 =2ag cosθ+sinθ\n4\n2 B1 N2L\nu\nm\n \nAt B, tension in string is zero, so\nmgsinθ=\n2\na\n(u2=4agsinθ\n)\nu2 DM1\nEliminate\ntanθ=2 OE A1\nmu2 B1 N2L\nAt A, T −mgcosθ=\na\n9 5 M1 A1 Substitute to find T.\nT = mg ( = 4.02mg)\n5\nAlternative method for question 4\n2 *M1 Energy equation, with 2 KE terms and a two-part GPE term,\n1 mu2 − 1 m   u  =mg ( acosθ+asinθ) allow cos/sin mix.\n2 2 2\n3 u2 =2ag ( cosθ+sinθ) A1\n4\n© UCLES 2021 Page 9 of 16\n\n4 2 B1 N2L\nu\nm\n \nAt B, tension in string is zero, so\nmgsinθ=\n2\na\n(u2=4agsinθ)\nEliminate θ : u2 = 8ag 5 DM1 A1\n5\nmu2 B1 N2L\nAt A, T −mgcosθ=\na\n9 5 M1 A1 Substitute to find T.\nT = mg ( = 4.02mg)\n5\n8\n© UCLES 2021 Page 10 of 16",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p09.png",
        "../answer-assets/9231_s21_ms_33-p10.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q05",
      "question_id": "9231-2021-mj-33-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) dv B1 Use of SUVAT implies 0 marks.\nm =−mg−2mv\ndt\nN2L, must include .\nm\n( ) ( )\nln 5+v =−2t +A M1 Separate variables and integrate 3-term N2L, condone omission\nof constant.\n( )\nln 5+v =−2t +A A1 FT FT only sign error in N2L.\nt=0, v=20, A=ln25 M1 Use correct initial condition.\n2t =ln   5 2 + 5 v    , e2t = 5 2 + 5 v M1 Remove all logs.\nv = 25e−2t −5 A1\nAlternative method for question 5(a)\ndv B1 N2L, must include .\nm =−mg−2mv m\ndt\ndv M1\n+2v=−g : Integrating factor = e2t\ndt\n( )\nd ve2t M1 Integrate both sides, condone omission of constant.\n=−ge2t, ve2t =− g e2t (+A )\ndt 2\ng A1 FT FT only sign error in N2L.\nve2t =− e2t +A\n2\n© UCLES 2021 Page 11 of 16\n\n5(a) t =0, v=20, A=25 M1 Use correct initial condition.\ng A1\nve2t =− e2t +25, v=25e −2t −5\n2\n6\n\n5(b) x=− 25 e −2t −5t (+B ) M1 Use of SUVAT implies 0 marks.\nIntegrate their expression from part (a).\n2\nt=0, x=0, B=\n25 A1 FT\nFT only expressions of the form\nv=Pekt +Q\nfor P, Q non-zero.\n2\n25( )\nx= 1−e −2t −5t\n2\n2\n\n5(c) 1 M1 Use of SUVAT in part (a) or part (b) implies 0 marks.\nGreatest height when v = 0, so t = 0.8047…or ln5\nFind value of t, may be embedded.\n2\nx = 5.98 m A1 CWO\n2\n© UCLES 2021 Page 12 of 16",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p11.png",
        "../answer-assets/9231_s21_ms_33-p12.png"
      ]
    },
    {
      "id": "9231-2021-mj-33-q06",
      "question_id": "9231-2021-mj-33-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) Let velocities of A and B along line of centres after collision be v M1 Momentum, must include m, allow cos/sin mix.\n1\nand v .\n2\nmv + kmv = mucosθ .\n1 2\n1 M1 Restitution, consistent signs, correct way up.\nv −v = ucosθ\n2 1 3\n4ucosθ A1 AG shown convincingly.\nSolve: v 2 = 3 ( 1+k )\n3\n\n6(b) ( 3−k ) ucosθ B1 Or equivalent, may be unsimplified.\nv =\n1 3 ( 1+k )\nUse velocity of A with both components. B1 v2+( usinθ)2\nseen.\n1\n1 kmv 2 + 1 m ( v2 +( usinθ)2 ) = 3 × 1 mu2 M1 KE after = 30% KE before (all terms present).\n2 2 2 1 10 2 M0 if incorrect masses.\nSubstitute from part (a) and for θ . M1 Eliminate trigonometric terms, must be KE equation, in terms of\nk only.\n( 3−k )2 +16k=2 ( 1+k )2\n, k2 −6k −7=0\nM1 Obtain simplified quadratic equation in k .\nk =7 A1\n6\n© UCLES 2021 Page 13 of 16",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p13.png"
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    },
    {
      "id": "9231-2021-mj-33-q07",
      "question_id": "9231-2021-mj-33-q07",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) R2 M1\ny = 0 in trajectory equation: Rtanθ−g =0\n2u2\n( cosθ)2\n(\nR=)2u2sinθcosθ\nonly\nA1 A\nu\nn\n2\ny\nsi\ne\nn\nq\n2\nu\nθ\nivale\n2\nn\nu\nt s\n2\ni\nt\nn\na\ng\nn\nl\nθ\ne term expression, for example:\ng\n, , at least one intermediate line of\ng g sec2θ\nworking, not just quoting a result.\nSC B1 using SUVAT.\n2\n\n7(b) u2sinθcosθ M1 Or use SUVAT.\nx=their\nand substitute in trajectory equation.\ng\nu2 ( sinθ)2 A1 Single term.\nH =\n2g\n2\n\n7(c) 4H B1 AG\nUse R= and simplify: tanθ= 3, θ = 60°\n3\n1\n© UCLES 2021 Page 14 of 16\n\n7(d) dy gx M1 Differentiate with respect to x.\n=tanθ−\ndx u2\n( cosθ)2\nx M1 dy\ntanθ− =±1 used Use =±1 as limiting case.\n4 ( cosθ)2 dx\nx = 3 +1, x = 3 −1 A1\n3 −1< x < 3 +1 A1 Strict inequality, exact values.\nAlternative method for question 7(d)\n1 dy M1 Differentiate with respect to x.\ny= 3x− x2, = 3−x\n2 dx\ndy M1 dy\n=±1 used Use =±1 as limiting case.\ndx dx\nx = 3 +1 , x = 3 −1 A1\n3 −1< x < 3 +1 A1 Strict inequality, exact values.\n© UCLES 2021 Page 15 of 16\n\n7(d) Alternative method for question 7(d)\nv =±v M1 Used, both cases considered.\nWhen moving at 45° to horizontal,\nx y\nv = 40 cosθ , v = 40sinθ−10t M1\nx y\n1 ( ) 1 ( )\nt= 30− 10 , t= 30+ 10\n10 10\nx = 3 +1 , x = 3 −1 A1\n3 −1< x < 3 +1 A1 Strict inequality, exact values.\n4\n© UCLES 2021 Page 16 of 16",
      "source_pages": [
        14,
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_33-p14.png",
        "../answer-assets/9231_s21_ms_33-p15.png",
        "../answer-assets/9231_s21_ms_33-p16.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q01",
      "question_id": "9231-2021-mj-41-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 56.56 B1 Accept unsimplified.\nx = =8.08\n7\ns2 = 1 457.275− 56.562   = 0.045033  = 1351  M1 Accept unsimplified.\n \n6  7   30000\n8.08−8.22 M1 A1 1.74 – 1.75\nt = = − 1.745\ns2\n7\nTabular value 1.440 M1 Comparison with 1.440 and correct FT conclusion.\n1.745 > 1.440 , so reject H\n0\nThere is insufficient evidence to support the hypothesis that the A1 Correct conclusion, in context, following correct work. Level\n(population) mean is 8.22 OR sufficient evidence to suggest mean of uncertainty in language is used.\nis less than 8.22 OR sufficient evidence to suggest μ<8.22 OE\n6\n\n1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed.\n1\n© UCLES 2021 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q02",
      "question_id": "9231-2021-mj-41-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 M1 At least 3 correct to 4sf.\n72 (61.425) 42 (44.46) 52 (54.405) 68 (73.71)\n33 (43.575) 34 (31.54) 41 (38.595) 58 (52.29)\nCalculates E values (in brackets above) A1 All correct to 4sf (if 6.037 not found).\n( O−E )2 M1 Correct formula used.\nTest value:  = 1.8206 + 0.1361 + 0.1063 +0.4423 +\nE\n2.5664 + 0.1919 + 0.1499 + 0.6235\n= 6.037 A1 Correct to 3 sf.\nH : (driving) test success is independent of instructor B1\n0\nTabular value: 3 df 10% = 6.251 M1 Compare with correct tabular value and conclusion.\n6.04 < 6.251\nAccept H\n0\nSufficient evidence to suggest that (driving) test success is A1 Correct conclusion, in context, following correct work. Level\nindependent of instructor of uncertainty in language is used.\nAllow ±1 in the third significant figure of test value.\n7\n© UCLES 2021 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q03",
      "question_id": "9231-2021-mj-41-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)\n\n9 2\nx1.5 dx\nB1\nf\n(\nx\n)= 2\nx, correct expression as integrand.\n81 81\n0\n2 2 M1 Integrate, FT only on their PDF expression.\n× x2.5\n81 5\n12 A1\n=2.4\n5\n3\n\n3(b) 9 M1\n\n2\nx2\ndx−(\ntheir a\n)2\n81\n0\n2 93 6 A1\n× −2.42 =\n81 3 25\n2\n\n3(c) 1 M1\nG(y) = y6\n81\n2 M1\ng(y) = y5\n27\nFully correct including ‘for 0  y  39 and 0 otherwise’ A1 Condone 2.08\n3\n© UCLES 2021 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q04",
      "question_id": "9231-2021-mj-41-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 s 2 = 1  1596− 2822   = 0.11265  = 138  M1 One correct, unsimplified.\nx 49  50    1225  \ns y 2 = 5 1 9    1808− 3 6 2 0 82    = 0.25311    = 8 2 8 2 5 4   A1 Both correct to 3sf.\n0.11265 0.25312 M1 A1 May be implied.\ns2 = + =0.006471…\n50 60\n282 328 M1\n−\nz= 50 60\ntheir s\n[z] = 2.155 A1 Tolerance 2.15 – 2.16\nComparison with tabular value 1.96 OE M1 Comparison with 1.96 OE and conclusion.\n2.155 > 1.96\nReject H\n0\nSufficient evidence to reject population means are equal OR A1 Correct conclusion following correct work. Level of\nsufficient evidence to accept population means are not equal OR uncertainty in language is used.\nsufficient evidence to suggest μ ≠μ OE\nA B\n8\n© UCLES 2021 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p09.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q05",
      "question_id": "9231-2021-mj-41-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) Underlying distribution or population of differences is unknown B1 Not known to be normal.\nUnderlying distributions or population of scores for both X and Y Condone ‘scores for X and Y cannot be assumed to be normally\nare unknown distributed’.\n1\n\n5(b) H :difference of population medians=0 B1 Correct hypotheses, allow m but not μ or mean.\n0\nH :difference of population median≠0 B1 ‘population’ included.\n1\nDiff: 4 −1 −21 3 −7 −9 −5 8 −10 −12 −11 M1 Differences, allow one error.\nRank: 3 −1 −11 2 −5 −7 −4 6 −8 −10 −9 M1 Signed ranks, allow one sign error.\n[ ]\nQ= −55, P=11 A1 P = 11\nCritical tabular value = 13 M1 Comparison with 13 and correct ft conclusion.\n‘11’ < 13 so reject H\n0\nSufficient evidence of a preference for one of the uniforms OE A1 Correct conclusion, in context, following correct work, except\npossibly second B1.\nLevel of uncertainty in language is used.\n7\n© UCLES 2021 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p10.png"
      ]
    },
    {
      "id": "9231-2021-mj-41-q06",
      "question_id": "9231-2021-mj-41-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 3 1 M1 At least 2 correct probabilities used in a polynomial.\nP(3R, 2R, 1R, 0R) = , , , 0\n5 5 5\nG ( t )= 1 t+ 3 t2 + 1 t3 A1\nX 5 5 5\n2\n\n6(b) 1 6 9 M1 One correct probability and all three adding to 1 used in a\nP(2H, 1H, 0H) = , ,\npolynomial.\n16 16 16\nG ( t )= 9 + 6 t+ 1 t2 A1\nY 16 16 16\n2\n\n6(c) Attempt to multiply results from part (a) and part (b) M1\nObtain polynomial M1\n9 33 28 9 1 A1 Accept exact equivalent decimals.\nt+ t2 + t3 + t4 + t5\n80 80 80 80 80\n3\n\n6(d) E(Z) = G ' ( 1 )= 1 ( 9+66+84+36+5 ) M1 Differentiate and put t = 1.\nZ 80\n200 A1\n= =2.5\n80\nG ' ' ( 1 )= 1 ( 66+168+108+20 )=4.525= 362 M1\nZ 80   80  \n© UCLES 2021 Page 11 of 12\n\n6(d)\nVar (Z) = G\n′′(\n1\n)+G ′(\n1\n)− (\nG'\n(\n1\n))2 M1 Use result.\n2 A1\n362 200 200 31\n+ −   = 0.775=  \n80 80  80  40\n5\n© UCLES 2021 Page 12 of 12",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_41-p11.png",
        "../answer-assets/9231_s21_ms_41-p12.png"
      ]
    },
    {
      "id": "9231-2021-mj-42-q01",
      "question_id": "9231-2021-mj-42-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 56.56 B1 Accept unsimplified.\nx = =8.08\n7\ns2 = 1 457.275− 56.562   = 0.045033  = 1351  M1 Accept unsimplified.\n \n6  7   30000\n8.08−8.22 M1 A1 1.74 – 1.75\nt = = − 1.745\ns2\n7\nTabular value 1.440 M1 Comparison with 1.440 and correct FT conclusion.\n1.745 > 1.440 , so reject H\n0\nThere is insufficient evidence to support the hypothesis that the A1 Correct conclusion, in context, following correct work. Level\n(population) mean is 8.22 OR sufficient evidence to suggest mean of uncertainty in language is used.\nis less than 8.22 OR sufficient evidence to suggest μ<8.22 OE\n6\n\n1(b) Underlying distribution is normal/ population is normal B1 X is normal distributed.\n1\n© UCLES 2021 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_42-p06.png"
      ]
    },
    {
      "id": "9231-2021-mj-42-q02",
      "question_id": "9231-2021-mj-42-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 M1 At least 3 correct to 4sf.\n72 (61.425) 42 (44.46) 52 (54.405) 68 (73.71)\n33 (43.575) 34 (31.54) 41 (38.595) 58 (52.29)\nCalculates E values (in brackets above) A1 All correct to 4sf (if 6.037 not found).\n( O−E )2 M1 Correct formula used.\nTest value:  = 1.8206 + 0.1361 + 0.1063 +0.4423 +\nE\n2.5664 + 0.1919 + 0.1499 + 0.6235\n= 6.037 A1 Correct to 3 sf.\nH : (driving) test success is independent of instructor B1\n0\nTabular value: 3 df 10% = 6.251 M1 Compare with correct tabular value and conclusion.\n6.04 < 6.251\nAccept H\n0\nSufficient evidence to suggest that (driving) test success is A1 Correct conclusion, in context, following correct work. Level\nindependent of instructor of uncertainty in language is used.\nAllow ±1 in the third significant figure of test value.\n7\n© UCLES 2021 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_42-p07.png"
      ]
    },
    {
      "id": "9231-2021-mj-42-q03",
      "question_id": "9231-2021-mj-42-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)\n\n9 2\nx1.5 dx\nB1\nf\n(\nx\n)= 2\nx, correct expression as integrand.\n81 81\n0\n2 2 M1 Integrate, FT only on their PDF expression.\n× x2.5\n81 5\n12 A1\n=2.4\n5\n3\n\n3(b) 9 M1\n\n2\nx2\ndx−(\ntheir a\n)2\n81\n0\n2 93 6 A1\n× −2.42 =\n81 3 25\n2\n\n3(c) 1 M1\nG(y) = y6\n81\n2 M1\ng(y) = y5\n27\nFully correct including ‘for 0  y  39 and 0 otherwise’ A1 Condone 2.08\n3\n© UCLES 2021 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_42-p08.png"
      ]
    },
    {
      "id": "9231-2021-mj-42-q04",
      "question_id": "9231-2021-mj-42-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 s 2 = 1  1596− 2822   = 0.11265  = 138  M1 One correct, unsimplified.\nx 49  50    1225  \ns y 2 = 5 1 9    1808− 3 6 2 0 82    = 0.25311    = 8 2 8 2 5 4   A1 Both correct to 3sf.\n0.11265 0.25312 M1 A1 May be implied.\ns2 = + =0.006471…\n50 60\n282 328 M1\n−\nz= 50 60\ntheir s\n[z] = 2.155 A1 Tolerance 2.15 – 2.16\nComparison with tabular value 1.96 OE M1 Comparison with 1.96 OE and conclusion.\n2.155 > 1.96\nReject H\n0\nSufficient evidence to reject population means are equal OR A1 Correct conclusion following correct work. Level of\nsufficient evidence to accept population means are not equal OR uncertainty in language is used.\nsufficient evidence to suggest μ ≠μ OE\nA B\n8\n© UCLES 2021 Page 9 of 12",
      "source_pages": [
        9
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-42-q05",
      "question_id": "9231-2021-mj-42-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) Underlying distribution or population of differences is unknown B1 Not known to be normal.\nUnderlying distributions or population of scores for both X and Y Condone ‘scores for X and Y cannot be assumed to be normally\nare unknown distributed’.\n1\n\n5(b) H :difference of population medians=0 B1 Correct hypotheses, allow m but not μ or mean.\n0\nH :difference of population median≠0 B1 ‘population’ included.\n1\nDiff: 4 −1 −21 3 −7 −9 −5 8 −10 −12 −11 M1 Differences, allow one error.\nRank: 3 −1 −11 2 −5 −7 −4 6 −8 −10 −9 M1 Signed ranks, allow one sign error.\n[ ]\nQ= −55, P=11 A1 P = 11\nCritical tabular value = 13 M1 Comparison with 13 and correct ft conclusion.\n‘11’ < 13 so reject H\n0\nSufficient evidence of a preference for one of the uniforms OE A1 Correct conclusion, in context, following correct work, except\npossibly second B1.\nLevel of uncertainty in language is used.\n7\n© UCLES 2021 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_42-p10.png"
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    },
    {
      "id": "9231-2021-mj-42-q06",
      "question_id": "9231-2021-mj-42-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 3 1 M1 At least 2 correct probabilities used in a polynomial.\nP(3R, 2R, 1R, 0R) = , , , 0\n5 5 5\nG ( t )= 1 t+ 3 t2 + 1 t3 A1\nX 5 5 5\n2\n\n6(b) 1 6 9 M1 One correct probability and all three adding to 1 used in a\nP(2H, 1H, 0H) = , ,\npolynomial.\n16 16 16\nG ( t )= 9 + 6 t+ 1 t2 A1\nY 16 16 16\n2\n\n6(c) Attempt to multiply results from part (a) and part (b) M1\nObtain polynomial M1\n9 33 28 9 1 A1 Accept exact equivalent decimals.\nt+ t2 + t3 + t4 + t5\n80 80 80 80 80\n3\n\n6(d) E(Z) = G ' ( 1 )= 1 ( 9+66+84+36+5 ) M1 Differentiate and put t = 1.\nZ 80\n200 A1\n= =2.5\n80\nG ' ' ( 1 )= 1 ( 66+168+108+20 )=4.525= 362 M1\nZ 80   80  \n© UCLES 2021 Page 11 of 12\n\n6(d)\nVar (Z) = G\n′′(\n1\n)+G ′(\n1\n)− (\nG'\n(\n1\n))2 M1 Use result.\n2 A1\n362 200 200 31\n+ −   = 0.775=  \n80 80  80  40\n5\n© UCLES 2021 Page 12 of 12",
      "source_pages": [
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        12
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-43-q01",
      "question_id": "9231-2021-mj-43-q01",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 1 1522  M1 13\ns2 = 2313−\n\n=0.28889 , accept unsimplified.\n45\n9 10\n \nH : μ = 14.8 H : μ > 14.8 B1 If μ not used, ‘population mean’ required.\n0 1\nMust see 14.8, must be = and >.\n152 M1 Using unbiased estimate.\n−14.8\n[ t= ] 10 p-value is 0.0215.\ns/ 10\n[ ]\nt= 2.35 A1 Rounds to 2.35.\nTabular value = 1.833 M1 Comparison with 1.833 and correct FT conclusion.\n‘2.35’ > 1.833\nReject H\n0\nSufficient evidence to accept Farmer B’s claim oe A1 Correct conclusion, in context, following correct work. Level\nof uncertainty in language is used.\nAllow ±1 difference in third significant figure of their t for this\nmark.\n6\n© UCLES 2021 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-43-q02",
      "question_id": "9231-2021-mj-43-q02",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 H : difference in population medians = 0 (m = m ) B1 Correct hypotheses, allow m . Do not allow μ or mean.\n0 a b d\nH : difference in population medians ≠ 0 (m ≠ m ) B1 ‘population’ included.\n1 a b\nDiff: 12 −14 −1 3 −13 8 −5 15 6 M1 Allow one error.\nRank: 6 −8 −1 2 −7 5 −3 9 4 M1 Allow one error.\nSmaller sum 19 A1 (Other sum is 26.)\nCritical tabular value = 8 M1 Comparison with 8 and correct FT conclusion.\n‘19’ > 8 so accept H\n0\nThere is insufficient evidence that marks differ A1 Correct conclusion, in context, following correct work, except\npossibly the hypotheses. Level of uncertainty in language is\nused.\n7\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_43-p07.png"
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    },
    {
      "id": "9231-2021-mj-43-q03",
      "question_id": "9231-2021-mj-43-q03",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3 89 67.2 B1 Implied by 0.1 in the CI formula.\nx = =1.78 y = =1.68\n50 40\nσ x 2 = 4 1 9    159.4− 8 5 9 0 2    = 0.02    = 5 1 0    M1 One variance correct unsimplified.\nσ y 2 = 3 1 9    113.1− 67 4 . 0 22    = 0.0052308    = 3 1 2 7 50    A1 Both correctly calculated.\n0.02 0.0052308  69  M1 Unsimplified expression must be seen, 40 and 50 in correct\nσ2 = + =   places.\n50 40 130000\n[σ2 =] 0.0005308 A1 May be implied by correct final answer.\nCI : 1.78 – 1.68 ± z × 0.0005308 M1 Correct form for CI with a z-value.\nwith z = 1.96 A1\n0.10 ± 0.0452 or [0.0548, 0.145] A1 Allow either form.\n8 SC Assuming equal variances award B1M0A0, M1A1 (for use\nof pooled variance formula, gives 0.0135), M1A1A0.\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s21_ms_43-p08.png"
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    },
    {
      "id": "9231-2021-mj-43-q04",
      "question_id": "9231-2021-mj-43-q04",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 4t M1 kt\nG’(t) = Differentiate to obtain OE.\n(\n3−2t2\n)2 (\n3−2t2\n)2\nso E(X) = G’(1) = 4 A1 CAO WWW\nG’’(t) =\n12+24t2\nor 4\n(\n3−2t2\n)−2\n+32t2\n(\n3−2t2\n)−3 M1 OE. Differentiate, allow only numerical or sign slips.\n(\n3−2t2\n)3\n( ( ))2 M1 Substitute their values into correct formula, dependent on\nVar(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16\nattempt at G' '(t).\n[Var(X)] = 24 A1 CAO WWW\n5\n\n4(b) G ( t )= 1 = ( 3−2t2 )−1 M1 Expand given expression or give expression for the term in t4 .\nX 3−2t2\n1 2 4 \n= 1+ t2 + t4 +….\n3 3 9 \nP(X = 4) = their coefficient of t4 M1 Found from legitimate method.\n4 A1 WWW\n[P(X = 4) =]\n27\n3\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2021-mj-43-q05",
      "question_id": "9231-2021-mj-43-q05",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) (Mean of distribution = 2.4) M1 Allow 1.52.\nVariance = 2.25\nMean approx. equal to variance (so Poisson might be suitable) A1 Must have ‘approximately’ oe, not equality.\nMust have 2.25.\n2\n\n5(b) Po(2.5) leads to frequencies: M1 At least 3 correct to 3sf.\n14.77[5], 36.938, 46.173, 38.477, 24.048, 12.024, 5.010, 2.554 A1 All correct to 4sf.\nCombine last two values: 7.564 M1 FT their table values if final figure is less than 5.\n( O−E )2 M1 Apply correct formula to their frequencies.\n =2.6227+0.4198+0.00065+1.4529+0.04570\nE\n+1.3148+0.7845\n6.64 A1 Correct to 3sf.\n\n5(b) H : distribution fits the data B1 Must mention data and distribution .\n0\nTabular value, 5 degrees of freedom, is 9.236 M1 Compare with correct tabular value and conclusion.\n‘6.64’ < 9.236, so accept H\n0\nInsufficient evidence to show that data does not follow a Poisson A1 Correct conclusion, in context, following correct work.\ndistribution Level of uncertainty in language is used.\nAllow ±1 difference in third significant figure.\n8\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-May-June/ms/9231_s21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-mj-43-q06",
      "question_id": "9231-2021-mj-43-q06",
      "subject": "9231",
      "year": 2021,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 M1 Integrate both parts (all powers of x increase by 1), allow M1 if\nx 0x<1,\n no constant term.\n8\nF(x)=  A1 Both parts correct, with domains.\n1  1  1\n  8x− x2  − 1  x  8. Any equivalent form, for example: 2 x− 1 x2 − 1 .\n28 2  7 7 56 7\nF(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1\nAlternative method for question 6(a)\n1 M1 Integrate both parts (all powers of x increase by 1), allow M1 if\nx 0x<1,\n no constant term.\n8\nF(x)=\n A1 Both parts correct, with domains.\n  1− 5 1 6 ( 8−x )2 1  x  8. Any equivalent form, for example: 1 ( 16x−x2 −8 ) .\n56\nF(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1\n3\n\n6(b) 5 1  1  1 5 *M1 Correct method and attempt to simplify, ft their part (a).\nF(a) = leading to 8a− a2\n\n− =\n7 28 2  7 7\na2 −16a+48=0 DM1 Obtain quadratic and solve.\n[ ]\na=4 a=12\na = 4 A1 Correct single answer, WWW.\n3\n© UCLES 2021 Page 11 of 13\n\n6(c) 1 M1 Substitute x = y3 into both parts of their F(x) from part (a).\ny3 0  y <1,\n A1 Correct functions in terms of y, any equivalent form.\n8\nG(y)=\n\n1 ( ) 1\n 16y3 − y6 − 1 y 2.\n56 7\n0  y < 1 and 1  y  2seen in CDF or PDF A1 Condone  instead of <.\n3 M1 Differentiate their G(y), all powers of y decrease by 1.\ny2 0  y< 1,\n A1 Must include ‘otherwise’.\n8\n\n 3 ( )\ng(y)=  8y2 − y5 1  y<2,\n28\n\n\n0 otherwise.\n\n\n5\n© UCLES 2021 Page 12 of 13\n\n6(c) Alternative method for question 6(c)\n1 1 −2 M1\ny= x3, dy= x 3dx, dx=3y2dy\n3\n0x<1, f ( x ) dx= 1 dx= 3 y2dy M1\n8 8\ng\n(\ny\n)= 3\ny2\nA1\n8\n1x8, f ( x ) dx= 1 ( 8−x ) dx= 1 ( 8−y3 ) 3y2dy A1\n28 28\ng ( y )= 3 ( 8y2 −y5 )\n28\n0  y < 1 and 1  y  2 seen in CDF or PDF A1\n5\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
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        12,
        13
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-may-june/9231_s21_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s21_ms_43-p13.png"
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    },
    {
      "id": "9231-2021-on-11-q01",
      "question_id": "9231-2021-on-11-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1\nb=−(α+β+γ)=−3\nB1\n5=32 −2(αβ+βγ+γα) M1 A1 Uses formula for sum of squares.\nc=2 A1\n6−3(5)+2(3)+3d =0 M1 Uses original equation or formula for sum of cubes.\n[Equation is x3 −3x2 +2x+1=0 ] A1\nd =1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_11-p06.png"
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    },
    {
      "id": "9231-2021-on-11-q02",
      "question_id": "9231-2021-on-11-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) n ( ) M1 A1 Expands and substitutes formulae.\n r3+3r2 +2r = 1n2(n+1)2 +1n(n+1)(2n+1)+n(n+1)\nAt least two formulae used for M1.\n4 2\nr=1\n( )\n= 1n(n+1) n2 +n+4n+2+4 = 1n(n+1)(n+2)(n+3) A1\n4 4\n3\n© UCLES 2021 Page 6 of 14\n\n2(b) 1 1 1 1 M1 A1 Finds partial fractions.\n= − +\nr(r+1)(r+2) 2r r+1 2(r+2)\nn 1 M1 A1 Shows enough terms for cancelation to be clear.\n = 1 f(1)− f(2)+1 f(3)\nr(r+1)(r+2) 2 2\nr=1\n+1 f(2)− f(3)+1 f(4) 1\n2 2 Where f(x)=\n+1 f(3)− f(4)+1 f(5) r\n2 2\n\n+1 f(n−1)− f(n)+1 f(n+1)\n2 2\n+1 f(n)− f(n+1)+1 f(n+2)\n2 2\n1 1 1 A1 ISW\n= − +\n4 2 ( n+1 ) 2 ( n+2 )\n5\n\n2(c) 1 B1 FT FT from their answer to part (b).\n4\n1\n© UCLES 2021 Page 7 of 14",
      "source_pages": [
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        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_11-p07.png"
      ]
    },
    {
      "id": "9231-2021-on-11-q03",
      "question_id": "9231-2021-on-11-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) a =23 leading to lna =3ln2 so true when n=2. B1 Checks base case. Must be n =2 and exact.\n2 2\nAssume that lna ≥3k−1ln2. B1 States inductive hypothesis.\nk\n 1  M1 Applies result given in part (a).\nlna k+1 =3lna k + a  >3lna k\n \nk\n≥ 3k ln 2 M1 A1 Applies inductive hypothesis and writes in required\nform.\nSo true when n=k+1. By induction, true for all integers n ≥ 2. A1 States conclusion, all previous marks must be gained.\n6\n\n3(b)  1  M1 Applies results given in parts (a) and (b).\n3lna +  −lna >2lna\nn a n n\n \nn\n>2×3n−1ln2=3n−1ln4 A1 AG\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
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    },
    {
      "id": "9231-2021-on-11-q04",
      "question_id": "9231-2021-on-11-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) One-way stretch [scale factor 3 parallel to the x-axis] followed by a rotation, B2 Award B1 if given in the wrong order.\n[anticlockwise centred at the origin, through an angle θ.]\n2\n© UCLES 2021 Page 8 of 14\n\n4(b) 3cosθ −sinθ  B1\nM=\n \n3sinθ cosθ\n\n3cosθ −sinθ x 3xcosθ− ysinθ  B1 x X\n   =   Transforms   to  .\n3sinθ cosθ y 3xsinθ+ ycosθ  y Y \n3xsinθ+mxcosθ=3mxcosθ−m2xsinθ M1 A1 Uses y=mx and Y =mX .\nm2sinθ−2mcosθ+3sinθ=0 A1\n4cos2θ−12sin2θ=0 M1 A1 Sets discriminant equal to 0.\n1 M1 Uses an appropriate trigonometric identity.\ntanθ=± or\n3\n( )\n4cos2θ−12 1−cos2θ =0 leading to 16cos2θ−12=0 leading to\ncosθ=± 3\n2\nθ= 1π and θ= 5π A1 Allow 30° and 150°\n6 6\n9\n© UCLES 2021 Page 9 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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        "../answer-assets/9231_w21_ms_11-p09.png"
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    },
    {
      "id": "9231-2021-on-11-q05",
      "question_id": "9231-2021-on-11-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k  −3 M1 A1 Finds vector perpendicular to Π.\n \n1 0 1 = 2\n \n \n2 3 0 3\n \n−3(−2)+2(3)+3(3)=21 M1 Substitutes point on Π.\n−3x+2y+3z=21 A1\n4\n\n5(b)  2  B1 Forms general point on line (given as a single vector).\n \n−3\n \n 5+t \n\n−3(2)+2(−3)+3(5+t)=21 leading to t=6 M1 Substitutes into the equation forΠ.\n 2  A1\n \n−3\n \n \n11\n3\n\n5(c) 0  −3 M1 Uses dot product of k and their normal.\n 0    2  = 1 22cosαleading to cosα= 3 A1 FT\n   \n    22\n1 3\n   \nAcute angle between l and Π is 90−α=39.8 ° A1\n3\n© UCLES 2021 Page 10 of 14\n\n5(d) 0  2   −2 B1 Finds direction vector from P to plane.\n     \n0 − −3 = 3\n     \n     \n7  5   2 \n −2  −3 M1 A1 Uses dot product of their direction and normal vectors.\n1     18\n3  2 = =3.84\n   \n22 22\n   \n2 3\n   \n3",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2021-on-11-q06",
      "question_id": "9231-2021-on-11-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) 2 ( c(c)−(1+s)s )=0 M1 dr\nFinds and sets equal to 0.\ndθ\n1−s−2s2 =0 A1\nsinθ=−1, sinθ= 1 leading to θ=1π M1 A1 Solves quadratic in sinθ.\n2 6\n( ) ( )\n3 3, 1π A1 Allow 2.60, 1π 3sf\n2 6 6\n5\n\n6(b) B1 π\nCorrect shape. Tangential to θ= at pole, r strictly\n2\nπ\nincreasing to then decreasing.\n6\nB1 Correct position. All in first quadrant.\nθ=0\n2\n© UCLES 2021 Page 11 of 14\n\n6(c) 1 1 2 π 4c2(1+s)2 dθ=2 1 2 π c2 +2sc2 +s2c2 dθ M1 A1 Uses 1\n2\nr2 dθ with correct limits.\n2\n0 0\n2 1 2 π 1 ( cos2θ+1 )+2sinθcos2θ+1sin22θ dθ M1 Uses double angle formulae on first and last terms.\n2 4\n0\n1π A1 Obtains integrable form.\n2 cos2θ+1+4sinθcos2θ+ 1 −1cos4θ dθ\n4 4\n0\n5θ+1sin2θ−4cos3θ− 1 sin4θ 1 2 π M1 Integrates.\n4 2 3 16  0\n5π+4 A1\n8 3\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2021-on-11-q07",
      "question_id": "9231-2021-on-11-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=1, x=−1 B1 Vertical asymptotes.\ny =0 B1 Horizontal asymptote.\n2\n\n7(b) dy (4−4x2)(4)−(4x+5)(−8x) M1 dy\n= Finds .\ndx ( 4−4x2 ) 2 dx\n16x2 +40x+16=0 M1 Sets equal to 0 and forms equation.\n( ) ( )\n−2,1 , −1,1 A1 A1\n4 2\n4\n© UCLES 2021 Page 12 of 14\n\n7(c) B1 Axes and asymptotes.\nB1 Correct shape and position.\nx\n4x(cid:3397)5\ny(cid:3404)\n4-4x(cid:2870)\n( )\n(−5,0), 0,5 B1 States coordinates of intersections with axes.\n4 4\n3\n© UCLES 2021 Page 13 of 14\n1-(cid:3404)x 1(cid:3404)x\n\n7(d) B1 FT FT from sketch in part (c).\nB1 Correct shape at infinity.\n4x(cid:3397)5\ny(cid:3404)(cid:3628) (cid:3628)\n4-4x(cid:2870)\nx\n4x+5 5 4x+5 5 M2 Finds critical points, award M1 for each case.\n= or =−\n4−4x2 4 4−4x2 4\n5x2 +4x=0 or 5x2 −4x−10=0\nx=−4, x=0 or x= 2−3 6, x= 2+3 6 A1 A0 if –1.07, 1.87\n5 5 5 5 5\n2−3 6<x<−4, 0<x< 2+3 6 , x≠±1 A1 FT Condone exclusion of x=±1 from the range.\n5 5 5 5 5\n6\n© UCLES 2021 Page 14 of 14\n1-(cid:3404)x\n1(cid:3404)x",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2021-on-12-q01",
      "question_id": "9231-2021-on-12-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "1(a) Enlargement, scale factor 6. B1\n1\n\n1(b) detA=6−8=−2 M1 Finds detA.\nArea\n=2×13=26cm2 A1\n2\n\n1(c)  2 −4 M1 Finds A −1.\nA −1 =−1  \n2  −2 3 \n 2 −4  −6 12 A1 AEF\nB=−3\n\n=\n  Could be solved by equations.\n −2 3   6 −9\n2\n\n1(d) 3 4x 3x+4y B1 X\n   =   Finds  \n2 2y 2x+2y Y \n3x+4y= x 2x+4y=0 M1 x X \n leading to  Uses   =   to form simultaneous equations.\n2x+2y= y 2x+ y=0 y Y \ny=−2xleading to −6x=0 leading tox=0, y=0 M1 A1 2 4\nSolves equations or states thatdet  ≠0, AG.\n2 1\n4\n© UCLES 2021 Page 6 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-12-q02",
      "question_id": "9231-2021-on-12-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 dy M1 A1 Differentiates once using the product rule.\n=axeax +eax =(ax+1)eax so true when n=1.\ndx\ndky ( ) B1 States inductive hypothesis.\nAssume that = akx+kak−1 eax.\ndxk\ndk+1y ( ) ( ) ( ) M1 A1 Differentiates kth derivative.\n=a akx+kak−1 eax +eax ak = ak+1+(k+1)ak eax\ndxk+1\nSo true when n=k +1. By induction, true for all positive integers n. A1 States conclusion.\n6\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-12-q03",
      "question_id": "9231-2021-on-12-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) n B1 Separates logarithms into correct form using a\n( lnr−2ln(r+1)+ln(r+2) )\ndifference.\nr=1 Or as logarithm of product.\nln1−2ln2+ln3 M1 A1 Shows enough terms to make cancellation clear.\nln2−2ln3+ln4\nln3−2ln4+ln5\n\nln(n−1)−2lnn+ln(n+1)\nlnn−2ln(n+1)+ln(n+2)\nn+2 A1 AG\n[ln1]−ln2−ln(n+1)+ln(n+2)=ln .\n2(n+1)\n4\n© UCLES 2021 Page 7 of 16\n\n3(b) S =−ln2 B1 States sum to infinity. AEF\nS n −S =ln    n n + +1 2   <0.01 leading to n+2<e0.01(n+1) M1 Forms inequality\nLeast value of n is 99 A1 CAO\n3\nQuestion Answer Marks Guidance",
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2021-on-12-q04",
      "question_id": "9231-2021-on-12-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) (−2)2 −2(3) M1 Uses formula for sum of squares.\n−2 A1\n2\n\n4(b) α3+β3+γ3 =−2(−2)−3(−2)−3(3) M1 Uses original equation or formula for sum of cubes.\n1 A1 AG\n2\n© UCLES 2021 Page 8 of 16\n\n4(c) (α+r )3 =α3 +3α2r+3αr2 +r3 B1 Expands.\nn ( ) n ( ) M1 A1 Collects like terms and uses results from parts (a) and\n (α+r )3 +(β+r )3 +(γ+r )3 = 1+3(−2)r+3(−2)r2 +3r3\n(b).\nr=1 r=1\n( ) ( )\nn−6 1n(n+1) −6 1n(n+1)(2n+1) + 3n2(n+1)2 M1 Applies formulae from MF19.\n2 6 4\nn−3n(n+1)−n(n+1)(2n+1)+ 3n2(n+1)2\n4\nn+1n(n+1) (−12−4(2n+1)+3n(n+1) ) M1 A1 Simplifies.\n4\n( )\nn+1n(n+1) 3n2 −5n−16\n4\n6\n© UCLES 2021 Page 9 of 16",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2021-on-12-q05",
      "question_id": "9231-2021-on-12-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Correct symmetrical shape, closed loop.\n1\n\n5(b) Line l parallel to initial line and correct side of pole. B1\n2=3sinθ+2sin2θ M1\nForms quadratic in sinθ. Or in r r=3+\n4\nr\nsinθ= 1 M1 Solves for sinθ.\n2\n( )\n4,1π A1 SC1 For finding both angles correctly\n6\n( )\n4,5π\nA1\n6\n5\n© UCLES 2021 Page 10 of 16\n\n5(c) 2×1 π 6 ( 3+2sinθ)2 dθ M1 Finds the part of the required area enclosed by the\n2 −π curved outer edge and two line segments from the pole.\n2\nLimits must be correct.\n π 6 9+12sinθ+2 ( 1−cos2θ) dθ M1 Uses double angle formula and integrates.\n−π\n2\n[ 11θ−12cosθ−sin2θ ] 6 π = 22π−13 3 A1 A1\n−π 3 2\n2\n22π−13 3+(4cosπ)×2 M1 Adds area of triangle.\n3 2 6\n22π− 5 3 A1\n3 2\n6\n© UCLES 2021 Page 11 of 16",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2021-on-12-q06",
      "question_id": "9231-2021-on-12-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x=3 B1 States vertical asymptote.\n9 M1 A1 Finds oblique asymptote.\ny=x+3+ leading toy=x+3\nx−3\n3\n\n6(b) yx−3y=x2 leading tox2 −yx+3y=0 M1 A1 Forms quadratic in x.\ny2 −4(3y)<0 leading to y2 −12y<0 M1 Uses that discriminant is negative.\n0< y<12 A1 AG\n4\n© UCLES 2021 Page 12 of 16\n\n6(c) B1 Axes and asymptotes.\nB1 Branches correct.\n𝑥(cid:2870)\n𝑦 (cid:3404)\n𝑥(cid:3398)3\n2\n© UCLES 2021 Page 13 of 16\n3\n(cid:3404)\n𝑥\n\n6(d)(i) B1 FT FT from sketch in (c).\n𝑥(cid:2870)\n𝑦 (cid:3404) (cid:4708) (cid:4708)\n𝑥(cid:3398)3 B1 Correct shape at infinity.\nB1 Correct shape of y= x −3.\nB1 Correct intercepts with axes (may be seen on graph).\nx\n4\n\n6(d)(ii) c ⩽ –3 B1\n1\n© UCLES 2021 Page 14 of 16\n3\n(cid:3404)\n𝑥",
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    },
    {
      "id": "9231-2021-on-12-q07",
      "question_id": "9231-2021-on-12-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 17,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k  2   1  M1 A1 Finds common perpendicular.\n   \nn= 2 2 0 = −2 ~ −1\n   \n0 −1 1  −2   −1 \n 1  A1\n \nr. −1 =0\n \n\n\n−1 \n\n3\n\n7(b) 1 1 B1 Divides by magnitude of the normal to Π\n=\n0.267\n12 +32 +22 14\n1\n\n7(c)  1   1  M1 Takes dot product of normal vectors.\n −1    −3  = 3 14cosαleading tocosα= 6 A1 FT\n   \n3 14\n\n\n−1 \n\n\n\n−2 \n\n22.2° A1 Accept 0.388 radians. Mark final answer.\n3\n© UCLES 2021 Page 15 of 16\n\n7(d) i j k  −20  5  M1 A1 Finds direction of common perpendicular.\n   \n2 1 −7 = 12 ~ −3\n   \n−2 −3 1   −4     1  \n\n 2  2−2μ  2−2μ−2λ  M1 A1 Finds PQ.\n  \n     \nOP=λ 1 ,OQ= 2−3μ leading toPQ= 2−3μ−λ\n     \n\n\n−7 \n\n\n\nμ \n\n\n\nμ+7λ \n\n\n2−2μ−2λ   −2 2−2μ−2λ   5  M1 Uses that dot product of PQ with line direction is zero,\n        \n2−3μ−λ  −3 =0 or 2−3μ−λ =k −3\n        or, alternatively, PQ is a multiple of the common\n  μ+7λ     1     μ+7λ     1   perpendicular (parameter k not 1).\n14μ+14λ=10 A1 Deduces one equation.\n2−2μ−2λ   2  A1 Deduces second equation.\n   \n2−3μ−λ  1 =014μ+54λ=6\n   \n\n μ+7λ \n\n\n\n−7 \n\n 2  M1 A1 Solves for λ and substitutes into OP.\nλ=− 1 leading to O  P  =− 1  1 \n \n10 10\n\n\n−7 \n\n −0.2  5  B1 FT FT using their common perpendicular.\n   \nr= −0.1 +k −3\n   \n   \n 0.7   1 \n10\n© UCLES 2021 Page 16 of 16",
      "source_pages": [
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_12-p15.png",
        "../answer-assets/9231_w21_ms_12-p16.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q01",
      "question_id": "9231-2021-on-13-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1\nb=−(α+β+γ)=−3\nB1\n5=32 −2(αβ+βγ+γα) M1 A1 Uses formula for sum of squares.\nc=2 A1\n6−3(5)+2(3)+3d =0 M1 Uses original equation or formula for sum of cubes.\n[Equation is x3 −3x2 +2x+1=0 ] A1\nd =1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q02",
      "question_id": "9231-2021-on-13-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) n ( ) M1 A1 Expands and substitutes formulae.\n r3+3r2 +2r = 1n2(n+1)2 +1n(n+1)(2n+1)+n(n+1)\nAt least two formulae used for M1.\n4 2\nr=1\n( )\n= 1n(n+1) n2 +n+4n+2+4 = 1n(n+1)(n+2)(n+3) A1\n4 4\n3\n© UCLES 2021 Page 6 of 14\n\n2(b) 1 1 1 1 M1 A1 Finds partial fractions.\n= − +\nr(r+1)(r+2) 2r r+1 2(r+2)\nn 1 M1 A1 Shows enough terms for cancelation to be clear.\n = 1 f(1)− f(2)+1 f(3)\nr(r+1)(r+2) 2 2\nr=1\n+1 f(2)− f(3)+1 f(4) 1\n2 2 Where f(x)=\n+1 f(3)− f(4)+1 f(5) r\n2 2\n\n+1 f(n−1)− f(n)+1 f(n+1)\n2 2\n+1 f(n)− f(n+1)+1 f(n+2)\n2 2\n1 1 1 A1 ISW\n= − +\n4 2 ( n+1 ) 2 ( n+2 )\n5\n\n2(c) 1 B1 FT FT from their answer to part (b).\n4\n1\n© UCLES 2021 Page 7 of 14",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p06.png",
        "../answer-assets/9231_w21_ms_13-p07.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q03",
      "question_id": "9231-2021-on-13-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) a =23 leading to lna =3ln2 so true when n=2. B1 Checks base case. Must be n =2 and exact.\n2 2\nAssume that lna ≥3k−1ln2. B1 States inductive hypothesis.\nk\n 1  M1 Applies result given in part (a).\nlna k+1 =3lna k + a  >3lna k\n \nk\n≥ 3k ln 2 M1 A1 Applies inductive hypothesis and writes in required\nform.\nSo true when n=k+1. By induction, true for all integers n ≥ 2. A1 States conclusion, all previous marks must be gained.\n6\n\n3(b)  1  M1 Applies results given in parts (a) and (b).\n3lna +  −lna >2lna\nn a n n\n \nn\n>2×3n−1ln2=3n−1ln4 A1 AG\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p08.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q04",
      "question_id": "9231-2021-on-13-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) One-way stretch [scale factor 3 parallel to the x-axis] followed by a rotation, B2 Award B1 if given in the wrong order.\n[anticlockwise centred at the origin, through an angle θ.]\n2\n© UCLES 2021 Page 8 of 14\n\n4(b) 3cosθ −sinθ  B1\nM=\n \n3sinθ cosθ\n\n3cosθ −sinθ x 3xcosθ− ysinθ  B1 x X\n   =   Transforms   to  .\n3sinθ cosθ y 3xsinθ+ ycosθ  y Y \n3xsinθ+mxcosθ=3mxcosθ−m2xsinθ M1 A1 Uses y=mx and Y =mX .\nm2sinθ−2mcosθ+3sinθ=0 A1\n4cos2θ−12sin2θ=0 M1 A1 Sets discriminant equal to 0.\n1 M1 Uses an appropriate trigonometric identity.\ntanθ=± or\n3\n( )\n4cos2θ−12 1−cos2θ =0 leading to 16cos2θ−12=0 leading to\ncosθ=± 3\n2\nθ= 1π and θ= 5π A1 Allow 30° and 150°\n6 6\n9\n© UCLES 2021 Page 9 of 14",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p08.png",
        "../answer-assets/9231_w21_ms_13-p09.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q05",
      "question_id": "9231-2021-on-13-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k  −3 M1 A1 Finds vector perpendicular to Π.\n \n1 0 1 = 2\n \n \n2 3 0 3\n \n−3(−2)+2(3)+3(3)=21 M1 Substitutes point on Π.\n−3x+2y+3z=21 A1\n4\n\n5(b)  2  B1 Forms general point on line (given as a single vector).\n \n−3\n \n 5+t \n\n−3(2)+2(−3)+3(5+t)=21 leading to t=6 M1 Substitutes into the equation forΠ.\n 2  A1\n \n−3\n \n \n11\n3\n\n5(c) 0  −3 M1 Uses dot product of k and their normal.\n 0    2  = 1 22cosαleading to cosα= 3 A1 FT\n   \n    22\n1 3\n   \nAcute angle between l and Π is 90−α=39.8 ° A1\n3\n© UCLES 2021 Page 10 of 14\n\n5(d) 0  2   −2 B1 Finds direction vector from P to plane.\n     \n0 − −3 = 3\n     \n     \n7  5   2 \n −2  −3 M1 A1 Uses dot product of their direction and normal vectors.\n1     18\n3  2 = =3.84\n   \n22 22\n   \n2 3\n   \n3",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p10.png",
        "../answer-assets/9231_w21_ms_13-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q06",
      "question_id": "9231-2021-on-13-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) 2 ( c(c)−(1+s)s )=0 M1 dr\nFinds and sets equal to 0.\ndθ\n1−s−2s2 =0 A1\nsinθ=−1, sinθ= 1 leading to θ=1π M1 A1 Solves quadratic in sinθ.\n2 6\n( ) ( )\n3 3, 1π A1 Allow 2.60, 1π 3sf\n2 6 6\n5\n\n6(b) B1 π\nCorrect shape. Tangential to θ= at pole, r strictly\n2\nπ\nincreasing to then decreasing.\n6\nB1 Correct position. All in first quadrant.\nθ=0\n2\n© UCLES 2021 Page 11 of 14\n\n6(c) 1 1 2 π 4c2(1+s)2 dθ=2 1 2 π c2 +2sc2 +s2c2 dθ M1 A1 Uses 1\n2\nr2 dθ with correct limits.\n2\n0 0\n2 1 2 π 1 ( cos2θ+1 )+2sinθcos2θ+1sin22θ dθ M1 Uses double angle formulae on first and last terms.\n2 4\n0\n1π A1 Obtains integrable form.\n2 cos2θ+1+4sinθcos2θ+ 1 −1cos4θ dθ\n4 4\n0\n5θ+1sin2θ−4cos3θ− 1 sin4θ 1 2 π M1 Integrates.\n4 2 3 16  0\n5π+4 A1\n8 3\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p11.png",
        "../answer-assets/9231_w21_ms_13-p12.png"
      ]
    },
    {
      "id": "9231-2021-on-13-q07",
      "question_id": "9231-2021-on-13-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=1, x=−1 B1 Vertical asymptotes.\ny =0 B1 Horizontal asymptote.\n2\n\n7(b) dy (4−4x2)(4)−(4x+5)(−8x) M1 dy\n= Finds .\ndx ( 4−4x2 ) 2 dx\n16x2 +40x+16=0 M1 Sets equal to 0 and forms equation.\n( ) ( )\n−2,1 , −1,1 A1 A1\n4 2\n4\n© UCLES 2021 Page 12 of 14\n\n7(c) B1 Axes and asymptotes.\nB1 Correct shape and position.\nx\n4x(cid:3397)5\ny(cid:3404)\n4-4x(cid:2870)\n( )\n(−5,0), 0,5 B1 States coordinates of intersections with axes.\n4 4\n3\n© UCLES 2021 Page 13 of 14\n1-(cid:3404)x 1(cid:3404)x\n\n7(d) B1 FT FT from sketch in part (c).\nB1 Correct shape at infinity.\n4x(cid:3397)5\ny(cid:3404)(cid:3628) (cid:3628)\n4-4x(cid:2870)\nx\n4x+5 5 4x+5 5 M2 Finds critical points, award M1 for each case.\n= or =−\n4−4x2 4 4−4x2 4\n5x2 +4x=0 or 5x2 −4x−10=0\nx=−4, x=0 or x= 2−3 6, x= 2+3 6 A1 A0 if –1.07, 1.87\n5 5 5 5 5\n2−3 6<x<−4, 0<x< 2+3 6 , x≠±1 A1 FT Condone exclusion of x=±1 from the range.\n5 5 5 5 5\n6\n© UCLES 2021 Page 14 of 14\n1-(cid:3404)x\n1(cid:3404)x",
      "source_pages": [
        12,
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_13-p12.png",
        "../answer-assets/9231_w21_ms_13-p13.png",
        "../answer-assets/9231_w21_ms_13-p14.png"
      ]
    },
    {
      "id": "9231-2021-on-21-q01",
      "question_id": "9231-2021-on-21-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 ex tanx+sec2x B1 Finds first derivative.\n( )\nex tanx+2sec2x+2sec2xtanx M1 A1 Finds second derivative.\ny(0)=0, y'(0)=1, y''(0)=2 M1 Evaluates derivatives at x=0.\ny=x+x2 A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-21-q02",
      "question_id": "9231-2021-on-21-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 (λ+1 )(λ−1 )(λ−3 )=λ3−3λ2 −λ+3[=0] M1 A1 Finds characteristic equation.\nA3 −3A2 −A+3I=0A4 −3A3 −A2 +3A=0 M1 A1 Substitutes A and multiplies through by A.\n( )\nA4 =3 3A2 +A−3I +A2 −3A=10A2 −9I M1 A1 Substitutes A3.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-21-q03",
      "question_id": "9231-2021-on-21-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 3xy2y'+ y3 B1 Differentiates xy3.\n( )\n−4 x3y'+3x2y =0 B1 Differentiates x3y.\n( ) ( )\n3(−1)(1)y'+13−4 (−1)3y'+3(−1)2(1) =0 leading to y'=11 B1 Substitutes (−1,1), AG. y' 3xy2 −4x3 =12x2y− y3\n3\n© UCLES 2021 Page 6 of 13\n\n3(b) 3xy2 −4x3 y''+ y' 6xyy'+3y2 −12x2 B1 B1 Differentiates 3xy2 −4x3 y'.\n( )\n+3y2y'−12 x2y'+2xy =0 B1 Differentiates y3−12x2y.\n3xy2y''−4x3y''+6xy ( y' )2 +6y2y'−24x2y'−24xy=0\n( )( ) ( )( )\n3xy2 −4x3 12x2y'+24xy−3y2y' − 12x2y− y3 6xyy'+3y2 −12x2\ny''=\n(\n3xy2 −4x3\n)2\ny''+11(−75)+3(11)−12(9)=0 M1 Substitutes (–1, 1)\ny''=900 A1\n5\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2021-on-21-q04",
      "question_id": "9231-2021-on-21-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) N lnr ln2 N lnr M1 A1 Compares with sum of the areas of the rectangles.\n = +\nr=1 r2 4 r=3 r2 M1 for writing out sum, A1 for considering N lnr .\nr2\nr=3\nln2 Nlnx M1 Compares with integral.\n< + dx\n4 2 x2\nNlnx  lnx+1 N M1 A1 Finds integral.\n dx= −\n2 x2   x  \n2\nN lnr ln2  lnN +1 ln2+1 2+3ln2 1+lnN M1 A1 Inserts limits, AG. A1 requires second M1.\n < +\n\n− +\n\n= −\nr2 4  N 2  4 N\nr=1\n7\n\n4(b) N lnr N−1lnr lnN Nlnx lnN M1 A1 Compares with integral. Lower limit of 1 scores M0.\n = + > dx+\nr=1 r2 r=2 r2 N2 2 x2 N2 Accept  N lnr > N+1lnx .\nr2 2 x2\nr=1\nln2+1 lnN +1 lnN A1 ln2+1 ln(N +1)+1\n= − + Accept − .\n2 N N2 2 N +1\n3\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-21-q05",
      "question_id": "9231-2021-on-21-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5 m2 −2m+1=0m=1 M1 Auxiliary equation.\ny=ex ( Ax+B ) A1 Complementary function. Accept ‘y =’ missing.\ny= psinx+qcosx, y'= pcosx−qsinx, y''=−psinx−qcosx M1 A1 Particular integral and its derivatives.\n−p+2q+ p=0 −q−2p+q=4 M1 Substitutes and equates coefficients.\np=−2 q=0 A1\ny=ex ( Ax+B )−2sinx A1\ny'= Aex +ex ( Ax+B )−2cosx M1 Differentiates.\nB=−4 A+B−2=3 A=9 M1 A1 Forms simultaneous equations using initial conditions.\ny=ex ( 9x−4 )−2sinx A1\n11\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_21-p09.png"
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    },
    {
      "id": "9231-2021-on-21-q06",
      "question_id": "9231-2021-on-21-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) ( cos5θ+isin5θ)=( c+is )5 M1 Uses binomial theorem.\nsin5θ=s5 −10c2s3 +5c4s A1\ns5 −10 ( 1−s2 ) s3 +5 ( 1−s2 )2 s M1 Uses c2 =1−s2 or cot2θ=cosec2θ−1 after dividing\nnumerator and denominator by s5.\n16s5 −20s3 +5s A1\n1 cosec5θ M1 Divides simplified numerator and denominator by s5.\ncosec5θ= ×\n16s5 −20s3 +5s cosec5θ\ncosec5θ A1 AG\n5cosec4θ−20cosec2θ+16\n6\n\n6(b) ( ) x5 M1 Relates with equation in part (a).\nx5 =2 5x4 −20x2 +16 leading to =2\n5x4 −20x2 +16\ncosec5θ=2leading to sin5θ= 1 M1 Solves sin5θ= 1.\n2 2\n( )\nx=cosec 1 π A1 Gives one correct solution.\n30\n( ) ( ) ( ) ( )\ncosec 5 π ,cosec − 7 π ,cosec −11π ,cosec 13π A1 Gives four other solutions. Allow different values of q\n30 30 30 30\nas long as all five solutions are found.\n9.56, 2, –1.49, –1.09, 1.02\n4\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_21-p10.png"
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    },
    {
      "id": "9231-2021-on-21-q07",
      "question_id": "9231-2021-on-21-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) dy 1 x2 −x x2 −1 B1 Divides through by x2−1.\n+ y=\ndx x2 −1 x2 −1\n1 M1 A1 Finds integrating factor. M1 for correct form\n dx\ne\n√(x2−1) =ecosh−1x\n1\na dx\ne\n√(x2−1)\n, where a is a non-zero constant.\nx+ x2 −1 A1 AG\n4\n( ( ))\n\n7(b) d x M1 A1 Correct form on LHS and simplifies RHS.\ny x+ x2 −1 =\ndx x2 −1\n( )\nM1 A1 Integrates RHS. For M1, needs to integrate non-zero\ny x+ x2 −1 = x2 −1+C\nx\nmultiple of .\nx2 −1\n2= 3+C leading toC= 5 M1 Substitutes initial conditions.\n4 4\nx2 −1+ 5 M1 A1 Divides through by coefficient of y.\ny= 4\nx+ x2 −1\n7\n© UCLES 2021 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_21-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-21-q08",
      "question_id": "9231-2021-on-21-q08",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) coshA= 1 eA+e −A B1 Writes in exponential form\n2\n( ) ( )\n2cosh2 A= 1 e2A +2+e −2A = 1 e2A +e −2A +1=cosh2A+1 M1 A1 Expands, AG.\n2 2\n3\n\n8(b)(i) dx dy B1\n=4sinh2t+3 =3sinh2t−4\ndt dt\n( 4sinh2t+3 )2 +( 3sinh2t−4 )2 =25(sinh22t+1)=25cosh22t M1 A1 Expands and applies cosh2 A=sinh2 A+1\n2π 1 2 ( 2cosh2t+3t )( 5cosh2t ) dt  =π 1 2 20cosh22t+30tcosh2tdt  A1 Correct formula for surface area, AG.\n \n−1  −1 \n2 2\n4\n\n8(b)(ii) 20 1 2 cosh22tdt =10 1 2 cosh4t+1dt M1 Applies cosh2 A= 1\n2\n( cosh2A+1 ) .\n−1 −1\n2 2\n101sinh4t+t 1 2 =10 ( 1sinh2+1 ) M1 A1 Integrates.\n \n4 −1 2\n2\n30\n1\n2 tcosh2tdt=30\n\n 1\n[\ntsinh2t\n]1\n2 −1\n1\n2 sinh2tdt\n\n\nM1 A1 Integrates by parts.\n−1 2 2 −1 2 2 −1 2 \n15  tsinh2t−1cosh2t  1 2 =0 A1 Accept  1 2 tcosh2tdt =0 since tcosh2t is odd.\n2 −1\n2 −1\n2\n( ( ) )\n10π 1 e2 −e −2 +1 A1 64.82457...\n4\n© UCLES 2021 Page 12 of 13\n\n8(b)(ii) Alternative method for question 8(b)(ii)\n10π 1 2 ( 2cosh2t+3t ) cosh2tdt = M1 A1 I u nt = eg 2 r c a o te s s h b 2 y t + pa 3 r t t , s . u '=4sinh2t+3\n−1\n2\n10πcosh2tsinh2t+(3tsinh2t) 1 2 −5π 1 2 sinh2t ( 4sinh2t+3 ) dt\nv'=cosh2t, v= 1\n2\nsinh2t\n 2  −1 −1\n2 2\n10π  1sinh4t  1 2 −20π 1 2 sinh22tdt−  15π 1 2 sinh2tdt   = M1 A1 Applies 2sinh22t =cosh4t−1\n2 −1 −1  −1 \n2 2 2\n10π  1sinh4t  1 2 −10π 1 2 cosh4t−1dt−  15π 1 2 sinh2tdt  \n2 −1 −1  −1 \n2 2 2\n( ) 1 M1 A1 Integrates\nπ5sinh4t−10sinh4t+10t− 15cosh2t 2\n2 4 2  −1\n2\nπ5sinh4t+10t− ( 15cosh2t )  1 2 =π ( 5sinh2+10 ) =10π ( 1 ( e2 −e −2 ) +1 ) A1 64.82457...\n2 2  −1 4\n2\n7\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_21-p13.png"
      ]
    },
    {
      "id": "9231-2021-on-22-q01",
      "question_id": "9231-2021-on-22-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) sinh x2 +cosh x2 M1 Combines using correct power series. Or sensible\nattempt at all four derivatives of y.\n1+x2 +1x4 A1\n2\n2\n\n1(b) 1×4!=12 B1\n2\n1\n\n1(c)  1 21+x2 + 1x4 dx M1 Substitutes their power series, must be at least a+bx2.\n2\n0\n1 A1\n=x+1x3 + 1 x52 = 523\n 3 10  0 960\n2\n© UCLES 2021 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p06.png"
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    },
    {
      "id": "9231-2021-on-22-q02",
      "question_id": "9231-2021-on-22-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 e ln x4+5 =x4 +5 M1 A1 Finds integrating factor.\nd ( ( )) M1 Writes in correct form.\ny x4 +5 =6x5 +30x\ndx\ny(x4 +5)=x6 +15x2 +C A1\n6=16+C M1 Substitutes initial conditions.\nx6 +15x2 −10 M1 A1 Division through by coefficient of y.\ny=\nx4 +5\n7\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p07.png"
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    },
    {
      "id": "9231-2021-on-22-q03",
      "question_id": "9231-2021-on-22-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)  1 1−x2dx< 1 + 1 ( 1− 1 ) ++ 1 ( 1−(n−1)2 ) M1 A1 Forms the sum of the areas of the rectangles. Three\n0\nn n n2 n n2 terms, including last for A1. Allow use of sigma.\n1 n−1 (n−1)n(2n−1) M1 n−1\n1− r2 =1− Applies r2 =1(n−1)n(2n−1).\nn3 6n3 6\nr=1 r=1\n6n2 −(n−1)(2n−1) 4n2 +3n−1 A1 AG. [Can get this A1 without previous A1.]\n=\n6n2 6n2\n4\n© UCLES 2021 Page 7 of 14\n\n3(b)  1 1−x2dx> 1 ( 1− 1 ) + 1 ( 1−22 ) ++ 1 1− ( n−1 )2 M1 A1 Forms the sum of the areas of appropriate rectangles.\n0\nn n2 n n2 n n2\nThree terms, including last for A1. Allow \nn\n1\n(\n1− r2\n)\n.\nn n2\nr=1\nn−1 1 n−1 n−1 (n−1)n(2n−1) M1 n−1\n− r2 = − Applies r2 =1(n−1)n(2n−1).\nn n3 n 6n3 6\nr=1 r=1\n6n(n−1)−(n−1)(2n−1) 4n2 −3n−1 A1 4n2 +3n−1 1\n= SC: − scores 2/4.\n6n2 6n2 6n2 n\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2021-on-22-q04",
      "question_id": "9231-2021-on-22-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) ei2 5 kπ, k =0,1,2,3,4 B2 OE. B1 for 1 correct fifth root of unity.\nB2 for exactly 5 distinct, correct roots.\n2\n\n4(b) ( c+is )4 =c4 +4c3(is)+6c2(is)2 +4c(is)3+(is)4 M1 A1 Uses binomial expansion. Can also go RHS to LHS\nusing 2cosθ=z+z −1.\ncos4θ=c4 −6c2s2 +s4 =c4 −6c2(1−c2)+(1−c2)2 M1 Applies s2 =1−c2 or 2cosθ=z+z −1.\n8cos4θ−8cos2θ+1 A1 AG. Can get final A1 without first.\nSC: Using double angle formulae scores 1/4.\n4\n© UCLES 2021 Page 8 of 14\n\n4(c) 8x9 −8x7 +x5 −8x4 +8x2 −1=(x5 −1)(8x4 −8x2 +1) B1\ncos4θ=0 leading to4θ= 1(2k+1)π M1 A1 Solves cos4θ=0. A1 for 4θ= 1π,3π,5π,7π OE.\n2 2 2 2 2\ncos0,cos1π,cos3π,cos5π,cos7π A1 Gives exactly 5 distinct, real roots.\n8 8 8 8\nAccept 1,±cos1π,±cos3π.\n8 8\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_22-p09.png"
      ]
    },
    {
      "id": "9231-2021-on-22-q05",
      "question_id": "9231-2021-on-22-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) x=3−2t −2 +2t2 y =4+ 3t −2 −3t2 B1 Differentiates x and y with respect to t.\n2 2\n4t4 +12t2 + 4 − 12 +1+ 23 + 9t4 + 12 + 9 −12t2 M1 Expands x2 + y2. Accept 25+ 25 ( t −2 −t2 )2 .\n4\nt4 t2 2 4 t2 4t4\n25t4 + 25t −4 + 25 = 25 ( t2 +t −2 )2 M1 A1 Factorises x2 + y2. AG.\n4 4 2 4\n5 2 t2 +t −2 dt = 51t3 −t −1 2 = 85 M1 A1 Applies correct formula for arc length.\n2 1 23  1 12\n6\n© UCLES 2021 Page 9 of 14\n\n5(b) dy y 4+ 3t −2 B1 dy\n= = 2 Finds .\ndx x 3−2t −2 dx\n( )( ) ( )( )\nd d t     4 3 + −2 3 2 t t − − 2 2     = 3−2t −2 −3 ( t 3 − − 3 2 − t −2 4 )2 + 3 2 t −2 4t −3 B1 Differentiates d d y x with respect to t.\nd2y d 4+ 3t −2  dt −25t −3 M1 A1 Applies chain rule.\n=  2  × = =−25 when t =1.\ndx2 dt   3−2t −2   dx ( 3−2t −2 )3\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p09.png",
        "../answer-assets/9231_w21_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2021-on-22-q06",
      "question_id": "9231-2021-on-22-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) (λ−1 ) (λ−2)(λ+3)=λ3−7λ+6[=0] M1 A1 Finds characteristic equation.\nP2 −7I+6P −1=0 M1 Substitutes P and multiplies through by P −1.\n1 18 24 1 −3 −4 M1 A1\nP2 =  0 4 −6  leading toP −1 =  0 1 1 \n   2 \n 0 0 9     0 0 −1 3 \n5\n© UCLES 2021 Page 10 of 14\n\n6(b) 4 0 0 M1 Applies A=PDP −1.\nA=P  0 5 0  P −1\n \n \n0 0 6\n \n4 30 36 1 −3 −4 1 6 6 4 −12 −16 M1 A1 Multiplies two adjacent matrices.\n     \n0 10 36 0 1 1 or 0 2 6 0 5 5\n  2    2 \n 0 0 −18     0 0 −1 3     0 0 −3     0 0 −2  \n4 3 2  A1 A1 linked to first M1.\n \n0 5 −2\n \n \n0 0 6\n \n4\n\n6(c) 1 6  6  B1 Accept any non-zero multiples of these.\n     \nEigenvectors: 0 , 2 , 6\n     \n 0    0     −3  \nEigenvalues: 64, 125, 216 B1\n2\n© UCLES 2021 Page 11 of 14",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p10.png",
        "../answer-assets/9231_w21_ms_22-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-22-q07",
      "question_id": "9231-2021-on-22-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) dy dw B1 dw 1 dy 2\n=x2 +2xw = − y\ndx dx dx x2 dx x3\nd2y d2w dw B1 d2w 1 d2y 4 dy 6\n=x2 +4x +2w = − + y\ndx2 dx2 dx dx2 x2 dx2 x3 dx x4\nd2y dy d2w dw dw M1 Uses substitution to find w-x equation, AG.\n+4 +5y=x2 +4x +2w+4x2 +8xw+5x2w\ndx2 dx dx2 dx dx\nd2w ( )dw ( ) A1 AG\n5x2 + 4x2 +4x + 5x2 +8x+2 w=5x2 +4x+2\ndx2 dx\n4\n\n7(b) m2 +4m+5=0 leading tom=−2±i M1 Auxiliary equation.\ny=e −2x ( Acosx+Bsinx ) A1 Complimentary function. Allow “y =” missing. Don’t\nallow e.g. e −2x ( Acosx+Bisinx ) .\ny= px2 +qx+r , y'=2px+q, y''=2p B1 Particular integral and its derivatives.\n2p+8px+4q+5px2 +5qx+5r=5px2 +(8p+5q)x+(2p+4q+5r) M1 Substitutes and equates coefficients.\n5p=5 8p+5q=4 2p+4q+5r=2\np=1 q=−4 r = 16 A1\n5 25\nx2w=e −2x ( Acosx+Bsinx )+x2 −4x+16 M1 Substitutes for y and find w in terms of x.\n5 25\nw= ( xex )−2( Acosx+Bsinx )+1−(5x) −1+ ( 5x )−2 A1\n4 4\n7\n© UCLES 2021 Page 12 of 14",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p12.png"
      ]
    },
    {
      "id": "9231-2021-on-22-q08",
      "question_id": "9231-2021-on-22-q08",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) ex −e −x 2 B1 Writes in exponential form.\ntanhx= sechx=\nex +e −x ex +e −x\nex −e −x  2 ( ex +e −x )2 − ( ex −e −x )2 4 M1 Writes over common denominator.\n1−\n ex +e −x  \n=\n( ex +e −x )2\n=\n( ex +e −x )2\nsech2 x A1 AG\n3\n\n8(b) u2 du= 1tanh3x+C M1 A1 Uses substitution correctly. Allow by inspection.\n3\n2\n\n8(c) I = ln3 sechn−2 xsech2 xtanh2 xdx B1 Separates into correct structure.\nn\n0\n1sechn−2 xtanh3x ln3 +1(n−2) ln3 sechn−2 xtanh4 xdx M1 Uses integration by parts correctly.\n3  0 3 0\n1 ( 3 )n−2( 4 )3 + 1(n−2) ( I −I ) M1 A1 Uses 1−sech2x=tanh2x.\n3 5 5 3 n−2 n\n( 3+n−2 ) I = ( 3 )n−2( 4 )3 +(n−2)I leading to A1 AG\nn 5 5 n−2\n(n+1)I =\n(\n3\n)n−2(\n4\n)3\n+(n−2)I\nn 5 5 n−2\n5\n© UCLES 2021 Page 13 of 14\n\n8(d) I = 1 ( 4 )3 = 64 B1\n2 3 5 375\n(4+1)I =\n(\n3\n)2(\n4\n)3\n+2I leading to I =\n4928\n=0.105\nM1 A1 Applies reduction formula with n=4.\n4 5 5 2 4 46875\n3\n© UCLES 2021 Page 14 of 14",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_22-p13.png",
        "../answer-assets/9231_w21_ms_22-p14.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q01",
      "question_id": "9231-2021-on-23-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 ex tanx+sec2x B1 Finds first derivative.\n( )\nex tanx+2sec2x+2sec2xtanx M1 A1 Finds second derivative.\ny(0)=0, y'(0)=1, y''(0)=2 M1 Evaluates derivatives at x=0.\ny=x+x2 A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q02",
      "question_id": "9231-2021-on-23-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 (λ+1 )(λ−1 )(λ−3 )=λ3−3λ2 −λ+3[=0] M1 A1 Finds characteristic equation.\nA3 −3A2 −A+3I=0A4 −3A3 −A2 +3A=0 M1 A1 Substitutes A and multiplies through by A.\n( )\nA4 =3 3A2 +A−3I +A2 −3A=10A2 −9I M1 A1 Substitutes A3.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q03",
      "question_id": "9231-2021-on-23-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 3xy2y'+ y3 B1 Differentiates xy3.\n( )\n−4 x3y'+3x2y =0 B1 Differentiates x3y.\n( ) ( )\n3(−1)(1)y'+13−4 (−1)3y'+3(−1)2(1) =0 leading to y'=11 B1 Substitutes (−1,1), AG. y' 3xy2 −4x3 =12x2y− y3\n3\n© UCLES 2021 Page 6 of 13\n\n3(b) 3xy2 −4x3 y''+ y' 6xyy'+3y2 −12x2 B1 B1 Differentiates 3xy2 −4x3 y'.\n( )\n+3y2y'−12 x2y'+2xy =0 B1 Differentiates y3−12x2y.\n3xy2y''−4x3y''+6xy ( y' )2 +6y2y'−24x2y'−24xy=0\n( )( ) ( )( )\n3xy2 −4x3 12x2y'+24xy−3y2y' − 12x2y− y3 6xyy'+3y2 −12x2\ny''=\n(\n3xy2 −4x3\n)2\ny''+11(−75)+3(11)−12(9)=0 M1 Substitutes (–1, 1)\ny''=900 A1\n5\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p06.png",
        "../answer-assets/9231_w21_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q04",
      "question_id": "9231-2021-on-23-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) N lnr ln2 N lnr M1 A1 Compares with sum of the areas of the rectangles.\n = +\nr=1 r2 4 r=3 r2 M1 for writing out sum, A1 for considering N lnr .\nr2\nr=3\nln2 Nlnx M1 Compares with integral.\n< + dx\n4 2 x2\nNlnx  lnx+1 N M1 A1 Finds integral.\n dx= −\n2 x2   x  \n2\nN lnr ln2  lnN +1 ln2+1 2+3ln2 1+lnN M1 A1 Inserts limits, AG. A1 requires second M1.\n < +\n\n− +\n\n= −\nr2 4  N 2  4 N\nr=1\n7\n\n4(b) N lnr N−1lnr lnN Nlnx lnN M1 A1 Compares with integral. Lower limit of 1 scores M0.\n = + > dx+\nr=1 r2 r=2 r2 N2 2 x2 N2 Accept  N lnr > N+1lnx .\nr2 2 x2\nr=1\nln2+1 lnN +1 lnN A1 ln2+1 ln(N +1)+1\n= − + Accept − .\n2 N N2 2 N +1\n3\n© UCLES 2021 Page 8 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p08.png"
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    },
    {
      "id": "9231-2021-on-23-q05",
      "question_id": "9231-2021-on-23-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5 m2 −2m+1=0m=1 M1 Auxiliary equation.\ny=ex ( Ax+B ) A1 Complementary function. Accept ‘y =’ missing.\ny= psinx+qcosx, y'= pcosx−qsinx, y''=−psinx−qcosx M1 A1 Particular integral and its derivatives.\n−p+2q+ p=0 −q−2p+q=4 M1 Substitutes and equates coefficients.\np=−2 q=0 A1\ny=ex ( Ax+B )−2sinx A1\ny'= Aex +ex ( Ax+B )−2cosx M1 Differentiates.\nB=−4 A+B−2=3 A=9 M1 A1 Forms simultaneous equations using initial conditions.\ny=ex ( 9x−4 )−2sinx A1\n11\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p09.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q06",
      "question_id": "9231-2021-on-23-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) ( cos5θ+isin5θ)=( c+is )5 M1 Uses binomial theorem.\nsin5θ=s5 −10c2s3 +5c4s A1\ns5 −10 ( 1−s2 ) s3 +5 ( 1−s2 )2 s M1 Uses c2 =1−s2 or cot2θ=cosec2θ−1 after dividing\nnumerator and denominator by s5.\n16s5 −20s3 +5s A1\n1 cosec5θ M1 Divides simplified numerator and denominator by s5.\ncosec5θ= ×\n16s5 −20s3 +5s cosec5θ\ncosec5θ A1 AG\n5cosec4θ−20cosec2θ+16\n6\n\n6(b) ( ) x5 M1 Relates with equation in part (a).\nx5 =2 5x4 −20x2 +16 leading to =2\n5x4 −20x2 +16\ncosec5θ=2leading to sin5θ= 1 M1 Solves sin5θ= 1.\n2 2\n( )\nx=cosec 1 π A1 Gives one correct solution.\n30\n( ) ( ) ( ) ( )\ncosec 5 π ,cosec − 7 π ,cosec −11π ,cosec 13π A1 Gives four other solutions. Allow different values of q\n30 30 30 30\nas long as all five solutions are found.\n9.56, 2, –1.49, –1.09, 1.02\n4\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p10.png"
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    },
    {
      "id": "9231-2021-on-23-q07",
      "question_id": "9231-2021-on-23-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) dy 1 x2 −x x2 −1 B1 Divides through by x2−1.\n+ y=\ndx x2 −1 x2 −1\n1 M1 A1 Finds integrating factor. M1 for correct form\n dx\ne\n√(x2−1) =ecosh−1x\n1\na dx\ne\n√(x2−1)\n, where a is a non-zero constant.\nx+ x2 −1 A1 AG\n4\n( ( ))\n\n7(b) d x M1 A1 Correct form on LHS and simplifies RHS.\ny x+ x2 −1 =\ndx x2 −1\n( )\nM1 A1 Integrates RHS. For M1, needs to integrate non-zero\ny x+ x2 −1 = x2 −1+C\nx\nmultiple of .\nx2 −1\n2= 3+C leading toC= 5 M1 Substitutes initial conditions.\n4 4\nx2 −1+ 5 M1 A1 Divides through by coefficient of y.\ny= 4\nx+ x2 −1\n7\n© UCLES 2021 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_23-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-23-q08",
      "question_id": "9231-2021-on-23-q08",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) coshA= 1 eA+e −A B1 Writes in exponential form\n2\n( ) ( )\n2cosh2 A= 1 e2A +2+e −2A = 1 e2A +e −2A +1=cosh2A+1 M1 A1 Expands, AG.\n2 2\n3\n\n8(b)(i) dx dy B1\n=4sinh2t+3 =3sinh2t−4\ndt dt\n( 4sinh2t+3 )2 +( 3sinh2t−4 )2 =25(sinh22t+1)=25cosh22t M1 A1 Expands and applies cosh2 A=sinh2 A+1\n2π 1 2 ( 2cosh2t+3t )( 5cosh2t ) dt  =π 1 2 20cosh22t+30tcosh2tdt  A1 Correct formula for surface area, AG.\n \n−1  −1 \n2 2\n4\n\n8(b)(ii) 20 1 2 cosh22tdt =10 1 2 cosh4t+1dt M1 Applies cosh2 A= 1\n2\n( cosh2A+1 ) .\n−1 −1\n2 2\n101sinh4t+t 1 2 =10 ( 1sinh2+1 ) M1 A1 Integrates.\n \n4 −1 2\n2\n30\n1\n2 tcosh2tdt=30\n\n 1\n[\ntsinh2t\n]1\n2 −1\n1\n2 sinh2tdt\n\n\nM1 A1 Integrates by parts.\n−1 2 2 −1 2 2 −1 2 \n15  tsinh2t−1cosh2t  1 2 =0 A1 Accept  1 2 tcosh2tdt =0 since tcosh2t is odd.\n2 −1\n2 −1\n2\n( ( ) )\n10π 1 e2 −e −2 +1 A1 64.82457...\n4\n© UCLES 2021 Page 12 of 13\n\n8(b)(ii) Alternative method for question 8(b)(ii)\n10π 1 2 ( 2cosh2t+3t ) cosh2tdt = M1 A1 I u nt = eg 2 r c a o te s s h b 2 y t + pa 3 r t t , s . u '=4sinh2t+3\n−1\n2\n10πcosh2tsinh2t+(3tsinh2t) 1 2 −5π 1 2 sinh2t ( 4sinh2t+3 ) dt\nv'=cosh2t, v= 1\n2\nsinh2t\n 2  −1 −1\n2 2\n10π  1sinh4t  1 2 −20π 1 2 sinh22tdt−  15π 1 2 sinh2tdt   = M1 A1 Applies 2sinh22t =cosh4t−1\n2 −1 −1  −1 \n2 2 2\n10π  1sinh4t  1 2 −10π 1 2 cosh4t−1dt−  15π 1 2 sinh2tdt  \n2 −1 −1  −1 \n2 2 2\n( ) 1 M1 A1 Integrates\nπ5sinh4t−10sinh4t+10t− 15cosh2t 2\n2 4 2  −1\n2\nπ5sinh4t+10t− ( 15cosh2t )  1 2 =π ( 5sinh2+10 ) =10π ( 1 ( e2 −e −2 ) +1 ) A1 64.82457...\n2 2  −1 4\n2\n7\n© UCLES 2021 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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        "../answer-assets/9231_w21_ms_23-p13.png"
      ]
    },
    {
      "id": "9231-2021-on-31-q01",
      "question_id": "9231-2021-on-31-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 3mgx B1 Their tensions equated to obtain a quadratic equation,\nT =\nCAO.\na\n4mga B1\nT =\n3 ( a+x )\n9x2 +9ax−4a2 =0 leading to ( 3x−a )( 3x+4a )=0 M1\n1 A1\nx= a\n3\n4",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p06.png"
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    },
    {
      "id": "9231-2021-on-31-q02",
      "question_id": "9231-2021-on-31-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) Separate variables and integrate: M1 A1 Must include logs. Condone missing modulus.\ndv 1−2t2 \n=  dt so ln v =lnt−t2 +c\nv t\n \nv = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 A1 CAO.\n3\n© UCLES 2021 Page 6 of 12\n\n2(b) −Ate −t2 1−2t2\n( )\nM1 Substituting their answer to part (a) into given formula\na= =−Ae −t2 1−2t2\nt\nt =1, a=5 ( A=5e ) M1 Use initial condition.\n( )\nForce = 5me1−t2 2t2 −1 A1 Use N2L, correct work only.\nAlternative method for question 2(b)\n( )\nv 1−2t2 M1 Use initial condition.\na= substitute t=1, a=5 so v=−5 Use N2L, correct work only.\nt\nSubstituting in their answer to part (a) so ( A=5e ) M1\n( )\nForce = 5me1−t2 2t2 −1 A1\n3\n© UCLES 2021 Page 7 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p06.png",
        "../answer-assets/9231_w21_ms_31-p07.png"
      ]
    },
    {
      "id": "9231-2021-on-31-q03",
      "question_id": "9231-2021-on-31-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3\nLoss in EPE =\n1\n×\n12mge2\n−\n1\n×\n12mg\n×\n\n\ne−\na\n\n2 \n\n=\n2mg(\n6e−a\n)\n\nB1 E ither term correct.\n2 a 2 a  3  3 \n1 mga B1\nGain in KE = mv2 and Gain in GPE =\n2 3\nGain in KE + Gain in GPE = Loss in EPE M1 KE, GPE and at least one EPE term.\n1 mv2 + mga = 2mg( 6e−a ) A1 All terms correct.\n2 3 3\nSimplify to a linear equation in e. M1\n1 A1\ne= a\n2\n6\n© UCLES 2021 Page 8 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p08.png"
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    },
    {
      "id": "9231-2021-on-31-q04",
      "question_id": "9231-2021-on-31-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Attempt at moments with three terms.\nArea Centre of mass from AD\n3\nSquare 9a2 a\n2\n3\nCDF ah a\n2\n3 1\nBEC ah 3a− h\n2 3\nResulting AEFC 9a2 −3ah x\nTaking moments about AD: A1 Two terms correct.\n( )  3  3  3  1  A1 All correct.\n9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h \n 2  2  2  3 \n27a2 −12ah+h2\n\n9a−h A1 AEF\nx =\n6 ( 3a−h )  \n=\n6  \ny =x B1 By symmetry or equal to their x.\n5\n\n4(b) For equilibrium, x ⩽ 3a – h B1 Accept strict inequality.\n27a2 – 12ah + h2 ⩽ 6(3a – h)2\n27a2– 24ah + 5h2 ⩾ 0 M1 Homogeneous 3-term quadratic inequality.\n9 A1 CAO.\nh ⩽ a\n5\n3\n© UCLES 2021 Page 9 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p09.png"
      ]
    },
    {
      "id": "9231-2021-on-31-q05",
      "question_id": "9231-2021-on-31-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5 At A: ↑usinθ−8g →ucosθ M1 Both.\nusinθ−8g A1\ntanα=\nucosθ\nAt B: ↑usinθ−32g →ucosθ M1 Both.\nusinθ−32g A1\ntanβ=\nucosθ\nusinθ−8g usinθ−32g B1 Perpendicular directions, so tanα×tanβ=−1.\n× =−1\nucosθ ucosθ\nu2 −320u+25600=0 M1 Simplify to a quadratic in u.\nu=160 A1\n7\n© UCLES 2021 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p10.png"
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    },
    {
      "id": "9231-2021-on-31-q06",
      "question_id": "9231-2021-on-31-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) 5ag B1 N2L\nAt A T −mgcosθ=m×\nA a\nv2 B1 N2L\nAt B T +mgcosθ=m×\nB a\n1 1 M1 Energy equation with correct number of terms.\nm×5ag− mv2 =mga×2cosθ\n2 2\nv2 =5ag−4gacosθ A1 Accept multiplied by m and/or divided by a.\nUse ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ.\n2 A1 CAO\ncosθ=\n5\n6\n\n6(b) Greatest speed at lowest point M1 Energy equation including lowest point, correct number\n− 1 m×5ag+ 1 mV2 =mga×( 1−cosθ) of terms.\n2 2\n31ag A1 FT Ft their cosθ from part (a).\nV =\n5\n2\n© UCLES 2021 Page 11 of 12",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-31-q07",
      "question_id": "9231-2021-on-31-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) ucosα=vcosβ M1\neusinα=vsinβ M1\nDivide: tanβ=etanα A1 AG. Must see divide OE.\n3\n\n7(b) vsinβ=wcosγ (=eusinα) M1\nevcosβ=wsinγ (=eucosα) M1\nDivide: tanγ=1/tanα: γ=90°−α *A1\nAfter second rebound, direction of motion is parallel to initial path. DB1\n4\n\n7(c) Final KE = 1 m ( ( eusinα)2 +( eucosα)2 )  = 1 me2u2  M1 Energy expression in terms of u.\n \n2  2 \n1 1 1 1 A1\nSo me2u2 = × mu2 giving e=\n2 9 2 3\nPart (a) gives tan\n( 90−α)=etanα\nM1\nSo tanα= 3 A1\n4\n© UCLES 2021 Page 12 of 12",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_31-p12.png"
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    },
    {
      "id": "9231-2021-on-32-q01",
      "question_id": "9231-2021-on-32-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) Velocity: →ucosα B1\n↑usinα−gT B1 Allow 10 for g. Must be T.\n2\n\n1(b) ucosα sinα M1 FT Allow missing minus sign on RHS for M1.\n=− oe\nusinα−gT cosα FT from (a).\nu A1\nT =\ngsinα\n2\n\n1(c) u B1 AG\nsinα<1 giving T >\ng\n1\n© UCLES 2021 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p06.png"
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    },
    {
      "id": "9231-2021-on-32-q02",
      "question_id": "9231-2021-on-32-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 At the collision of P and Q: ( m+km ) v=kmu M1 Momentum conserved, allow missing k on RHS.\nA1\nk 4ga\nSo v=\n( 1+k )\n2 B1\n1 5mg a  mga\nEPE = × × =\n   \n2 a 5  10 \n2 M1 Energy equation, LHS correct, EPE dimensionally correct.\nLoss in KE = Gain in EPE: 1 m ( k+1 ) v2 = 1 × 5mg ×   a \n2 2 a 5\nSubstitute for v and rearrange to form quadratic equation in k M1\n20k2 =1+k\n1 A1\nk =\n4\n6\n© UCLES 2021 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p07.png"
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    },
    {
      "id": "9231-2021-on-32-q03",
      "question_id": "9231-2021-on-32-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3(a) T =3mg and Tcosθ=mg M1 Must see both of these separately.\n1 A1 At least one step of working, AG.\nCombining, cosθ=\n3\n2\n\n3(b)\n(cosθ=\na−x\n,where x = AR)\nB1 \nsinθ=\n8\n\nx  3 \n \n3 1 a\nAR= a or BR= a or radius =\n4 4 2\nmv2 M1\nTsinθ=\nr\nCombining to find an equation in v2, a and g only. DM1\nv2 =2ga, v= 2ga A1\n4\n© UCLES 2021 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p08.png"
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    },
    {
      "id": "9231-2021-on-32-q04",
      "question_id": "9231-2021-on-32-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Attempt at moments, 3 terms.\nVolume Centre of mass from AB\n2 3\nHemisphere πa3 a\n3 8\nCylinder πka(\na\n)2\nka\n2 2\n2\n2 a\nRemainder πa3 −πka   x\n3 2\nTaking moments about AB: A1 Any 2 terms correct.\n2 a 2 2 3   a 2 ka A1 All correct.\n πa3 −πka   ×x =  πa3× a  − πka   × \n3 2  3 8   2 2 \n   \n( )\n3a 2−k2 A1 Shown convincingly, AG.\nx =\n2 ( 8−3k )\n4\n\n4(b) x B1\ntanθ=\na\n( )\n3 2−k2\n7\n=\n2 ( 8−3k ) 18\n27k2 −21k+2=0 M1 Rearrange to form quadratic.\n2 1 A1 Both answers correct.\nk = and k =\n3 9\n3\n© UCLES 2021 Page 9 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-32-q05",
      "question_id": "9231-2021-on-32-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) Let speeds of A and B along line of centres after collision be v and M1 Momentum with masses correct.\n1\nv\n2\n3 3  u\nmv + mv =− mucos60°+mu =\n1 2 2 2   4  \nv −v =− 2(−ucos60°−u ) (=u) M1 Restitution, with consistent signs on LHS.\n2 1 3\n 1  1 A1\n  v 1 =− 2 u  v 2 = 2 u\n3 B1\nPerpendicular to line of centres, speed of B is usin60°= u\n2\nDirection of B is now 60° above line of centres. M1\nAngle of deflection is 60°. A1 FT FT (120° – their direction of B angle)\n6\n\n5(b) 1 1 3m 5 B1\nKE before = mu2 + . u2 = mu2\n2 2 2 4\n1 u 2 1 3m  u 2  3u 2  7  B1 FT FT only their speeds from (a)\nKE after = m  + .    +       = mu2 \n2 2 2 2 2  2    8 \n \n3 B1\nLoss in KE = mu2\n8\n3\n© UCLES 2021 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p10.png"
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    },
    {
      "id": "9231-2021-on-32-q06",
      "question_id": "9231-2021-on-32-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) dv 128 M1 Separate variables and integrate, + c not required for M1.\n2v =8x−\ndx x3\nv2 =4x2 +64x −2 +c A1 OE.\nx = 8, v = ‒15 and c = ‒32 M1 Use initial condition.\nv2 =\n4 (\nx4 −8x2 +16\n)\nor 4x2 −32+\n64 A1 Correct expression for v2, AEF.\nx2 x2\nv2 =\n4 (\nx2 −4\n)2\ngiving v=−\n2(\nx2 −4\n) A1 Convincingly shown, e.g. v is negative initially, AG.\nx2 x\n5\n\n6(b) 1 ln(x2 −4)=−2t (+A ) M1 Use v= dx and integrate.\n2 dt\n1 DM1 Use initial condition.\nt=0, x=8, A= ln60\n2\n1 x2 −4 x2 −4 M1 Remove log.\nln  =−2t giving =e −4t\n2  60  60\nx= 4+60e −4t A1 CAO\n4\n© UCLES 2021 Page 11 of 12",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-32-q07",
      "question_id": "9231-2021-on-32-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "7(a) Coordinates of A: x=asin60, y=a−acos60 B1\n( ) 2 M1 Substitute their (x, y) into correct trajectory equation.\na 3\ng\na\n=\na 3\n3−\n22\n2 2 1\n2V2.\n4\nRearrange to find V2. M1\n3 3 A1\nV2 = ag, V = ag\n2 2\n4\n\n7(b) 1 mu2 − 1 mV2 =mga ( 1+cos60 ) M1 Energy equation.\n2 2\n9 A1 u is the speed at P.\nu2 = ag\n2\nm M1 N2L\nT −mg = u2\na\n11 A1\nT = mg\n2\n4\n© UCLES 2021 Page 12 of 12",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_32-p12.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q01",
      "question_id": "9231-2021-on-33-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 3mgx B1 Their tensions equated to obtain a quadratic equation,\nT =\nCAO.\na\n4mga B1\nT =\n3 ( a+x )\n9x2 +9ax−4a2 =0 leading to ( 3x−a )( 3x+4a )=0 M1\n1 A1\nx= a\n3\n4",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q02",
      "question_id": "9231-2021-on-33-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) Separate variables and integrate: M1 A1 Must include logs. Condone missing modulus.\ndv 1−2t2 \n=  dt so ln v =lnt−t2 +c\nv t\n \nv = Ate −t2 , −v= Ate −t2 , v=−Ate −t2 A1 CAO.\n3\n© UCLES 2021 Page 6 of 12\n\n2(b) −Ate −t2 1−2t2\n( )\nM1 Substituting their answer to part (a) into given formula\na= =−Ae −t2 1−2t2\nt\nt =1, a=5 ( A=5e ) M1 Use initial condition.\n( )\nForce = 5me1−t2 2t2 −1 A1 Use N2L, correct work only.\nAlternative method for question 2(b)\n( )\nv 1−2t2 M1 Use initial condition.\na= substitute t=1, a=5 so v=−5 Use N2L, correct work only.\nt\nSubstituting in their answer to part (a) so ( A=5e ) M1\n( )\nForce = 5me1−t2 2t2 −1 A1\n3\n© UCLES 2021 Page 7 of 12",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p06.png",
        "../answer-assets/9231_w21_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q03",
      "question_id": "9231-2021-on-33-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3\nLoss in EPE =\n1\n×\n12mge2\n−\n1\n×\n12mg\n×\n\n\ne−\na\n\n2 \n\n=\n2mg(\n6e−a\n)\n\nB1 E ither term correct.\n2 a 2 a  3  3 \n1 mga B1\nGain in KE = mv2 and Gain in GPE =\n2 3\nGain in KE + Gain in GPE = Loss in EPE M1 KE, GPE and at least one EPE term.\n1 mv2 + mga = 2mg( 6e−a ) A1 All terms correct.\n2 3 3\nSimplify to a linear equation in e. M1\n1 A1\ne= a\n2\n6\n© UCLES 2021 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p08.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q04",
      "question_id": "9231-2021-on-33-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Attempt at moments with three terms.\nArea Centre of mass from AD\n3\nSquare 9a2 a\n2\n3\nCDF ah a\n2\n3 1\nBEC ah 3a− h\n2 3\nResulting AEFC 9a2 −3ah x\nTaking moments about AD: A1 Two terms correct.\n( )  3  3  3  1  A1 All correct.\n9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h \n 2  2  2  3 \n27a2 −12ah+h2\n\n9a−h A1 AEF\nx =\n6 ( 3a−h )  \n=\n6  \ny =x B1 By symmetry or equal to their x.\n5\n\n4(b) For equilibrium, x ⩽ 3a – h B1 Accept strict inequality.\n27a2 – 12ah + h2 ⩽ 6(3a – h)2\n27a2– 24ah + 5h2 ⩾ 0 M1 Homogeneous 3-term quadratic inequality.\n9 A1 CAO.\nh ⩽ a\n5\n3\n© UCLES 2021 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p09.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q05",
      "question_id": "9231-2021-on-33-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5 At A: ↑usinθ−8g →ucosθ M1 Both.\nusinθ−8g A1\ntanα=\nucosθ\nAt B: ↑usinθ−32g →ucosθ M1 Both.\nusinθ−32g A1\ntanβ=\nucosθ\nusinθ−8g usinθ−32g B1 Perpendicular directions, so tanα×tanβ=−1.\n× =−1\nucosθ ucosθ\nu2 −320u+25600=0 M1 Simplify to a quadratic in u.\nu=160 A1\n7\n© UCLES 2021 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p10.png"
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    },
    {
      "id": "9231-2021-on-33-q06",
      "question_id": "9231-2021-on-33-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) 5ag B1 N2L\nAt A T −mgcosθ=m×\nA a\nv2 B1 N2L\nAt B T +mgcosθ=m×\nB a\n1 1 M1 Energy equation with correct number of terms.\nm×5ag− mv2 =mga×2cosθ\n2 2\nv2 =5ag−4gacosθ A1 Accept multiplied by m and/or divided by a.\nUse ratio of tensions = 9 : 5 M1 Use ratio and simplify to an expression in cosθ.\n2 A1 CAO\ncosθ=\n5\n6\n\n6(b) Greatest speed at lowest point M1 Energy equation including lowest point, correct number\n− 1 m×5ag+ 1 mV2 =mga×( 1−cosθ) of terms.\n2 2\n31ag A1 FT Ft their cosθ from part (a).\nV =\n5\n2\n© UCLES 2021 Page 11 of 12",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p11.png"
      ]
    },
    {
      "id": "9231-2021-on-33-q07",
      "question_id": "9231-2021-on-33-q07",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) ucosα=vcosβ M1\neusinα=vsinβ M1\nDivide: tanβ=etanα A1 AG. Must see divide OE.\n3\n\n7(b) vsinβ=wcosγ (=eusinα) M1\nevcosβ=wsinγ (=eucosα) M1\nDivide: tanγ=1/tanα: γ=90°−α *A1\nAfter second rebound, direction of motion is parallel to initial path. DB1\n4\n\n7(c) Final KE = 1 m ( ( eusinα)2 +( eucosα)2 )  = 1 me2u2  M1 Energy expression in terms of u.\n \n2  2 \n1 1 1 1 A1\nSo me2u2 = × mu2 giving e=\n2 9 2 3\nPart (a) gives tan\n( 90−α)=etanα\nM1\nSo tanα= 3 A1\n4\n© UCLES 2021 Page 12 of 12",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_33-p12.png"
      ]
    },
    {
      "id": "9231-2021-on-41-q01",
      "question_id": "9231-2021-on-41-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 78.5 B1 157\ns2 = =8.7222\n9 18\ns2 M1 Correct expression with a t value.\nCI: 25.6±t\n10\nWith t = 1.833 A1 With correct t value.\n25.6±1.71 ( 2 ) or [23.9, 27.3] A1 Accept in either form, ISW. Accept inequality form.\n4\n\n1(b) 25.6−k M1\nt\ns2\n10\nWith t = −1.383 A1 Allow 1.383.\n26.9 A1 CAO\n3\n© UCLES 2021 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_41-p06.png"
      ]
    },
    {
      "id": "9231-2021-on-41-q02",
      "question_id": "9231-2021-on-41-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) ìï1+x -1≤x≤0, M1 Differentiation attempted.\nï\nïï\nf(x)=í1-x 0<x≤1,\nA1 All correct, including 0 otherwise.\nï\nï\nï0 otherwise.\nïî\n2\n\n2(b)  1 1 1  1 7 1 M1 Can do by integration with correct limits.\nP − X = F −F − = −\n     \n 2 2 2  2 8 8\n3 A1\n4\n2\n\n2(c) 0 1 M1\nx2 ( 1+x ) dx +x2 ( 1−x ) dx\n−1 0\n1 1  1 1 \nx3 + x4 + x3 − x4\n   \n3 4  3 4 \n1 1 1 A1\n+ =\n12 12 6\n2\n© UCLES 2021 Page 7 of 14\n\n2(d) 0 1 2 M1 Complete method with their mean squared explicit.\nx4 ( 1+x ) dx +x4 ( 1−x ) dx− \n\n1\n\n6\n−1 0\n1 1  1 1  1\nx5 + x6 + x5 − x6 −\n   \n5 6  5 6  36\n1 1 7 A1\n− =\n15 36 180\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_41-p07.png",
        "../answer-assets/9231_w21_ms_41-p08.png"
      ]
    },
    {
      "id": "9231-2021-on-41-q03",
      "question_id": "9231-2021-on-41-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) B1 3 dp or better.\nx 0 1 2 3 4 5 6 or\nmore\nO 35 48 43 15 6 3 0\nE 40.874 57.704 35.641 12.579 2.775 0.392 0.0352\np = 57.704\nq = 0.035 or 0.036 B1 Accept numbers which round to 0.035 or 0.036.\n2\n© UCLES 2021 Page 8 of 14\n\n3(b) Combine frequencies less than 5: last 4 columns give 24 / 15.7812 M1 Allow last 2 or 3 columns combined for this M1.\n( O−E )2 M1\n =0.8444+1.6319+1.5199+4.2803\nE\n8.28 A1 Accept 8.27 – 8.28.\nSC: 25.60, if values not combined, scores M1A0.\nSC: 14.96 if last 3 combined, scores M1A0.\nSC: if last 2 combined, scores M1A0.\nH : B(8, 0.15) fits the data B1 Must mention distribution and data.\n0\n3 degrees of freedom, tabular value = 7.815. M1 Compare their value with 7.815 and correct FT conclusion\n8.28 > 7.815 3 columns combined compared with 9.488 and correct\nReject H . conclusion\n0\nInsufficient evidence to support manager’s claim. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\n6\n© UCLES 2021 Page 9 of 14",
      "source_pages": [
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        9
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
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    },
    {
      "id": "9231-2021-on-41-q04",
      "question_id": "9231-2021-on-41-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 H :μ −μ =2 and H :μ −μ >2 B1 Or use of μ.\n0 B A 1 B A d\nDifferences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error.\nd =32, d2 =272 M1 Sample mean and variance.\n1 322   848\nd =3.2, s2 = 272−  =18.84  = \n9  10   45 \n3.2−2 M1 A1\nt = =0.874\ns2\n10\nCompare with tabular value 1.833: 0.874 < 1.833. M1 Compare their value with 1.833 and conclusion.\nAccept H .\n0\nInsufficient evidence to support claim. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\nAssumption: population differences are normally distributed B1\n8\n© UCLES 2021 Page 10 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-41-q05",
      "question_id": "9231-2021-on-41-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 3 2 1 6 B1 At least 2 probabilities correct.\nP(3, 6, 9) = × × =\n9 8 7 504\n3 2 6 108\nP(Two of 3, 6, 9) = × × ×3=\n9 8 7 504\n3 6 5 270\nP(one of 3, 6, 9) = × × ×3=\n9 8 7 504\n6 5 4 120\nP(none of 3, 6, 9) = × × =\n9 8 7 504\nG ( t )= 20 + 45 t+ 18 t2 + 1 t3 M1 Attempt with at least 3 probabilities in a polynomial,\nX 84 84 84 84 A1 CAO.\n3\n\n5(b) 12 M1\nP(both even ) =\n72\n40\nP(one even) =\n72\n20\nP(no even) =\n72\nG ( t )= 5 + 10 t+ 3 t2 A1 CAO\nY 18 18 18\n2\n© UCLES 2021 Page 11 of 14\n\n5(c) 1 20 45 18 1   5 10 3  M1 Method and attempt to multiply.\n+ t+ t2 + t3 + t+ t2\n   \n151284 84 84 84  18 18 18 \nM1 Multiplication to obtain single quintic polynomial.\n1 A1 CAO\n(100+425t+600t2 +320t3 +64t4 +3t5)\n1512\n3\n\n5(d) G'(1) = 1 ( 425+1200+960+256+15 ) M1\n1512\n17 A1 CWO\n9\n2\n© UCLES 2021 Page 12 of 14",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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        "../answer-assets/9231_w21_ms_41-p12.png"
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    },
    {
      "id": "9231-2021-on-41-q06",
      "question_id": "9231-2021-on-41-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 M1 Attempt at ranking.\n51 1 72 3\n55 2 86 5\n75 4 170 10\n98 6 175 11\n137 7 220 12\n152 8 260 17\n167 9 262 18\n235 13 302 19\n238 14 311 20\n242 15 320 21\n256 16 333 22\n351 23\nTotal ranks: 95 A1\nH :m =m and H :m ≠m B1 Allow words, must include ‘population’ median.\n0 x y 1 x y\nUse normal approximation with attempts at mean and variance. M1\nMean =132, variance = 264 A1\n95.5−132 M1 Allow no or wrong continuity correction for M1 only.\n264\n© UCLES 2021 Page 13 of 14\n\n6 – 2.246 A1 CAO\nCritical value is – 1.96. M1 Compare their value with – 1.96.\n– 2.246 < – 1.96 reject H . Or area comparison 0.0123 or 0.0124 with 0.025 and FT\n0\nconclusion.\nThere is sufficient evidence of a difference in levels. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\n9\n© UCLES 2021 Page 14 of 14",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_41-p14.png"
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    },
    {
      "id": "9231-2021-on-42-q01",
      "question_id": "9231-2021-on-42-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1 1  7522  304 M1 One correct unsimplified.\ns 2 = 14320−  =4.677 =\nx 39 40  65\n1  15482  1308 A1 Both correct to at least 3sf.\ns 2 = 40200−  =4.434 =\ny 59 60  295\n4.6769 4.4339 M1\ns2 = +\n40 60\n 3659  A1\n=\n0.1908  \n 19175\n752 1548 M1A1 Correct form with a z-value with 1.96 s can be unsimplified.\nCI = \n−\n\n±1.96s\n 40 60 \n− 7.0 ±0.856 = [−7.86, −6.14] A1 Accept in either form, ISW.\nAccept [6.14, 7.86].\nAccept inequality form.\n7\n© UCLES 2021 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-42-q02",
      "question_id": "9231-2021-on-42-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) P(130x<140) = P\n\n130−125 ≤Z< 140−125\n\n M1 Must be seen.\n 12 12  Must see 0.4167 and 1.25, may be implied by 0.8944 and 0.6616\nP( 0.4167<Z <1.25) Accept 0.417.\n0.8944−0.6616=0.2328 A1 AG\nMultiply by 60 leads to 13.97 Accept 0.6615. Need to see 0.2328 or 0.2329 or 13.968 or\n( 0.8944−0.6616 )×60.\n2\n\n2(b) M1 Combine first two values.\n0 3 15 23 11 8\n1.12 5.22 13.97 19.38 13.97 6.34\nTest stat = 1.7588 + 0.0761 + 0.6743 + 0.6309 + 0.4352 M1\n3.58 A1 Accept 3.57 – 3.58 .\nSC: 3.88, if values not combined, scores M1A0.\nH :distribution fits data B1 Must mention distribution and data.\n0\nN(125, 122) is a good model for the data oe\n4 degrees of freedom, so tabular value = 9.488 M1 Compare their value with 9.488 (9.49) and correct FT conclusion.\n3.58 < 9.488 Or, if values not combined, compare their value with 11.07 and\nAccept H . correct FT conclusion for M1A0.\n0\nThere is sufficient evidence that the normal distribution fits the A1 Correct conclusion, in context, following correct work. Level of\ndata uncertainty in language is used.\nThere is sufficient evidence that the claim is supported by the data\n6\n© UCLES 2021 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_42-p07.png"
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    },
    {
      "id": "9231-2021-on-42-q03",
      "question_id": "9231-2021-on-42-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 1  1  2  1  *M1 Correct expression integrated and equated to 1.\nTotal prob = 1, so   a+ x  dx+  2a− x  dx=1 Powers of x correct.\n 5   5 \n0 1\n 1   1 \nax+ x2 + 2ax− x2 = 1\n   \n 10   10 \n1 DM1 Limits used and attempt to solve.\n3a− =1\n5\n2 A1\nSolve to give a =\n5\n3\n\n3(b)  1 1( 2+x ) x2 dx+ 2 1( 4−x ) x2 dx M1 C orrect expressions integrated, FT their a.\n5 5\n0 1\n12 1  14 1 \nx3+ x4 + x3− x4\n   \n53 4  53 4 \n13 A1 11 67\n+\n10 60 60\n2\n© UCLES 2021 Page 8 of 13\n\n3(c) ìï 0 x<0 M1 Integration of their PDF.\nï\nï\nï ï ï ï ïï 1 5 æ ç ç çè 2x+ 1 2 x2 ö ÷ ÷ ÷ ø 0x<1, A1 M iddle 2 parts correct.\nF(x)=í\nï ï\nï ï ï 1\n5 çè\næ ç ç4x- 1\n2\nx2-1\n÷ ø\nö ÷ ÷ 1x2,\nï\nï\nï 1 x>2.\nïî\n=0 ( x<0 ) and =1 (x>2) A1 First and last parts correct and all domains correct, with no gaps.\n3\n© UCLES 2021 Page 9 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_42-p08.png",
        "../answer-assets/9231_w21_ms_42-p09.png"
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    },
    {
      "id": "9231-2021-on-42-q04",
      "question_id": "9231-2021-on-42-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Attempt at ranking.\n12 1 28 4\n18 2 31 6\n25 3 35 9\n29 5 36 10\n32 7 37 11\n33 8 41 13\n40 12 48 15\n46 14 49 16\nTest statistic: 52 A1\nH :m =m and H :m ≠m B1 Allow in words but ‘population’ must be included.\n0 x y 1 x y\nCritical value for (8, 8) is 49. *B1 Allow 51 if clearly one-tail test in hypotheses.\n52 > 49 DM1 Compare their calculated value with 49 and correct FT conclusion.\nAccept H\n0\nInsufficient evidence of difference in medians. A1 Correct conclusion, in context, following correct work. Level of\nuncertainty in language is used.\n6\n\n4(b) Not a paired sample. B1\nUnderlying distribution/population not (known to be) normal B1 B0 for ‘data is not normally distributed’\nUnderlying distribution/population unknown B0 for ‘marks are not normally distributed’\n2\n© UCLES 2021 Page 10 of 13",
      "source_pages": [
        10
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w21_ms_42-p10.png"
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    },
    {
      "id": "9231-2021-on-42-q05",
      "question_id": "9231-2021-on-42-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) ( 1+4+9+16 ) k =1 so k = 1 B1\n30\n1\n\n5(b) G ( t )= 1 t+ 4 t2 + 9 t3 + 16 t4 M1A1 Using their k in a polynomial, at least two terms correct for their k.\nX 30 30 30 30\n2\n\n5(c)  1 4 9 16 1 1 1  M1 Method and attempt to multiply.\n\nt+ t2 + t3+ t4\n\n+ t+ t2\n\n30 30 30 30 4 2 4 \n1 ( ) M1A1 Multiplication to obtain single polynomial of order 6.\nt+6t2 +18t3 +38t4 +41t5 +16t6\n120\n3\n\n5(d) Given: G'(1) = 13/3 M1 Differentiate twice.\n1\nG''(t) =\n(12+108t+456t2 +820t3 +480t4)\n120\n2 M1 Use correct formula.\n13 13\n−\nVar(X) = G''(1) +  \n3  3 \n1876 13 169 107 A1 CAO\n+ − =\nor 1.19\n120 3 9 90\n3\n© UCLES 2021 Page 11 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2021-on-42-q06",
      "question_id": "9231-2021-on-42-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 H : μ = μ and H : μ > μ B1\n0 A B 1 A B\nx = 21.2 x2 = 47.92 M1 A1\n1 21.22  124\ns 2 = 47.92−  =0.33067 =\nx 9 10  375\n1  24.482  9527\ns 2 = 53.75−  =0.34644 =\ny 11 12  27500\n9×0.33067+11×0.34644 M1 A1\ns2 = = 0.3393\n10+12−2\n2.12−2.04 M1 A1 Accept 0.321 – 0.322.\nt = =0.321\n1 1\ns +\n10 12\nCritical value: 1.325 M1 Compare their value with 1.325 and correct FT conclusion.\n0.321 < 1.325 accept H .\n0\nInsufficient evidence that mean of A is greater than mean of B A1 Correct conclusion, in context, following correct work. Level of\nuncertainty in language is used.\nEITHER: Distributions are normal and equal variances B1 Assumptions consistent with method used.\nOR: Distributions are normal Accept ‘population is normal’.\n‘Distribution of difference of means is normal’.\nAlternative method for question 6\nH : μ = μ and H : μ > μ B1\n0 A B 1 A B\n© UCLES 2021 Page 12 of 13\n\n9231/42 Cambridge International AS & A Level – Mark Scheme October/November 2021\nPUBLISHED\nQuestion Answer Marks Guidance\nx = 21.2 x2 = 47.92 M1 For one correct unsimplified.\n1 21.22  124\ns 2 = 47.92−  =0.33067 = A1 For both correct.\nx 9 10  375\n1  24.482  9527\ns 2 = 53.75−  =0.34644 =\ny 11 12  27500\n0.33067 0.34644 M1 A1\ns2 = + = 0.061936\n10 12\nt = 2.12−2.04 =0.321 ( 5 ) M1 A1 Accept 0.321 – 0.322.\ns\nCritical value: 1.325 M1 Compare their value with 1.325 and correct FT conclusion.\n0.321 < 1.325 accept H .\n0\nInsufficient evidence that mean of A is greater than mean of B. A1 Correct conclusion, in context, following correct work. Level of\nuncertainty in language is used.\nEITHER: Distributions are normal and equal variances B1 Assumptions consistent with method used.\nOR: Distributions are normal. Accept ‘population is normal’.\n‘Distribution of difference of means is normal’.\n10\n© UCLES 2021 Page 13 of 13",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_42-p13.png"
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    },
    {
      "id": "9231-2021-on-43-q01",
      "question_id": "9231-2021-on-43-q01",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 78.5 B1 157\ns2 = =8.7222\n9 18\ns2 M1 Correct expression with a t value.\nCI: 25.6±t\n10\nWith t = 1.833 A1 With correct t value.\n25.6±1.71 ( 2 ) or [23.9, 27.3] A1 Accept in either form, ISW. Accept inequality form.\n4\n\n1(b) 25.6−k M1\nt\ns2\n10\nWith t = −1.383 A1 Allow 1.383.\n26.9 A1 CAO\n3\n© UCLES 2021 Page 6 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-43-q02",
      "question_id": "9231-2021-on-43-q02",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) ìï1+x -1≤x≤0, M1 Differentiation attempted.\nï\nïï\nf(x)=í1-x 0<x≤1,\nA1 All correct, including 0 otherwise.\nï\nï\nï0 otherwise.\nïî\n2\n\n2(b)  1 1 1  1 7 1 M1 Can do by integration with correct limits.\nP − X = F −F − = −\n     \n 2 2 2  2 8 8\n3 A1\n4\n2\n\n2(c) 0 1 M1\nx2 ( 1+x ) dx +x2 ( 1−x ) dx\n−1 0\n1 1  1 1 \nx3 + x4 + x3 − x4\n   \n3 4  3 4 \n1 1 1 A1\n+ =\n12 12 6\n2\n© UCLES 2021 Page 7 of 14\n\n2(d) 0 1 2 M1 Complete method with their mean squared explicit.\nx4 ( 1+x ) dx +x4 ( 1−x ) dx− \n\n1\n\n6\n−1 0\n1 1  1 1  1\nx5 + x6 + x5 − x6 −\n   \n5 6  5 6  36\n1 1 7 A1\n− =\n15 36 180\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w21_ms_43-p08.png"
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    },
    {
      "id": "9231-2021-on-43-q03",
      "question_id": "9231-2021-on-43-q03",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) B1 3 dp or better.\nx 0 1 2 3 4 5 6 or\nmore\nO 35 48 43 15 6 3 0\nE 40.874 57.704 35.641 12.579 2.775 0.392 0.0352\np = 57.704\nq = 0.035 or 0.036 B1 Accept numbers which round to 0.035 or 0.036.\n2\n© UCLES 2021 Page 8 of 14\n\n3(b) Combine frequencies less than 5: last 4 columns give 24 / 15.7812 M1 Allow last 2 or 3 columns combined for this M1.\n( O−E )2 M1\n =0.8444+1.6319+1.5199+4.2803\nE\n8.28 A1 Accept 8.27 – 8.28.\nSC: 25.60, if values not combined, scores M1A0.\nSC: 14.96 if last 3 combined, scores M1A0.\nSC: if last 2 combined, scores M1A0.\nH : B(8, 0.15) fits the data B1 Must mention distribution and data.\n0\n3 degrees of freedom, tabular value = 7.815. M1 Compare their value with 7.815 and correct FT conclusion\n8.28 > 7.815 3 columns combined compared with 9.488 and correct\nReject H . conclusion\n0\nInsufficient evidence to support manager’s claim. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\n6\n© UCLES 2021 Page 9 of 14",
      "source_pages": [
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      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-43-q04",
      "question_id": "9231-2021-on-43-q04",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 H :μ −μ =2 and H :μ −μ >2 B1 Or use of μ.\n0 B A 1 B A d\nDifferences: 5 8 2 0 4 7 −2 1 10 −3 M1 Allow one error.\nd =32, d2 =272 M1 Sample mean and variance.\n1 322   848\nd =3.2, s2 = 272−  =18.84  = \n9  10   45 \n3.2−2 M1 A1\nt = =0.874\ns2\n10\nCompare with tabular value 1.833: 0.874 < 1.833. M1 Compare their value with 1.833 and conclusion.\nAccept H .\n0\nInsufficient evidence to support claim. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\nAssumption: population differences are normally distributed B1\n8\n© UCLES 2021 Page 10 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2021-on-43-q05",
      "question_id": "9231-2021-on-43-q05",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 3 2 1 6 B1 At least 2 probabilities correct.\nP(3, 6, 9) = × × =\n9 8 7 504\n3 2 6 108\nP(Two of 3, 6, 9) = × × ×3=\n9 8 7 504\n3 6 5 270\nP(one of 3, 6, 9) = × × ×3=\n9 8 7 504\n6 5 4 120\nP(none of 3, 6, 9) = × × =\n9 8 7 504\nG ( t )= 20 + 45 t+ 18 t2 + 1 t3 M1 Attempt with at least 3 probabilities in a polynomial,\nX 84 84 84 84 A1 CAO.\n3\n\n5(b) 12 M1\nP(both even ) =\n72\n40\nP(one even) =\n72\n20\nP(no even) =\n72\nG ( t )= 5 + 10 t+ 3 t2 A1 CAO\nY 18 18 18\n2\n© UCLES 2021 Page 11 of 14\n\n5(c) 1 20 45 18 1   5 10 3  M1 Method and attempt to multiply.\n+ t+ t2 + t3 + t+ t2\n   \n151284 84 84 84  18 18 18 \nM1 Multiplication to obtain single quintic polynomial.\n1 A1 CAO\n(100+425t+600t2 +320t3 +64t4 +3t5)\n1512\n3\n\n5(d) G'(1) = 1 ( 425+1200+960+256+15 ) M1\n1512\n17 A1 CWO\n9\n2\n© UCLES 2021 Page 12 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9231_w21_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2021-on-43-q06",
      "question_id": "9231-2021-on-43-q06",
      "subject": "9231",
      "year": 2021,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 M1 Attempt at ranking.\n51 1 72 3\n55 2 86 5\n75 4 170 10\n98 6 175 11\n137 7 220 12\n152 8 260 17\n167 9 262 18\n235 13 302 19\n238 14 311 20\n242 15 320 21\n256 16 333 22\n351 23\nTotal ranks: 95 A1\nH :m =m and H :m ≠m B1 Allow words, must include ‘population’ median.\n0 x y 1 x y\nUse normal approximation with attempts at mean and variance. M1\nMean =132, variance = 264 A1\n95.5−132 M1 Allow no or wrong continuity correction for M1 only.\n264\n© UCLES 2021 Page 13 of 14\n\n6 – 2.246 A1 CAO\nCritical value is – 1.96. M1 Compare their value with – 1.96.\n– 2.246 < – 1.96 reject H . Or area comparison 0.0123 or 0.0124 with 0.025 and FT\n0\nconclusion.\nThere is sufficient evidence of a difference in levels. A1 Correct conclusion, in context, following correct work.\nLevel of uncertainty in language is used.\n9\n© UCLES 2021 Page 14 of 14",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2021-oct-nov/9231_w21_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-mj-11-q01",
      "question_id": "9231-2022-mj-11-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 1 1 1  M1 A1 Finds partial fractions.\n   \n(ar1)(ara1) aar1 ara1\nn 1 1 1 1 1 1 1 1  M1 Writes at least three complete terms,\n         including first and last.\n(ar1)(ara1) aa1 2a1 2a1 3a1 an1 ana1\nr1\nn 1 1 1 1  A1 OE ISW\n \n\n\n\n(ar1)(ara1) aa1 ana1\nr1\n4\n\n1(b)  1 1 1 M1 A1 Finds sum to infinity, forms quadratic.\n   leading to a2 a60\n(ar1)(ara1) a2 a 6\nr1\na2 A1 CAO\n3\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-mj-11-q02",
      "question_id": "9231-2022-mj-11-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "2(a) AB 8i7j5k AC3j3k B1 Finds direction vectors of two lines in the\nplane.\ni j k  6  1 M1 A1 Finds normal to the plane ABC.\n   \n8 7 5  24 ~ 4\n   \n   \n0 3 3 24 4\n(4)4(4)4(1) 8x4y4z8 M1 A1 Substitutes point.\nAlternative method for question 2(a)\nSetting up 3 equations using the points given M1\nx4y4z 8 A1 A1 A1 A1\n5\n\n2(b) 8 M1 A1 Divides their constant by magnitude of\n1.39\ntheir normal vector.\n12 42 42\n8\nCAO\n33\n2\n\n2(c)  2  B1 Equation of line OD.\n \nrt 3\n \n \n3\n2t12t12t 8 M1 Substitutes into equation of plane.\n8,12,12 A1\n3\n© UCLES 2022 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2022-mj-11-q03",
      "question_id": "9231-2022-mj-11-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) u 4 (given) B1 States base case.\n1\nAssume that u 4 for some positive integer k. B1 States inductive hypothesis.\nk\nu 2 u 12 u 2 7u 12 M1 Considers u 4, puts over common\nThen u 4 k k 4 k k k1\nk1 2u 2u denominator.\nk k\nu 3u 4 A1\nu 4  k k 0\nk1 2u\nk\nHence, by induction, u 4 for all positive integers n. A1\nn\n5\n\n3(b) u 2 u 12 u 4(u 3) M1 A1 Considers u u .\nu u  n n  n n n1 n\nn1 n 2u 2u\nn n\nSo u 4u u 0 A1 Uses u 4.\nn n1 n n\n3\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
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    },
    {
      "id": "9231-2022-mj-11-q04",
      "question_id": "9231-2022-mj-11-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) yx3 x y1 3 B1 Substitutes.\n2y15y2\n3 60 leading to\n5y2\n3\n62y1125y6y23 M1 Cubes to eliminate radical.\n216y3216y2 53y80 A1\nAlternative method for question 4(a)\n(2x36)3 (5x2)3 M1 8x9 72x6 125x6 216x3 2160\nyx3 leading to x3  y1 B1 Substitutes.\n216y3216y2 53y80 A1\n3\n\n4(b) 333 1 33 33 33  53 B1 FT Using their answer to part (a).\n216\n6 6 6 12 2   53  M1 6 6 6 (3 3 3)2 \n216\n2(33 33 33)\n6 6 6 161 A1\n108\n3\n\n4(c) 216S 216S 53S 24 M1 Using their\n9 6 3\n2169 2166 533 80\nS  399 133 A1\n9 216 72\n2\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
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      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2022-mj-11-q05",
      "question_id": "9231-2022-mj-11-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) 30 M1 A1 Finds discriminant or roots of x2 x1,\nor completes square.\ny2 B1 Horizontal asymptote.\n3\n\n5(b) dy (x2 x1)(4x1)(2x2 x1)(2x1) M1 dy\n Finds .\ndx  x2 x1 2 dx\n3x2 6x0 M1 Sets equal to 0 and forms equation.\n0,1 2,3 A1 A1 One point correct, or both x values.\nOther point correct.\nAlternative method for question 5(b)\n2x2 x1y(x2 x1)0 M1 Forms quadratic equation\nFinds discriminant (y1)2 4(y1)(y2)AND states y exists if discriminant ≥ 0 M1\nOR does not exist if discriminant < 0\nFinds (0, -1) and (-2, 3) A1\nExplains why they are stationary values. A1 Double xroots for y1 andy3\nor no vertical asymptote etc\n4\n© UCLES 2022 Page 10 of 15\n\n5(c) B1 Axes and asymptote.\nB1 Correct shape and position.\n1,0, 1,0, (0,1) B1 States coordinates of intersections with\n2\naxes.\n3\n\n5(d) B1 FT FT from sketch in part (c). There must be\ncusps on the x axis.\n0k1 B1\n2\n© UCLES 2022 Page 11 of 15",
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2022-mj-11-q06",
      "question_id": "9231-2022-mj-11-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) B1 Correct domain, r strictly increasing.\nB1 Correct gradient when 0 and 2.\nMaximum distance of C from the pole is 1. B1 1\n4 Must be exact. π\n2\n3\n\n6(b) 1 2 tan11 d M1 Uses 1r2d with correct limits.\n2\n2 2\n0\n1tan11 2  2  d M1 A1 Applies integration by parts.\n2 2  0 02 4\n1tan11ln(2 4) 2 A1\n2 2  0\n11ln81ln4 11ln2 A1 Must be exact.\n4 2 2 4 2\n5\n© UCLES 2022 Page 12 of 15\n\n6(c)\nx\n tan111\n2 cos\nB1 Uses xrcos\n2\ndx   tan111 2sincos  tan111 2  2 4 1 0 M1 A1 Sets derivative equal to zero.\nd 2 2\n 2 4  tan11sincos0 A1 AG\n2\n 0.62 4  tan10.3sin0.6cos0.60.108 and B1 Shows sign change.\n 0.72 4  tan10.35sin0.7cos0.70.209\n5\n© UCLES 2022 Page 13 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s22_ms_11-p13.png"
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    },
    {
      "id": "9231-2022-mj-11-q07",
      "question_id": "9231-2022-mj-11-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 17,
      "status": "available",
      "reason": null,
      "text": "7(a) k 6 4 6 4 k M1 A1 Sets detA0. SOI\n2 3 012k600\n8 9 7 9 7 8\nk 5 A1\n3\n\n7(b) 1 k 6 9k48 3k16 M1 A1 Finds one correct entry.\n \n6012k 8 9 6012k 45k\n1 2 3 6 1 A1\n \n6012k 8 9 6012k 2(5k)\n1 2 3 123k 4k A1\n \n6012k k 6 6012k 45k\n4\n© UCLES 2022 Page 14 of 15\n\n7(c) 1 2 3 M1 A1 Finds BA.\n1 0 0  1 2 3\nBA  4 k 6   \n0 1 0\n\n4 k 6\n7 8 9\na d M1 Sets C as 32 matrix.\n1 2 3   a2b3c d 2e3f  2 1\n  b e    \n4 k 6  4akb6c 4d ke6f  k 4\nc f \n0 1 A1 Accept any a, b, c, d, e and f such that\nC  1 0   a2b3c d 2e3f  2 1\n    .\n 0 0   4akb6c 4d ke6f  k 4\n4\n\n7(d) 2 1x 2x y  B1 x X\n    Transforms   to  .\nk 4y kx4y y Y \nkx4mxm2xmx M1 A1 Uses y mx and Y mX .\nm2 2mk 0 A1\n44k 0 M1 Sets discriminant positive.\nk1 k8 A1\n6\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2022-mj-12-q01",
      "question_id": "9231-2022-mj-12-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 1 1 1  M1 A1 Finds partial fractions.\n   \n(ar1)(ara1) aar1 ara1\nn 1 1 1 1 1 1 1 1  M1 Writes at least three complete terms,\n         including first and last.\n(ar1)(ara1) aa1 2a1 2a1 3a1 an1 ana1\nr1\nn 1 1 1 1  A1 OE ISW\n \n\n\n\n(ar1)(ara1) aa1 ana1\nr1\n4\n\n1(b)  1 1 1 M1 A1 Finds sum to infinity, forms quadratic.\n   leading to a2 a60\n(ar1)(ara1) a2 a 6\nr1\na2 A1 CAO\n3\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_12-p06.png"
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    },
    {
      "id": "9231-2022-mj-12-q02",
      "question_id": "9231-2022-mj-12-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "2(a) AB 8i7j5k AC3j3k B1 Finds direction vectors of two lines in the\nplane.\ni j k  6  1 M1 A1 Finds normal to the plane ABC.\n   \n8 7 5  24 ~ 4\n   \n   \n0 3 3 24 4\n(4)4(4)4(1) 8x4y4z8 M1 A1 Substitutes point.\nAlternative method for question 2(a)\nSetting up 3 equations using the points given M1\nx4y4z 8 A1 A1 A1 A1\n5\n\n2(b) 8 M1 A1 Divides their constant by magnitude of\n1.39\ntheir normal vector.\n12 42 42\n8\nCAO\n33\n2\n\n2(c)  2  B1 Equation of line OD.\n \nrt 3\n \n \n3\n2t12t12t 8 M1 Substitutes into equation of plane.\n8,12,12 A1\n3\n© UCLES 2022 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_12-p07.png"
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    },
    {
      "id": "9231-2022-mj-12-q03",
      "question_id": "9231-2022-mj-12-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) u 4 (given) B1 States base case.\n1\nAssume that u 4 for some positive integer k. B1 States inductive hypothesis.\nk\nu 2 u 12 u 2 7u 12 M1 Considers u 4, puts over common\nThen u 4 k k 4 k k k1\nk1 2u 2u denominator.\nk k\nu 3u 4 A1\nu 4  k k 0\nk1 2u\nk\nHence, by induction, u 4 for all positive integers n. A1\nn\n5\n\n3(b) u 2 u 12 u 4(u 3) M1 A1 Considers u u .\nu u  n n  n n n1 n\nn1 n 2u 2u\nn n\nSo u 4u u 0 A1 Uses u 4.\nn n1 n n\n3\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_12-p08.png"
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    },
    {
      "id": "9231-2022-mj-12-q04",
      "question_id": "9231-2022-mj-12-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) yx3 x y1 3 B1 Substitutes.\n2y15y2\n3 60 leading to\n5y2\n3\n62y1125y6y23 M1 Cubes to eliminate radical.\n216y3216y2 53y80 A1\nAlternative method for question 4(a)\n(2x36)3 (5x2)3 M1 8x9 72x6 125x6 216x3 2160\nyx3 leading to x3  y1 B1 Substitutes.\n216y3216y2 53y80 A1\n3\n\n4(b) 333 1 33 33 33  53 B1 FT Using their answer to part (a).\n216\n6 6 6 12 2   53  M1 6 6 6 (3 3 3)2 \n216\n2(33 33 33)\n6 6 6 161 A1\n108\n3\n\n4(c) 216S 216S 53S 24 M1 Using their\n9 6 3\n2169 2166 533 80\nS  399 133 A1\n9 216 72\n2\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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        "../answer-assets/9231_s22_ms_12-p09.png"
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    },
    {
      "id": "9231-2022-mj-12-q05",
      "question_id": "9231-2022-mj-12-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) 30 M1 A1 Finds discriminant or roots of x2 x1,\nor completes square.\ny2 B1 Horizontal asymptote.\n3\n\n5(b) dy (x2 x1)(4x1)(2x2 x1)(2x1) M1 dy\n Finds .\ndx  x2 x1 2 dx\n3x2 6x0 M1 Sets equal to 0 and forms equation.\n0,1 2,3 A1 A1 One point correct, or both x values.\nOther point correct.\nAlternative method for question 5(b)\n2x2 x1y(x2 x1)0 M1 Forms quadratic equation\nFinds discriminant (y1)2 4(y1)(y2)AND states y exists if discriminant ≥ 0 M1\nOR does not exist if discriminant < 0\nFinds (0, -1) and (-2, 3) A1\nExplains why they are stationary values. A1 Double xroots for y1 andy3\nor no vertical asymptote etc\n4\n© UCLES 2022 Page 10 of 15\n\n5(c) B1 Axes and asymptote.\nB1 Correct shape and position.\n1,0, 1,0, (0,1) B1 States coordinates of intersections with\n2\naxes.\n3\n\n5(d) B1 FT FT from sketch in part (c). There must be\ncusps on the x axis.\n0k1 B1\n2\n© UCLES 2022 Page 11 of 15",
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2022-mj-12-q06",
      "question_id": "9231-2022-mj-12-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) B1 Correct domain, r strictly increasing.\nB1 Correct gradient when 0 and 2.\nMaximum distance of C from the pole is 1. B1 1\n4 Must be exact. π\n2\n3\n\n6(b) 1 2 tan11 d M1 Uses 1r2d with correct limits.\n2\n2 2\n0\n1tan11 2  2  d M1 A1 Applies integration by parts.\n2 2  0 02 4\n1tan11ln(2 4) 2 A1\n2 2  0\n11ln81ln4 11ln2 A1 Must be exact.\n4 2 2 4 2\n5\n© UCLES 2022 Page 12 of 15\n\n6(c)\nx\n tan111\n2 cos\nB1 Uses xrcos\n2\ndx   tan111 2sincos  tan111 2  2 4 1 0 M1 A1 Sets derivative equal to zero.\nd 2 2\n 2 4  tan11sincos0 A1 AG\n2\n 0.62 4  tan10.3sin0.6cos0.60.108 and B1 Shows sign change.\n 0.72 4  tan10.35sin0.7cos0.70.209\n5\n© UCLES 2022 Page 13 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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        "../answer-assets/9231_s22_ms_12-p13.png"
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    },
    {
      "id": "9231-2022-mj-12-q07",
      "question_id": "9231-2022-mj-12-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 17,
      "status": "available",
      "reason": null,
      "text": "7(a) k 6 4 6 4 k M1 A1 Sets detA0. SOI\n2 3 012k600\n8 9 7 9 7 8\nk 5 A1\n3\n\n7(b) 1 k 6 9k48 3k16 M1 A1 Finds one correct entry.\n \n6012k 8 9 6012k 45k\n1 2 3 6 1 A1\n \n6012k 8 9 6012k 2(5k)\n1 2 3 123k 4k A1\n \n6012k k 6 6012k 45k\n4\n© UCLES 2022 Page 14 of 15\n\n7(c) 1 2 3 M1 A1 Finds BA.\n1 0 0  1 2 3\nBA  4 k 6   \n0 1 0\n\n4 k 6\n7 8 9\na d M1 Sets C as 32 matrix.\n1 2 3   a2b3c d 2e3f  2 1\n  b e    \n4 k 6  4akb6c 4d ke6f  k 4\nc f \n0 1 A1 Accept any a, b, c, d, e and f such that\nC  1 0   a2b3c d 2e3f  2 1\n    .\n 0 0   4akb6c 4d ke6f  k 4\n4\n\n7(d) 2 1x 2x y  B1 x X\n    Transforms   to  .\nk 4y kx4y y Y \nkx4mxm2xmx M1 A1 Uses y mx and Y mX .\nm2 2mk 0 A1\n44k 0 M1 Sets discriminant positive.\nk1 k8 A1\n6\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2022-mj-13-q01",
      "question_id": "9231-2022-mj-13-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) B1 Axes and asymptotes.\nB1 Branches correct.\nx\n2\n© UCLES 2022 Page 6 of 17\n\n1(b) B1 FT FT from sketch in part (a), symmetrical about x0.\nB1 Correct shape at x0(reflection not turning point).\nx 1 M1 Finds critical point(s).\n2 leading to x 1\nx 1 3\n1x1, 1 x1 A1\n3 3\n4\n© UCLES 2022 Page 7 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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        "../answer-assets/9231_s22_ms_13-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-13-q02",
      "question_id": "9231-2022-mj-13-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) 10 B1 SOI\n2 2 2 52 205 M1 A1 Uses 2 2 2 2 2.\nAlternative method for question 2(a)\nPut yx2 and form cubic not involving x M1 A1 y(y10)2 25y2 20y4(need not be simplified).\ny35y2 120y40\n2 2 2 (5)5 A1\n3\n\n2(b) 1   M1 A1 Finds determinant.\n1     1\n 1     12 2 2 2\n 1  1  \n  1\n12 2 2 2152(2)0 M1 A1 Substitutes and uses 2\n4\n© UCLES 2022 Page 8 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2022-mj-13-q03",
      "question_id": "9231-2022-mj-13-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) x1 B1 States vertical asymptote.\nx1(axa1)a a M1 Finds oblique asymptote.\ny axa1\nx1 x1\nyaxa1 A1\n3\n© UCLES 2022 Page 9 of 17\n\n3(b) ax2 x1 yx y leading to ax2 (y1)x(y1)0 M1 A1 Forms a three term quadratic in x.\ny12\n4a(y1)0\nM1 Correct inequality using discriminant.\n1 y14a A1 AG\nClear method to reach given answer\neg (y1)(y14a)0\nAlternative method for question 3(b)\nax2 x1 yx y leading to ax2 (y1)x(y1)0 M1 A1 Forms three term quadratic in x.\ny12\n4a(y1)0\nM1 Correct inequality using discriminant. Must be clear\nlooking for where values of yexist.\n1 y14a A1 AG Clear method to reach given answer\neg (y1)(y14a)0\nSecond alternative method for question 3(b)\ndy (x1)(2ax1)(ax2 x1) M1\n 0\ndx (x1)2\nx0or x2 A1\nProves (0, 1) is local maximum and (2,4a1) is local minimum. M1 d2y\nEither or gradient either side\ndx2\nReference to position of asymptote or two distinct branches to justify A1 AG\n1 y14a\n4\n© UCLES 2022 Page 10 of 17\n\n3(c) B1 FT Asymptotes identified. FT from part 3a.\nB1 Two branches in correct position relative to correct\nasymptotes and of correct shape. Good asymptotic\nbehaviour.\nB1 Lower branch in all four quadrants from correct\nasymptotes.\n3\n© UCLES 2022 Page 11 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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      ]
    },
    {
      "id": "9231-2022-mj-13-q04",
      "question_id": "9231-2022-mj-13-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) n M1 A1 Shows enough complete terms to make pattern of\nerx e2x 2ex 1  e3x 2e2x ex cancelling clear\nr1\ne4x 2e3x e2x GP method.\ne5x 2e4x e3x  e2x 2ex 1  n erx   e2x 2ex 1  ex (ex)n 1\n ex 1\nr1\ne(n1)x 2enx e(n1)x\ne(n2)x 2e(n1)x enx\nex e2x e(n1)x e(n2)x A1 OE\nex(ex 1)(enx 1)\n3\n\n4(b) x0 B1 Accept x0.\nenx 0as n leading to u u u ex e2x M1 A1 Must see enx 0[as n] agreeing with their set of\n1 2 3\nx\n3\n\n4(c) n n M1* Uses laws of logarithms correctly.\nlnu (rxln(ex 1)2)\nr\nr1 r1\nn dM1 A1 n\nlnu  1xn(n1)nln(ex 1)2 Applies r  1n(n1).\nr 2 2\nr1 r1\nAEF for nln(ex 1)2\n3\n© UCLES 2022 Page 12 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-mj-13-q05",
      "question_id": "9231-2022-mj-13-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) A shear in the x-direction. B1 Accept ‘shear’.\n1\n\n5(b) 1 a B1 States base case.\nA  so true when n1.\n0 1\n1 ka B1 States inductive hypothesis.\nAssume that it is true for nk, so Ak  .\n0 1 \n1 ka1 a 1 aak 1 (k1)a M1 A1 Multiplies Ak with A.\nThen Ak1     \n0 1 0 1 0 1  0 1 \nIf it is true for [n1 and] nk then it is also true for nk1. Hence, A1 Everything correct and states conclusion.\n[by induction,] true for all positive integers.\n5\n\n5(c) 1 nab b bn bn M1 A1 Uses formula given in part (b).\nAnB   \n0 1  1 a 1 a  1 a a 1 \nbn bnx (bn)(x y) B1 x X \n    Transforms   by multiplying matrices to find  .\n 1 a 1 a y  1 a (x y)  y Y \n1(1m)m(bn)(1m) M1 Uses ymx and Y mX\na\ny x A1\ny 1 x A1\na(bn)\n6\n© UCLES 2022 Page 13 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-mj-13-q06",
      "question_id": "9231-2022-mj-13-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) xrcos yrsin x2  y2 r2 B1 SOI\nr21sincosr21 1sin2a M1 Eliminates both xand yand uses the double angle\n2\nformula.\n2a A1 AG\nr2 \n2sin2\n3\n\n6(b) B1 Polar graph with curve in correct domain.\nGraph needs gradient 0and r0 for all  in the\ndomain.\nB1 π\nr strictly decreasing such that r at  is greater than\n4\nhalf r at 0. Concave graph.\nπ\nGradient at 0 and must not be vertical or\n4\nhorizontal, respectively.\n2\n© UCLES 2022 Page 14 of 17\n\n6(c) 1π B1 AG\n1π 4 1tan2\nR 14 r2 da d\n2 0   2  1tan2  2tan\n0\ndt M1 Applies given substitution.\nt tan leading to sec2\nd\ndt M1 Applies sec21tan2\nsec21t2\nd\n 1 1 A1\nR 1a dt\n2  t2 t1\n0\nt2 t1(t 1)2  3 B1 Completes the square.\n2 4\n1 M1 A1 Applies formula.\nR 1a   1 dt  1 atan1  2 t 1  1\n2  (t 1)2  3 3  3 3  0\n0 2 4\nπa A1\n6 3\n8\n© UCLES 2022 Page 15 of 17",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2022-mj-13-q07",
      "question_id": "9231-2022-mj-13-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 18,
      "status": "available",
      "reason": null,
      "text": "7(a) AB4ijk CDj(3)k B1 Finds direction vectors of the two lines.\ni j k  2  M1 A1 Finds common perpendicular.\n  This may also be done by setting up simultaneous\n4 1 1  124\n  equations and solving them.\n \n0 1 3  4 \n5  2  M1 A1 Uses formula for perpendicular distance.\n1    \n2  124\n   \n(2)2 (412)2 16   \n 4   4 \nM1 Sets equal to 3 and forms in quadratic in .\n303 39(10)2 9  172 100164 \n172 100164\n162 80640 leading to 2 540 A1 AG\n7\n\n7(b)(i) r7i4jks4ijkt5i3j2k M1 A1 OE\nM1 for using a correct point and attempting to find\nrelevant direction vectors.\n2\n\n7(b)(ii) i j k  8  M1 A1 Finds normal to the plane  .\n2\n \n4 1 1  25\n \n \n5 3 5  7 \n8(7)25(4)7(1) leading to 8x25y7z163 M1 A1 Substitutes point.\n4\n© UCLES 2022 Page 16 of 17\n\n7(c) i j k  5  M1 A1 Finds normal to the plane  .\n1\n \n4 1 1  13\n \n \n5 3 2  7 \n 5   8  M1 A1 Uses dot product of normal vectors.\n    414 cos0.9776176...\n13  25  243 738cos leading to cos\n   \n243 738\n   \n 7   7 \n12.1 A1 Mark final answer. Accept 0.212c\n5\n© UCLES 2022 Page 17 of 17",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2022-mj-21-q01",
      "question_id": "9231-2022-mj-21-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 s    e 3 2  9 e 3 2 d 5  e 3 4 d M1 A1 Forms s correctly.\n16 4\n0 0\ns 5e 3 4    5  e 3 4 1  M1 A1 Integrates.\n3  3\n0\n 4ln1 3s A1\n3 5\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q02",
      "question_id": "9231-2022-mj-21-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) coshx 1  ex ex sinhx 1  ex ex B1\n2 2\n1  ex ex2 1 1  e2x e2x cosh2x M1 A1 Expands, AG.\n2 2\n3\n\n2(b) 2sinh2 xksinhx10 M1 A1 Applies identity.\nk2 80 M1 A1 Sets discriminant positive.\nA1\nk  8, k  8\n5\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q03",
      "question_id": "9231-2022-mj-21-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) z2 z10 leading to m1 i 3 M1 Auxiliary equation.\n2 2\nx e1 2 t  Asin 3tBcos 3t  A1 Complimentary function.\n2 2\nx pt2 qtr leading to x2ptq leading to x2p B1 Particular integral and its derivatives.\n2p2ptq pt2 qtr t2 1 M1 Substitutes and equates coefficients. PI must have the\ncorrect form.\nr 1, q  2, p 1 A1\nxe1 2 t Asin 3tBcos 3t  t2 2t1 A1\n2 2\n6\n\n3(b) e1 2 t 0 as t  M1\nx2 A1 FT\n2\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q04",
      "question_id": "9231-2022-mj-21-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)  1 2xdx 1 2N 1  1 2N 2  1 2 N N 1  1 2 N N M1 A1 Forms the sum of the areas of the rectangles.\nN N N N\n0\n 1  N  2N 1 n  2N 1 M1 A1 Applies  N rn  r  rN 1  , AG.\nN\nn1\nN  2N 1 1 \nn1\nr1\n4\n\n4(b)  1 2xdx 1  1 2N 1  1 2 N N 2  1 2 N N 1 M1 A1 Forms the sum of the areas of appropriate rectangles.\nN N N N\n0\n\n1\n\nN1\n2N 1\nn\n\n1 M1 A1\nApplies \nN1\nrn \nrN 1\n.\nN\nn0\nN  2N 1 1 \nn0\nr1\n4\n\n4(c) 1 M1 c\nN  2 2 N 1 N 1   N  2 1 N 1 1   N 1 104 leading to N 104 Simplifies U n L n to n .\nLeast value of N is 10001 A1\n2\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p08.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q05",
      "question_id": "9231-2022-mj-21-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) d B1 Differentiates (x1)y correctly.\n(x1)y(x1)y' y\ndx\nd  x y13 3x y12y'1 B1 Differentiates x y13 correctly.\ndx\n(1)y'3y'10 y'3 B1 Substitutes (0,0), AG.\n4\n3\n\n5(b) x1y''2y' B1 Differentiates (x1)y' y.\n3x y12 y''61 y'2x y10 B1 B1 Differentiates 3(x y1)21 y'.\nExpect\n6x y11 y6x y11 yy\ninstead of 61 y2x y1.\ny''233y''612 0 M1 Substitutes (0,0).\n4 4\ny'' 9 A1\n32\ny''(0) M1 A1 Finds Maclaurin’s series.\ny  y(0) y'(0)x x2 3 x 9 x2\n2! 4 64\n7\n© UCLES 2022 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p09.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q06",
      "question_id": "9231-2022-mj-21-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 dy dv B1\nvx\ndx dx\n dv M1 Substitutes and derives first order separable equation.\nx\n\nvx\n\nvx 9x2 x2v2\n dx\ndv A1\nx  v2 9\ndx\n1 1 M1 Separates variables and integrates both sides.\n dv dx\nv2 9 x\nv A1\nsinh1\n \nlnxC\n3\nC 0 B1 Substitutes initial conditions.\n  M1 Converts to logarithmic form.\nln v  v2 1 lnx\n3 9\nA1\ny y2 9x2 3x2\n9x4 6x2y y2 y2 9x2 0 M1 Squares to eliminate radical.\ny  3(x2 1) A1\n2\n10\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p10.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q07",
      "question_id": "9231-2022-mj-21-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) sin7Imcis7 M1 Uses binomial theorem.\nsin7s7 21c2s535c4s37c6s A1\nsin7s7 21  1s2 s5 35  1s22 s37  1s23 s M1 Uses c2 1s2.\nsin7s7 21  1s2 s5 35  12s2 s4 s37  13s2 3s4 s6 s\nsin764s7 112s556s37s A1\n1 M1 Divides numerator and denominator by s7.\ncosec7\n64s7 112s5 56s37s\ncosec7 A1 AG\ncosec7 CWO\n7cosec656cosec4112cosec264\n6\n\n7(b) x7 M1 A1 Relates with equation in part (a).\nx7 2  7x6 56x4 112x2 64  leading to 2\n7x6 56x4 112x2 64\ncosec72 leading to sin7 1 M1 Solves sin7 1.\n2 2\nxcosec 1 π A1 Gives one correct solution.\n42\nxcosecqπ, q19,11, 7 , 5 ,13,17 A1 Gives six other solutions. Allow different values of q as\n42 42 42 42 42 42\nlong as all seven solutions are found.\n5\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p11.png"
      ]
    },
    {
      "id": "9231-2022-mj-21-q08",
      "question_id": "9231-2022-mj-21-q08",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) 3 a 0 M1 Sets determinant equal to zero.\n5 1 0 010a60\n1 3 2\na3 A1\n5\n2\n\n8(b) Eigenvalues of A are 3, 1 and 2. B1 Lower diagonal matrix or characteristic equation.\ni j k  4  M1 A1 Uses vector product (or equations) to find corresponding\n  eigenvectors.\n3: 5 4 0  5\n \n \n1 3 1 19\n \ni j k  0   0  A1\n   \n1: 4 0 0  12  1\n   \n   \n1 3 3 12 1\n   \ni j k 0 0 A1\n   \n2: 1 0 0  0  0\n   \n   \n1 3 0 3 1\n   \n 4 0 0 9 0 0 M1 A1 Or correctly matched permutations of columns.\n   \nThus P  5 1 0 and D 0 1 0\n   \n   \n19 1 1 0 0 4\n   \n7\n© UCLES 2022 Page 12 of 13\n\n8(c) 312342 60 B1 Characteristic equation.\nA6IA34A2 M1 Substitutes for A and makes A6I the subject.\nA6I2   A3 4A22  A4A4I2 M1 A1 Squares and factorises.\n4\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_21-p12.png",
        "../answer-assets/9231_s22_ms_21-p13.png"
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    },
    {
      "id": "9231-2022-mj-22-q01",
      "question_id": "9231-2022-mj-22-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 s    e 3 2  9 e 3 2 d 5  e 3 4 d M1 A1 Forms s correctly.\n16 4\n0 0\ns 5e 3 4    5  e 3 4 1  M1 A1 Integrates.\n3  3\n0\n 4ln1 3s A1\n3 5\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p06.png"
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    },
    {
      "id": "9231-2022-mj-22-q02",
      "question_id": "9231-2022-mj-22-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) coshx 1  ex ex sinhx 1  ex ex B1\n2 2\n1  ex ex2 1 1  e2x e2x cosh2x M1 A1 Expands, AG.\n2 2\n3\n\n2(b) 2sinh2 xksinhx10 M1 A1 Applies identity.\nk2 80 M1 A1 Sets discriminant positive.\nA1\nk  8, k  8\n5\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p06.png"
      ]
    },
    {
      "id": "9231-2022-mj-22-q03",
      "question_id": "9231-2022-mj-22-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) z2 z10 leading to m1 i 3 M1 Auxiliary equation.\n2 2\nx e1 2 t  Asin 3tBcos 3t  A1 Complimentary function.\n2 2\nx pt2 qtr leading to x2ptq leading to x2p B1 Particular integral and its derivatives.\n2p2ptq pt2 qtr t2 1 M1 Substitutes and equates coefficients. PI must have the\ncorrect form.\nr 1, q  2, p 1 A1\nxe1 2 t Asin 3tBcos 3t  t2 2t1 A1\n2 2\n6\n\n3(b) e1 2 t 0 as t  M1\nx2 A1 FT\n2\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-22-q04",
      "question_id": "9231-2022-mj-22-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)  1 2xdx 1 2N 1  1 2N 2  1 2 N N 1  1 2 N N M1 A1 Forms the sum of the areas of the rectangles.\nN N N N\n0\n 1  N  2N 1 n  2N 1 M1 A1 Applies  N rn  r  rN 1  , AG.\nN\nn1\nN  2N 1 1 \nn1\nr1\n4\n\n4(b)  1 2xdx 1  1 2N 1  1 2 N N 2  1 2 N N 1 M1 A1 Forms the sum of the areas of appropriate rectangles.\nN N N N\n0\n\n1\n\nN1\n2N 1\nn\n\n1 M1 A1\nApplies \nN1\nrn \nrN 1\n.\nN\nn0\nN  2N 1 1 \nn0\nr1\n4\n\n4(c) 1 M1 c\nN  2 2 N 1 N 1   N  2 1 N 1 1   N 1 104 leading to N 104 Simplifies U n L n to n .\nLeast value of N is 10001 A1\n2\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p08.png"
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    },
    {
      "id": "9231-2022-mj-22-q05",
      "question_id": "9231-2022-mj-22-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) d B1 Differentiates (x1)y correctly.\n(x1)y(x1)y' y\ndx\nd  x y13 3x y12y'1 B1 Differentiates x y13 correctly.\ndx\n(1)y'3y'10 y'3 B1 Substitutes (0,0), AG.\n4\n3\n\n5(b) x1y''2y' B1 Differentiates (x1)y' y.\n3x y12 y''61 y'2x y10 B1 B1 Differentiates 3(x y1)21 y'.\nExpect\n6x y11 y6x y11 yy\ninstead of 61 y2x y1.\ny''233y''612 0 M1 Substitutes (0,0).\n4 4\ny'' 9 A1\n32\ny''(0) M1 A1 Finds Maclaurin’s series.\ny  y(0) y'(0)x x2 3 x 9 x2\n2! 4 64\n7\n© UCLES 2022 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p09.png"
      ]
    },
    {
      "id": "9231-2022-mj-22-q06",
      "question_id": "9231-2022-mj-22-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 dy dv B1\nvx\ndx dx\n dv M1 Substitutes and derives first order separable equation.\nx\n\nvx\n\nvx 9x2 x2v2\n dx\ndv A1\nx  v2 9\ndx\n1 1 M1 Separates variables and integrates both sides.\n dv dx\nv2 9 x\nv A1\nsinh1\n \nlnxC\n3\nC 0 B1 Substitutes initial conditions.\n  M1 Converts to logarithmic form.\nln v  v2 1 lnx\n3 9\nA1\ny y2 9x2 3x2\n9x4 6x2y y2 y2 9x2 0 M1 Squares to eliminate radical.\ny  3(x2 1) A1\n2\n10\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2022-mj-22-q07",
      "question_id": "9231-2022-mj-22-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) sin7Imcis7 M1 Uses binomial theorem.\nsin7s7 21c2s535c4s37c6s A1\nsin7s7 21  1s2 s5 35  1s22 s37  1s23 s M1 Uses c2 1s2.\nsin7s7 21  1s2 s5 35  12s2 s4 s37  13s2 3s4 s6 s\nsin764s7 112s556s37s A1\n1 M1 Divides numerator and denominator by s7.\ncosec7\n64s7 112s5 56s37s\ncosec7 A1 AG\ncosec7 CWO\n7cosec656cosec4112cosec264\n6\n\n7(b) x7 M1 A1 Relates with equation in part (a).\nx7 2  7x6 56x4 112x2 64  leading to 2\n7x6 56x4 112x2 64\ncosec72 leading to sin7 1 M1 Solves sin7 1.\n2 2\nxcosec 1 π A1 Gives one correct solution.\n42\nxcosecqπ, q19,11, 7 , 5 ,13,17 A1 Gives six other solutions. Allow different values of q as\n42 42 42 42 42 42\nlong as all seven solutions are found.\n5\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p11.png"
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    },
    {
      "id": "9231-2022-mj-22-q08",
      "question_id": "9231-2022-mj-22-q08",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) 3 a 0 M1 Sets determinant equal to zero.\n5 1 0 010a60\n1 3 2\na3 A1\n5\n2\n\n8(b) Eigenvalues of A are 3, 1 and 2. B1 Lower diagonal matrix or characteristic equation.\ni j k  4  M1 A1 Uses vector product (or equations) to find corresponding\n  eigenvectors.\n3: 5 4 0  5\n \n \n1 3 1 19\n \ni j k  0   0  A1\n   \n1: 4 0 0  12  1\n   \n   \n1 3 3 12 1\n   \ni j k 0 0 A1\n   \n2: 1 0 0  0  0\n   \n   \n1 3 0 3 1\n   \n 4 0 0 9 0 0 M1 A1 Or correctly matched permutations of columns.\n   \nThus P  5 1 0 and D 0 1 0\n   \n   \n19 1 1 0 0 4\n   \n7\n© UCLES 2022 Page 12 of 13\n\n8(c) 312342 60 B1 Characteristic equation.\nA6IA34A2 M1 Substitutes for A and makes A6I the subject.\nA6I2   A3 4A22  A4A4I2 M1 A1 Squares and factorises.\n4\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_22-p12.png",
        "../answer-assets/9231_s22_ms_22-p13.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q01",
      "question_id": "9231-2022-mj-23-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 z3 7 37i14ei1 6 π B1 Converts 7 37i to exponential form.\nz 14 1 3ei 1 1 8 π M1 A1 Finds one root.\n1\nz 14 1 3ei 1 1 8 1π, z 14 1 3ei 1 1 8 3π A1FT A1FT Finds other two roots. FT on modulus. Others given\n2 3 scores A1 A0.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q02",
      "question_id": "9231-2022-mj-23-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) d B1 Finds first derivative.\nlncosxtanx\ndx\nd B1 Finds second derivative.\ntanxsec2x\ndx\ny''(0) 1 M1 A1 Evaluates second derivative at zero. Accept 1x2.\n 2\n2! 2\n4\n© UCLES 2022 Page 6 of 15\n\n2(b) 1π M1 Forms correct integral\n4 1tan2 x dx\n0\n1π A1\n4 secx dx\n0\nln  tan1x 1π 1 4 π M1 Integrates (answer given in List MF19).\n 2 4  Accept ln(secxtanx).\n0\nlntan3π0.881 A1\nAccept ln\n\n21\n\n.\n8\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p06.png",
        "../answer-assets/9231_s22_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q03",
      "question_id": "9231-2022-mj-23-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) 6 9 5 M1 Sets determinant equal to zero.\n5 8 5 0\n1 1 2\n3760 M1 Expands.\n1,2,3 A1 A1 Award A1A0 for 2 correct solutions, A0A0 for 1\ncorrect solution.\n4\n© UCLES 2022 Page 7 of 15\n\n3(b) A37A6I0 B1 States that A satisfies its characteristic equation.\nA2 7I6A1 0 M1 Multiplies through by A1.\nA1 1A2  7I A1\n6 6\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p07.png",
        "../answer-assets/9231_s22_ms_23-p08.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q04",
      "question_id": "9231-2022-mj-23-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) dy 1 B1\n1\ndt 1t2\ndx 1 B1\n1\ndt 1t2\ndy dy dt 1t2 1 1t2 M1 Uses chain rule.\n   \ndx dt dx 1t2 t2\n 1t2 1  1t2  1t2 A1 AG\n 1t2  \n t2  t2\n \n4\n© UCLES 2022 Page 8 of 15\n\n4(b) t2  2tt  1t21 2  2t  1t2  1t2  M1 A1 Applies product or quotient rule.\nd dy  \n\n \ndtdx t4\nt2 1t21 2 2  1 1t2  M1 A1 Uses chain rule. Simplification not required for A1.\nd2y\n \n 1t2\ndx2 t5\n \nt2 22 t2 1 1t2\n\nt5\n80 when t 3. A1\n3 4\n5\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p08.png",
        "../answer-assets/9231_s22_ms_23-p09.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q05",
      "question_id": "9231-2022-mj-23-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5 dy 7y 1 B1 Divides through by x(x7).\n \ndx x(x7) x7\n7 1 1 7 7 B1 Converts to partial fractions or completes the\n  or \nx(x7) x x7 x(x7) x 72  49 square.\n2 4\nx M1 A1 Finds integrating factor. Accept non-zero multiple\ne7x1(x7)1dx elnxln(x7) \nx\nx7\nof .\nx7\nd  yx  x 1 7 M1 Correct form on LHS and attempt to integrate RHS.\ndx  x7    x72  x7  x72 RHS must be of the form a  b .\nx7 x72\nyx A1\nln(x7)7(x7)1 (C)\nx7\n7 ln8 7 C M1 Finds C, substitutes into their expression.\n8 8\n 7 x7 7 M1 A1 Divides their expression through by coefficient of y.\ny 1  ln  \n x  8  x\n9\n© UCLES 2022 Page 10 of 15",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2022-mj-23-q06",
      "question_id": "9231-2022-mj-23-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a)  1 ln1xdx1ln1 11ln1 21ln1 n M1 A1 Forms the sum of the areas of the rectangles.\nn n n n n n\n0\n 1nln1 lnn1lnn2ln2n 1ln 2n! lnn M1 A1 Applies laws of logarithms, AG.\nn n n n!\n4\n\n6(b)  1 ln1xdx1ln1 11ln1 21ln1 n1 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n rectangles.\n0\n 1(n1)ln1 lnn1lnn2ln2n1 M1 Applies laws of logarithms.\nn n\n1ln 2n1! n1lnn A1 Equivalent answers: 1ln(2n)!lnn1ln2;\nn n! n n n! n\n1ln(2n1)!lnn.\nn (n1)!\n4\n\n6(c) ln2 M1 c\nU L  Simplifies their U L to .\nn n n n n n\nln2 A1 AG\n0 as n\nn\n2\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2022-mj-23-q07",
      "question_id": "9231-2022-mj-23-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) 4m2 10 leading to m1, M1 Auxiliary equation.\n2\ny Ae1 2 x Be 1 2 x A1 Complementary function. Allow \"y\" missing.\nk 3 leading to k 3 M1 A1 Substitutes yk.\ny Ae1 2 x Be 1 2 x 3 A1 FT General solution. Must have y or sensible\nequivalent, such as y . Don’t allow GS .\nGS\ny'1 Ae1 2 x  1Be 1 2 x B1 Derivative of general solution.\n2 2\nAB0 1 A1B2 leading to A2,B2 M1 Substitutes initial conditions into general solution,\n2 2\nsolves simultaneous equations.\ny2e 1 2 x 2e1 2 x 3 A1 Particular solution. Must have y or sensible\nequivalent, such as y . Don’t allow PS .\nPS\n8\n\n7(b) y0 leading to 4sinh1x3 leading to 1xsinh13 M1 Makes x the subject.\n2 2 4\ny must be of the form aebx a ebx c.\n1 2\nsinh13 ln  3  9 1  ln2 M1 Uses logarithmic form.\n4 4 16 y must be of the form aebx a ebx c.\n1 2\nx2ln2ln4 A1\n3\n© UCLES 2022 Page 12 of 15",
      "source_pages": [
        12
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_23-p12.png"
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    },
    {
      "id": "9231-2022-mj-23-q08",
      "question_id": "9231-2022-mj-23-q08",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "8(a) cosn1 M1 A1 Allow with \"C\"missing.\nsincosnd (C)\nn1\n2\n\n8(b) 1π M1 A1 Integrates by parts using part (a).\n1π  sinm1cosn12 m1 1π\n2 sinm1sincosnd   2 cosn2sinm2d Allow with limits missing.\n0  n1  n1 0\n0\n m1  1 2 π cosnsinm2(1sin2)d M1 Applies cos21sin2.\nn1 0\nm1 m1 M1 Combines with LHS.\nI  I  I\nm,n n1 m2,n n1 m,n\nnm m1 m1 A1 AG.\nI  I leading to I  I .\nn1 m,n n1 m2,n m,n mn m2,n\n© UCLES 2022 Page 13 of 15\n\n8(b) Alternative method for question 8(b)\n 1 2 π sinm2sin2cosnd 1 2 π sinm2  1cos2  cosnd M1 Applies sin21cos2\n0 0\n1π 1π\n2 sinm2cosnd2 sinm2cosn2d\n0 0\n1π\nI 2 sinm2coscosn1d\nm2,n\n0\n 1 \n1\n2\nπ\nn1 1π\nM1 A1 ucosn1u'(n1)cosnsin\nI  sinm1cosn1  2 cosnsinmd\nm2,n  m1   0 m1 0 v'sinm2cosv 1 sinm1\nm1\nn1  n1 M1 A1 Combines with LHS.\nI m,n I m2,n  m1 I m,n    1 m1  I m,n I m2,n AG\n5\n\n8(c) zz1 2cos B1\n zz15   z5 z5 5  z3 z3 10  zz1 M1 A1 Expands and groups. A0 if no grouping shown.\n25cos52cos552cos3102cos M1 Substitutes zn zn 2cosn.\ncos5 1 cos5 5 cos3 5cos A1\n16 16 8\n5\n© UCLES 2022 Page 14 of 15\n\n8(d) I  1I M1 Applies reduction formula in part (b).\n2,5 7 0,5\n1π 1π M1 Applies part (c).\n1I  12 cos5d 12 1 cos5 5 cos35cosd\n7 0,5 7 7 16 16 8\n0 0\n1π A1\n 11 sin5 5 sin3 5sin2\n780 48 8  0\n 8 A1\n105\n4\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
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        15
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_23.pdf?download=true",
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    },
    {
      "id": "9231-2022-mj-31-q01",
      "question_id": "9231-2022-mj-31-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) Tcos7.5, Tsin10 B1\n1 B1\nT   7.52 102 2 = 12.5 N\n2\n\n1(b) 50x B1\nHooke’s law: T  , x0.2\n0.8\n10 M1\nx0.8sin1\n12.5\n4 A1\nVertical distance 0.8\n5\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_31-p06.png"
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    },
    {
      "id": "9231-2022-mj-31-q02",
      "question_id": "9231-2022-mj-31-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1 1 M1 A1 Energy equation.\n mv2  m3gamgacoscos\n2 2\nm B1 N2L\nT mgcos v2\na\n4 M1 Combine to find cos.\nagcos3ga2gacos2ga\n5\n7 A1\ncos\n15\n5\n© UCLES 2022 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_31-p06.png"
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    },
    {
      "id": "9231-2022-mj-31-q03",
      "question_id": "9231-2022-mj-31-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "3 dv 4000 400 M1 A1 Integrate.\n ; v A\ndt 5t43 5t42 Constant of integration needed for A1.\nt 0, v25 A25250 M1 Find constant.\ndx M1 Integrate and find constant.\nv :\nx805t41\nB\ndt\nx0, t0 B20\n80  100t  A1\nx 20  \n5t4  5t4\n5\n© UCLES 2022 Page 7 of 14",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_31-p07.png"
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    },
    {
      "id": "9231-2022-mj-31-q04",
      "question_id": "9231-2022-mj-31-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Moments equation, condone missing ends of cylinder.\nArea Centre of mass from AB\nOne expression on the RHS correct.\n1\nCylinder 2πah2πa2 h\n2\nShell 2π2a2 ha\nMoments about AB A1 A1 One correct expression on RHS correct scores A1.\nx  2πah2πa2 2π2a2 2π2a2 ha  2πah2πa2  1 h  \n2 \n2h10ax h2 ah8ah8a2 A1\nh2 9ah8a2\nx \n2h5a\n4\n\n4(b) a B1\ntan\nx\nh2 9ah8a2 3 M1 Form inequality and rearrange to quadratic, condone\nx   a equation.\n2h5a 2\nh2 6ah7a20\nhah7a0 M1 Attempt to solve, condone equation.\n7a ha A1\n4\n© UCLES 2022 Page 8 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_s22_ms_31-p08.png"
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    },
    {
      "id": "9231-2022-mj-31-q05",
      "question_id": "9231-2022-mj-31-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5 For A: Tsinmr2 M1 N2L horizontal.\nr3asin B1 Correct expression for radius.\nT m3a2 A1\n3 M1 N2L horizontal\nSimilarly, for B: Tcos mrk22\n4\n3 A1\nT  mak22\n4\n3 M1 Equate expressions for T.\nm3a2  mak22\n4\nk2 4, k 2 A1\n© UCLES 2022 Page 9 of 14\n\n5 Alternative method for question 5\nFor A: Tcosmg, Tsinmr2 M1 N2L horizontal and vertical.\nr3asin B1 Correct expression for radius.\n5 5 g A1 Combine to obtain expression for 2.\nT m3a2  mg,  2  \n4 12 a\n3 M1 N2L horizontal.\nSimilarly, for B: Tcos mrk22\n4\n3 A1\nT  mak22\n4\n5 3 5g M1 Substitute for T and .\nmg  mak2\n4 4 12a\nk2 4, k 2 A1\n7\n© UCLES 2022 Page 10 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2022-mj-31-q06",
      "question_id": "9231-2022-mj-31-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v and w be speeds after collision: M1 Momentum along line of centres.\nmvkmwmucos\n1 M1 NEL consistent signs.\nwv ucos\n2\n3ucos A1 AG\nAdd to give\n21k Convincing working.\nSubstitute back or re-solve: A1 Accept without modulus sign.\n2kucos\nv\n21k\n4\n© UCLES 2022 Page 11 of 14\n\n6(b) 2kucos 2 B1 For speed of A (SOI).\n(usin)2  \n 21k \n \nEqual KE after collision: M1 Equate KEs.\n1 3ucos 2 1  2kucos 2\nkm    m(usin)2   \n2  21k 2   21k  \n   \n \n9kcos2 41k2sin2 2k2cos2\n \n2 M1\nUse tan :\n3\n16\n 12kk2\n9\n 44kk2\n81k\n25k2 85k520 leading to 5k45k130 M1 Obtain quadratic and attempt to solve.\n4 13 A1\nk  or\n5 5\n5\n© UCLES 2022 Page 12 of 14",
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2022-mj-31-q07",
      "question_id": "9231-2022-mj-31-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) For Q: xucosT B1\n35 B1\nFor P: x cosT 1\n2\n35 M1 Equate and attempt to rearrange.\nCollision, so cosT 1ucosT\n2\n35 3 2 A1 AG\n T 1u T\nShown convincingly.\n2 5 5\n4uT 21 5T 1\n4\n\n7(b) Vertical motion to collision: M1 A1 M1 for both expressions, one correct.\n1\nFor Q: yusinT  gT2\n2\n35 1\nFor P: y sinT 1 g(T 1)2\n2 2\n1 1 35 4 1 M1 Equate and attempt to solve\nEquate: u T  gT2   T 1 g(T 1)2\n5 2 2 5 2\n14T 1 1 g  T2 2T 1T2  21 T 1\n2 4\n16T 3621T 21, 1 55T A1\nT 3\n4\n© UCLES 2022 Page 13 of 14\n\n7(c) x42 B1\ny 24 M1\ny24 (or 24 m below O) A1 Correct sign or in words.\n3\n© UCLES 2022 Page 14 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s22_ms_31-p14.png"
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    },
    {
      "id": "9231-2022-mj-32-q01",
      "question_id": "9231-2022-mj-32-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) Tcos7.5, Tsin10 B1\n1 B1\nT   7.52 102 2 = 12.5 N\n2\n\n1(b) 50x B1\nHooke’s law: T  , x0.2\n0.8\n10 M1\nx0.8sin1\n12.5\n4 A1\nVertical distance 0.8\n5\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_32-p06.png"
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    },
    {
      "id": "9231-2022-mj-32-q02",
      "question_id": "9231-2022-mj-32-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1 1 M1 A1 Energy equation.\n mv2  m3gamgacoscos\n2 2\nm B1 N2L\nT mgcos v2\na\n4 M1 Combine to find cos.\nagcos3ga2gacos2ga\n5\n7 A1\ncos\n15\n5\n© UCLES 2022 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_32-p06.png"
      ]
    },
    {
      "id": "9231-2022-mj-32-q03",
      "question_id": "9231-2022-mj-32-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "3 dv 4000 400 M1 A1 Integrate.\n ; v A\ndt 5t43 5t42 Constant of integration needed for A1.\nt 0, v25 A25250 M1 Find constant.\ndx M1 Integrate and find constant.\nv :\nx805t41\nB\ndt\nx0, t0 B20\n80  100t  A1\nx 20  \n5t4  5t4\n5\n© UCLES 2022 Page 7 of 14",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_32-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-32-q04",
      "question_id": "9231-2022-mj-32-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Moments equation, condone missing ends of cylinder.\nArea Centre of mass from AB\nOne expression on the RHS correct.\n1\nCylinder 2πah2πa2 h\n2\nShell 2π2a2 ha\nMoments about AB A1 A1 One correct expression on RHS correct scores A1.\nx  2πah2πa2 2π2a2 2π2a2 ha  2πah2πa2  1 h  \n2 \n2h10ax h2 ah8ah8a2 A1\nh2 9ah8a2\nx \n2h5a\n4\n\n4(b) a B1\ntan\nx\nh2 9ah8a2 3 M1 Form inequality and rearrange to quadratic, condone\nx   a equation.\n2h5a 2\nh2 6ah7a20\nhah7a0 M1 Attempt to solve, condone equation.\n7a ha A1\n4\n© UCLES 2022 Page 8 of 14",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_32-p08.png"
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    },
    {
      "id": "9231-2022-mj-32-q05",
      "question_id": "9231-2022-mj-32-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5 For A: Tsinmr2 M1 N2L horizontal.\nr3asin B1 Correct expression for radius.\nT m3a2 A1\n3 M1 N2L horizontal\nSimilarly, for B: Tcos mrk22\n4\n3 A1\nT  mak22\n4\n3 M1 Equate expressions for T.\nm3a2  mak22\n4\nk2 4, k 2 A1\n© UCLES 2022 Page 9 of 14\n\n5 Alternative method for question 5\nFor A: Tcosmg, Tsinmr2 M1 N2L horizontal and vertical.\nr3asin B1 Correct expression for radius.\n5 5 g A1 Combine to obtain expression for 2.\nT m3a2  mg,  2  \n4 12 a\n3 M1 N2L horizontal.\nSimilarly, for B: Tcos mrk22\n4\n3 A1\nT  mak22\n4\n5 3 5g M1 Substitute for T and .\nmg  mak2\n4 4 12a\nk2 4, k 2 A1\n7\n© UCLES 2022 Page 10 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_32-p09.png",
        "../answer-assets/9231_s22_ms_32-p10.png"
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    },
    {
      "id": "9231-2022-mj-32-q06",
      "question_id": "9231-2022-mj-32-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v and w be speeds after collision: M1 Momentum along line of centres.\nmvkmwmucos\n1 M1 NEL consistent signs.\nwv ucos\n2\n3ucos A1 AG\nAdd to give\n21k Convincing working.\nSubstitute back or re-solve: A1 Accept without modulus sign.\n2kucos\nv\n21k\n4\n© UCLES 2022 Page 11 of 14\n\n6(b) 2kucos 2 B1 For speed of A (SOI).\n(usin)2  \n 21k \n \nEqual KE after collision: M1 Equate KEs.\n1 3ucos 2 1  2kucos 2\nkm    m(usin)2   \n2  21k 2   21k  \n   \n \n9kcos2 41k2sin2 2k2cos2\n \n2 M1\nUse tan :\n3\n16\n 12kk2\n9\n 44kk2\n81k\n25k2 85k520 leading to 5k45k130 M1 Obtain quadratic and attempt to solve.\n4 13 A1\nk  or\n5 5\n5\n© UCLES 2022 Page 12 of 14",
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2022-mj-32-q07",
      "question_id": "9231-2022-mj-32-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) For Q: xucosT B1\n35 B1\nFor P: x cosT 1\n2\n35 M1 Equate and attempt to rearrange.\nCollision, so cosT 1ucosT\n2\n35 3 2 A1 AG\n T 1u T\nShown convincingly.\n2 5 5\n4uT 21 5T 1\n4\n\n7(b) Vertical motion to collision: M1 A1 M1 for both expressions, one correct.\n1\nFor Q: yusinT  gT2\n2\n35 1\nFor P: y sinT 1 g(T 1)2\n2 2\n1 1 35 4 1 M1 Equate and attempt to solve\nEquate: u T  gT2   T 1 g(T 1)2\n5 2 2 5 2\n14T 1 1 g  T2 2T 1T2  21 T 1\n2 4\n16T 3621T 21, 1 55T A1\nT 3\n4\n© UCLES 2022 Page 13 of 14\n\n7(c) x42 B1\ny 24 M1\ny24 (or 24 m below O) A1 Correct sign or in words.\n3\n© UCLES 2022 Page 14 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2022-mj-33-q01",
      "question_id": "9231-2022-mj-33-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 M1 Attempt at moments equation with all necessary terms.\nArea Distance from Oy\nOther options possible for RHS of moments equation, for\nTriangle OCD 6 2 example:\n(1) OAC: 30  6 and ABC: 12  9\nRectangle (2) OBC: 12  4 and OAB: 30 8\n24 6\nDEBC (3) Subtraction: 607.5611213\nTriangle BAE 12 11\nTrapezium\n42 x\nOCBA\nwhere D is point with coordinates (3, 0) and E is a point with\ncoordinates (9, 0).\nParts that would give correct total area 42 B1\nMoments about Oy A1 Correct equation.\n42x 622461211 (=288)\n288 A1\nx  6.86\n42\n4\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p06.png"
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    },
    {
      "id": "9231-2022-mj-33-q02",
      "question_id": "9231-2022-mj-33-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 4 B1 EPE term correct\nmg\n1 . 3 x2\n2 a\n1 B1 Work term correct\nmgx\n3\nLoss in KE = gain in EPE + work done against friction M1 Energy equation with 3 terms, allow sign error.\n4\nmg\n1 mv2  1  3 x2  1 mgx\n2 2 a 3\n2 M1\ng\n1  1 ga 3 x2  1 gx\n2 4 a 3\nObtain and attempt to solve a 3-term quadratic equation.\n16x2 8ax3a2 0\n4xa4x3a0\n1 A1\nx a\n4\n5\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2022-mj-33-q03",
      "question_id": "9231-2022-mj-33-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Components of velocity : 25cos 25sin2g B1\nM1 A1 Expression for speed or square of speed.\nSpeed =\n25cos2 25sin2g2\n25cos2 25sin2g2 152 M1 Attempt to solve and find value for sin\n625100gsin4g2 225\n800 4 A1\nsin \n1000 5\n5\n\n3(b) 225sin B1\nTime of flight   = 4 (s)\n g \n225sin M1 Any equivalent method.\nRange = 25cos\ng\nRange = 60 (m) A1 CWO\nAlternative method for question 3(b)\n4 1 B1 Equation of trajectory..\ny x x2\n3 45\nSubstitute y0 and solve M1\n60 (m) A1\n3\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2022-mj-33-q04",
      "question_id": "9231-2022-mj-33-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 1 M1 Energy equation with all necessary terms, GPE terms must be\nmv2  mu2 mgacoscos\nresolved, allow sin/cos mix, allow sign error.\n2 2\n1 1 A1 2u may be substituted later.\nm2u2  mu2 mgacoscos\n3 2\n2 2 Implied by  ag gacoscos\n2 3\nm B1 N2L\nAt A, T mgcos u2\na\nm B1 N2L and use of tension (10T).\nAlso, 10T mgcos 4u2\na\nUse all three (two N2L and energy) equations to find T in terms of M1 Might see\nm and g only. 3m 2\n9T mgcoscos  ga\na 3\ncoscos1\n10coscos4\n1 A1\nT  mg\n3\n6\n\n4(b) 1 4m 2 M1 Any appropriate method to obtain cos.\nSubstitute back, 10 mgmgcos  ga\n3 a 3\n2 A1\ncos\n3\n2\n© UCLES 2022 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p09.png"
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    },
    {
      "id": "9231-2022-mj-33-q05",
      "question_id": "9231-2022-mj-33-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) 4v dv   4ex 12  ex B1\ndx\n1 1 M1 Expression of the correct form.\nv2  e2x 3exA\n2 2\n9 A1\nv4, x0, A\n2\nv2 e2x 6ex 9  3ex2 A1 AG\nMust see the factorisation.\n13ex Condone lack of justification for taking positive square root.\nv3ex \nex\n4\n\n5(b) dx 13ex ex M1* A1 Integration to obtain ln term\n so  dx1 dt\nCorrect answer with constant of integration\ndt ex 3ex 1\n1 ln  3ex 1  tB\n3\n1 DM1 Find the constant and substitute into their general solution.\nt 0, x0, B ln4\n3\n3ex 1\n3t ln\n4\n4 1 A1 OE\nxln e3t  \n3 3\n4\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p10.png"
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    },
    {
      "id": "9231-2022-mj-33-q06",
      "question_id": "9231-2022-mj-33-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v be speed of rebound from 1st collision: B1 Energy loss.\n1 2 1 2\nEnergy loss: mv2   mu2, v2  u2\n2 5 2 5\nvcosucos B1 Both.\nvsineusin\nCombine to form equation in e only M1 v2 ucos 2 eusin 2\n2 1 4\n e2\n5 5 5\n1 A1\ne\n2\n4\n\n6(b) tanetan, so tan1,  45 B1\nFor 2nd collision M1 Both.\nwcosvcos18060 May be implied by the A1.\nwsinevsin18060\ntanetan120their  M1 Divide to find .\n61.8 A1\n4\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p11.png"
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    },
    {
      "id": "9231-2022-mj-33-q07",
      "question_id": "9231-2022-mj-33-q07",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) Frictional force =  normal reaction at D and E B1 F R, F N\nAB BC\nMoments about B, NaRaWa(sincos) B1 One moments equation about any point involving all relevant\nforces, resolved if necessary (AEF).\nMoments about centre, F aF aWa(cossin)\nAB BC\nMoments about D, F aNaWacossin\nBC\nMoments about E, RaF aWacossin\nAB\nParallel to AB, N F WsinWsin B1 Two resolutions: all relevant terms, different frictional forces\nAB\nPerpendicular to AB, F RWcosWcos [Vertical: RcosF BC cosNsinF AB sinW W\nBC\nHorizontal: F sinF cosRsin Ncos]\nBC AB\nAlternative approach using two moments equations can earn the\nB1B1\n1 M1 Combine appropriate equations.\nN R 1N 1R\n2\n 1   1  M1 Collect terms to obtain ratio/fraction in terms of  only (CWO),\nN1 1 R1 1 \n 2   2  any equivalent simplified form.\n1 1  1 1 \nN  R  \n2 2  2 2 \nR :N 1  :1  A1\n6\n© UCLES 2022 Page 12 of 13\n\n7(b) N R M1 Must include  terms.\nDivide resolution equations: tan\nN R\n1 M1 FT their answer to part (a).\nN\n3\nUse R2N: tan\n7\nN\n3\n1 A1\ntan\n7\n3\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_33-p12.png",
        "../answer-assets/9231_s22_ms_33-p13.png"
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    },
    {
      "id": "9231-2022-mj-41-q01",
      "question_id": "9231-2022-mj-41-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) Differences: M1 Attempt at differences, allow one error.\n0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4\nd 5.6 d2 6.74 M1\n1 5.62 \ns2  6.74 0.4004\n9 10\n \nH :   0.4 and H :   0.4 B1 Allow use of \n0 B A 1 B A d\n0.560.4 M1\nt  (0.800)\n0.4004\n10\nt =0.7996 (0.800) A1\nCritical value = t (9) = 1.383 M1 Compare calculated value with 1.383 and correct FT conclusion.\n0.90\nCompare: ‘0.7996’ < 1.383\nAccept H\n0\nInsufficient evidence to support manager’s claim. A1 Correct conclusion, in context.\nLevel of uncertainty in language is used.\n7\n\n1(b) Distribution of population differences is normal B1\n1\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_41-p06.png"
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    },
    {
      "id": "9231-2022-mj-41-q02",
      "question_id": "9231-2022-mj-41-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) G ' (t) = 0.20.74t0.9t2 0.36t3 M1 PGF differentiated twice.\nY\nG \"(t) = 0.741.8t1.08t2\nY\nG '(1) = 2.2 A1\nY\nG \"(1) = 3.62\nY\nVar(Y) = G \"(1) + G '(1) – (G '(1))2 M1 Correct formula used and attempt at Var(Y).\nY Y Y\n3.62 + 2.2 – 2.22\n0.98 A1\n4\n\n2(b)\nG (t) =\n abtct22 M1* Correct method.\nY\nSo a2 0.04, a0.2 depM1 Attempt to find all coefficients.\nc2 0.09, c0.3\n0.1\n2ab0.2, b 0.5\n0.2\nG X (t) = 0.20.5t0.3t2 A1 Note: G = 1 1t223t2 M1, accept without 1 .\nY\n100 100\n1 1\nSo G = 1t23t dep M1, must have .\nX\n10 10\nFinal answer A1.\n3\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_41-p07.png"
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    },
    {
      "id": "9231-2022-mj-41-q03",
      "question_id": "9231-2022-mj-41-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 1  1  B1 Integrate and equate to 1, no errors seen.\nk2x2  x3k 6x x2 1\n \n3  2 \n 8  40\nk 8 3618122  k 1\n \n 3  3\n3\nk \n40\n1\n\n3(b) 2 6 M1* Attempt at E(X 2), correct limits, integrated.\nE(X 2) = k  4x3 x4 dxk  6x2 x3 dx =\n0 2\n 1   1 \nk x4  x5 k 2x3 x4 = k9.6967.92\n   \n 5   4 \nVar(X) = E(X2) – E2(X) = their (7.92)2.52 depM1 Use of correct formula, with their E(X2).\n1.67 A1\n3\n© UCLES 2022 Page 8 of 13\n\n3(c) 2 m 1 M1* Correct expression and attempt to integrate.\nkx4xdxk6x\n2\n0 2\n2 2 B1 May be implied\nkx4xdx\n5\n0\n2  1  1 depM1 First integral and 1 may be seen combined as 1 .\nk  6x x2   2 10\n5  2  2\n1 1 Obtain quadratic equation.\n6m m2 122\n2 10k\n68\nm2 12m 0\n3\n2 A1 Single answer.\nm6 30\n3\n2\nm6 30 2.35\n3\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_41-p08.png",
        "../answer-assets/9231_s22_ms_41-p09.png"
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    },
    {
      "id": "9231-2022-mj-41-q04",
      "question_id": "9231-2022-mj-41-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) p 16.03 B1\nq 1.70 B1 Condone 1.7\n2\n© UCLES 2022 Page 9 of 13\n\n4(b) M1 Add last 2 columns (or 3 columns: 13.06).\n0 1 2 3 4 5 6 7\n12 20 36 32 13 6 1 0\n9.85 24.63 30.78 25.65 16.03 8.02 3.34 1.70\nCombine last two columns: O value = 1, E value = 5.04\nCalculate values of\nOE2\n:\nM1 Calculate values.\nE\n0.4693 0.8704 0.8853 1.5720\n0.5727 0.5088 3.2384\nTest statistic = 8.12 A1\nH : data fits a Poisson distribution with mean 2.5 B1 Need data and distribution.\n0\nH : data does not fit a Poisson distribution with mean 2.5 e.g Data fits the given distribution.\n1\nNumber of butterflies per square metre fits Po(2.5).\nCritical value of chi-squared = 10.64 M1 Compare their test statistic with 10.64.\nCompare: ‘8.12’ < 10.64, accept H Note: allow 9.236 if 3 columns combined\n0\nor 12.02 if none combined.\nThere is sufficient evidence to support the scientist’s claim / there A1 Correct conclusion, in context.\nis sufficient evidence to suggest that the data fits a Poisson Level of uncertainty in language is used.\ndistribution with mean 2.5\n6\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_41-p09.png",
        "../answer-assets/9231_s22_ms_41-p10.png"
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    },
    {
      "id": "9231-2022-mj-41-q05",
      "question_id": "9231-2022-mj-41-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) x44.4, x2 247.48 B1\n1 44.42   53  M1\ns2  247.48 0.1514\n\n\n\n7  8   350\nM1 Correct formula with a t value.\n44.4 0.1514\nCI: 1.895\n8 8\n1.895 B1 1.895 used.\n5.550.2615.29, 5.81 A1 Either form.\n5\n\n5(b) H : 5.9 and H : 5.9 B1\n0 1\n5.555.9 M1\nt  2.544\ns2\n8\nCritical value t (7) = 2.365 M1 Compare calculated value with 2.365 and correct FT\n0.975\nCompare ‘– 2.544’ < – 2.365, reject H . conclusion.\n0\nSufficient evidence to support Raman’s claim. A1 Correct conclusion, in context.\nLevel of uncertainty in language is used.\n4\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_41-p11.png"
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    },
    {
      "id": "9231-2022-mj-41-q06",
      "question_id": "9231-2022-mj-41-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 Differences/rank: M1 A1 Attempt at differences, allow 4 errors.\n0.2 – 0.5 –1.6 –2.9 2.5 8.5 1.2 5.2\n2 –5 –11 –18 17 24 9 21\n–0.1 0.8 7.8 1.9 –0.3 –1.8 –0.7 –1.4\n–1 7 23 13 –3 –12 –6 –10\n–0.9 2.1 2.3 –5.4 –2.0 3.4 4.7 0.4\n–8 15 16 –22 –14 19 20 4\nQ = 110 P = 190 A1 Either sum correct.\nH : population median =18.0 and H : population median 18.0 B1 Must be ‘population’, allow m.\n0 1\n1 1 B1 Normal approximation: mean.\nMean = nn1 2425150\n4 4\n1 1 B1\nVariance = nn12n1 2425491225\n24 24\n110.5150 M1 Attempt at test statistic, allow incorrect or missing cc, FT their\n110.\n1225\n1.129 A1\n© UCLES 2022 Page 12 of 13\n\n6 10% 2-tail, critical value is 1.645 M1 Compare with 1.645, correct FT conclusion.\n1.1291.645 or 0.1296 > 0.05, accept H .\n0\nInsufficient evidence to suggest that median is not 18.0/ A1 Correct conclusion, in context.\nInsufficient evidence to reject teacher’s claim. Level of uncertainty in language is used.\n10\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s22_ms_41-p13.png"
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    },
    {
      "id": "9231-2022-mj-42-q01",
      "question_id": "9231-2022-mj-42-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) Differences: M1 Attempt at differences, allow one error.\n0.6 1.3 0.0 0.7 0.6 0.6 1.8 –0.2 –0.2 0.4\nd 5.6 d2 6.74 M1\n1 5.62 \ns2  6.74 0.4004\n9 10\n \nH :   0.4 and H :   0.4 B1 Allow use of \n0 B A 1 B A d\n0.560.4 M1\nt  (0.800)\n0.4004\n10\nt =0.7996 (0.800) A1\nCritical value = t (9) = 1.383 M1 Compare calculated value with 1.383 and correct FT conclusion.\n0.90\nCompare: ‘0.7996’ < 1.383\nAccept H\n0\nInsufficient evidence to support manager’s claim. A1 Correct conclusion, in context.\nLevel of uncertainty in language is used.\n7\n\n1(b) Distribution of population differences is normal B1\n1\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_42-p06.png"
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    },
    {
      "id": "9231-2022-mj-42-q02",
      "question_id": "9231-2022-mj-42-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) G ' (t) = 0.20.74t0.9t2 0.36t3 M1 PGF differentiated twice.\nY\nG \"(t) = 0.741.8t1.08t2\nY\nG '(1) = 2.2 A1\nY\nG \"(1) = 3.62\nY\nVar(Y) = G \"(1) + G '(1) – (G '(1))2 M1 Correct formula used and attempt at Var(Y).\nY Y Y\n3.62 + 2.2 – 2.22\n0.98 A1\n4\n\n2(b)\nG (t) =\n abtct22 M1* Correct method.\nY\nSo a2 0.04, a0.2 depM1 Attempt to find all coefficients.\nc2 0.09, c0.3\n0.1\n2ab0.2, b 0.5\n0.2\nG X (t) = 0.20.5t0.3t2 A1 Note: G = 1 1t223t2 M1, accept without 1 .\nY\n100 100\n1 1\nSo G = 1t23t dep M1, must have .\nX\n10 10\nFinal answer A1.\n3\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-mj-42-q03",
      "question_id": "9231-2022-mj-42-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 1  1  B1 Integrate and equate to 1, no errors seen.\nk2x2  x3k 6x x2 1\n \n3  2 \n 8  40\nk 8 3618122  k 1\n \n 3  3\n3\nk \n40\n1\n\n3(b) 2 6 M1* Attempt at E(X 2), correct limits, integrated.\nE(X 2) = k  4x3 x4 dxk  6x2 x3 dx =\n0 2\n 1   1 \nk x4  x5 k 2x3 x4 = k9.6967.92\n   \n 5   4 \nVar(X) = E(X2) – E2(X) = their (7.92)2.52 depM1 Use of correct formula, with their E(X2).\n1.67 A1\n3\n© UCLES 2022 Page 8 of 13\n\n3(c) 2 m 1 M1* Correct expression and attempt to integrate.\nkx4xdxk6x\n2\n0 2\n2 2 B1 May be implied\nkx4xdx\n5\n0\n2  1  1 depM1 First integral and 1 may be seen combined as 1 .\nk  6x x2   2 10\n5  2  2\n1 1 Obtain quadratic equation.\n6m m2 122\n2 10k\n68\nm2 12m 0\n3\n2 A1 Single answer.\nm6 30\n3\n2\nm6 30 2.35\n3\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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        "../answer-assets/9231_s22_ms_42-p09.png"
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    },
    {
      "id": "9231-2022-mj-42-q04",
      "question_id": "9231-2022-mj-42-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) p 16.03 B1\nq 1.70 B1 Condone 1.7\n2\n© UCLES 2022 Page 9 of 13\n\n4(b) M1 Add last 2 columns (or 3 columns: 13.06).\n0 1 2 3 4 5 6 7\n12 20 36 32 13 6 1 0\n9.85 24.63 30.78 25.65 16.03 8.02 3.34 1.70\nCombine last two columns: O value = 1, E value = 5.04\nCalculate values of\nOE2\n:\nM1 Calculate values.\nE\n0.4693 0.8704 0.8853 1.5720\n0.5727 0.5088 3.2384\nTest statistic = 8.12 A1\nH : data fits a Poisson distribution with mean 2.5 B1 Need data and distribution.\n0\nH : data does not fit a Poisson distribution with mean 2.5 e.g Data fits the given distribution.\n1\nNumber of butterflies per square metre fits Po(2.5).\nCritical value of chi-squared = 10.64 M1 Compare their test statistic with 10.64.\nCompare: ‘8.12’ < 10.64, accept H Note: allow 9.236 if 3 columns combined\n0\nor 12.02 if none combined.\nThere is sufficient evidence to support the scientist’s claim / there A1 Correct conclusion, in context.\nis sufficient evidence to suggest that the data fits a Poisson Level of uncertainty in language is used.\ndistribution with mean 2.5\n6\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s22_ms_42-p10.png"
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    },
    {
      "id": "9231-2022-mj-42-q05",
      "question_id": "9231-2022-mj-42-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) x44.4, x2 247.48 B1\n1 44.42   53  M1\ns2  247.48 0.1514\n\n\n\n7  8   350\nM1 Correct formula with a t value.\n44.4 0.1514\nCI: 1.895\n8 8\n1.895 B1 1.895 used.\n5.550.2615.29, 5.81 A1 Either form.\n5\n\n5(b) H : 5.9 and H : 5.9 B1\n0 1\n5.555.9 M1\nt  2.544\ns2\n8\nCritical value t (7) = 2.365 M1 Compare calculated value with 2.365 and correct FT\n0.975\nCompare ‘– 2.544’ < – 2.365, reject H . conclusion.\n0\nSufficient evidence to support Raman’s claim. A1 Correct conclusion, in context.\nLevel of uncertainty in language is used.\n4\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_42-p11.png"
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    },
    {
      "id": "9231-2022-mj-42-q06",
      "question_id": "9231-2022-mj-42-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 Differences/rank: M1 A1 Attempt at differences, allow 4 errors.\n0.2 – 0.5 –1.6 –2.9 2.5 8.5 1.2 5.2\n2 –5 –11 –18 17 24 9 21\n–0.1 0.8 7.8 1.9 –0.3 –1.8 –0.7 –1.4\n–1 7 23 13 –3 –12 –6 –10\n–0.9 2.1 2.3 –5.4 –2.0 3.4 4.7 0.4\n–8 15 16 –22 –14 19 20 4\nQ = 110 P = 190 A1 Either sum correct.\nH : population median =18.0 and H : population median 18.0 B1 Must be ‘population’, allow m.\n0 1\n1 1 B1 Normal approximation: mean.\nMean = nn1 2425150\n4 4\n1 1 B1\nVariance = nn12n1 2425491225\n24 24\n110.5150 M1 Attempt at test statistic, allow incorrect or missing cc, FT their\n110.\n1225\n1.129 A1\n© UCLES 2022 Page 12 of 13\n\n6 10% 2-tail, critical value is 1.645 M1 Compare with 1.645, correct FT conclusion.\n1.1291.645 or 0.1296 > 0.05, accept H .\n0\nInsufficient evidence to suggest that median is not 18.0/ A1 Correct conclusion, in context.\nInsufficient evidence to reject teacher’s claim. Level of uncertainty in language is used.\n10\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_42-p12.png",
        "../answer-assets/9231_s22_ms_42-p13.png"
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    },
    {
      "id": "9231-2022-mj-43-q01",
      "question_id": "9231-2022-mj-43-q01",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 94.5 B1\ns2  13.5\n7\ns2 M1 Correct formula with a t value, must be 33.8.\nCI: 33.82.365\n8\nB1 2.365 used.\n33.83.072 30.7, 36.9 A1 Either form.\n4\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_43-p06.png"
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    },
    {
      "id": "9231-2022-mj-43-q02",
      "question_id": "9231-2022-mj-43-q02",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 H : the size of shell is independent of the location B1\n0\nH : the size of shell is not independent of the location\n1\nM1 A1 Attempt at calculation of E values.\nO 68 69 96 81\nE 60.29 77.87 100.48 75.36\nO 28 55 64 39\nE 35.71 46.13 59.52 44.64\nOE2 M1 Attempt at Chi-squared contributions.\nCalculate values of : Accuracy of 0.1.\nE\n0.9859 1.0104 0.1997 0.4221\n1.6646 1.7055 0.3372 0.7126\nTest statistic = 7.04 A1 Correct total to 3sf.\nTabular value for 3df = 6.251 M1 Compare with 6.251 and correct FT conclusion.\n7.04 > 6.251 May be implied by FT correct conclusion in words.\nReject H\no\nInsufficient evidence to suggest size of shell is independent of A1 Correct conclusion, in context.\nlocation/Evidence suggests that size of shell is dependent on Level of uncertainty in language is used.\n(beach) location\n7\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
        7
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      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_43-p07.png"
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    },
    {
      "id": "9231-2022-mj-43-q03",
      "question_id": "9231-2022-mj-43-q03",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 6 5 1 M1 Obtain two correct equations and attempt to solve.\nG (t) =  t t2\nX\n12 12 12\n6\nP(0 heads) = 1a1b\n12\n5\nP(1 head) = a1b1ab\n12\n1\nP(2 heads) = ab\n12\n 7 1 \nab , ab \n 12 12\nSolve to give A1 Correct value for a.\n1  1\na b \n3  4\n2\n\n3(b) 2 M1 Square their G (t).\n 6 5 1  X\nG Y (t) =   t t2 \n12 12 12 \n1  3660t37t2 10t3t4 M1 Obtain quartic polynomial from 3-term G X (t).\n144\nA1\n3\n© UCLES 2022 Page 8 of 13\n\n3(c) G '(t) = 1  6074t30t2 4t3 M1 Differentiate their G Y (t) twice.\nY\n144\nG \"(t) =\n1  7460t12t2\nY\n144\nVar(Y) = G \" (1) + G ' (1) – (G ' (1))2 M1 Use correct formula and attempt at Var(Y).\nY Y Y\n1 7 49 59\n146   0.819\n144 6 36 72\n0.819 A1\n3\nQuestion Answer Marks Guidance",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2022-mj-43-q04",
      "question_id": "9231-2022-mj-43-q04",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) E( X) x 3  1 1   dx   3 3 x0.5 x1.5 dx   M1\n8 x2   8 \n \n1\n32  A1\nx1.5 2x0.5\n \n83 \n3 2 2  1  A1\n2 3  2  1 3 ( = 1.37)\n \n8 3 3  2\n3\n© UCLES 2022 Page 9 of 13\n\n4(b) 3 1 M1 Integrate.\nFx x \n8 x\n3 1  M1 Change variable throughout in their CDF.\nGy  y  \n8\ny\n\n \ngy 3  y0.5  y1.5 , 1y9 A1 Differentiate to find g(y).\n16\n0 otherwise A1 1y9 and 0 otherwise, dependent on second M1.\n4\n\n4(c) 3 1 M1* Correct method.\nFp0.4 so p 0.4\n8 p\n15p2 16p150 M1 dep Obtain quadratic and solve.\n5\n5p33p50, p\n3\n25 A1\n40th percentile of Y is\n9\n3\n© UCLES 2022 Page 10 of 13",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2022-mj-43-q05",
      "question_id": "9231-2022-mj-43-q05",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5 Signs: + – – + + + + – + + + and use Binomial M1 11 signs (values not needed) allow one error.\nConsider B(11, 0.5) with P(X 8) or P(X 3) M1 Use of correct binomial.\nCompare 0.227 with 0.1 or 0.113 with 0.05 A1\nDifferences: M1 Allow one error.\n0.06 –0.05 –0.03 0.08 0.04 0.02\n0.10 –0.15 0.09 0.01 0.12\nSigned Ranks: M1 Allow one swap.\n6 -5 -3 7 4 2\n9 -11 8 1 10\n(P = 47) Q = 19 A1\nCritical value from table = 13 B1\n19 > 13 M1 FT their (19) compared with 13 and correct FT conclusion\nAccept H\n0\nInsufficient evidence to support median not being 5.50 A1 BOTH conclusions correct from correct working, in context.\nLevel of uncertainty in language.\n9\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_43-p11.png"
      ]
    },
    {
      "id": "9231-2022-mj-43-q06",
      "question_id": "9231-2022-mj-43-q06",
      "subject": "9231",
      "year": 2022,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) 1  11202  M1 40\ns 2  31400 1.0256\nx 39 40 39\n \n1  13702 \ns\ny\n2 \n49\n37600\n50\n1.2653 62\nBoth required\n  49\n1.0256 1.2653 M1 2434\ns2  \n40 50 47775\n1.03 1.27\nCondone s2   or 0.05115.\n40 50\ns2 0.050947 A1 May be implied by 2.66 as z value.\nH :   0 and H :   0 B1\n0 a b 1 a b\n1120 1370 M1\n\n40 50 0.6\nz \ns 0.050947\nz = 2.66 A1\nCritical value of z = 2.326 M1 Compare their z value with 2.326 and correct FT conclusion.\n2.66 > 2.326\nReject H\n0\nSufficient evidence to support difference in population means oe A1 Correct conclusion, in context.\nLevel of uncertainty in language is used.\n8\n© UCLES 2022 Page 12 of 13\n\n6(b) 0.60.25 M1 Condone 1.54.\nz\ns\nz 1 .55 A1\n0.9392 – 0.9395 or 0.0605 – 0.0608 A1\n6.05 A1 Condone 6.05 – 6.08, accept \n4\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-May-June/ms/9231_s22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-may-june/9231_s22_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s22_ms_43-p12.png",
        "../answer-assets/9231_s22_ms_43-p13.png"
      ]
    },
    {
      "id": "9231-2022-on-11-q01",
      "question_id": "9231-2022-on-11-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) 2+=−b B1\n2 +2=0\n=−2 leading to −4+=−b M1 Solves simultaneous equations, or, express b and d\nin terms of  and .\n= 1b, =−2b A1 3\n3 3 b=3=−  and d =−2.\n2\n2=−d leading to 4 b3 =−d leading to4b3+27d =0 M1 A1 Substitutes into third equation, AG.\n27\n5\n\n1(b) 3b=b2 leading to b=3 M1 A1 Uses 22 +2 =(2+)2 −2 ( 2 +2 ) or\nsubstitutes for , in terms of b.\nd=−4 A1\n3\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p06.png"
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    },
    {
      "id": "9231-2022-on-11-q02",
      "question_id": "9231-2022-on-11-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 72 +97−50=96 is divisible by 48 B1 Checks base case.\nAssume that 72k +97k −50 is divisible by 48 for some positive integer k. B1 States inductive hypothesis.\nThen 72k+2 +97k+1−50=(48+1)72k +(96+1)97k −50 M1 A1 Separates 72k +97k −50 or considers difference.\nis divisible by 48 because 4872k +9697k is divisible by 48. A1\nHence, by induction, true for every positive integer n. A1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p07.png"
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    },
    {
      "id": "9231-2022-on-11-q03",
      "question_id": "9231-2022-on-11-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 8r3+12r2 +6r+1− ( 8r3−12r2 +6r−1 ) =24r2 +2 M1 Expands.\nn M1 A1 A1 Sums both sides and uses method of differences\n24r2 +2n=(2n+1)3−1\nwith sufficient complete terms.\nr=1\nn A1 AG.\n24r2 =8n3+12n2 +4n=4n(n+1)(2n+1)\nr=1\n5\n© UCLES 2022 Page 7 of 15\n\n3(b) 2n n 2n n M1 Relates with sum of squares.\nS =r2 +2(2r)2 =r2 +8r2\n2n\nr=1 r=1 r=1 r=1\nS = 1(2n)(2n+1)(4n+1)+ 8n(n+1)(2n+1) M1 n\n2n 6 6 Applies r2 =1n(n+1)(2n+1).\n6\nr=1\nS = 1n(2n+1)(4n+1+4n+4)= 1n(2n+1)(8n+5) A1 n\n2n 3 3 r = 1n(n+1).\n2\nr=1\nAlternative for 3(b)\nn n n M1 Relates with sum of squares.\nS =(2r−1)2 +3(2r)2 =( 4r2 −4r+1+12r2)\n2n\nr=1 r=1 r=1\n16n(n+1)(2n+1) n(n+1) M1 Applies\n−4 +n\n6 2 n n\nr2 =1n(n+1)(2n+1) and r = 1n(n+1)\n6 2\nr=1 r=1\nn( 16n2 +18n+5 ) = n (2n+1)(8n+5) A1\n3 3\n3\n\n3(c) 16 B1 FT\n3\n1\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p07.png",
        "../answer-assets/9231_w22_ms_11-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-11-q04",
      "question_id": "9231-2022-on-11-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) i j k −4  2  M1 A1 Finds vector perpendicular to Π.\n   \n−1 2 1 = 2 ~ −1\n   \n   \n3 2 −1 −8  4 \n2(3)−(−2)+4(1)=12 M1 Substitutes point on Π.\n2x−y+4z=12 A1\n4\n\n4(b) 0  2  M1 A1 FT Uses dot product of j+k and their normal.\n    3\n1 −1 = 2 21cosleading to cos= May be implied by angle 62.4°\n   \n42\n   \n1  4 \nAcute angle between l and Π is 90−=27.6 A1 0.48 radians\n3\n\n4(c)  2  2+2t M1 A1 Forms OFusing parameter.\n   \nOF =OP+t −1 = 3−t\n   \n   \n 4  1+4t\n21t+5=12 M1 Substitutes into equation of plane.\n8 A1\n1 \nOF = 8\n \n3\n \n7\n4\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p09.png"
      ]
    },
    {
      "id": "9231-2022-on-11-q05",
      "question_id": "9231-2022-on-11-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) Shear (in the x-direction) followed by a rotation (anticlockwise about the origin B2 Award B1 if given in the wrong order.\nthrough 45°).\n2\n\n5(b)  k+1 k−1  k+1 k−1 M1 A1 Finds M−1.\n− −\n   \n1 2 2 2 2\nM−1 =  = \n1 1 1   1 1 \n− −\n   \n 2 2   2 2 \n2\n\n5(c) 1 2 1(k−1) 2 B1\nM=2 2 \n1 2 1(k+1) 2 \n \n2 2\n1 k−1x x+(k−1)y B1 x X\n1 2  = 1 2  Transforms   to  .\n2 1 k+1y 2 \n\nx+(k+1)y\n\n y Y \nx+m(k+1)x=m ( x+m(k−1)x ) M1 A1 Uses y=mx and Y =mX .\n(k−1)m2 −km−1=0 A1\nk2 +4k−40 M1 Sets discriminant negative.\n( ) ( ) A1\n2 −1− 2 k2 −1+ 2\n© UCLES 2022 Page 10 of 15\n\n5(c) Alternative method for question 5(c)\n1 2 1(k−1) 2 B1\nM=2 2 \n1 2 1(k+1) 2 \n \n2 2\nB1 Sets up to find characteristic equation.\n1 2− 1(k−1) 2\n[M - I =] 2 2 =0\n1 2 1(k+1) 2−\n2 2\n( 1 2− )( 1(k+1) 2− ) −1(k−1) 21 2=0 M1 A1 Evaluates determinant.\n2 2 2 2\n22− 2(k+2)+2=0 A1\nk2 +4k−40 M1 Sets discriminant negative.\n( ) ( ) A1\n2 −1− 2 k2 −1+ 2\n7\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p10.png",
        "../answer-assets/9231_w22_ms_11-p11.png"
      ]
    },
    {
      "id": "9231-2022-on-11-q06",
      "question_id": "9231-2022-on-11-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 + y2 =r2 x=rcos y=rsin B1 Used.\nr4 =36r2( cos2−sin2 ) =36r2cos2 M1 Substitutes and applies cos2=cos2−sin2 .\nr2 =36cos2 A1 AG.\n3\n\n6(b) B1 Closed curve.\nCorrect position and symmetrical about initial line.\nB1 Single correct loop.\n6 B1 States maximum distance or labels sketch.\n3\n© UCLES 2022 Page 12 of 15\n\n6(c) 18  4 cos2d=9sin2 4 M1 Forms 1 2 r2 d\n− 4 − 4\n18 A1\n2\n\n6(d) 1 B1\ny=6cos22sin\ncos 1 2 2cos−cos−1 2 2sin2sin=0 M1 A1 Sets dy =0.\nd\n2tan2 M1 Applies suitable trigonometric identity.\ncos2cos−sin2sin=0 leading to 1=\n1−tan2\n=1π A1\n6\n3 2 A1\n2\n6\n© UCLES 2022 Page 13 of 15",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p12.png",
        "../answer-assets/9231_w22_ms_11-p13.png"
      ]
    },
    {
      "id": "9231-2022-on-11-q07",
      "question_id": "9231-2022-on-11-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) x= 2 B1 Vertical asymptote.\n5\n(5x−2)(x+ 2)+ 4 M1 A1 Oblique asymptote.\ny= 5 5 leading to y= x+ 2\n5x−2 5\n3\n\n7(b) dy (5x−2)(10x)−(5x2)(5) M1 dy\n= Finds .\ndx (5x−2)2 dx\n5x2 −4x=0 M1 Sets equal to 0 and forms quadratic equation.\n(0,0), (4,8) A1 A1\n5 5\n4\n\n7(c) B1 Axes and asymptotes.\nB1 Correct upper branch and asymptotic behaviour.\nB1 Correct lower branch.\n3\n© UCLES 2022 Page 14 of 15\n\n7(d) B1 FT FT from sketch in part (c).\nB1 Correct shape as x tends to infinity.\n5x2 5x2 M2 Finds critical points, award M1 for each case.\n=2 or =−2\n5x−2 5x−2\n5x2 −10x+4=0 or 5x2 +10x−4=0\nx=−1−3 5, x=−1+3 5 or x=1−1 5, x=1+1 5 A1 Must be exact.\n5 5 5 5\n−1−3 5x−1+3 5, 1−1 5x1+1 5 A1 FT Follow through on use of decimals.\n5 5 5 5\n6\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_11-p14.png",
        "../answer-assets/9231_w22_ms_11-p15.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q01",
      "question_id": "9231-2022-on-12-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) n n M1 Uses List of formulae (MF19).\nr(r+2)=r2 +2r=1n(n+1)(2n+1)+n(n+1)\n6\nr=1 r=1\n1n(n+1)(2n+7) A1 At least all like terms collected and\n6\nfractions simplified.\n2\n\n1(b) 1 11 1  M1 A1 Expresses in partial fractions.\n=  − \nr(r+2) 2r r+2\nn 2 1 1 1 M1 Writes enough terms.\n =1− + −\nr(r+2) 3 2 4\nr=1\n1 1 1 1\n+ − + −\n3 5 4 6\n1 1 1 1\n+ − + −\nn−1 n+1 n n+2\nn 1 13 1 1  A1 OE\n = − −\n \nr(r+2) 22 n+1 n+2\nr=1\n4\n\n1(c) 3 B1 FT FT their part (b) of correct form.\n4\n1\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p06.png"
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    },
    {
      "id": "9231-2022-on-12-q02",
      "question_id": "9231-2022-on-12-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) y=x−2 leading tox= y−1 2 B1 Correct substitution.\ny−2+3y−1+2y−1 2 +6=0 leading to 1+3y+2y 3 2 +6y2 =0 M1 Obtains an equation not involving\nradicals.\n36y4 +32y3 +21y2 +6y+1=0 A1\n1 1 1 1 B1 FT\n+ + + =−32 =−8\n2 2 2 2 36 9\n4\n\n2(b) =6 B1 SOI\n1 1 1 1 222 +222 +222 +222 M1 Relates to coefficients.\n+ + + =\n2 2 2 2 2222\n222 +222222 +222 =−32 A1\n3\n\n2(c) 2 M1 Uses formula for sum of squares.\n1 1 1 1  1 1 1 1 \n+ + + = + + + \n4 4 4 4 2 2 2 2\n\n−2 ( −2−2 +−2−2 +a−2−2 +−2−2 +−2−2 +−2−2)\n=\n( −8)2\n−2\n(21)\n9 36\n− 61 A1\n162\n2\n© UCLES 2022 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p07.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q03",
      "question_id": "9231-2022-on-12-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) Shear [in the y-direction] followed by a stretch, [parallel to the y-axis, scale factor k]. B2 Award B1 if given in the wrong order.\n2\n\n3(b) 1 0 −1  1 0 1 0 −1 1 0 B1\n  =  ,   = \nk 1 −k 1 0 k\n\n0 1\n\nk\n 1 01 0 B1 Correct order.\nM−1=\n \n−k 1 0 1\n\nk\n2\n\n3(c)  1 0 B1\nM= \nk2 k\n 1 0x  x  B1 x X\n  =  Transforms   to  \nk2 ky k2x+ky y Y \nk2x+ky= y M1 X x\nSets  = \nY  y\nk2 A1 AG.\ny= x\n1−k\n4\n\n3(d) 1= k M1 Uses DEF = detM ABC\nk=−1 A1\n2\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q04",
      "question_id": "9231-2022-on-12-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4 d B1 Checks base case using product rule.\n(xf(x))=xf '(x)+ f(x)=xf '(x)+(2(1)−1)f(x)\ndx\nd2k−1 B1 States inductive hypothesis.\nAssume true for n=k, so (xf(x))=xf '(x)+(2k−1)f(x)\ndx2k−1\nd2k M1 A1 Differentiates once.\nThen (xf(x))=xf(x)+2kf '(x)\ndx2k\nd2k+1 M1 A1 Differentiates again.\n(xf(x))=xf '(x)+ f(x)+2kf(x)\ndx2k+1\n=xf '(x)+(2k+1)f(x)\nSo, it is also true for n=k+1. Hence, by induction, true for all positive integers. A1 States conclusion.\n7\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p09.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q05",
      "question_id": "9231-2022-on-12-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Correct position and domain and r strictly\nincreasing.\nB1 Decreasing positive gradient.\n=0\n(a,0) (2a,1π) B1 Can be labelled on their sketch.\n4\n3\n\n5(b) y=(2a)sin (1π ) M1 Uses y=rsin\n4\na 2 A1\n2\n© UCLES 2022 Page 10 of 15\n\n5(c) 1a2 1 4 π sec4d M1 Forms 1\n2\nr2d with correct limits.\n2\n0\n=1a2 1 4 π sec2 ( tan2+1 ) d M1 Applies sec2=tan2+1 to obtain an\n2 0 integrable form.\n1π A1\n= 1a21tan3+tan4\n2 3  0\n= 2a2 A1\n3\n4\n\n5(d) x2 + y2 =r2 x=rcos M1 Substitutes to eliminate either r or .\nr2cos2=ar leading to x2 =a x2+y2 leading tox4 =a2( x2 +y2) M1 Completes substitution to eliminate both\nr and  and makes y or y2 the subject.\nA1 AEF\ny= a−2x4−x2 =x a−2x2−1\n3\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p10.png",
        "../answer-assets/9231_w22_ms_12-p11.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q06",
      "question_id": "9231-2022-on-12-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) 0 2 −2 B1 Finds a vector joining any point of l 1 to\n      any point of l .\n2 − 0 = 2 2\n     \n     \n6 1  5 \n. i j k  2  M1 A1 Finds common perpendicular.\n \n1 2 −2 = −4\n \n \n1 −1 2 −3\n−2  2  M1 A1 Uses formula for shortest distance.\n1     27\n2 −4 = =5.01\n   \n29     29\n 5  −3\n5\n\n6(b)(i) r=2i+k+s(i−j+2k)+t(2i−4j−3k) B1 FT their common perpendicular from part\n(a).\n1\n\n6(b)(ii) i j k −14 M1 A1FT Finds normal to the plane  . FT their\n2\n \n1 2 −2 = −1 common perpendicular from part (a).\n \n \n2 −4 −3  −8 \n14(0)+(2)+8(6)=5014x+ y+8z=50 M1 A1 Substitutes point.\n4\n© UCLES 2022 Page 12 of 15\n\n6(c) i j k 11 M1 A1 FT Finds normal to the plane  . FT their\n1\n \n1 −1 2 = 7 part (b)(i).\n \n \n2 −4 −3 −2\n11 14 M1 A1 Uses dot product of normal vectors.\n    145\n7 1 = 174 261cos leading to cos=\n   \n174 261\n   \n−2  8 \n47.1 A1 0.822 radians\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p12.png",
        "../answer-assets/9231_w22_ms_12-p13.png"
      ]
    },
    {
      "id": "9231-2022-on-12-q07",
      "question_id": "9231-2022-on-12-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1 B1 States vertical asymptote.\n(x+1)(x−2)+2 M1 Finds oblique asymptote.\ny=\nx+1\ny=x−2 A1\n3\n\n7(b) dy M1 A1 Differentiates and sets derivative equal to\n=1−2(x+1)−2 =0(x+1)2 =2\n0.\ndx\n( ) ( ) A1 A1\n−1+ 2,−3+2 2 , −1− 2,−3−2 2\n4\n© UCLES 2022 Page 13 of 15\n\n7(c) B1 Axes and asymptotes labelled.\nB1 Upper branch with (0,0) and (1,0) stated\nor clear on scale.\nB1 Lower branch correct and all asymptotic\napproaches correct.\n3\n© UCLES 2022 Page 14 of 15\n\n7(d) B1 FT FT from sketch in part (c).\nx2 −x x2 −x M2 Finds critical points, award M1 for each\n=6 or =−6\ncase.\nx+1 x+1\nx2 −7x−6=0 or x2 +5x+6=0\nx=−3,−2, 7−1 73, 7+1 73 A1\n2 2 2 2\n−3x−2 and 7−1 73x7+1 73 A1\n2 2 2 2\n5\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
        13,
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_12-p13.png",
        "../answer-assets/9231_w22_ms_12-p14.png",
        "../answer-assets/9231_w22_ms_12-p15.png"
      ]
    },
    {
      "id": "9231-2022-on-13-q01",
      "question_id": "9231-2022-on-13-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) 2+=−b B1\n2 +2=0\n=−2 leading to −4+=−b M1 Solves simultaneous equations, or, express b and d\nin terms of  and .\n= 1b, =−2b A1 3\n3 3 b=3=−  and d =−2.\n2\n2=−d leading to 4 b3 =−d leading to4b3+27d =0 M1 A1 Substitutes into third equation, AG.\n27\n5\n\n1(b) 3b=b2 leading to b=3 M1 A1 Uses 22 +2 =(2+)2 −2 ( 2 +2 ) or\nsubstitutes for , in terms of b.\nd=−4 A1\n3\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_13-p06.png"
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    },
    {
      "id": "9231-2022-on-13-q02",
      "question_id": "9231-2022-on-13-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 72 +97−50=96 is divisible by 48 B1 Checks base case.\nAssume that 72k +97k −50 is divisible by 48 for some positive integer k. B1 States inductive hypothesis.\nThen 72k+2 +97k+1−50=(48+1)72k +(96+1)97k −50 M1 A1 Separates 72k +97k −50 or considers difference.\nis divisible by 48 because 4872k +9697k is divisible by 48. A1\nHence, by induction, true for every positive integer n. A1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_13-p07.png"
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    },
    {
      "id": "9231-2022-on-13-q03",
      "question_id": "9231-2022-on-13-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 8r3+12r2 +6r+1− ( 8r3−12r2 +6r−1 ) =24r2 +2 M1 Expands.\nn M1 A1 A1 Sums both sides and uses method of differences\n24r2 +2n=(2n+1)3−1\nwith sufficient complete terms.\nr=1\nn A1 AG.\n24r2 =8n3+12n2 +4n=4n(n+1)(2n+1)\nr=1\n5\n© UCLES 2022 Page 7 of 15\n\n3(b) 2n n 2n n M1 Relates with sum of squares.\nS =r2 +2(2r)2 =r2 +8r2\n2n\nr=1 r=1 r=1 r=1\nS = 1(2n)(2n+1)(4n+1)+ 8n(n+1)(2n+1) M1 n\n2n 6 6 Applies r2 =1n(n+1)(2n+1).\n6\nr=1\nS = 1n(2n+1)(4n+1+4n+4)= 1n(2n+1)(8n+5) A1 n\n2n 3 3 r = 1n(n+1).\n2\nr=1\nAlternative for 3(b)\nn n n M1 Relates with sum of squares.\nS =(2r−1)2 +3(2r)2 =( 4r2 −4r+1+12r2)\n2n\nr=1 r=1 r=1\n16n(n+1)(2n+1) n(n+1) M1 Applies\n−4 +n\n6 2 n n\nr2 =1n(n+1)(2n+1) and r = 1n(n+1)\n6 2\nr=1 r=1\nn( 16n2 +18n+5 ) = n (2n+1)(8n+5) A1\n3 3\n3\n\n3(c) 16 B1 FT\n3\n1\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_13-p07.png",
        "../answer-assets/9231_w22_ms_13-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-13-q04",
      "question_id": "9231-2022-on-13-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) i j k −4  2  M1 A1 Finds vector perpendicular to Π.\n   \n−1 2 1 = 2 ~ −1\n   \n   \n3 2 −1 −8  4 \n2(3)−(−2)+4(1)=12 M1 Substitutes point on Π.\n2x−y+4z=12 A1\n4\n\n4(b) 0  2  M1 A1 FT Uses dot product of j+k and their normal.\n    3\n1 −1 = 2 21cosleading to cos= May be implied by angle 62.4°\n   \n42\n   \n1  4 \nAcute angle between l and Π is 90−=27.6 A1 0.48 radians\n3\n\n4(c)  2  2+2t M1 A1 Forms OFusing parameter.\n   \nOF =OP+t −1 = 3−t\n   \n   \n 4  1+4t\n21t+5=12 M1 Substitutes into equation of plane.\n8 A1\n1 \nOF = 8\n \n3\n \n7\n4\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_13-p09.png"
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    },
    {
      "id": "9231-2022-on-13-q05",
      "question_id": "9231-2022-on-13-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) Shear (in the x-direction) followed by a rotation (anticlockwise about the origin B2 Award B1 if given in the wrong order.\nthrough 45°).\n2\n\n5(b)  k+1 k−1  k+1 k−1 M1 A1 Finds M−1.\n− −\n   \n1 2 2 2 2\nM−1 =  = \n1 1 1   1 1 \n− −\n   \n 2 2   2 2 \n2\n\n5(c) 1 2 1(k−1) 2 B1\nM=2 2 \n1 2 1(k+1) 2 \n \n2 2\n1 k−1x x+(k−1)y B1 x X\n1 2  = 1 2  Transforms   to  .\n2 1 k+1y 2 \n\nx+(k+1)y\n\n y Y \nx+m(k+1)x=m ( x+m(k−1)x ) M1 A1 Uses y=mx and Y =mX .\n(k−1)m2 −km−1=0 A1\nk2 +4k−40 M1 Sets discriminant negative.\n( ) ( ) A1\n2 −1− 2 k2 −1+ 2\n© UCLES 2022 Page 10 of 15\n\n5(c) Alternative method for question 5(c)\n1 2 1(k−1) 2 B1\nM=2 2 \n1 2 1(k+1) 2 \n \n2 2\nB1 Sets up to find characteristic equation.\n1 2− 1(k−1) 2\n[M - I =] 2 2 =0\n1 2 1(k+1) 2−\n2 2\n( 1 2− )( 1(k+1) 2− ) −1(k−1) 21 2=0 M1 A1 Evaluates determinant.\n2 2 2 2\n22− 2(k+2)+2=0 A1\nk2 +4k−40 M1 Sets discriminant negative.\n( ) ( ) A1\n2 −1− 2 k2 −1+ 2\n7\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2022-on-13-q06",
      "question_id": "9231-2022-on-13-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 + y2 =r2 x=rcos y=rsin B1 Used.\nr4 =36r2( cos2−sin2 ) =36r2cos2 M1 Substitutes and applies cos2=cos2−sin2 .\nr2 =36cos2 A1 AG.\n3\n\n6(b) B1 Closed curve.\nCorrect position and symmetrical about initial line.\nB1 Single correct loop.\n6 B1 States maximum distance or labels sketch.\n3\n© UCLES 2022 Page 12 of 15\n\n6(c) 18  4 cos2d=9sin2 4 M1 Forms 1 2 r2 d\n− 4 − 4\n18 A1\n2\n\n6(d) 1 B1\ny=6cos22sin\ncos 1 2 2cos−cos−1 2 2sin2sin=0 M1 A1 Sets dy =0.\nd\n2tan2 M1 Applies suitable trigonometric identity.\ncos2cos−sin2sin=0 leading to 1=\n1−tan2\n=1π A1\n6\n3 2 A1\n2\n6\n© UCLES 2022 Page 13 of 15",
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2022-on-13-q07",
      "question_id": "9231-2022-on-13-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) x= 2 B1 Vertical asymptote.\n5\n(5x−2)(x+ 2)+ 4 M1 A1 Oblique asymptote.\ny= 5 5 leading to y= x+ 2\n5x−2 5\n3\n\n7(b) dy (5x−2)(10x)−(5x2)(5) M1 dy\n= Finds .\ndx (5x−2)2 dx\n5x2 −4x=0 M1 Sets equal to 0 and forms quadratic equation.\n(0,0), (4,8) A1 A1\n5 5\n4\n\n7(c) B1 Axes and asymptotes.\nB1 Correct upper branch and asymptotic behaviour.\nB1 Correct lower branch.\n3\n© UCLES 2022 Page 14 of 15\n\n7(d) B1 FT FT from sketch in part (c).\nB1 Correct shape as x tends to infinity.\n5x2 5x2 M2 Finds critical points, award M1 for each case.\n=2 or =−2\n5x−2 5x−2\n5x2 −10x+4=0 or 5x2 +10x−4=0\nx=−1−3 5, x=−1+3 5 or x=1−1 5, x=1+1 5 A1 Must be exact.\n5 5 5 5\n−1−3 5x−1+3 5, 1−1 5x1+1 5 A1 FT Follow through on use of decimals.\n5 5 5 5\n6\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-21-q01",
      "question_id": "9231-2022-on-21-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 ex B1 Finds first derivative.\ny'=\n1+ex\nex B1 Finds second derivative.\ny''=\n( 1+ex)2\ny(0)=ln2, y'(0)= 1, y''(0)= 1 B1 Evaluates at x=0.\n2 4\ny=y(0)+y'(0)x+ 1 y''(0)x2+ M1 Allow 2! missing.\n2!\nln2+1x+1x2 A1 Decimal used for ln2 scores A0.\n2 8\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-21-q02",
      "question_id": "9231-2022-on-21-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) 1 −1 2 M1 A1 Shows that determinant is zero.\n1 −1 −3 =−7−3+10−2(0)=0\n1 −1 7\n2\n\n2(b) x− y+ 2z= 4, M1 A1 M1 for row operations or eliminating one\nx− y− 3z= −5, z= 9, x− y= 2 variable.\n5 5 A1 for deriving one correct equation with two\nx− y+ 7z= 13,\nunknowns.\nThe three planes form a sheaf. B1 All three planes have the same common line.\n3\n© UCLES 2022 Page 6 of 13\n\n2(c) x− y+ 2z= 4, B1 Derives contradiction.\nx− y− 3z= a, z= 9 ,3z+a=4−2za=−5\n5\nx− y+ 7z= 13,\nThe three planes form a triangular prism. B1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2022-on-21-q03",
      "question_id": "9231-2022-on-21-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 x=et −t2 B1\ny=2te 1 2 t B1\n( et −t2)2 +4t2et = e2t +2t2et +t4 =et +t2 M1 A1 Finds x2 + y2.\n2 2 M1 A1\n et +t2dt =et +1t3 =e2 + 5\n0  3  0 3\n6\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2022-on-21-q04",
      "question_id": "9231-2022-on-21-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "4(a) coshx= 1 ( ex +e−x) sinhx= 1 ( ex −e−x) B1\n2 2\n1 ( e2x +2+e−2x −e2x +2−e−2x) =1 M1 A1 Expands, AG. Clear LHS to RHS for A1.\n4\n3\n© UCLES 2022 Page 7 of 13\n\n4(b) d ( tan−1sinhx ) = coshx M1 A1 Applies d ( tan−1u ) = 1 .\ndx sinh2x+1 du u2 +1\ncoshx A1 AG.\n= =sechx\ncosh2x\n3\ny\n\n4(c) B1 Correct shape, symmetrical about x=0.\ny=0 B1 Accept labels on their sketch.\n2\n\n4(d) n M1 A1 Compares sum with integral. Limits correct for\nn\nsechr sechxdx A1.\n0\nr=1\nn A1 AG.\n=tan−1sinhx =tan−1sinhn\n \n0\n3\n\n4(e) 1 B1\nπ\n2\n1\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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        "../answer-assets/9231_w22_ms_21-p08.png"
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    },
    {
      "id": "9231-2022-on-21-q05",
      "question_id": "9231-2022-on-21-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 2m2+2m+1=0m=−11i M1 Correct auxiliary equation.\n2 2\ny=e−1\n2\nx(Acos1x+Bsin1x) A1 Complementary function. Accept “y=”\n2 2\nmissing.\ny= px2 +qx+r leading toy'=2px+q leading to y''=2p B1 Particular integral and its derivatives.\np=4 4p+q=3 4p+2q+r=3 M1 Substitutes and equates coefficients.\nq=−13 r=13 A1\ny=e−1 2 x(Acos1x+Bsin1x)+4x2 −13x+13 A1 General solution.\n2 2\ny'=e−1\n2\nx(−1 Asin1x+1Bcos1x)−1e−1\n2\nx(Acos1x+Bsin1x)+8x−13 M1 Full use of product rule. y must be of form\n2 2 2 2 2 2 2 shown in the general solution above with p0.\nA+13=0 1B−1 A−13=0 A=−13,B=13 M1 A1 Uses initial conditions to derive two linear\n2 2\nequations in two unknowns.\ny=e−1\n2\nx(−13cos1x+13sin1x)+4x2−13x+13 A1\n2 2\n10\n© UCLES 2022 Page 9 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2022-on-21-q06",
      "question_id": "9231-2022-on-21-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Eigenvalues of A are 2, 5 and −2. B1 Lower diagonal matrix or characteristic\nequation.\ni j k −12 1 M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\n=2: 0 3 7 = 0 ~ 0\n   \n   \n0 0 −4  0  0\ni j k −21  1  A1\n   \n=5: −3 −3 −7 = 21 −1\n   \n   \n0 0 7  0   0 \ni j k  28   1  A1\n   \n=−2: 4 −3 −7 = −28 −1\n   \n   \n0 7 7  28   1 \n1 1 1  32 0 0  M1 A1 Or correctly matched permutations of columns.\n    Their eigenvectors must be non-zero and\nThus P= 0 −1 −1 and D= 0 3125 0\n    correctly matched to their eigenvalues raised to\n   \n0 0 1   0 0 −32 the fifth power for M1.\n7\n\n6(b) (−2)(−5)(+2)=3−52−4+20=0 B1 Characteristic equation.\nA3 =5A2 +4A−20I M1 Substitutes for A and makes A3 the subject.\nTolerate I missing.\nA4 =5A3+4A2 −20A=5 ( 5A2 +4A−20I ) +4A2 −20A M1 Multiplies by A and substitutes for A3.\nA4 =29A2 −100I A1 Has Iin answer or states the values of a and b.\n4\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2022-on-21-q07",
      "question_id": "9231-2022-on-21-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) wn −1 B1\nw−1\n1\n\n7(b) (1+itan)k =seck(cos+isin)k =seck(cosk+isink) M1 A1 Applies de Moivre’s theorem, AG.\n2\n\n7(c) n−1 (1+itan)n −1 M1 A1 Applies part (a).\n(1+itan)k\n=\nitan\nk=0\n−cotsecn(icosn−sinn)+icot A1\nn−1 M1 A1 Takes imaginary part, AG.\nsecksink=cot ( 1−secncosn )\nk=0\n5\n\n7(d) 6m−1 M1 A1 Sets = 1π.\n=1π 2ksin (1kπ ) =1 3 ( 1−26mcos(2mπ)) =1 3 ( 1−26m) 3\n3 3 3 3\nk=0\n2\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2022-on-21-q08",
      "question_id": "9231-2022-on-21-q08",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) du B1\n=−2(−1) leading to (−1)d=−1du\nd 2\n −1  1 M1 A1 Applies substitution. For M1, integrand must be\n  1−(−1)2 d=−1 2   u du=− u +C=− 1−(−1)2 +C of the form k u , where k0 is constant. For\nA1, allow “+C” missing.\n3\n\n8(b) dy y B1 Divides through by .\n− =sin−1(−1)\nd \ne\n−−1d\n=e−ln=−1\nM1 A1 Finds integrating factor.\nd ( −1y ) =sin−1(−1) M1 Correct form on LHS using their integrating\nfactor.\nd\n  M1 A1 Integrates sin−1(−1).\n−1y=sin−1(−1)− d\n\n 1−(−1)2\n   −1  1 M1 Applies part (a) and uses\n d=  d+  d 1\n 1−(−1)2  1−(−1)2  1−(−1)2 \n1−x2\ndx=sin−1x+C.\n−1y=sin−1(−1)+ 1−(−1)2 −sin−1(−1)+C A1\n© UCLES 2022 Page 12 of 13\n\n8(b) 1=1+C M1 Substitutes initial conditions into their\nexpression in y and .\ny=(−1)sin−1(−1)+ 1−(−1)2 M1 A1 Divides through by their integrating factor.\n11\n© UCLES 2022 Page 13 of 13",
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      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-22-q01",
      "question_id": "9231-2022-on-22-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 2 3 M1 A1 Evaluates determinant or solves system of\nequations (eliminate at least one variable for M1):\nk 4 6 =6k−12\n7 8 9\n2 4 k−8\nx=− , y= , z= .\nk−2 k−2 3(k−2)\nk2 A1\n3\n\n1(b) Two parallel planes. Other plane not parallel. B1 SC k =2,two parallel planes.\nParallel planes not identical. B1 Accept diagram.\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_22-p06.png"
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    },
    {
      "id": "9231-2022-on-22-q02",
      "question_id": "9231-2022-on-22-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) d B1 Differentiates (x+1)y correctly.\n((x+1)y)=(x+1)y'+ y\ndx\nd ( y2)\n=2yy'\nB1 Differentiates y2correctly.\ndx\n(1)y'−2+2(−2)y'=0 leading toy'=−2 B1 Substitutes (0,−2), AG.\n3\n3\n© UCLES 2022 Page 6 of 15\n\n2(b) (x+1)y''+2y' B1 Differentiates (x+1)y'+ y.\n+2yy''+2(y')2\n=0\nB1 Differentiates 2yy'.\ny''+2\n( −2)\n+2(−2)y''+2\n( −2)2\n=0\nM1 Substitutes (0,−2) into their expression with y''.\n3 3\ny''=− 4 A1\n27\nAlternative method for question 2(b)\n−y (x+1+2y)(−y')+ y(1+2y') B1 B1 B1 for (x+1+2y)(−y') in numerator.\ny'= y''=\nx+1+2y (x+1+2y)2 B1 for y(1+2y') in numerator.\n(0+1+2(−2))(2)−2 ( 1+2(−2)) M1 Substitutes (0,−2) and y'=−2 into their\ny''= 3 3 3\n(0+1+2(−2))2 expression.\ny''=− 4 A1\n27\n4\n© UCLES 2022 Page 7 of 15",
      "source_pages": [
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        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-22-q03",
      "question_id": "9231-2022-on-22-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) M1\n1+ ( ex −1e−x)2 = 1+e2x −1+ 1 e−2x = e2x +1+ 1 e−2x dy 2\n4 2 16 2 16 Finds 1+  \ndx\nA1\n( ex +1e−x)2 =ex +1e−x\n4 4\n2π 1( ex + 1e−x)( ex + 1e−x) dx M1 Correct formula for surface area with limits.\n4 4\n0\n1 A1 Expanded integrand.\n2π e2x + 1 + 1 e−2x dx\n2 16\n0\n1 M1 Integrates and substitutes limits. Integrand must be\n21e2x + 1x− 1 e−2x\n2 2 32  0 of the form ae2x +be−2x +c, with a and b\nnonzero.\n2π ( 1e2 − 1 e−2 + 1 ) =π ( e2 − 1 e−2 + 1 ) A1\n2 32 32 16 16\n6\n\n3(b)  x2  1 (−x)2  M1 Uses expansion of ex or finds first and second\n1+x+ + + 1−x+ + \n2! 4 2! derivatives ex −1e−x and ex +1e−x.\n    4 4\n5+3x+5x2 A1\n4 4 8\n2\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-22-q04",
      "question_id": "9231-2022-on-22-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4 dx 4 B1 Divides through by4t2 −1 .\n+ x=1\ndt 4t2 −1\n4 2 2 4 4 1 B1 Writes as partial fractions or a form to which the\n= − or = =\n4t2 −1 2t−1 2t+1 4t2 −1 (2t)2 −1 t2 −(1)2 formula from MF19 can be applied.\n2\ne\n4 ( 4t2−1 )−1 dt =eln(2t−1)−ln(2t+1)\n=\n2t−1 M1 A1 Finds integrating factor.\n2t+1\nd  2t−1 2t−1 2t+1−2 M1 Correct form on LHS and attempt to integrate RHS.\nx = =\nat+b\ndt 2t+1 2t+1 2t+1\nRHS must be of the form with a,b,c,d 0.\nct+d\n2t−1 A1\nx =t−ln(2t+1)+C\n2t+1\n1=1−ln3+C M1 Substitutes into their expression in x and t.\n2t+1  3  M1 A1 After integrating, divides their expression through\nx=  t+ln   by their coefficient of x. (Their coefficient must\n2t−1 2t+1\ndepend on t.)\n9\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-22-q05",
      "question_id": "9231-2022-on-22-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 1,i B1\nAccept exponential form\ne0,eiπ 2,eiπ,ei3\n2\nπ\nor\ntrigonometric form. Four values specified.\n1\n© UCLES 2022 Page 9 of 15\n\n5(b) cos4=Re(cos+isin)4 =cos4−6cos2sin2+sin4 M1 A1 Expands and takes real part.\ncos4−6cos2 ( 1−cos2 ) + ( 1−2cos2+cos4 ) M1 Applies sin2=1−cos2.\n8cos4−8cos2+1 A1 AG. Can award final A1 without previous A1 if\npenultimate line seen.\nSC. Applying cos2A=2cos2 A−1 twice scores\n1/4.\n4\n\n5(c) ( 8x4 −8x2 +1 )4 = 9 , x=cos M1 Applies identify given in (b).\n16\ncos4=1 3 A1 Need  and exact value for A1.\n2\n4=1π+2kπ 4=5π+2kπ M1 Solves cos4=1 3 or cos4=−1 3\n6 6 2 2\ncos ( 1 π ) A1 Gives one correct solution.\n24\ncos(11π),cos(13π),cos(23π), A1 Gives other solutions. Accept\n24 24 24\ncos ( 1 π ) ,cos(11π),cos(5 π),cos(7 π)\ncos(5 π),cos(7 π),cos(17π),cos(19π),\n24 24 24 24\n24 24 24 24\n5\n© UCLES 2022 Page 10 of 15",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_22-p09.png",
        "../answer-assets/9231_w22_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2022-on-22-q06",
      "question_id": "9231-2022-on-22-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 1 1 1 n 1 M1A1 Forms sum of the areas of the rectangles. At least\nSum= + + + = three terms, including first and last for A1.\n   ( n2 +2n ) r2 +2r\nr=2 SC Areas of these three rectangles written down but\nsum not clearly formed scores B1.\nn M1 Compares with integral.\n 1\n\n   ( x2 +2x )\ndx\n1\nx2+2x=(x+1)2\n−1\nB1 Completes the square.\nn M1 A1 Uses formula. Correct limits for A1.\n 1 n\n\ndx=\n\ncosh−1(x+1)\n\n (x+1)2 −1 1\n1\n ln ( x+1+ x2 +2x ) n M1 A1 Uses logarithmic form of cosh−1.\n \n1\nln ( n+1+ n2 +2n ) −ln ( 2+ 3 ) M1 Inserts limits.\n n 1 ln ( n+1+ n2 +2n ) −ln ( 2+ 3 ) +1 3 A1 Adds term from r=1.\nAG\nr2 +2r 3\nr=1\n10\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-22-q07",
      "question_id": "9231-2022-on-22-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) Ae=e leading to e=A-1e M1 Multiples by A-1on LHS.\n−1e=A-1e A1\n2\n\n7(b) i j k  9   1  M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvector.\n=−1: 3 0 3 = 36 ~ 4\n   \n   \n15 −3 3 −9 −1\n2\n\n7(c)  2 0 30  0  B1\n    \n15 −4 3 1 = −4 =−4\n    \n    \n 3 0 20  0 \n 2 0 31  5  B1\n    \n15 −4 3 2 = 10 =5\n    \n    \n 3 0 21  5 \n2\n\n7(d)  1 0 1 −1 0 0 M1 A1 Or correctly matched permutations of columns.\n    M1 for their eigenvectors (all three must be non-\nThus P= 4 1 2 and D= 0 −1 0\n   4  zero) correctly matched to reciprocals of their\n\n−1 0 1\n\n\n\n 0 0\n1\n eigenvalues.\n5\n2\n© UCLES 2022 Page 12 of 15\n\n7(e) (+1)(+4)(−5)=3−21−20=0 B1 Characteristic equation.\nA3−21A−20I=0 M1 Substitutes for A. Tolerate missing I.\n20A−1 =A2 −21I leading toA−1= 1 A2 −21I M1 A1 Multiplies both sides of equation byA−1, need I for\n20 20\nA1.\n4",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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        "../answer-assets/9231_w22_ms_22-p13.png"
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    },
    {
      "id": "9231-2022-on-22-q08",
      "question_id": "9231-2022-on-22-q08",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) dy du B1\n=sinhu\ndx dx\nd2y d2u du 2 B1\n=sinhu +coshu\n \ndx2 dx2 dx\nd2y dy d2u du 2 du M1 Uses substitution to find y-x equation, AG.\n+ −2y=sinhu +coshu +sinhu −2coshu\n \ndx2 dx dx2 dx dx\nd2u du du 2 A1 AG\ncosh2u−1 + +coshu\n \n−2coshu=4e−x\ndx2 dx\n\ndx\n© UCLES 2022 Page 13 of 15\n\n8(a) Alternative for 8(a)\ndu dy B1\n=(sinhu)−1\ndx dx\nd2u  1 d2y dy coshu du B1\n=  −  \ndx2 sinhudx2 dxsinh2udx\nd2u  1 d2y dy 2  coshu \n= −\n     \ndx2 sinhudx2 dx sinh3u\nsinhu\n\n\n\n\n1 \n\nd2y\n−\n\n\ndy\n\n2 \n\ncoshu  +(sinhu)−1 dy\n+\ncoshu \n\ndy\n\n2\n−2coshu\nM1 U ses substitution to find y-x equation, AG.\n sinhudx2 dx sinh3u dx sinh2udx\n \nd2y dy 2  coshu  dy coshu dy 2\n− + + −2coshu\n     \ndx2 dx sinh2u dx sinh2udx\nd2y dy A1 AG\n+ −2y=4e−x\ndx2 dx\n4\n© UCLES 2022 Page 14 of 15\n\n8(b) m2 +m−2=0m=1,−2 M1 Auxiliary equation.\ny= Aex +Be−2x A1 Complimentary function. Allow ‘y=’ missing.\ny=ke−x y'=−ke−x y''=ke−x B1 Particular integral and its derivatives.\nke−x −ke−x −2ke−x =4e−x M1 Substitutes and equates coefficients.\nk=−2 A1 WWW\ny=coshu= Aex +Be−2x −2e−x A1 Must have ‘y=’ or ‘coshu=’ .\ndu B1\ny'=sinhu = Aex −2Be−2x +2e−x\ndx\nA+B−2= 5 A−2B+2=4 A= 28, B= 5 M1 A1 Substitutes initial conditions and forms\n3 9 9\nsimultaneous equations. For M1, allow substitution\ninto their y and y' if two linear equations in two\nunknowns derived.\nu=cosh−1( 28ex +5e−2x −2e−x) A1 Substitutes for y and finds u in terms of x.\n9 9\n10\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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        "../answer-assets/9231_w22_ms_22-p15.png"
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    },
    {
      "id": "9231-2022-on-23-q01",
      "question_id": "9231-2022-on-23-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 ex B1 Finds first derivative.\ny'=\n1+ex\nex B1 Finds second derivative.\ny''=\n( 1+ex)2\ny(0)=ln2, y'(0)= 1, y''(0)= 1 B1 Evaluates at x=0.\n2 4\ny=y(0)+y'(0)x+ 1 y''(0)x2+ M1 Allow 2! missing.\n2!\nln2+1x+1x2 A1 Decimal used for ln2 scores A0.\n2 8\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-23-q02",
      "question_id": "9231-2022-on-23-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) 1 −1 2 M1 A1 Shows that determinant is zero.\n1 −1 −3 =−7−3+10−2(0)=0\n1 −1 7\n2\n\n2(b) x− y+ 2z= 4, M1 A1 M1 for row operations or eliminating one\nx− y− 3z= −5, z= 9, x− y= 2 variable.\n5 5 A1 for deriving one correct equation with two\nx− y+ 7z= 13,\nunknowns.\nThe three planes form a sheaf. B1 All three planes have the same common line.\n3\n© UCLES 2022 Page 6 of 13\n\n2(c) x− y+ 2z= 4, B1 Derives contradiction.\nx− y− 3z= a, z= 9 ,3z+a=4−2za=−5\n5\nx− y+ 7z= 13,\nThe three planes form a triangular prism. B1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w22_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2022-on-23-q03",
      "question_id": "9231-2022-on-23-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 x=et −t2 B1\ny=2te 1 2 t B1\n( et −t2)2 +4t2et = e2t +2t2et +t4 =et +t2 M1 A1 Finds x2 + y2.\n2 2 M1 A1\n et +t2dt =et +1t3 =e2 + 5\n0  3  0 3\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p07.png"
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    },
    {
      "id": "9231-2022-on-23-q04",
      "question_id": "9231-2022-on-23-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "4(a) coshx= 1 ( ex +e−x) sinhx= 1 ( ex −e−x) B1\n2 2\n1 ( e2x +2+e−2x −e2x +2−e−2x) =1 M1 A1 Expands, AG. Clear LHS to RHS for A1.\n4\n3\n© UCLES 2022 Page 7 of 13\n\n4(b) d ( tan−1sinhx ) = coshx M1 A1 Applies d ( tan−1u ) = 1 .\ndx sinh2x+1 du u2 +1\ncoshx A1 AG.\n= =sechx\ncosh2x\n3\ny\n\n4(c) B1 Correct shape, symmetrical about x=0.\ny=0 B1 Accept labels on their sketch.\n2\n\n4(d) n M1 A1 Compares sum with integral. Limits correct for\nn\nsechr sechxdx A1.\n0\nr=1\nn A1 AG.\n=tan−1sinhx =tan−1sinhn\n \n0\n3\n\n4(e) 1 B1\nπ\n2\n1\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p07.png",
        "../answer-assets/9231_w22_ms_23-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-23-q05",
      "question_id": "9231-2022-on-23-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 2m2+2m+1=0m=−11i M1 Correct auxiliary equation.\n2 2\ny=e−1\n2\nx(Acos1x+Bsin1x) A1 Complementary function. Accept “y=”\n2 2\nmissing.\ny= px2 +qx+r leading toy'=2px+q leading to y''=2p B1 Particular integral and its derivatives.\np=4 4p+q=3 4p+2q+r=3 M1 Substitutes and equates coefficients.\nq=−13 r=13 A1\ny=e−1 2 x(Acos1x+Bsin1x)+4x2 −13x+13 A1 General solution.\n2 2\ny'=e−1\n2\nx(−1 Asin1x+1Bcos1x)−1e−1\n2\nx(Acos1x+Bsin1x)+8x−13 M1 Full use of product rule. y must be of form\n2 2 2 2 2 2 2 shown in the general solution above with p0.\nA+13=0 1B−1 A−13=0 A=−13,B=13 M1 A1 Uses initial conditions to derive two linear\n2 2\nequations in two unknowns.\ny=e−1\n2\nx(−13cos1x+13sin1x)+4x2−13x+13 A1\n2 2\n10\n© UCLES 2022 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p09.png"
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    },
    {
      "id": "9231-2022-on-23-q06",
      "question_id": "9231-2022-on-23-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Eigenvalues of A are 2, 5 and −2. B1 Lower diagonal matrix or characteristic\nequation.\ni j k −12 1 M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\n=2: 0 3 7 = 0 ~ 0\n   \n   \n0 0 −4  0  0\ni j k −21  1  A1\n   \n=5: −3 −3 −7 = 21 −1\n   \n   \n0 0 7  0   0 \ni j k  28   1  A1\n   \n=−2: 4 −3 −7 = −28 −1\n   \n   \n0 7 7  28   1 \n1 1 1  32 0 0  M1 A1 Or correctly matched permutations of columns.\n    Their eigenvectors must be non-zero and\nThus P= 0 −1 −1 and D= 0 3125 0\n    correctly matched to their eigenvalues raised to\n   \n0 0 1   0 0 −32 the fifth power for M1.\n7\n\n6(b) (−2)(−5)(+2)=3−52−4+20=0 B1 Characteristic equation.\nA3 =5A2 +4A−20I M1 Substitutes for A and makes A3 the subject.\nTolerate I missing.\nA4 =5A3+4A2 −20A=5 ( 5A2 +4A−20I ) +4A2 −20A M1 Multiplies by A and substitutes for A3.\nA4 =29A2 −100I A1 Has Iin answer or states the values of a and b.\n4\n© UCLES 2022 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2022-on-23-q07",
      "question_id": "9231-2022-on-23-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) wn −1 B1\nw−1\n1\n\n7(b) (1+itan)k =seck(cos+isin)k =seck(cosk+isink) M1 A1 Applies de Moivre’s theorem, AG.\n2\n\n7(c) n−1 (1+itan)n −1 M1 A1 Applies part (a).\n(1+itan)k\n=\nitan\nk=0\n−cotsecn(icosn−sinn)+icot A1\nn−1 M1 A1 Takes imaginary part, AG.\nsecksink=cot ( 1−secncosn )\nk=0\n5\n\n7(d) 6m−1 M1 A1 Sets = 1π.\n=1π 2ksin (1kπ ) =1 3 ( 1−26mcos(2mπ)) =1 3 ( 1−26m) 3\n3 3 3 3\nk=0\n2\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p11.png"
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    },
    {
      "id": "9231-2022-on-23-q08",
      "question_id": "9231-2022-on-23-q08",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) du B1\n=−2(−1) leading to (−1)d=−1du\nd 2\n −1  1 M1 A1 Applies substitution. For M1, integrand must be\n  1−(−1)2 d=−1 2   u du=− u +C=− 1−(−1)2 +C of the form k u , where k0 is constant. For\nA1, allow “+C” missing.\n3\n\n8(b) dy y B1 Divides through by .\n− =sin−1(−1)\nd \ne\n−−1d\n=e−ln=−1\nM1 A1 Finds integrating factor.\nd ( −1y ) =sin−1(−1) M1 Correct form on LHS using their integrating\nfactor.\nd\n  M1 A1 Integrates sin−1(−1).\n−1y=sin−1(−1)− d\n\n 1−(−1)2\n   −1  1 M1 Applies part (a) and uses\n d=  d+  d 1\n 1−(−1)2  1−(−1)2  1−(−1)2 \n1−x2\ndx=sin−1x+C.\n−1y=sin−1(−1)+ 1−(−1)2 −sin−1(−1)+C A1\n© UCLES 2022 Page 12 of 13\n\n8(b) 1=1+C M1 Substitutes initial conditions into their\nexpression in y and .\ny=(−1)sin−1(−1)+ 1−(−1)2 M1 A1 Divides through by their integrating factor.\n11\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_23-p12.png",
        "../answer-assets/9231_w22_ms_23-p13.png"
      ]
    },
    {
      "id": "9231-2022-on-31-q01",
      "question_id": "9231-2022-on-31-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 2v2 M1\nUse F = ma: 20= OR 20=20.62\n0.6\n50 A1\nv2 =6 OR 2 =\n3\n60v 60 A1 FT 38.9848….\nNumber of revolutions per min = OR\n0.62π 2π\nso 39(.0) revolutions\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p06.png"
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    },
    {
      "id": "9231-2022-on-31-q02",
      "question_id": "9231-2022-on-31-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) Loss in KE = Gain in EPE, so B1 EPE terms correct.\n1 1 v 2 1 4mg 1  2 1  2\nmv2 − m  =   a − a  M1 All 4 terms and no extras.\n2 2 2 2 a  2  4  \n \n3 4mg 3 M1 Simplify.\nmv2 =  a2\n4 a 16\nv2 =ag, v= ag A1\n4\n\n2(b) 4mg 1 M1\nHooke’s law: tension =  a (= 2mg)\na 2\n2mg A1 Accept -2g.\nAcceleration = =2g\nm\n2\n© UCLES 2022 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p06.png"
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    },
    {
      "id": "9231-2022-on-31-q03",
      "question_id": "9231-2022-on-31-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) Let F and R be friction and normal reaction at A M1 Correct terms, allow sign errors and cos/sin mix.\nTake moments about A, for rod\nN3a=W2acos+kW4acos\n4 A1 At least one intermediate line of working.\n3N =(2+4k)W\n5\n8\nN = W(1+2k) AG\n15\n2\n\n3(b) Ncos+R=W +kW B1 Resolve (to include R) for rod.\n6 B1 Both.\n→F=Nsin and F = R\n7\n28 21 M1 Find R or N .\nso R= W(1+2k) or R= W(1+k)\n75 45\nEliminate to find k M1 Complete method.\n1 A1\nk =\n3\n5\n© UCLES 2022 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p07.png"
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    },
    {
      "id": "9231-2022-on-31-q04",
      "question_id": "9231-2022-on-31-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) m dv =50−2v2 dv =4 ( 25−v2) B1 N2L\ndt dt\n1 1 1 M1 Separate variables and use partial fractions.\n + dv=4dt\n10 5−v 5+v\n1 M1 A1 Integrate into log terms.\n(−ln(5−v)+ln(5+v))=4t+A\n(Note: formula on MF19).\n10\n1 M1 Use initial condition.\nUse t=0, v=3 to give A= ln4\n10\n1 5+v 5+v M1 Rearrange to make v the subject.\n4t = ln leading to =e40t\n10 4(5−v) 20−4v\n5\n( 4−e−40t) A1\nv=\n4+e−40t\n7\n\n4(b) As t→, v→5 B1\n1\n© UCLES 2022 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-31-q05",
      "question_id": "9231-2022-on-31-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 1 B1 Energy equation.\nmv2 − mu2 =mgacos\n2 2\n 1 \nkag = ag+2agcos\n \n 3 \nm B1 N2L at B.\nT −mgcos= v2\na\n11 m 11\nSo mg−mgcos= .kag, −cos=k\n6 a 6\nSolve simultaneously. M1\n4 1 A1 Both.\nk = , cos=\n3 2\n4\n\n5(b) Initial speed  = kagsin B1\nUse v2 =u2 +2as: 0= ( kagsin )2 −2gs M1\n1 A1\ns= a\n2\n1 1 A1 FT\nHeight above lowest point = s+a−acos = a+a− a=a\n2 2\n4\n© UCLES 2022 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p09.png"
      ]
    },
    {
      "id": "9231-2022-on-31-q06",
      "question_id": "9231-2022-on-31-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let speed of A after collision be →v and speed of B M1\nA\nperpendicular to line of centres be v\n5\nAlong line of centres: mu−km ucos=mv Momentum.\n8 A\n5  M1 NEL\nNEL: 0−v\nA\n=e\n8\nu cos+u\n\n5 25  M1 Solve.\nSo u− kucos=−  u cos+u\n8 38 \nSubstitute for cos, to give k=4 A1\n4\n\n6(b) 5 3 B1\nv = usin = u\nB 8 8 Velocity perpendicular to line of centres\nv =−u B1 FT\nA\n2 M1 NOTE: KE before and after for A is unchanged.\n1 1 5  1 25 41\nKE before = mu2 + km\n\nu\n\n= mu2 + mu2 = mu2\n2 2 8  2 32 32\nBoth.\n1 1 1 9 25\nKE after = mv 2 + kmv 2 = mu2 +2m u2 = mu2\n2 A 2 B 2 64 32\n41 25 1 A1\nLoss = mu2  − = mu2\n32 32 2\n4\n© UCLES 2022 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p10.png"
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    },
    {
      "id": "9231-2022-on-31-q07",
      "question_id": "9231-2022-on-31-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) 1 M1\n0=Vsin75t− gt2\n2\n2V A1 AG\nt = sin75\ng\n2\n\n7(b) Vertical component of velocity is unchanged. B1\n1\n© UCLES 2022 Page 11 of 12\n\n7(c)  15  B1\nHorizontally to wall, →15=Vcos75t;  t = \n Vcos75\neVcos75 B1 Speed after rebound.\n15 M1 Time back to O\nT =\neVcos75  3 \nt = T\n 5 \nVertically for whole flight: M1\n2V\nt+T = sin75\ng\n15 15 2V A1\n+ = sin75\nVcos75 eVcos75 g\nV2cos75sin75=20g M1\nMultiply by 2: V2sin150=40g, V 2 =80g A1\n( )\nV =4 5g =8.94 g\n7\n© UCLES 2022 Page 12 of 12",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_31-p11.png",
        "../answer-assets/9231_w22_ms_31-p12.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q01",
      "question_id": "9231-2022-on-32-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 m B1 N2L\nWhen string goes slack, mgcos= v2, v2 =agcos\nMay include T, but B1 not awarded until T = 0.\na\n1 1 B1 Energy equation.\nm.3ag− mv2 =mg(acos+acos)\n2 2\n 2 M1 Combine.\nSo u2 −agcos=2agcos+ \n 3\n4 A1\nu2 − ag\n3 5\ncos= =\n3ag 9\n4\n© UCLES 2022 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p06.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q02",
      "question_id": "9231-2022-on-32-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2\nArea Distance from AB\nABC 27a2 2a\n9 1\nABD ax x\n2 3\n9\nShape ADC 27a2 − ax x\n2\nTaking moments about AB M1 Moments equation with 3 terms.\n 9  9 1\nx 27a2 − ax= 27a2 2a− ax  x\nA1 At least 2 terms correct.\n 2  2 3\n 54a3− 3 ax2  A1 All correct.\n \n2\nx = \n9\n 27a2 − ax \n 2 \nFor equilibrium, x x, B1 Use correct condition: allow strict inequality.\nCan be implied by correct final answer x 3a.\n3  9 \n54a3− ax2 x27a2 − ax\n2  2 \n54a2 −27ax+3x2 0 M1 Simplify and attempt to solve a quadratic inequality or equation.\n(x−3a)(x−6a) 0\n(0 ) x 3a [only] A1 CAO\n© UCLES 2022 Page 7 of 15\n\n2 Alternative method for question 2\nTaking moments with B as origin. M1\n1 1 A2\nx = (0+x+6a)=2a+ x\n3 3\n1 B1 Allow strict inequality.\nFor equilibrium, x x, so x 2a+ x\n3\n(0 ) x 3a M1\nA1\n6\n© UCLES 2022 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p07.png",
        "../answer-assets/9231_w22_ms_32-p08.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q03",
      "question_id": "9231-2022-on-32-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 16 B1\nMge\n3 3\nIn equilibrium, =4Mg, e= a\na 4\nIn subsequent motion, M1 Energy equation with GPE and KE terms correct and at least one\nLoss in GPE = gain in EPE + gain in KE EPE term.\nDimensionally correct.\n6Mga 1 16 Mg  3a 2 1 B1 EPE correct.\n= . . .a2 −  + .6Mv2\n4 2 3 a    4    2 A1 All correct.\n3Mga 8Mg 7 M1 Attempt to find v in terms of a and g.\n= . a2 +3Mv2 etc\n2 3a 16\nga 1 A1\n=3v2, v= ga\n3 3\n6\n© UCLES 2022 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p09.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q04",
      "question_id": "9231-2022-on-32-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) dv 500 1 B1 Sight of m or 5 is required.\n5v = − v2\ndx v 2\n10v2dv M1 Separate variables and attempt to integrate into a log term.\n=dx\n1000−v3\n− 10 ln ( 1000−v3) =x(+A) A1\n3\n10 M1 Evaluate constant: correct initial condition used.\nx=0, v=5, A=− ln875\n3\n10 875 M1 Make v the subject: correct use of logs.\nx= ln\n3 1000−v3\n1 A1 1\nv=(1000−875e−0.3x3 v=5(8−7e−0.3x3 : A0 if eln terms.\n   \n6\n\n4(b) Maximum value of v is 10 B1 No FT: result can be found from initial equation.\n1\n© UCLES 2022 Page 10 of 15",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p10.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q05",
      "question_id": "9231-2022-on-32-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) → x=ucos t B1 Result quoted from MF19 scores 0/4.\n1 B1\n y=usin t− gt2\n2\n2 M1\nx 1  x \nEliminate t: y=usin  − g  \ncos 2 ucos \ngx2 A1 Must be an intermediate line of working.\ny=xtan− (1+tan2)\nAG\n2u2\n4\n\n5(b) 4 302  4 2 M1 Substituting values correctly.\n20=30 −10 1+  \n3 2u2  3 \n \nu2 =625, u=25 A1\nSubstitute back into trajectory equation, M1 Obtain a 3-term quadratic.\ng302 36\n20=30tan − sec2 =30tan − (1+tan2 )\n2.252 5\n18tan2−75tan+68=0\n4 M1\nOne solution is , (3tan−4)(6tan−17)=0\n3\n17 A1\nGiving tan=\n6\n5\n© UCLES 2022 Page 11 of 15",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p11.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q06",
      "question_id": "9231-2022-on-32-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) 6 B1 May be implied.\nT = mg\n7\nTsin=mr2 =mhtan2 B1 Allow r for radius.\nRadius of circle = htan B1\n6g\n[So 2 = cos]\n7h\n9 6g M1 Second scenario, equivalent result .\nIn second scenario, 2 = cos\n4 7h\n6g 4 6g A1 Combine convincingly to obtain given result.\nEquate, cos=  cos giving\n7h 9 7h\n4\ncos= cos AG\n9\n5\n© UCLES 2022 Page 12 of 15\n\n6(b) First scenario: N+Tcos=mg\n1 B1 Both.\nSecond scenario, N +Tcos=mg\n2\n6 12 M1 12cos−6cos=7\nEquate: mg− mgcos=2mg− mgcos\n7 7\n3 A1\ncos=\n4\n5 A1\nN = mg\n7\n4\n© UCLES 2022 Page 13 of 15",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p12.png",
        "../answer-assets/9231_w22_ms_32-p13.png"
      ]
    },
    {
      "id": "9231-2022-on-32-q07",
      "question_id": "9231-2022-on-32-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) Let v, w be speeds of A and B along line of centres after collision M1\n1 1\nmv+ mw=mucos− m.2ucos\nMomentum: masses correct, opposite signs on RHS.\n2 2\nw−v=e(2ucos+ucos) M1 NEL: LHS signs must be consistent with momentum equation,\nsame sign for both terms on RHS.\n+=90, so cos=sin M1 Solve to find an expression of the correct form.\nUse this fact and solve to find w\n2 1 13  A1\nw= u sin+ cos\n3 4 8 \n4\n© UCLES 2022 Page 14 of 15\n\n7(b) Perpendicular to line of centres, speed of B is B1\n2usin=2ucos\nAfter, velocity of B makes angle  with line of centres, so B1\n2ucos\ntan=\nw\nsin 2ucos M1* Obtain homogeneous equation in cos and sin or an equation in tan\n= giving\ncos 2 1 13 \nu sin+ cos\n3 4 8 \n1 13 DM1 Obtain quadratic and solve to find values of tan\n3(cos)2\n=\n(sin)2\n+ sincos\n4 8\n2(tan)2 +13tan−24=0, (2tan−3)(tan+8)=0\n3 A1\ntan=\n2\n5\n© UCLES 2022 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_32-p14.png",
        "../answer-assets/9231_w22_ms_32-p15.png"
      ]
    },
    {
      "id": "9231-2022-on-33-q01",
      "question_id": "9231-2022-on-33-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 2v2 M1\nUse F = ma: 20= OR 20=20.62\n0.6\n50 A1\nv2 =6 OR 2 =\n3\n60v 60 A1 FT 38.9848….\nNumber of revolutions per min = OR\n0.62π 2π\nso 39(.0) revolutions\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p06.png"
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    },
    {
      "id": "9231-2022-on-33-q02",
      "question_id": "9231-2022-on-33-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) Loss in KE = Gain in EPE, so B1 EPE terms correct.\n1 1 v 2 1 4mg 1  2 1  2\nmv2 − m  =   a − a  M1 All 4 terms and no extras.\n2 2 2 2 a  2  4  \n \n3 4mg 3 M1 Simplify.\nmv2 =  a2\n4 a 16\nv2 =ag, v= ag A1\n4\n\n2(b) 4mg 1 M1\nHooke’s law: tension =  a (= 2mg)\na 2\n2mg A1 Accept -2g.\nAcceleration = =2g\nm\n2\n© UCLES 2022 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p06.png"
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    },
    {
      "id": "9231-2022-on-33-q03",
      "question_id": "9231-2022-on-33-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) Let F and R be friction and normal reaction at A M1 Correct terms, allow sign errors and cos/sin mix.\nTake moments about A, for rod\nN3a=W2acos+kW4acos\n4 A1 At least one intermediate line of working.\n3N =(2+4k)W\n5\n8\nN = W(1+2k) AG\n15\n2\n\n3(b) Ncos+R=W +kW B1 Resolve (to include R) for rod.\n6 B1 Both.\n→F=Nsin and F = R\n7\n28 21 M1 Find R or N .\nso R= W(1+2k) or R= W(1+k)\n75 45\nEliminate to find k M1 Complete method.\n1 A1\nk =\n3\n5\n© UCLES 2022 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2022-on-33-q04",
      "question_id": "9231-2022-on-33-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) m dv =50−2v2 dv =4 ( 25−v2) B1 N2L\ndt dt\n1 1 1 M1 Separate variables and use partial fractions.\n + dv=4dt\n10 5−v 5+v\n1 M1 A1 Integrate into log terms.\n(−ln(5−v)+ln(5+v))=4t+A\n(Note: formula on MF19).\n10\n1 M1 Use initial condition.\nUse t=0, v=3 to give A= ln4\n10\n1 5+v 5+v M1 Rearrange to make v the subject.\n4t = ln leading to =e40t\n10 4(5−v) 20−4v\n5\n( 4−e−40t) A1\nv=\n4+e−40t\n7\n\n4(b) As t→, v→5 B1\n1\n© UCLES 2022 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p08.png"
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    },
    {
      "id": "9231-2022-on-33-q05",
      "question_id": "9231-2022-on-33-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 1 B1 Energy equation.\nmv2 − mu2 =mgacos\n2 2\n 1 \nkag = ag+2agcos\n \n 3 \nm B1 N2L at B.\nT −mgcos= v2\na\n11 m 11\nSo mg−mgcos= .kag, −cos=k\n6 a 6\nSolve simultaneously. M1\n4 1 A1 Both.\nk = , cos=\n3 2\n4\n\n5(b) Initial speed  = kagsin B1\nUse v2 =u2 +2as: 0= ( kagsin )2 −2gs M1\n1 A1\ns= a\n2\n1 1 A1 FT\nHeight above lowest point = s+a−acos = a+a− a=a\n2 2\n4\n© UCLES 2022 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p09.png"
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    },
    {
      "id": "9231-2022-on-33-q06",
      "question_id": "9231-2022-on-33-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let speed of A after collision be →v and speed of B M1\nA\nperpendicular to line of centres be v\n5\nAlong line of centres: mu−km ucos=mv Momentum.\n8 A\n5  M1 NEL\nNEL: 0−v\nA\n=e\n8\nu cos+u\n\n5 25  M1 Solve.\nSo u− kucos=−  u cos+u\n8 38 \nSubstitute for cos, to give k=4 A1\n4\n\n6(b) 5 3 B1\nv = usin = u\nB 8 8 Velocity perpendicular to line of centres\nv =−u B1 FT\nA\n2 M1 NOTE: KE before and after for A is unchanged.\n1 1 5  1 25 41\nKE before = mu2 + km\n\nu\n\n= mu2 + mu2 = mu2\n2 2 8  2 32 32\nBoth.\n1 1 1 9 25\nKE after = mv 2 + kmv 2 = mu2 +2m u2 = mu2\n2 A 2 B 2 64 32\n41 25 1 A1\nLoss = mu2  − = mu2\n32 32 2\n4\n© UCLES 2022 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p10.png"
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    },
    {
      "id": "9231-2022-on-33-q07",
      "question_id": "9231-2022-on-33-q07",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) 1 M1\n0=Vsin75t− gt2\n2\n2V A1 AG\nt = sin75\ng\n2\n\n7(b) Vertical component of velocity is unchanged. B1\n1\n© UCLES 2022 Page 11 of 12\n\n7(c)  15  B1\nHorizontally to wall, →15=Vcos75t;  t = \n Vcos75\neVcos75 B1 Speed after rebound.\n15 M1 Time back to O\nT =\neVcos75  3 \nt = T\n 5 \nVertically for whole flight: M1\n2V\nt+T = sin75\ng\n15 15 2V A1\n+ = sin75\nVcos75 eVcos75 g\nV2cos75sin75=20g M1\nMultiply by 2: V2sin150=40g, V 2 =80g A1\n( )\nV =4 5g =8.94 g\n7\n© UCLES 2022 Page 12 of 12",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_33-p11.png",
        "../answer-assets/9231_w22_ms_33-p12.png"
      ]
    },
    {
      "id": "9231-2022-on-41-q01",
      "question_id": "9231-2022-on-41-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1  1625  1  16252  M1 775\n  x = 50 =32.5   s x 2 = 49   53200− 50   =7.908 98\n 1854  1  18542  A1 3057\n  y = 60 =30.9   s y 2 = 59   57900− 60   =10.3627 295\nBoth correct\n7.908 10.3627 M1 A1\ns2 = + [=0.33087]\n50 60\nCI: (32.5−30.9)1.96s M1 A1 Correct form with a z-value. Correct with 1.96.\n1.61.1274=0.473, 2.73 A1 At least 3sf.\n7 Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7\n© UCLES 2022 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p06.png"
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    },
    {
      "id": "9231-2022-on-41-q02",
      "question_id": "9231-2022-on-41-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) Using B(8, 0.6) B1 One correct.\np=0.098, q=41.804, r =31.353\nB1 Other two correct.\nIf B0B0 scored, SCB1 if all 3 correct but not rounded to 3dp.\n2\n\n2(b) H : the data fits the binomial distribution B(8, 0.6) B1\n0\nH : the data does not fit the binomial distribution B(8, 0.6)\n1\nAdd first 3 columns: O value 8, E value 7.471 M1\nAND last two columns: O value 11, E value 15.956 Both.\nChi-squared sum: M1\n0.0374 + 1.5817 + 2.9655 + 0.8058 + 0.9139 + 1.5394 Method must be seen.\n7.84 A1 If M0 awarded SCB1 for 7.84.\nCritical value, 5df = 9.236 M1 Compare calculated value with 9.236 OR\n7.84 < 9.236 13.36 if no columns added.\nAccept H 12.02 adding only last 2 columns.\n0\n10.64 adding only first 3 columns.\nThere is insufficient evidence to reject the director’s claim. A1 Correct work only except possibly B1 for hypotheses, in context,\nlevel of uncertainty in language. ‘prove’ scores A0\n6\n© UCLES 2022 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p07.png"
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    },
    {
      "id": "9231-2022-on-41-q03",
      "question_id": "9231-2022-on-41-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Differences: M1\n4 -8 3 1 –7 11 19 6 10 –13 14 5 –9 12 Attempt at signed differences, allow 2 errors.\nSigned Ranks: M1\n3 –7 2 1 –6 10 14 5 9 –12 13 4 –8 11 Ranks, allow 4 errors.\n[P = 72, Q = 33] T = 33 A1 33 clearly identified.\nH : difference in population medians = 0 B1\n0\nH : difference in population medians > 0 Correct, allow m for this mark.\n1\nCritical tabular value = 25 B1\n33 > 25 M1 Compares their value of sum of ranks with their 25 and\naccept H conclusion.\n0\nInsufficient evidence to support claim A1 All correct except possibly first B1, in context, level of\nuncertainty in language.\n‘Prove’ scores A0.\n7\n\n3(b) Underlying/population distribution of differences is unknown (not B1\nknown to be normal)\n1\n© UCLES 2022 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p08.png"
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    },
    {
      "id": "9231-2022-on-41-q04",
      "question_id": "9231-2022-on-41-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 1 1 1 2 1 1 6 M1\nP(0 H) =   = P(1 H) =   3=\nAt least 2 probabilities correct.\n3 3 3 27 3 3 3 27\n1 2 2 12 2 2 2 8\nP(2 H) =   3= P(3 H) =   =\n3 3 3 27 3 3 3 27\n1 6 12 8 M1A1 Correct form, ft their probabilities.\nG (t) = + t+ t2 + t3\nX\n27 27 27 27\n1 2 M1\nOR: PGF for one coin is + t\n3 3\n3 M1A1\n1 2 \nFor 3 coins, PGF is  + t \n3 3 \n3\n\n4(b) 1 M1 Correct method.\nG (t) = ( 1+2t+t2)( 1+6t+12t2 +8t3)\nZ\n108\n1 M1 Obtain quintic polynomial.\n( 1+8t+25t2 +38t3+28t4 +8t5)\n108\n1 2 25 19 7 2 A1\n+ t+ t2 + t3+ t4 + t5\n108 27 108 54 27 27\n3\n© UCLES 2022 Page 9 of 12\n\n4(c) 1 M1\nG ’ (t) = ( 8+50t+114t2 +112t3+40t4)\nZ Differentiate and evaluate at t = 1\n108\n1\nE(Z) = (8+50+114+112+40)\n108\n324 A1\n=3\n108\nAlternative method for questions 4(c)\n2 2 2 1 1 M1 Summing expected values for each coin\n+ + + +\n3 3 3 2 2\n3 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p09.png",
        "../answer-assets/9231_w22_ms_41-p10.png"
      ]
    },
    {
      "id": "9231-2022-on-41-q05",
      "question_id": "9231-2022-on-41-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 3 M1\nF(U) = 0.75, so 1−\n(12−U)2\n=\n144 4\n(12−U)2\n=36, 1 2−U =6,\nA1\nU =6 only\n2\n© UCLES 2022 Page 10 of 12\n\n5(b) 1 B1 Differentiate given PDF.\n[f(x) =] (12−x)\n72\nE(X2) = 1  2 1 ( 12x2 −x3) dx = 1   4x3− 1 x4   [= 24] M1 Integrate x2 f (x) with their f (x) .\n72 72 4 \n0\nE ( X4) = 1  2 1 ( 12x4 −x5) dx= 1   12 x5 − 1 x6   [= 1382.4] M1 Integrate x4 f (x) with their f (x) .\n72 72 5 6 \n0\nVar ( X2) =E ( X4) −E2( X2) [= 1382.4−242] M1 Using correct formula with numerical values possibly\nunsimplified.\n1382.4−576=806(.4) A1\n5\n\n5(c)  0 y0, M1\nChange the variable: 1−\n1 (\n12−\ny2)2\n\n144\nG(y)=   1− 1 ( 12− y2)2 0 y 12, Allow domain unchanged at this stage.\n144\n\n 1 y  12.\n1 (\n12y−\ny3) M1 Differentiate their G(y) .\n36\n\n\n1 (\n12y−\ny3)\n0 y 12\nA1 Fully correct.\ng(y)=36\n  0 otherwise\n© UCLES 2022 Page 11 of 12",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p10.png",
        "../answer-assets/9231_w22_ms_41-p11.png"
      ]
    },
    {
      "id": "9231-2022-on-41-q06",
      "question_id": "9231-2022-on-41-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 1  6.242  M1 19\ns 2 = 3.26− =0.001382\nx 11 12  13750\n19\n1 5.12 \ns 2 = 2.6124− =0.001267 15000\ny 9 10 \nA1 Both correct.\n110.001382+90.001267 M1 A1 Find pooled variance.\ns2 = =0.001330\n12+10−2\n6.24 5.1 M1 A1 Use correct formula for t.\n−\n12 10\nt = =0.640\n1 1\ns +\n12 10\nH :  = H :   B1\n0 a b 1 a b\nCritical value is 2.086 (t (20)) M1 Compare their calculated value with 2.086 and conclusion.\n0.975\n0.640 < 2.086\nAccept H\n0\nThere is insufficient evidence to support inspector’s claim A1 All correct except possibly B1, in context, level of uncertainty in\nlanguage.\n‘Prove’ scores A0\n9 Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9\n© UCLES 2022 Page 12 of 12",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_41-p12.png"
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    },
    {
      "id": "9231-2022-on-42-q01",
      "question_id": "9231-2022-on-42-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 M1\ns2\n2.02=x+1.833\n10\ns2\n1.78=x−1.833\n10\n2.02+1.78\nAdd, x = =1.90 Add.\n2\nx=19 A1\nM1 Allow 1.372, 1.383, 1.812 instead of 1.833.\ns2\nSubtract: 2.02−1.78=21.833\n10\ns2 =0.042859 A1 May be implied.\ns = 0.20702\n1\n(x)2 M1\nBut s2 = (x2 − )\n9 10\n192 A1 36.486\nx2 =9s2 + =36.5\n10\nUsing 1.645: maximum M1A1 M0A0 M1A0 3/6\n6\n© UCLES 2022 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_42-p06.png"
      ]
    },
    {
      "id": "9231-2022-on-42-q02",
      "question_id": "9231-2022-on-42-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 H : performance is independent of region B1\n0\nH : performance is not independent of region\n1\nM1 A1 Calculate expected frequencies, allow 1 error.\n62 (53.9) 41 (44.1) 44 (49)\n102 (106.7) 94 (87.3) 95 (97)\n56 (59.4) 45 (48.6) 61 (54)\nTest stat = 1.217 + 0.218 + 0.510 + 0.207 + 0.514 + 0.041 + 0.195 M1 Calculate test statistic.\n+ 0.267 + 0.907\n4.08 A1 SCB2 for 4.08 if totally unsupported.\nSCM1A1 B1 for 4.08 supported only by expected frequencies.\nTabular value, 4df = 7.779 M1 Compare with 7.779 and conclusion.\n1.08 < 7.779\nAccept H\n0\nInsufficient evidence that performance is not independent of A1 Correct conclusion in context, following correct work, level of\nregion uncertainty in language.\n‘Prove’ gives A0.\n7\n© UCLES 2022 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-42-q03",
      "question_id": "9231-2022-on-42-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3 1  75.52  M1 239\ns 2 = 115.2− =0.02439\nx 49 50 9800\n \n259\n1  116.82 \ns 2 = 172.6− =0.02623 9875\ny 79 80\n \nA1 Both correct\n0.02439 0.02623 M1 A1 May be implied by 1.75 for z\ns2 = + =0.00081565\n50 80\n75.5 116.8 M1\n−\n50 80\nz=\ns\n1.75 A1\nCompare with 1.645: 1.75 >1.645 M1\nReject null hypothesis Using areas, 0.04 < 0.05.\nSufficient evidence to suggest that population mean in country X A1 Correct conclusion in context, following correct work, level of\nis greater than population mean in country Y uncertainty in language.\n‘Prove’ gives A0.\n8 Pooled variance: z=1.74\nM1A1 M0A0M0A0 M1A0 max 3/8\n© UCLES 2022 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2022-on-42-q04",
      "question_id": "9231-2022-on-42-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 2 B1 Integration and k + 3k/2 = 1 oe seen.\nk  3 2\nkdx+kxdx= kx+ x2 , k+ k =1, k =\n \n2  2 5\n0 1\n1\n\n4(b)  kx 0 x1, M1 1 1\n CDF or integration with x2 + seen.\nF(x)=k 1 5 5\nx2 + 1 x 2\n\n2 5\nk 1 3 M1 Method for UQ\nUQ: x2 + = ,\n2 5 4\n 3 11 1 A1 UQ\nx2 +1=5 , x2 = x= 11\n \n 4 4 2\n1 5 B1 LQ\nLQ: kx= , x=\n4 8\n1 5 A1 CAO\nIQR = 11− =1.03(3)\n2 8\n5\n© UCLES 2022 Page 9 of 13\n\n4(c) 1 2 k  k  17 M1 With correct limits.\nE(X) = kx dx+ kx2 dx= x2 + x3 =\n   \n2  3  15\n0 1\n1 2 k  k  49 M1 With correct limits.\nE(X2) = kx2 dx+kx3 dx= x3 + x4 =\n   \n3  4  30\n0 1\n2 M1 Using correct formula with numerical values.\n49 17\nVar(X) = −  \n30 15\n157 A1\n(=0.349)\n450\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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        "../answer-assets/9231_w22_ms_42-p10.png"
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    },
    {
      "id": "9231-2022-on-42-q05",
      "question_id": "9231-2022-on-42-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 9 6 1 M1 A1 2 probabilities correct, in a polynomial\nG (t) = + t+ t2\nX\n16 16 16\n2\n© UCLES 2022 Page 10 of 13\n\n5(b)  9 6 1 25 10 1  M1 Second PGF correct and multiplied by part (a).\nG Z (t) =  + t+ t2  + t+ t2 \n16 16 16 36 36 36 \n1 ( 225+240t+94t2 +16t3+t4) M1 Obtains a quartic polynomial.\n576\nA1 1\nOr 25 + 5 t+ 47 t2 + 1 t3+ 1 t4 Note: (t+3)2(t+5)2 scores M1M1A0.\n576\n64 12 288 36 576\n3\n\n5(c) G ’ (t) = 1 ( 240+188t+48t2 +4t3) M1 Differentiate twice.\nZ\n576\nG ’’ (t) =\n1 ( 188+96t+12t3)\nZ\n576\n1 5 25 M1 Use correct formula.\nVar (Z) = (188+96+12)+ −\n576 6 36\n47 A1\n72\n1 3 1 5 M1 One term correct.\nOR: Var(Z) = 2  +2 \nM1 Two terms present and added.\n4 4 6 6\n47 A1\n72\n3\n\n5(d) 1 B1 FT FT their power with greatest coefficient in part (b).\n1\n© UCLES 2022 Page 11 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w22_ms_42-p11.png"
      ]
    },
    {
      "id": "9231-2022-on-42-q06",
      "question_id": "9231-2022-on-42-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) M1 Wrong test: max B1 for hypotheses, B1B1 for correct mean and\n359 3 342 1\nvariance\n374 5 355 2\n398 7 361 4\nAttempt at rankings (allow up to 4 errors)\n415 10 384 6\n427 11 401 8\n452 14 414 9\n461 16 437 12\n482 18 443 13\n502 20 454 15\n512 22 472 17\n545 23 491 19\n612 24 506 21\nTest statistic = 127 A1 Clearly identified\nH : population medians are equal B1\n0\nH : population median for X is greater than population median for\n1\nY\n1 B1\nMean = 1225=150\n2\n1 B1\nVariance = 121225=300\n12\n© UCLES 2022 Page 12 of 13\n\n6(a) 127.5−150 M1 Allow incorrect or no continuity correction.\n300\n−1.299 A1\nCompare with −1.645: −1.299−1.645, or 0.097 > 0.05 M1 Valid comparison with 1.645 or 0.05 and reach correct ft\nAccept H conclusion.\n0\nInsufficient evidence to support manager’s claim A1 Correct conclusion in context, following correct work, level of\nuncertainty in language.\n‘Prove’ is A0.\n9\n\n6(b) Not appropriate/no, not the same people B1 OE\nNo and reason needed, e.g. individuals in the samples cannot be\npaired up.\n1\n© UCLES 2022 Page 13 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_42-p12.png",
        "../answer-assets/9231_w22_ms_42-p13.png"
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    },
    {
      "id": "9231-2022-on-43-q01",
      "question_id": "9231-2022-on-43-q01",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1  1625  1  16252  M1 775\n  x = 50 =32.5   s x 2 = 49   53200− 50   =7.908 98\n 1854  1  18542  A1 3057\n  y = 60 =30.9   s y 2 = 59   57900− 60   =10.3627 295\nBoth correct\n7.908 10.3627 M1 A1\ns2 = + [=0.33087]\n50 60\nCI: (32.5−30.9)1.96s M1 A1 Correct form with a z-value. Correct with 1.96.\n1.61.1274=0.473, 2.73 A1 At least 3sf.\n7 Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7\n© UCLES 2022 Page 6 of 12",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_43-p06.png"
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    },
    {
      "id": "9231-2022-on-43-q02",
      "question_id": "9231-2022-on-43-q02",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) Using B(8, 0.6) B1 One correct.\np=0.098, q=41.804, r =31.353\nB1 Other two correct.\nIf B0B0 scored, SCB1 if all 3 correct but not rounded to 3dp.\n2\n\n2(b) H : the data fits the binomial distribution B(8, 0.6) B1\n0\nH : the data does not fit the binomial distribution B(8, 0.6)\n1\nAdd first 3 columns: O value 8, E value 7.471 M1\nAND last two columns: O value 11, E value 15.956 Both.\nChi-squared sum: M1\n0.0374 + 1.5817 + 2.9655 + 0.8058 + 0.9139 + 1.5394 Method must be seen.\n7.84 A1 If M0 awarded SCB1 for 7.84.\nCritical value, 5df = 9.236 M1 Compare calculated value with 9.236 OR\n7.84 < 9.236 13.36 if no columns added.\nAccept H 12.02 adding only last 2 columns.\n0\n10.64 adding only first 3 columns.\nThere is insufficient evidence to reject the director’s claim. A1 Correct work only except possibly B1 for hypotheses, in context,\nlevel of uncertainty in language. ‘prove’ scores A0\n6\n© UCLES 2022 Page 7 of 12",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_43-p07.png"
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    },
    {
      "id": "9231-2022-on-43-q03",
      "question_id": "9231-2022-on-43-q03",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Differences: M1\n4 -8 3 1 –7 11 19 6 10 –13 14 5 –9 12 Attempt at signed differences, allow 2 errors.\nSigned Ranks: M1\n3 –7 2 1 –6 10 14 5 9 –12 13 4 –8 11 Ranks, allow 4 errors.\n[P = 72, Q = 33] T = 33 A1 33 clearly identified.\nH : difference in population medians = 0 B1\n0\nH : difference in population medians > 0 Correct, allow m for this mark.\n1\nCritical tabular value = 25 B1\n33 > 25 M1 Compares their value of sum of ranks with their 25 and\naccept H conclusion.\n0\nInsufficient evidence to support claim A1 All correct except possibly first B1, in context, level of\nuncertainty in language.\n‘Prove’ scores A0.\n7\n\n3(b) Underlying/population distribution of differences is unknown (not B1\nknown to be normal)\n1\n© UCLES 2022 Page 8 of 12",
      "source_pages": [
        8
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w22_ms_43-p08.png"
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    },
    {
      "id": "9231-2022-on-43-q04",
      "question_id": "9231-2022-on-43-q04",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 1 1 1 2 1 1 6 M1\nP(0 H) =   = P(1 H) =   3=\nAt least 2 probabilities correct.\n3 3 3 27 3 3 3 27\n1 2 2 12 2 2 2 8\nP(2 H) =   3= P(3 H) =   =\n3 3 3 27 3 3 3 27\n1 6 12 8 M1A1 Correct form, ft their probabilities.\nG (t) = + t+ t2 + t3\nX\n27 27 27 27\n1 2 M1\nOR: PGF for one coin is + t\n3 3\n3 M1A1\n1 2 \nFor 3 coins, PGF is  + t \n3 3 \n3\n\n4(b) 1 M1 Correct method.\nG (t) = ( 1+2t+t2)( 1+6t+12t2 +8t3)\nZ\n108\n1 M1 Obtain quintic polynomial.\n( 1+8t+25t2 +38t3+28t4 +8t5)\n108\n1 2 25 19 7 2 A1\n+ t+ t2 + t3+ t4 + t5\n108 27 108 54 27 27\n3\n© UCLES 2022 Page 9 of 12\n\n4(c) 1 M1\nG ’ (t) = ( 8+50t+114t2 +112t3+40t4)\nZ Differentiate and evaluate at t = 1\n108\n1\nE(Z) = (8+50+114+112+40)\n108\n324 A1\n=3\n108\nAlternative method for questions 4(c)\n2 2 2 1 1 M1 Summing expected values for each coin\n+ + + +\n3 3 3 2 2\n3 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w22_ms_43-p10.png"
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    },
    {
      "id": "9231-2022-on-43-q05",
      "question_id": "9231-2022-on-43-q05",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 3 M1\nF(U) = 0.75, so 1−\n(12−U)2\n=\n144 4\n(12−U)2\n=36, 1 2−U =6,\nA1\nU =6 only\n2\n© UCLES 2022 Page 10 of 12\n\n5(b) 1 B1 Differentiate given PDF.\n[f(x) =] (12−x)\n72\nE(X2) = 1  2 1 ( 12x2 −x3) dx = 1   4x3− 1 x4   [= 24] M1 Integrate x2 f (x) with their f (x) .\n72 72 4 \n0\nE ( X4) = 1  2 1 ( 12x4 −x5) dx= 1   12 x5 − 1 x6   [= 1382.4] M1 Integrate x4 f (x) with their f (x) .\n72 72 5 6 \n0\nVar ( X2) =E ( X4) −E2( X2) [= 1382.4−242] M1 Using correct formula with numerical values possibly\nunsimplified.\n1382.4−576=806(.4) A1\n5\n\n5(c)  0 y0, M1\nChange the variable: 1−\n1 (\n12−\ny2)2\n\n144\nG(y)=   1− 1 ( 12− y2)2 0 y 12, Allow domain unchanged at this stage.\n144\n\n 1 y  12.\n1 (\n12y−\ny3) M1 Differentiate their G(y) .\n36\n\n\n1 (\n12y−\ny3)\n0 y 12\nA1 Fully correct.\ng(y)=36\n  0 otherwise\n© UCLES 2022 Page 11 of 12",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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        "../answer-assets/9231_w22_ms_43-p11.png"
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    },
    {
      "id": "9231-2022-on-43-q06",
      "question_id": "9231-2022-on-43-q06",
      "subject": "9231",
      "year": 2022,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 1  6.242  M1 19\ns 2 = 3.26− =0.001382\nx 11 12  13750\n19\n1 5.12 \ns 2 = 2.6124− =0.001267 15000\ny 9 10 \nA1 Both correct.\n110.001382+90.001267 M1 A1 Find pooled variance.\ns2 = =0.001330\n12+10−2\n6.24 5.1 M1 A1 Use correct formula for t.\n−\n12 10\nt = =0.640\n1 1\ns +\n12 10\nH :  = H :   B1\n0 a b 1 a b\nCritical value is 2.086 (t (20)) M1 Compare their calculated value with 2.086 and conclusion.\n0.975\n0.640 < 2.086\nAccept H\n0\nThere is insufficient evidence to support inspector’s claim A1 All correct except possibly B1, in context, level of uncertainty in\nlanguage.\n‘Prove’ scores A0\n9 Pooled variance not used: M1A1 M0A0 M0A0 B1 M1A0 max 4/9\n© UCLES 2022 Page 12 of 12",
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      "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9231_w22_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2022-oct-nov/9231_w22_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-11-q01",
      "question_id": "9231-2023-mj-11-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 6 0 23 0 B1 States base case.\n2A   so true when n1.\n2 2 31 2\n23k 0 B1 States inductive hypothesis.\nAssume that it is true for nk, so 2Ak  .\n 3k 1 2 \n23k 03 0  23k1 0 M1A1 Multiplies 2Ak with A.\nThen 2Ak1    \n 3k 1 2  1 1   3k132 2 \nSo, it is also true for nk1. Hence, by induction, true for all positive integers. A1 States conclusion.\n5\n\n1(b)  3n 0 B1\ndetAn det 3n\n1  3n 1  1\n2 \n 2 0 \nOr a multiple of An 213n  seen.\n13n 23n\n\n 1 0  B1  2 0 \nAn 3n  OE An 213n  \n1  13n 3n 13n 23n \n2 \n2\n© UCLES 2023 Page 6 of 16",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p06.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q02",
      "question_id": "9231-2023-mj-11-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) (4)2 2(6) M1 Uses formula for sum of squares.\n4 A1\n2\n\n2(b) r2 2 2rr2 B1 Expands.\n n  r2 r2 r2  n  42(4)r3r2 M1 A1 Collects like terms and uses 4 and\ntheir result from part (a).\nr1 r1\n4n4nn1 1nn12n1 M1 Applies formulae from MF19.\n2\n4n2  1nn12n1 M1 A1 Simplifies.\n2\nn  4n 1  2n2 3n1 \n2\nn  n2 5n 1 \n2 2\n6\n© UCLES 2023 Page 7 of 16",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q03",
      "question_id": "9231-2023-mj-11-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) 1 1 1 1  M1 A1 Finds partial fractions.\n   \n(kr1)(krk1) kk(r1)1 kr1\nn 1 1 1 1 1 1 1  M1 Writes at least three complete terms, including\n  1      last.\n(kr1)(krk1) k k1 k1 2k1 k(n1)1 kn1\nr1\n1 1  A1 n\n1  OE e.g.\nk kn1 kn1\n4\n\n3(b) 1 B1\nk\n1\n\n3(c) n2 1 n2 1 n1 1 M1 Or applies the method of differences again.\n  \n(kr1)(krk1) (kr1)(krk1) (kr1)(krk1)\nrn r1 r1\n1 1  1  1 1 1  A1 n2 (n1)\nk \n1\nkn21\n\n \n1\nk(n1)1  \n\nk  k(n1)1\n\nkn21 \nOE e.g.\nkn2 1\n\nk(n1)1\n2\n© UCLES 2023 Page 8 of 16",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p08.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q04",
      "question_id": "9231-2023-mj-11-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "4(a)  a b2 cos sin M1 Uses correct matrix for rotation.\n  \n c2 a   sin cos\nb2 c2 which is impossible since b and c are real and b0. A1 AG\n2\n\n4(b)  a b2x axb2y B1 x X \n    Transforms   to  .\n c2 a  y  c2xay   y Y \nc2xamxm  axb2mx  M1 A1 Uses ymx and Y mX .\nc2 ammab2m2 c2 b2m2 A1\ny c x and yc x A1\nb b\n5\n\n4(c) 1 55 0 M1 A1* Award M1 if matrices correct but order is\nM   wrong.\n0 10 5\n5 52 DA1 Dep: previous A1\n \n0 5\n3\n\n4(d) 12detM M1 Using their M.\n300 cm2 A1FT\n2\n© UCLES 2023 Page 9 of 16",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p09.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q05",
      "question_id": "9231-2023-mj-11-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) 0 B1* Correct shape and domain, polar graph with r\nstrictly decreasing but condone if not strictly\ndecreasing close to π.\nDB1 Fully correct including\nshape at 0 correct\nshape at π correct\nl,0 B1 Identified as point furthest from the pole and\ngiven as coordinates.\n3\n\n5(b)  π 1 M1 Uses correct formula with correct limits.\n1 d\n2 2 1\n0\nπ M1 A1 1\n1 2   tan1  0 Integrates 2 1 .\n1tan1π0.631 A1\n2\n4\n© UCLES 2023 Page 10 of 16\n\n5(c) sin B1 Uses yrsin\ny\n2 1\ndy  2 1 1 2cos  2 1 1 2sin M1 A1 Sets derivative equal to zero.\n 0\nd 2 1\n 1 A1 AG.\n0 cot10\n \n 1  B1 Shows sign change (1sf or better).\n1.1 cot1.110.02…\n 1.1\n 1 \nand 1.2 cot1.210.209…\n 1.2\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p10.png",
        "../answer-assets/9231_s23_ms_11-p11.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q06",
      "question_id": "9231-2023-mj-11-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 B1 States vertical asymptote.\nx2 2x15x2x47 y x4 M1 A1 Finds oblique asymptote.\n3\n© UCLES 2023 Page 11 of 16\n\n6(b)\ndy\nx22x2  x2 2x15  M1 Differentiates.\n\ndx x22\n  A1 dy\nx2 4x110 or dy 1 7  Forms quadratic equation or simplifies .\ndx\n dx x22 \n \n42 4(1)(11)280 (or y'0 ) There are no turning points. A1 Correct conclusion.\n3\n© UCLES 2023 Page 12 of 16\n\n6(c) B1 Axes and asymptotes.\nB1 Branches correct.\n(0,7.5),5,0,(3,0) B1 States coordinates of intersections with axes.\n3\n© UCLES 2023 Page 13 of 16\n\n6(d) B1FT FT from sketch in (c).\nB1 Correct shape at infinity and on x axis.\n2\n\n6(e) x2 2x15 x2 2x15 M2 Finds critical points, award M1 for each case.\n15 or 15\nx2 2 x2 2\nx2 11x0 or x2 19x300\n2 2\nx0,11 or x12, 5 A1\n2 2\n12x0 or 5 x11 A1FT\n2 2\n4\n© UCLES 2023 Page 14 of 16",
      "source_pages": [
        11,
        12,
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p11.png",
        "../answer-assets/9231_s23_ms_11-p12.png",
        "../answer-assets/9231_s23_ms_11-p13.png",
        "../answer-assets/9231_s23_ms_11-p14.png"
      ]
    },
    {
      "id": "9231-2023-mj-11-q07",
      "question_id": "9231-2023-mj-11-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k 0 0 M1 A1 Finds common perpendicular.\n   \n1 1 1  2 ~ 1\n   \n   \n1 1 1 2 1\n(4)(3)7 yz7 M1 A1 Substitutes point.\n4\n\n7(b) 7 B1 FT\n \nStates point common to both planes e.g. 2 .\n \n \n5\ni j k  2  M1 A1FT Finds direction of line.\n \n0 1 1  5\n \n \n5 3 5  5 \n7  2  A1 OE.\n   \nr 2  5\n   \n   \n5  5 \n4\n\n7(c) aa77 M1 Uses correct formula for distance from A to .\n 2 1\n2\na1 A1\n2\n© UCLES 2023 Page 15 of 16\n\n7(d) M1 A1 Uses correct formula.\nbb\n 1 2\n2 (1b)2 2b2 2\nM1 Solves for b.\n2b 1b2 2b2 b2 2b10\nb1 2 A1 CAO\n4\n© UCLES 2023 Page 16 of 16",
      "source_pages": [
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_11-p15.png",
        "../answer-assets/9231_s23_ms_11-p16.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q01",
      "question_id": "9231-2023-mj-12-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 6 0 23 0 B1 States base case.\n2A   so true when n1.\n2 2 31 2\n23k 0 B1 States inductive hypothesis.\nAssume that it is true for nk, so 2Ak  .\n 3k 1 2 \n23k 03 0  23k1 0 M1A1 Multiplies 2Ak with A.\nThen 2Ak1    \n 3k 1 2  1 1   3k132 2 \nSo, it is also true for nk1. Hence, by induction, true for all positive integers. A1 States conclusion.\n5\n\n1(b)  3n 0 B1\ndetAn det 3n\n1  3n 1  1\n2 \n 2 0 \nOr a multiple of An 213n  seen.\n13n 23n\n\n 1 0  B1  2 0 \nAn 3n  OE An 213n  \n1  13n 3n 13n 23n \n2 \n2\n© UCLES 2023 Page 6 of 16",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p06.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q02",
      "question_id": "9231-2023-mj-12-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) (4)2 2(6) M1 Uses formula for sum of squares.\n4 A1\n2\n\n2(b) r2 2 2rr2 B1 Expands.\n n  r2 r2 r2  n  42(4)r3r2 M1 A1 Collects like terms and uses 4 and\ntheir result from part (a).\nr1 r1\n4n4nn1 1nn12n1 M1 Applies formulae from MF19.\n2\n4n2  1nn12n1 M1 A1 Simplifies.\n2\nn  4n 1  2n2 3n1 \n2\nn  n2 5n 1 \n2 2\n6\n© UCLES 2023 Page 7 of 16",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q03",
      "question_id": "9231-2023-mj-12-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) 1 1 1 1  M1 A1 Finds partial fractions.\n   \n(kr1)(krk1) kk(r1)1 kr1\nn 1 1 1 1 1 1 1  M1 Writes at least three complete terms, including\n  1      last.\n(kr1)(krk1) k k1 k1 2k1 k(n1)1 kn1\nr1\n1 1  A1 n\n1  OE e.g.\nk kn1 kn1\n4\n\n3(b) 1 B1\nk\n1\n\n3(c) n2 1 n2 1 n1 1 M1 Or applies the method of differences again.\n  \n(kr1)(krk1) (kr1)(krk1) (kr1)(krk1)\nrn r1 r1\n1 1  1  1 1 1  A1 n2 (n1)\nk \n1\nkn21\n\n \n1\nk(n1)1  \n\nk  k(n1)1\n\nkn21 \nOE e.g.\nkn2 1\n\nk(n1)1\n2\n© UCLES 2023 Page 8 of 16",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p08.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q04",
      "question_id": "9231-2023-mj-12-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "4(a)  a b2 cos sin M1 Uses correct matrix for rotation.\n  \n c2 a   sin cos\nb2 c2 which is impossible since b and c are real and b0. A1 AG\n2\n\n4(b)  a b2x axb2y B1 x X \n    Transforms   to  .\n c2 a  y  c2xay   y Y \nc2xamxm  axb2mx  M1 A1 Uses ymx and Y mX .\nc2 ammab2m2 c2 b2m2 A1\ny c x and yc x A1\nb b\n5\n\n4(c) 1 55 0 M1 A1* Award M1 if matrices correct but order is\nM   wrong.\n0 10 5\n5 52 DA1 Dep: previous A1\n \n0 5\n3\n\n4(d) 12detM M1 Using their M.\n300 cm2 A1FT\n2\n© UCLES 2023 Page 9 of 16",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p09.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q05",
      "question_id": "9231-2023-mj-12-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) 0 B1* Correct shape and domain, polar graph with r\nstrictly decreasing but condone if not strictly\ndecreasing close to π.\nDB1 Fully correct including\nshape at 0 correct\nshape at π correct\nl,0 B1 Identified as point furthest from the pole and\ngiven as coordinates.\n3\n\n5(b)  π 1 M1 Uses correct formula with correct limits.\n1 d\n2 2 1\n0\nπ M1 A1 1\n1 2   tan1  0 Integrates 2 1 .\n1tan1π0.631 A1\n2\n4\n© UCLES 2023 Page 10 of 16\n\n5(c) sin B1 Uses yrsin\ny\n2 1\ndy  2 1 1 2cos  2 1 1 2sin M1 A1 Sets derivative equal to zero.\n 0\nd 2 1\n 1 A1 AG.\n0 cot10\n \n 1  B1 Shows sign change (1sf or better).\n1.1 cot1.110.02…\n 1.1\n 1 \nand 1.2 cot1.210.209…\n 1.2\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p10.png",
        "../answer-assets/9231_s23_ms_12-p11.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q06",
      "question_id": "9231-2023-mj-12-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 B1 States vertical asymptote.\nx2 2x15x2x47 y x4 M1 A1 Finds oblique asymptote.\n3\n© UCLES 2023 Page 11 of 16\n\n6(b)\ndy\nx22x2  x2 2x15  M1 Differentiates.\n\ndx x22\n  A1 dy\nx2 4x110 or dy 1 7  Forms quadratic equation or simplifies .\ndx\n dx x22 \n \n42 4(1)(11)280 (or y'0 ) There are no turning points. A1 Correct conclusion.\n3\n© UCLES 2023 Page 12 of 16\n\n6(c) B1 Axes and asymptotes.\nB1 Branches correct.\n(0,7.5),5,0,(3,0) B1 States coordinates of intersections with axes.\n3\n© UCLES 2023 Page 13 of 16\n\n6(d) B1FT FT from sketch in (c).\nB1 Correct shape at infinity and on x axis.\n2\n\n6(e) x2 2x15 x2 2x15 M2 Finds critical points, award M1 for each case.\n15 or 15\nx2 2 x2 2\nx2 11x0 or x2 19x300\n2 2\nx0,11 or x12, 5 A1\n2 2\n12x0 or 5 x11 A1FT\n2 2\n4\n© UCLES 2023 Page 14 of 16",
      "source_pages": [
        11,
        12,
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p11.png",
        "../answer-assets/9231_s23_ms_12-p12.png",
        "../answer-assets/9231_s23_ms_12-p13.png",
        "../answer-assets/9231_s23_ms_12-p14.png"
      ]
    },
    {
      "id": "9231-2023-mj-12-q07",
      "question_id": "9231-2023-mj-12-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k 0 0 M1 A1 Finds common perpendicular.\n   \n1 1 1  2 ~ 1\n   \n   \n1 1 1 2 1\n(4)(3)7 yz7 M1 A1 Substitutes point.\n4\n\n7(b) 7 B1 FT\n \nStates point common to both planes e.g. 2 .\n \n \n5\ni j k  2  M1 A1FT Finds direction of line.\n \n0 1 1  5\n \n \n5 3 5  5 \n7  2  A1 OE.\n   \nr 2  5\n   \n   \n5  5 \n4\n\n7(c) aa77 M1 Uses correct formula for distance from A to .\n 2 1\n2\na1 A1\n2\n© UCLES 2023 Page 15 of 16\n\n7(d) M1 A1 Uses correct formula.\nbb\n 1 2\n2 (1b)2 2b2 2\nM1 Solves for b.\n2b 1b2 2b2 b2 2b10\nb1 2 A1 CAO\n4\n© UCLES 2023 Page 16 of 16",
      "source_pages": [
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_12-p15.png",
        "../answer-assets/9231_s23_ms_12-p16.png"
      ]
    },
    {
      "id": "9231-2023-mj-13-q01",
      "question_id": "9231-2023-mj-13-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 533233124 is divisible by 31. B1 Checks base case.\nAssume that 53k 32k 33 is divisible by 31 for some positive integer B1 States inductive hypothesis.\nk.\n53k332k133(1241)53k (311)32k 33 M1 A1 Separates 53k 32k 33 or considers difference.\nis divisible by 31 because 12453k 3132k is divisible by 31. A1\nHence, by induction, true for every positive integer n. A1\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_13-p06.png"
      ]
    },
    {
      "id": "9231-2023-mj-13-q02",
      "question_id": "9231-2023-mj-13-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) 1  1  M1 Substitutes formulae for r2 and r.\n6 nn12n1 6 nn1  5n\n6  2 \n2n36n2n A1\n2\n© UCLES 2023 Page 6 of 16\n\n2(b) 6r2 6r5 5 B1 Divides by denominator.\n6\nr2 r r(r1)\n1 1 1 B1 Finds partial fractions.\n \nrr1 r r1\nn 1 1 1 1 1 1 M1 Writes at least three complete terms, including the last\n 1     term, to show cancelation.\nrr1 2 2 3 n n1\nr1\nn 6r2 6r5 5 A1 6n2 n\n 6n5 OE e.g.\nr2 r n1 n1\nr1\n4\n\n2(c) 2n 6r2 6r5 n 6r2 6r5 M1 Or uses method of differences again.\n \nr2 r r2 r\nr1 r1\n5 5\n12n5 6n5\n2n1 n1\n5 5 5 5 A1 5n\n12n5 6n5 6n  Or 6n . OE, like terms collected.\n2n1 n1 2n1 n1 n12n1\n2\n© UCLES 2023 Page 7 of 16",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_13-p06.png",
        "../answer-assets/9231_s23_ms_13-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-13-q03",
      "question_id": "9231-2023-mj-13-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) y  x2 so, x y 1 2 B1 Correct substitution.\ny2  y2y 1 2 50 so,  2y 1 2 2   y2  y5 2 M1 Obtains an equation which eliminates radicals.\ny4 2y311y2 14y250 A1\n2 2 2 2 2 B1FT\n4\n\n3(b) 2222 25 B1FT\n1 1 1 1 222 222 222 222 M1 Relates to coefficients.\n   \n2 2 2 2 2222\n14 A1 CAO\n25\n3\n\n3(c)  2 M1 Uses formula for sum of squares or uses original equation.\n4 4 4 4  2 2 2 2\n \n2 22 22 22 22 22 22\n22 211\n18 A1\n2\n© UCLES 2023 Page 8 of 16",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_13-p08.png"
      ]
    },
    {
      "id": "9231-2023-mj-13-q04",
      "question_id": "9231-2023-mj-13-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "4(a) Rotation [anticlockwise] B1\nabout the origin through angle2. B1\nShear [in the x-direction] followed by a rotation [anticlockwise about the B1\norigin through angle2].\n3\n\n4(b) 1 1 B1\n1 k 1 k cos2 sin2  cos2 sin2\n    ,    \n0 1 0 1  sin2 cos2 sin2 cos2\n1 k cos2 sin2 B1 Correct order\nM1   \n0 1 sin2 cos2\n2\n\n4(c) cos21 kcos2sin2  B1\nMI \n sin2 ksin2cos21\ncos21(ksin2cos21)ksin2θcos2θsin22 0 M1 Evaluates detMI\n22cos2ksin20 A1 Brackets removed correctly and = 0\n4sin22ksincos M1 Uses 1cos22sin2 andsin22sincos or all\nnecessary double angle formulae.\ntan 1k A1\n2\n5\n© UCLES 2023 Page 9 of 16\n\n4(d)  1 3 3 B1\nM 2 2 .\n1 3 5 \n \n2 2\n 1 3 3x 1x 3 3y B1FT x X\n   1 2 2 3 2 5 2     y      1 2 2 3x 2 5 2 y    Transforms  y   to  Y  \n1 3  3 3  M1 X x\n x 3yx   x 3y0  x 3y0  Sets   \n2 2  2 2  Y  y\nand\n1 5  1 3 \n3x y y  3x y0  3x3y0\n \n2 2  2 2 \n3x3y0 A1 AG.\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2023-mj-13-q05",
      "question_id": "9231-2023-mj-13-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) r2 cos2sin2  a B1 Uses xrcos and/or yrsin.\nr2cos2a M1 Applies relevant double angle formulae.\nr2 asec2 A1 AG.\n3\n© UCLES 2023 Page 10 of 16\n\n5(b) B1* Initial line drawn. Correct domain and position, r strictly\nincreasing.\ndB1 Also sloping to right, concave on opposite side to pole,\ncorrect gradient when 0 and  π/4 .\nO\na B1\n3\n\n5(c) 1a1 1 2  sec2d M1 Uses 1 2 r2d with correct limits.\n2\n0\n1alntan 1π1 1 2 π or 1alntan2sec21 1 2 π M1 A1 Integrates.\n4  4  0 4   0\n1aln 3 1aln3 A1\n4 8\n4\n© UCLES 2023 Page 11 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2023-mj-13-q06",
      "question_id": "9231-2023-mj-13-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) AB2ij4k AC3i3k BCijk B1 Finds direction vectors of two lines in the plane.\ni j k  3   1  M1 A1FT Finds normal to the plane ABC.\n   \n2 1 4   ~ 2\n   \n   \n3 0 3     1 \n1(1)2(1)1(0)1x2yz1 M1 A1 Substitutes point. CAO\n5\n\n6(b) 1 1 M1 A1FT Divides by magnitude of normal vector.\n 0.408\nFT their (a).\n12 22 12 6\n2\n© UCLES 2023 Page 12 of 16\n\n6(c) 2 12 122 M1 A1 Finds PQ, where P is a point on OC and Q is a point on\n  \n     \nOP  ,OQ 1 PQ 1 AB.\n     \n     \n 3  4   43 \n\n122 2 M1* Uses that dot product of PQ with line direction is zero.\n   \n1  1 0\n   \n   \n 43   3 \n17141 dM1 Deduces one equation.\n122 2 dM1 Deduces second equation.\n   \n1  1 021171\n   \n   \n 43   4 \n\n2 dM1 A1 Solves for or  and substitutes into OP.\n4  4 \n OP 1\n \n5 5\n \n 3 \n2 1 A1 OE\n4   \nr 1 k 2\n   \n5\n   \n 3  0\n8\n© UCLES 2023 Page 13 of 16",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2023-mj-13-q07",
      "question_id": "9231-2023-mj-13-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "7(a) x3 B1 States vertical asymptote.\nx3(x5)16 16 M1 Finds oblique asymptote.\ny x5\nx3 x3\nyx5 A1\n3\n\n7(b) dy 16 M1 Differentiates and sets equal to zero.\n1 0(x3)2 16\ndx x32\nx1,7 A1 Finds x-coordinates\n(1,0),(7,16) A1 States coordinates of turning points.\n3\n© UCLES 2023 Page 14 of 16\n\n7(c) B1FT Axes and labelled asymptotes.\nB1 Upper branch correct.\nB1 Lower branch correct and no additional branches.\n3\n© UCLES 2023 Page 15 of 16\n\n7(d) B1FT Clear labels, axes and their vertical asymptote.\nx2 2x1\ny B1FT x2 2x1\nx3\ny correct, FT from their sketch in (c).\nx3\nB1 x2 2x1\nUpper branch of y2  (positive square root).\nx3\nx2 2x1\ny2 \nx3 B1FT x2 2x1\nLower branch of y2  (negative square root).\nx2 2x1 x3\ny\nFT from previous mark.\nx3\nx22x1\ny2\nx3\n4\n© UCLES 2023 Page 16 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2023-mj-21-q01",
      "question_id": "9231-2023-mj-21-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 2 3 M1 A1 Shows that determinant is zero or row operations to\nobtain e.g.\n4 5 6 59862497634875\nx 2y 3z 1,\n7 8 9\ny 2z 1,\n31290\n2\n\n1(b) x 2y 3z 1, M1 A1 Uses all three equations to reduce to one equation with\n4x 5y 6z 1, z11x2y y2x1 two unknowns. Reducing to two equations scores M1\n3 A0.\n7x 8y 9z 1,\nThe three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along\na common line.\n3\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-21-q02",
      "question_id": "9231-2023-mj-21-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 dy dz B1\n 1\ndx dx\ndz 13z M1 Substitutes and derives first order separable equation.\n1\ndx 3z1\ndz 13z 6z A1\n 1\ndx 3z1 3z1\n1 1z1dz1dx M1 Separates variables and integrates both sides.\n2 6\n1z1lnzxC1x1 y1ln(xy)C A1\n2 6 2 2 6\nC1 M1 Substitutes initial conditions into their expression.\n2\n1ln(x y)1xy1 0 A1 OE.\n6 2 2\n7\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_21-p07.png"
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    },
    {
      "id": "9231-2023-mj-21-q03",
      "question_id": "9231-2023-mj-21-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) zz1 2cos B1 Use of zz1 2cos.\n zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. M1 A0 for no clear grouping.\nCorrect substitution of zn cosnisinn for each\nterm scores M1 A1.\n2cos4 2cos442cos26 M1 Substitutes zn zn 2cosn. If zn cosnisinn\nused then must cancel sin.\ncos41cos44cos23 A1 AG. SC B1 expands cosisin4 and uses\n8\ntrigonometric identities.\n5\n\n3(b)  1 2 1x23 2dx 1 6 π cos4d 11sin42sin23 1 6 π M1 A1 Applies substitution (M1) gets to cos4d,\n \n0 0 8 4 0\nchanges limits, integration correct (A1).\n1  9 3π  A1\n16 4\n3\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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        "../answer-assets/9231_s23_ms_21-p08.png"
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    },
    {
      "id": "9231-2023-mj-21-q04",
      "question_id": "9231-2023-mj-21-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) d \nx\n 1x5n  5nx5 1x5n1\n\n 1x5n M1 A1 Uses the product rule to differentiate.\ndx\n5n  1x5 1  1x5n1   1x5n M1* Uses x5 x5 11.\n1 DM1 Integrates both sides using the limits given. Requires\n\nx\n 1x5n\n5nI 5nI I\n  n n1 n previous method mark.\n0\n2n (5n1)I 5nI (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG.\nn n1 n n1\nAlternative method for question 4(a)\nI \n1 1x5n\ndx\n\nx\n 1x5n 1\n5n\n1 x5 1x5n1\ndx\nM1 A1 Integrates by parts.\nn\n0\n \n0 0\nI \n\nx\n 1x5n 1\n5n\n1 1x5n\ndx5n\n1 1x5n1\ndx\nM1* Uses x5 x5 11.\nn  \n0 0 0\n1 DM1 Forms recursive formula. Requires previous method\nI \n\nx\n 1x5n\n5nI 5nI\nn   n n1 mark.\n0\n(5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG.\nn n1\n5\n© UCLES 2023 Page 9 of 15\n\n4(b) 1 B1\nI x1x6  7 or I 1\n1  6  0 6 0\n11I 22 10I I  47 M1 A1 Applies reduction formula.\n2 1 2 33\n16I 2315I I 323 A1\n3 2 3 176\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2023-mj-21-q05",
      "question_id": "9231-2023-mj-21-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 18 5 11 M1 Sets determinant equal to zero.\n8 6 4 0\n32 10 20\n6 4 8 4 8 6 A1 Expands determinant. Can use other rows/columns.\n18 5 11\n10 20 32 20 32 10\n342 20480 A1 AG\n(2)(4)(6)0 leading to 2,4,6 B1\n4\n© UCLES 2023 Page 10 of 15\n\n5(b) i j k  24   1  M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors. Must attempt to solve\n2: 16 5 11  24  1\n    equations for M1.\n   \n8 4 4  24   1 \ni j k 120 1 A1 A1\n   \n4: 8 10 4  0 ~ 0\n   \n   \n32 10 16 240 2\ni j k 20 1\n   \n6: 12 5 11  40 ~ 2\n   \n   \n8 0 4 40 2\n 1 1 1 32 0 0  M1 A1 Or correctly matched permutations of columns.\n    M0 if a column of zeros appears in P.\nThus P 1 0 2 and D 0 1024 0\n   \n   \n 1 2 2  0 0 7776\n6\n© UCLES 2023 Page 11 of 15",
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2023-mj-21-q06",
      "question_id": "9231-2023-mj-21-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6 m2 12m360 M1 Auxiliary equation.\nxe6tAtB A1 Complementary function. Allow “x” missing.\nx psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives.\np12q36p37 q12p36q0 M1 Substitutes and equates coefficients. Must have two\nunknowns.\np 35 q12 A1\n37 37\nxe6tAtB35sint12cost A1 FT Must have “x”. FT on CF.\n37 37\nx' Ae6t 6e6tAtB35cost12sint M1* Differentiates using product rule. Must have their PI\n37 37 differentiated.\nB12 A6B35 0A1 DM1 A1 Forms simultaneous equations using initial conditions.\n37 37\nxe6tt12 35sint12cost A1 Must have “x”.\n37 37 37\n11\n© UCLES 2023 Page 12 of 15",
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-21-q07",
      "question_id": "9231-2023-mj-21-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a)  x  1 M1 A1 Applies given substitution.\n dx 1 du\n x2 1 2 u\nx21C\nA1 Allow with “C ” missing. Answer must be in terms\nof x.\n3\n\n7(b)   B1\ncosh1r ln r r2 1\ncosh12cosh13cosh1N M1 Forms sum of the areas of the rectangles.\nN M1 Compares with integral with correct limits.\n cosh1xdx\n1\nN A1 Evaluates integral.\nN N  x\n cosh1xdxxcosh1x  dx\n \n1 1  x2 1\n1\nN     A1 AG.\nln r r2 1 Nln N  N2 1  N2 1\nr2\n5\n© UCLES 2023 Page 13 of 15\n\n7(c) (cosh11)cosh12cosh1(N1)  N cosh1xdx(or N cosh1xdx) M1 A1 Compares with integral with correct limits.\n1 2\n N ln  r r2 1  (N 1)ln  N  N2 1   N2 1 A1 Adds ln  N  N2 1  to both sides.\nr2\nor  N ln  r r2 1  (N 1)ln  N  N2 1   N2 12ln  2 3   3\nr2\nAlternative method for question 7(c)\n(cosh11)cosh12cosh1(N1)cosh1N M1 A1 Compares with integral with correct limits.\n N1 cosh1xdx  xln  x x2 1   x2 1  N1 (or  N1 cosh1xdx)\n1\n \n1 2\nN     A1\nln r r2 1 N 1ln N 1 N2 2N  N2 2N\nr2\nor N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3\nSecond alternative method for question 7(c)\ncosh1(11)cosh1(21)cosh1(N11) M1 A1 Compares with integral with correct limits.\n N cosh1x1dx x1ln  x1 x2 2x   x2 2x  N\n1\n \n1\nN 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 A1\n3\n© UCLES 2023 Page 14 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2023-mj-21-q08",
      "question_id": "9231-2023-mj-21-q08",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) 2 et et B1\nsecht  tanht \net et et et\n 2  2  et et2 4 e2t 2e2t 4  et et2 M1 A1 Expands, gets to e2t 2e2t 4 for M1, AG.\n1 et et     et et2   et et2   et et2  et et2\n3\n\n8(b) dy B1\n4tanh3tsech2t\ndt\ndx tanhtsech2ttanht  tanht  sech2t1  tanh3t  M1 A1 M1 sensible attempt at derivative of x.\ndt\ndy dy dt 4tanh3tsech2t M1 A1 dy dx\n   4sech2t Applies chain rule, must substitute their and\ndx dt dx tanh3t dt dt\nfor M1, AG.\n5\n\n8(c) d2y d dy dt 8sech2ttanht sech2t M1 A1 d2y d dy\n    8 Finds . For M1  csech2ttanht.AEF.\ndx2 dtdx dx tanh3t tanh2t dx2 dtdx\nAccept 8cosech2t.\n8\nsech2t\n9 8  1tanh2t   9tanh2ttanh2t  16\nM1 A1 Sets equal to 9\n2\n, uses identity from (a) or equivalent.\ntanh2t 2 2 25\nAccept sinh2t16 or cosh2t 25.\n9 9\nx,y 8 ln3,881 A1 A1 A1 for each correct coordinate. t ln3\n25 5 625\n6\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-22-q01",
      "question_id": "9231-2023-mj-22-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 2 3 M1 A1 Shows that determinant is zero or row operations to\nobtain e.g.\n4 5 6 59862497634875\nx 2y 3z 1,\n7 8 9\ny 2z 1,\n31290\n2\n\n1(b) x 2y 3z 1, M1 A1 Uses all three equations to reduce to one equation with\n4x 5y 6z 1, z11x2y y2x1 two unknowns. Reducing to two equations scores M1\n3 A0.\n7x 8y 9z 1,\nThe three planes form a sheaf. B1 Accept clear sketch or the three planes intersect along\na common line.\n3\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2023-mj-22-q02",
      "question_id": "9231-2023-mj-22-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 dy dz B1\n 1\ndx dx\ndz 13z M1 Substitutes and derives first order separable equation.\n1\ndx 3z1\ndz 13z 6z A1\n 1\ndx 3z1 3z1\n1 1z1dz1dx M1 Separates variables and integrates both sides.\n2 6\n1z1lnzxC1x1 y1ln(xy)C A1\n2 6 2 2 6\nC1 M1 Substitutes initial conditions into their expression.\n2\n1ln(x y)1xy1 0 A1 OE.\n6 2 2\n7\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_22-p07.png"
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    },
    {
      "id": "9231-2023-mj-22-q03",
      "question_id": "9231-2023-mj-22-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) zz1 2cos B1 Use of zz1 2cos.\n zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. M1 A0 for no clear grouping.\nCorrect substitution of zn cosnisinn for each\nterm scores M1 A1.\n2cos4 2cos442cos26 M1 Substitutes zn zn 2cosn. If zn cosnisinn\nused then must cancel sin.\ncos41cos44cos23 A1 AG. SC B1 expands cosisin4 and uses\n8\ntrigonometric identities.\n5\n\n3(b)  1 2 1x23 2dx 1 6 π cos4d 11sin42sin23 1 6 π M1 A1 Applies substitution (M1) gets to cos4d,\n \n0 0 8 4 0\nchanges limits, integration correct (A1).\n1  9 3π  A1\n16 4\n3\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_22-p08.png"
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    },
    {
      "id": "9231-2023-mj-22-q04",
      "question_id": "9231-2023-mj-22-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) d \nx\n 1x5n  5nx5 1x5n1\n\n 1x5n M1 A1 Uses the product rule to differentiate.\ndx\n5n  1x5 1  1x5n1   1x5n M1* Uses x5 x5 11.\n1 DM1 Integrates both sides using the limits given. Requires\n\nx\n 1x5n\n5nI 5nI I\n  n n1 n previous method mark.\n0\n2n (5n1)I 5nI (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG.\nn n1 n n1\nAlternative method for question 4(a)\nI \n1 1x5n\ndx\n\nx\n 1x5n 1\n5n\n1 x5 1x5n1\ndx\nM1 A1 Integrates by parts.\nn\n0\n \n0 0\nI \n\nx\n 1x5n 1\n5n\n1 1x5n\ndx5n\n1 1x5n1\ndx\nM1* Uses x5 x5 11.\nn  \n0 0 0\n1 DM1 Forms recursive formula. Requires previous method\nI \n\nx\n 1x5n\n5nI 5nI\nn   n n1 mark.\n0\n(5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG.\nn n1\n5\n© UCLES 2023 Page 9 of 15\n\n4(b) 1 B1\nI x1x6  7 or I 1\n1  6  0 6 0\n11I 22 10I I  47 M1 A1 Applies reduction formula.\n2 1 2 33\n16I 2315I I 323 A1\n3 2 3 176\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2023-mj-22-q05",
      "question_id": "9231-2023-mj-22-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 18 5 11 M1 Sets determinant equal to zero.\n8 6 4 0\n32 10 20\n6 4 8 4 8 6 A1 Expands determinant. Can use other rows/columns.\n18 5 11\n10 20 32 20 32 10\n342 20480 A1 AG\n(2)(4)(6)0 leading to 2,4,6 B1\n4\n© UCLES 2023 Page 10 of 15\n\n5(b) i j k  24   1  M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors. Must attempt to solve\n2: 16 5 11  24  1\n    equations for M1.\n   \n8 4 4  24   1 \ni j k 120 1 A1 A1\n   \n4: 8 10 4  0 ~ 0\n   \n   \n32 10 16 240 2\ni j k 20 1\n   \n6: 12 5 11  40 ~ 2\n   \n   \n8 0 4 40 2\n 1 1 1 32 0 0  M1 A1 Or correctly matched permutations of columns.\n    M0 if a column of zeros appears in P.\nThus P 1 0 2 and D 0 1024 0\n   \n   \n 1 2 2  0 0 7776\n6\n© UCLES 2023 Page 11 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2023-mj-22-q06",
      "question_id": "9231-2023-mj-22-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6 m2 12m360 M1 Auxiliary equation.\nxe6tAtB A1 Complementary function. Allow “x” missing.\nx psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives.\np12q36p37 q12p36q0 M1 Substitutes and equates coefficients. Must have two\nunknowns.\np 35 q12 A1\n37 37\nxe6tAtB35sint12cost A1 FT Must have “x”. FT on CF.\n37 37\nx' Ae6t 6e6tAtB35cost12sint M1* Differentiates using product rule. Must have their PI\n37 37 differentiated.\nB12 A6B35 0A1 DM1 A1 Forms simultaneous equations using initial conditions.\n37 37\nxe6tt12 35sint12cost A1 Must have “x”.\n37 37 37\n11\n© UCLES 2023 Page 12 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-22-q07",
      "question_id": "9231-2023-mj-22-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a)  x  1 M1 A1 Applies given substitution.\n dx 1 du\n x2 1 2 u\nx21C\nA1 Allow with “C ” missing. Answer must be in terms\nof x.\n3\n\n7(b)   B1\ncosh1r ln r r2 1\ncosh12cosh13cosh1N M1 Forms sum of the areas of the rectangles.\nN M1 Compares with integral with correct limits.\n cosh1xdx\n1\nN A1 Evaluates integral.\nN N  x\n cosh1xdxxcosh1x  dx\n \n1 1  x2 1\n1\nN     A1 AG.\nln r r2 1 Nln N  N2 1  N2 1\nr2\n5\n© UCLES 2023 Page 13 of 15\n\n7(c) (cosh11)cosh12cosh1(N1)  N cosh1xdx(or N cosh1xdx) M1 A1 Compares with integral with correct limits.\n1 2\n N ln  r r2 1  (N 1)ln  N  N2 1   N2 1 A1 Adds ln  N  N2 1  to both sides.\nr2\nor  N ln  r r2 1  (N 1)ln  N  N2 1   N2 12ln  2 3   3\nr2\nAlternative method for question 7(c)\n(cosh11)cosh12cosh1(N1)cosh1N M1 A1 Compares with integral with correct limits.\n N1 cosh1xdx  xln  x x2 1   x2 1  N1 (or  N1 cosh1xdx)\n1\n \n1 2\nN     A1\nln r r2 1 N 1ln N 1 N2 2N  N2 2N\nr2\nor N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3\nSecond alternative method for question 7(c)\ncosh1(11)cosh1(21)cosh1(N11) M1 A1 Compares with integral with correct limits.\n N cosh1x1dx x1ln  x1 x2 2x   x2 2x  N\n1\n \n1\nN 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 A1\n3\n© UCLES 2023 Page 14 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2023-mj-22-q08",
      "question_id": "9231-2023-mj-22-q08",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) 2 et et B1\nsecht  tanht \net et et et\n 2  2  et et2 4 e2t 2e2t 4  et et2 M1 A1 Expands, gets to e2t 2e2t 4 for M1, AG.\n1 et et     et et2   et et2   et et2  et et2\n3\n\n8(b) dy B1\n4tanh3tsech2t\ndt\ndx tanhtsech2ttanht  tanht  sech2t1  tanh3t  M1 A1 M1 sensible attempt at derivative of x.\ndt\ndy dy dt 4tanh3tsech2t M1 A1 dy dx\n   4sech2t Applies chain rule, must substitute their and\ndx dt dx tanh3t dt dt\nfor M1, AG.\n5\n\n8(c) d2y d dy dt 8sech2ttanht sech2t M1 A1 d2y d dy\n    8 Finds . For M1  csech2ttanht.AEF.\ndx2 dtdx dx tanh3t tanh2t dx2 dtdx\nAccept 8cosech2t.\n8\nsech2t\n9 8  1tanh2t   9tanh2ttanh2t  16\nM1 A1 Sets equal to 9\n2\n, uses identity from (a) or equivalent.\ntanh2t 2 2 25\nAccept sinh2t16 or cosh2t 25.\n9 9\nx,y 8 ln3,881 A1 A1 A1 for each correct coordinate. t ln3\n25 5 625\n6\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
        15
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_22-p15.png"
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    },
    {
      "id": "9231-2023-mj-23-q01",
      "question_id": "9231-2023-mj-23-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a)\nf '(x)\n 1x21\n2\nB1 Correct first derivative.\nf ''(x)x\n 1x23\n2\nB1 Correct second derivative.\nf\n'''(x)3x2 1x25\n2 \n 1x23\n2\nM1 Differentiates their f ''(x) using product rule.\nf(0)0 f '(0)1 f ''(0)0 f '''(0)1 M1 Evaluates their derivatives at x0. Must have\nattempted all three derivatives.\nsin1x x 1x3 A1 CWO.\n6\nAlternative method for question 1(a)\nsinyx f '(x)secy (B1) Finds first derivative.\nf\n''(x)tanysecy2 (B1) Finds second derivative.\nf '''(x)  2tan2 ysec2 ysec4 y  secy3sec5 y2sec3 y (M1) Differentiates tanysecy2 using product and chain\nrule.\nf(0)0 f '(0)1 f ''(0)0 f '''(0)1 (M1) Evaluates their derivatives at x0. Must have\nattempted all three derivatives.\nsin1x x 1x3 (A1) CWO.\n6\n5\n© UCLES 2023 Page 6 of 15\n\n1(b) 151 B1\n750\n1\nPartial\nQuestion Answer Guidance\nMarks",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_23-p06.png",
        "../answer-assets/9231_s23_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-23-q02",
      "question_id": "9231-2023-mj-23-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) 6m2 5m10m 1 , 1 M1 Auxiliary equation.\n2 3\nx Ae1 2 t Be1 3 t A1 Complementary function. Allow with ‘x =’ missing.\n2 B1 Particular integral and its derivatives.\nx pqtrt x'q2rtx''2r\n12r5q10rt pqtrt2 t2 10t13 M1 Substitutes and equates coefficients.\nr 1, q0, p1 A1\nx Ae1 2 t Be1 3 t t2 1 A1 Must have x.\n6\n\n2(b) xt2 1 B1FT Must have ‘x =’. Accept ‘x is approximately’ but do\nnot accept ‘x‘. FT on their PI.\n1\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_23-p07.png"
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    },
    {
      "id": "9231-2023-mj-23-q03",
      "question_id": "9231-2023-mj-23-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 zz1 2isinand zz1 2cos B1 Use of zz1 2isin and zz1 2cos\n zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups.\n zz14   z4 z4 4  z2 z2 6 M1 A1 Expands and groups. Only withhold one A1 mark\n(this or the previous) for no clear grouping.\n24cos4 2cos442cos26 M1 Substituteszn zn 2cosn once in LHS.\n\n24sin4 2cos442cos26\ncos44cos23 A1\ncot4\ncos44cos23\n7\nPartial\nQuestion Answer Guidance\nMarks",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_23-p08.png"
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    },
    {
      "id": "9231-2023-mj-23-q04",
      "question_id": "9231-2023-mj-23-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) d  4y3 12y2y' B1 Differentiates y3 correctly.\ndx\nd  x y6 6(x y)5(1 y') B1 Differentiates x y6 correctly.\ndx\n12  32 y'611 y'0 y' 1 B1 Substitutes (4,3), AG.\n17\n3\n© UCLES 2023 Page 8 of 15\n\n4(b) 2y2y''4yy2 B1 Differentiates y2y.\nx y5 y5x y41 y2 0 B1 B1 Differentiates (x y)51 y.\n17y12  1 2 5(1)4 182 017y 96 0 M1 Substitutes (4,3).\n17 17 17\n96 A1\ny\n289\n5\nPartial\nQuestion Answer Guidance\nMarks",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
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      ]
    },
    {
      "id": "9231-2023-mj-23-q05",
      "question_id": "9231-2023-mj-23-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) coshx 1  ex ex B1\n2\n2  ex ex2  1  e2x e2x 2  cosh2x1 M1 A1 Expands, AG.\n4 2\n3\n© UCLES 2023 Page 9 of 15\n\n5(b) e2tanhxdx e2lncoshx cosh2x M1 A1 Finds integrating factor.\nd  ycosh2x  cosh2x M1 Correct form on LHS, d yI for their integrating\ndx dx\nfactor I, and attempt to integrate their RHS.\nycosh2xcosh2xdx 1 cosh2x1dx 11 sinh2xx  C M1 A1 Uses cosh2 x 1cosh2x1. M1 A1 is for RHS.\n2 2 2 2\n1C M1 Finds C.\nysech2x 1 sinh2x 1 x1  M1 A1 Divides through by their coefficient of y.\n4 2\n8\n© UCLES 2023 Page 10 of 15",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-23-q06",
      "question_id": "9231-2023-mj-23-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a)  1 1x2 dx 1  1 02  1  1 1 2  1  1 n12 M1 A1 Forms the sum of the areas of the rectangles. Must\nhave first two terms and last term for A1. Accept\n0 n n n n n n\n1  n2\n1 for the last term. Accept written in\nn n\nsummation form with correct limits.\n1  1  n1 nr2  1  1  n1  n2 2nrr2 M1 A1 Setting up correct series so that formulae from\nMF19 can be applied. Can also recognise sum as\nn n3 n n3\nr1 r1 1 n\n 2 n1 1 n1 \nr2\n1 r r2  n3\n\nn2\nr1\nn3\nr1 \nr1\n1 n2n1 n2n1 n(n1)(2n1) 1 2n2 3n1 2n2 3n1 A1 AG. Must have gained previous accuracy mark.\n     \nn n3 n3 6n3 n 6n2 6n2\n5\n© UCLES 2023 Page 11 of 15\n\n6(b)  1 1x2 dx 1 112  1 122  1 1n12 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n rectangles. Must have first two and last terms for\n0\nA1. Accept 11 n2 for the last term.\nn n\nAccept written in summation form with correct\nlimits.\n1  n1  n2 2nrr2  n2n1  n2n1  n(n1)(2n1) M1 S M e F tt 1 in 9 g c u a p n c b o e r a re p c p t l i s e e d r . i e R s e s c o o g th n a i t s i f n o g rm su u m la e a s f rom\nn3 n3 n3 6n3\nr1 1 n1 1 n 1\n 2 n 1 n  r2  r2  without wrong working\n\n\n1 n2 \nr1\nr n3 \nr1\nr2 \n s\nn\nc\n3\nor\nr\ne\n\ns\n1\nM1 A\nn\n1\n3\nM\nr1\n1.\nn\n2n2 3n1 A1\n6n2\n4\n\n6(c) 1 M1 c\nU L  Simplifies U L to .\nn n n n n n\nl A1 AG.\n0 as n\nn\n2\n© UCLES 2023 Page 12 of 15",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2023-mj-23-q07",
      "question_id": "9231-2023-mj-23-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) I  4 3 1x21 2dxsinh1x 4 3 ln3 M1 A1 Recognises integral.\n1  \n0 0\n2\n\n7(b) d \nx\n 1x21\n2\nn nx2 1x21\n2\nn1\n\n 1x21\n2\nn M1 A1 Uses the product rule to differentiate.\ndx \nn  1x2 1  1x21 2 n1   1x21 2 n M1* Uses x2 1x2 1.\n4 DM1 Integrates both sides using the limits given. Requires\n x  1x21 2 n3 nI nI I previous method mark.\n  n n2 n\n \n0\n45n\nn1I nI n1I nI \n45n A1 Substitutes limits and rearranges. AG.\n3 3 n n2 n n2 3 3\n5\n\n7(c) 2 M1 Forms correct integral with correct limits.\ns3 14x2 dx\n0\n1  4 3 1u2 du 1 I A1 AG.\n2 2 1\n0\n45 M1 Applies reduction formula with n1.\n2I I \n1 1 3 3\n1 5 A1\ns ln3\n4 9\n4\n© UCLES 2023 Page 13 of 15",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_23-p13.png"
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    },
    {
      "id": "9231-2023-mj-23-q08",
      "question_id": "9231-2023-mj-23-q08",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) 12a (a1)2 2a2 7a1 M1 A1 Shows that determinant is non-zero or solves\nax1 2 \nequations.\na 6a 2a2 1a 1a 1a\n2\n0 1a 0 a(a1)0 or y\n1a\n0 2a 1\n2(2a) a1\nz 3\n1a 1a\nThree planes intersect at a single point. B1 Must be clear that there is just one point where all 3\nplanes intersect.\n3\n\n8(b) a 6a 2a2 M1 Equates determinant to zero or working to find\ncharacteristic equation.\n0 1a 0 0\n0 2a 1\na(1a)(1)0a,1a,1 A1 AG. Factorisation must be clear.\n2\n© UCLES 2023 Page 14 of 15\n\n8(c) i j k 2a2 a1 1 M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\na: 0 12a 0  0    0 \n   \n0 2a 1a\n\n0\n\n0\ni j k (2a)(2a2) 2 A1\n   \n1: a1 6a 2a2  0  0\n   \n   \n0 2a 0  (a1)(2a)   1 \ni j k 4a2 10a4 2 A1\n \n \n1a: 2a1 6a 2a2  2a2 5a2  1\n \n   \n0 2a 2a  2a2 5a2   1 \n \n1 2 2 a4 0 0  M1 A1 Correctly matched permutations of columns. Their\n    eigenvectors must be non-zero for M1.\nP  0 0 1  and D 0 1 0 \n 0 1 1     0 0 (1a)4  \n6\n\n8(d) a(1a)(1)3 (a2 a1)(a2 a)0 B1 Finds characteristic equation multiplied out.\nA4 (a2 a1)A2 (a2 a)A M1 A1 Multiplies through by A.\n3\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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        "../answer-assets/9231_s23_ms_23-p15.png"
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    },
    {
      "id": "9231-2023-mj-31-q01",
      "question_id": "9231-2023-mj-31-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 3mg B1 Correct EPE term seen\n2a2\n2a\n1 mv2 mg  3a 3 a   3mg 2a2 M1 D\non\ni\ne\nm\nK\nen\nE\ns\n,\ni o\no\nn\nn\na\ne\nl l\nE\ny\nP\nc\nE\no r\nt\nr\ne\ne\nr\nc\nm\nt e\na\nn\nn\ne\nd\nr g\na\ny\nt l\ne\ne\nq\na\nu\nst\na t\no\ni\nn\no\ne\nn .\nG\nM\nP\nu\nE\ns\n.\nt have\n2  4  2a\nAllow sign errors.\n15 A1 AEF\nv ag 2.74 ag\n \n2\n3\n\n1(b) 3mg M1 N2L and Hooke’s law\nT mg mA and T  2a\na\nAcceleration = 5g [upwards] A1 Allow 50 or 5g\n2\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_31-p06.png"
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    },
    {
      "id": "9231-2023-mj-31-q02",
      "question_id": "9231-2023-mj-31-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 Parallel to wall vcosucos M1 Both\nPerpendicular to wall vsineusin\n1 A1 AEF\nDividing, e\n2tan\nKE reduced by 20%, so 1mu2 cos2e2sin2   41mu2 M1 Dimensionally correct equation in u or v, but not\n2 5 2 both. Must have either  or , but not both.\nMust see 4 on the correct side of the equation.\n5\n4 A1\nEliminate e: cos\n5\n2 A1\ne\n3\n© UCLES 2023 Page 7 of 15\n\n2 Alternative method for question 2\nParallel to wall vcosucos M1 Both\nPerpendicular to wall vsineusin\n 5 2 5 5v 2 5v A1\nsin , cos  usin , ucos\n5 5 5e 5\n \n4v2 v2 A1 v2 1 \nu2   u2cos2u2sin2   5  5e2 AEF, e.g. 5   4 e2  \n1 4 1 2 v2  1  M1 Dimensionally correct equation in v.\n2\nmv2 \n 5\n\n2\nmu2 \n 5\nm\n5  \n4\ne2   Must have either  or , but not both.\n4 2\nMust see or on the correct side of the\n5 5\nequation.\n2 A1\ne\n3\n5\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s23_ms_31-p08.png"
      ]
    },
    {
      "id": "9231-2023-mj-31-q03",
      "question_id": "9231-2023-mj-31-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) m4ag M1 Allow cos instead of sin for M1. Do not award\nAt B, mgsin\nuntil tension 0 used. Mass must be seen. No sign\n5a\nerror.\n4 A1\nsin\n5\n2\n\n3(b) mu2 B1\nAt A, T mgcos\na\n1 1 4ag M1 A1 Energy equation with 4 terms, dimensionally\nEnergy mu2  m mgacossin\ncorrect. Mass must be present, allow sign errors.\n2 2 5\n1\nMust see in the KE terms.\n2\nSolve to find T M1 Complete method leading to an expression in mg\nfor T .\n21 A1 CWO\nT  mg\n5\n5\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_31-p09.png"
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    },
    {
      "id": "9231-2023-mj-31-q04",
      "question_id": "9231-2023-mj-31-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) [Mass is proportional to volume] M1 A1\nDistance of Distance of\nVolume centre of mass centre of mass\nfrom vertical axis from OC\n2 3\nHemisphere\n2a3\n2a  2a\n3 8\n1\nCylinder a2d a d\n2\nObject\n2\n2a3 a2d x y\n3\n  2 2a3 a2d  x  16 a32a a2da Moments equation, dimensionally correct, correct\n3  3 number of terms. Allow sign errors.\n32a2 3ad A1 AG. At least one line of intermediate working.\nSimplify to x \n16a3d\n  2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d M1 M nu o m m b e e n r t o s f e t q e u rm ati s o . n A , l d lo im w e s n i s g i n o n e a rr ll o y r s c . o rrect, correct\n3  3  8  2\n3  d2 8a2 A1 AEF\ny \n216a3d\n5\n© UCLES 2023 Page 10 of 15\n\n4(b) 2ax B1\nsin\n2a\n1 32a2 3ad M1 Remove fractions\n2a 2a\n6 16a3d\n5\n16a3d32a3d\n3\n8 A1\nd  a\n3\n3\n© UCLES 2023 Page 11 of 15",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_s23_ms_31-p10.png",
        "../answer-assets/9231_s23_ms_31-p11.png"
      ]
    },
    {
      "id": "9231-2023-mj-31-q05",
      "question_id": "9231-2023-mj-31-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) mg mg3  B1\nHooke’s law, T 1  a xa or T 2  a  4 xa \nmv2 mv2 mg M1 gxxa\nAlso, T = and equate  xa v2 \n1 x x a a\nDimensionally correct terms.\nmg   3x a   2m   1 v   2 M1 v2  3gx  3 xa  \n 4  2  2a 4 \nSimilarly: \na 3 1 3x\nx Must have v and on the RHS.\n4 2 4\nTheir dimensionally correct T .\n2\nEquate expressions for v2 and solve for x in terms of a. M1\nx4a A1 WWW\n5 SC B3 for answer of 4a using  instead of mg .\n\n5(b) a 2a M1 FT their expression for x.\n v2 or  v2 and substitute x4a, v 12ag\nxgxa 3 \n3xg xa\n4 \n1 A1 CAO\n2\n© UCLES 2023 Page 12 of 15",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_31-p12.png"
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    },
    {
      "id": "9231-2023-mj-31-q06",
      "question_id": "9231-2023-mj-31-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) dv M1\nv 6v v9 and attempt to separate variables and integrate\ndx\n2 v9 6xA A1\nx2, v72 A6 M1 Use initial condition to find constant.\nv9x12\n9\nA1 Correct, AEF.\n4\n\n6(b)    d d x t 9  x2 2x  , xx dx 2 9dt    1 2    1 x  x 1 2    dx9dt M1 S   e a pa  rate b va   r d ia x b  le d s t a n d write in the form\n x xc\n1  x  A1 Integrate, any correct form.\nln 9tB\n2 x2\n1 1 M1 Use initial condition to find constant.\nt 0, x2 B ln\n2 2\n 2x  2x M1 Take logarithms\n18t ln  e18t \nx2 x2\n2e18t 2 A1 Any correct form\nx or x\n2e18t 2e18t 1\n5\n© UCLES 2023 Page 13 of 15",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_31-p13.png"
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    },
    {
      "id": "9231-2023-mj-31-q07",
      "question_id": "9231-2023-mj-31-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) 800sin2 B1\nH 80sin2 or\ng\n40sin B1\nT 4sin or\ng\n2\n\n7(b) 1 1 M1 A1 No extra terms.\nBetween t T and t 3  H \n103T2\n4 2\n1 M1 Substitute their expressions for H and T from\nUse results from part (a)\n80sin2534sin2\npart (a) and obtain a quadratic equation in sin\n4\nwith no more than three terms.\n4sin28sin30\n1 A1 Single answer. NFWW.\nsin , 30\n2\nAlternative method for question 7 part (b)\n3 1 M1 A1 120 sin45\nH  y3403 sin 1032\n4 2\n3 M1 Substitute their expressions for H and T from\nUse results from (a): 80sin2120 sin45\npart (a) and obtain a quadratic equation in sin\n4\nwith no more than three terms.\n4sin28sin30\n1 A1 Single answer. NFWW.\nsin , 30\n2\n4\n© UCLES 2023 Page 14 of 15\n\n7(c) When t 3 speeds 40cos and 40sin103 B1\nSquare and add to find square of speed: v2   20 3 2 102 M1 Must be numerical.\nv2 1300, v10 13 [= 36.1] A1\n3\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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        "../answer-assets/9231_s23_ms_31-p15.png"
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    },
    {
      "id": "9231-2023-mj-32-q01",
      "question_id": "9231-2023-mj-32-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 3mg B1 Correct EPE term seen\n2a2\n2a\n1 mv2 mg  3a 3 a   3mg 2a2 M1 D\non\ni\ne\nm\nK\nen\nE\ns\n,\ni o\no\nn\nn\na\ne\nl l\nE\ny\nP\nc\nE\no r\nt\nr\ne\ne\nr\nc\nm\nt e\na\nn\nn\ne\nd\nr g\na\ny\nt l\ne\ne\nq\na\nu\nst\na t\no\ni\nn\no\ne\nn .\nG\nM\nP\nu\nE\ns\n.\nt have\n2  4  2a\nAllow sign errors.\n15 A1 AEF\nv ag 2.74 ag\n \n2\n3\n\n1(b) 3mg M1 N2L and Hooke’s law\nT mg mA and T  2a\na\nAcceleration = 5g [upwards] A1 Allow 50 or 5g\n2\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_32-p06.png"
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    },
    {
      "id": "9231-2023-mj-32-q02",
      "question_id": "9231-2023-mj-32-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 Parallel to wall vcosucos M1 Both\nPerpendicular to wall vsineusin\n1 A1 AEF\nDividing, e\n2tan\nKE reduced by 20%, so 1mu2 cos2e2sin2   41mu2 M1 Dimensionally correct equation in u or v, but not\n2 5 2 both. Must have either  or , but not both.\nMust see 4 on the correct side of the equation.\n5\n4 A1\nEliminate e: cos\n5\n2 A1\ne\n3\n© UCLES 2023 Page 7 of 15\n\n2 Alternative method for question 2\nParallel to wall vcosucos M1 Both\nPerpendicular to wall vsineusin\n 5 2 5 5v 2 5v A1\nsin , cos  usin , ucos\n5 5 5e 5\n \n4v2 v2 A1 v2 1 \nu2   u2cos2u2sin2   5  5e2 AEF, e.g. 5   4 e2  \n1 4 1 2 v2  1  M1 Dimensionally correct equation in v.\n2\nmv2 \n 5\n\n2\nmu2 \n 5\nm\n5  \n4\ne2   Must have either  or , but not both.\n4 2\nMust see or on the correct side of the\n5 5\nequation.\n2 A1\ne\n3\n5\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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        "../answer-assets/9231_s23_ms_32-p08.png"
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    },
    {
      "id": "9231-2023-mj-32-q03",
      "question_id": "9231-2023-mj-32-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) m4ag M1 Allow cos instead of sin for M1. Do not award\nAt B, mgsin\nuntil tension 0 used. Mass must be seen. No sign\n5a\nerror.\n4 A1\nsin\n5\n2\n\n3(b) mu2 B1\nAt A, T mgcos\na\n1 1 4ag M1 A1 Energy equation with 4 terms, dimensionally\nEnergy mu2  m mgacossin\ncorrect. Mass must be present, allow sign errors.\n2 2 5\n1\nMust see in the KE terms.\n2\nSolve to find T M1 Complete method leading to an expression in mg\nfor T .\n21 A1 CWO\nT  mg\n5\n5\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_32-p09.png"
      ]
    },
    {
      "id": "9231-2023-mj-32-q04",
      "question_id": "9231-2023-mj-32-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) [Mass is proportional to volume] M1 A1\nDistance of Distance of\nVolume centre of mass centre of mass\nfrom vertical axis from OC\n2 3\nHemisphere\n2a3\n2a  2a\n3 8\n1\nCylinder a2d a d\n2\nObject\n2\n2a3 a2d x y\n3\n  2 2a3 a2d  x  16 a32a a2da Moments equation, dimensionally correct, correct\n3  3 number of terms. Allow sign errors.\n32a2 3ad A1 AG. At least one line of intermediate working.\nSimplify to x \n16a3d\n  2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d M1 M nu o m m b e e n r t o s f e t q e u rm ati s o . n A , l d lo im w e s n i s g i n o n e a rr ll o y r s c . o rrect, correct\n3  3  8  2\n3  d2 8a2 A1 AEF\ny \n216a3d\n5\n© UCLES 2023 Page 10 of 15\n\n4(b) 2ax B1\nsin\n2a\n1 32a2 3ad M1 Remove fractions\n2a 2a\n6 16a3d\n5\n16a3d32a3d\n3\n8 A1\nd  a\n3\n3\n© UCLES 2023 Page 11 of 15",
      "source_pages": [
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        11
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_32-p10.png",
        "../answer-assets/9231_s23_ms_32-p11.png"
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    },
    {
      "id": "9231-2023-mj-32-q05",
      "question_id": "9231-2023-mj-32-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) mg mg3  B1\nHooke’s law, T 1  a xa or T 2  a  4 xa \nmv2 mv2 mg M1 gxxa\nAlso, T = and equate  xa v2 \n1 x x a a\nDimensionally correct terms.\nmg   3x a   2m   1 v   2 M1 v2  3gx  3 xa  \n 4  2  2a 4 \nSimilarly: \na 3 1 3x\nx Must have v and on the RHS.\n4 2 4\nTheir dimensionally correct T .\n2\nEquate expressions for v2 and solve for x in terms of a. M1\nx4a A1 WWW\n5 SC B3 for answer of 4a using  instead of mg .\n\n5(b) a 2a M1 FT their expression for x.\n v2 or  v2 and substitute x4a, v 12ag\nxgxa 3 \n3xg xa\n4 \n1 A1 CAO\n2\n© UCLES 2023 Page 12 of 15",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_32-p12.png"
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    },
    {
      "id": "9231-2023-mj-32-q06",
      "question_id": "9231-2023-mj-32-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) dv M1\nv 6v v9 and attempt to separate variables and integrate\ndx\n2 v9 6xA A1\nx2, v72 A6 M1 Use initial condition to find constant.\nv9x12\n9\nA1 Correct, AEF.\n4\n\n6(b)    d d x t 9  x2 2x  , xx dx 2 9dt    1 2    1 x  x 1 2    dx9dt M1 S   e a pa  rate b va   r d ia x b  le d s t a n d write in the form\n x xc\n1  x  A1 Integrate, any correct form.\nln 9tB\n2 x2\n1 1 M1 Use initial condition to find constant.\nt 0, x2 B ln\n2 2\n 2x  2x M1 Take logarithms\n18t ln  e18t \nx2 x2\n2e18t 2 A1 Any correct form\nx or x\n2e18t 2e18t 1\n5\n© UCLES 2023 Page 13 of 15",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_32-p13.png"
      ]
    },
    {
      "id": "9231-2023-mj-32-q07",
      "question_id": "9231-2023-mj-32-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) 800sin2 B1\nH 80sin2 or\ng\n40sin B1\nT 4sin or\ng\n2\n\n7(b) 1 1 M1 A1 No extra terms.\nBetween t T and t 3  H \n103T2\n4 2\n1 M1 Substitute their expressions for H and T from\nUse results from part (a)\n80sin2534sin2\npart (a) and obtain a quadratic equation in sin\n4\nwith no more than three terms.\n4sin28sin30\n1 A1 Single answer. NFWW.\nsin , 30\n2\nAlternative method for question 7 part (b)\n3 1 M1 A1 120 sin45\nH  y3403 sin 1032\n4 2\n3 M1 Substitute their expressions for H and T from\nUse results from (a): 80sin2120 sin45\npart (a) and obtain a quadratic equation in sin\n4\nwith no more than three terms.\n4sin28sin30\n1 A1 Single answer. NFWW.\nsin , 30\n2\n4\n© UCLES 2023 Page 14 of 15\n\n7(c) When t 3 speeds 40cos and 40sin103 B1\nSquare and add to find square of speed: v2   20 3 2 102 M1 Must be numerical.\nv2 1300, v10 13 [= 36.1] A1\n3\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2023-mj-33-q01",
      "question_id": "9231-2023-mj-33-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 1 1 M1 Energy equation, 4 terms, dimensionally correct,\nm.3ag mv2 mgacoscos\nmass must be present, allow sign errors, allow sin in\n2 2\nboth terms on RHS\nmv2 B1 N2L, may include tension initially but not awarded\nmgcos until tension = 0 used\na\n3 1 4  M1 Dependent on tension = 0 and on an energy\n2\nmag\n2\nm.agcosmga\n 5\ncos\n  equation, eliminate v2.\n3 7\ncos\n2 10\n7 A1 If no m in energy equation and no further errors,\ncos\naward SCB2 for correct final answer\n15\n4\n© UCLES 2023 Page 4 of 14",
      "source_pages": [
        4
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_33-p04.png"
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    },
    {
      "id": "9231-2023-mj-33-q02",
      "question_id": "9231-2023-mj-33-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 1 M1 Kinetic energy = elastic potential energy, 4 terms,\n1 m  v2  v2    2 mg     2 a   2    1 a   2  dimensionally correct, allow sign errors.\n2  4  a   3  3   \n1 A1 With v substituted.\n1  1  2 mg  2  2 1  2\nm  2ag ag    a    a  \n2  2  a  3  3  \n \n3 3  M1 Solve to find value for  dependent on energy\nSolve [ v2 g  a  ] equation with 3 or 4 terms\n4 9 \n9 A1 9\n SCB2 for  mg if given  not used\n2 2\n4\n© UCLES 2023 Page 5 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_33-p05.png"
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    },
    {
      "id": "9231-2023-mj-33-q03",
      "question_id": "9231-2023-mj-33-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) [Mass is proportional to area] B1 All correct for ABC and DEC.\nArea Centre of mass\nfrom AC\nABC 1 2a\n.6a.8a (24a2)\n2\nDEC 1 1\nx.5a x\n2 3\nADEB 5 x\n24a2  xa\n2\n 5  1 5 M1 All moment terms present, dimensionally correct,\nMoments [about AC] x  24a2  xa  24a22a x ax allow sign error.\n 2  3 2\nA1 All correct moments about AC.\n288a2 5x2 A1 AEF\nx \n348a5x\n4\n© UCLES 2023 Page 6 of 14\n\n3(b) 288a2 5x2 B1 FT FT their expression for x from part (a).\nOn the point of toppling about E: x x, x\n348a5x\nRearrange to 3-term quadratic: 10x2 144ax288a2 0 M1 Allow 3-term inequality.\n25x12ax12a0, A1 Single correct answer, no inequality, CWO.\n12\nx a\n5\n3\n© UCLES 2023 Page 7 of 14",
      "source_pages": [
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        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_33-p06.png",
        "../answer-assets/9231_s23_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-33-q04",
      "question_id": "9231-2023-mj-33-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) Let speeds of A and B along line of centres after collision be V and V M1 Allow sign errors, allow missing m.\nA B\nV V ucos30 (1)\nA B\nV V eucos30 (2) M1 Signs on LHS must be consistent with (1).\nA B\nSpeeds perpendicular to line of centres after collision are usin30 and 2u B1 SOI\nV V V 4V\nMoving in same direction, so A  B (3) B A\nusin30 2u\nUse V 4V in (1): 5V ucos30 M1 A complete method to find equation in e only\nB A A\nFrom (2): 3V eucos30 then\nA\nCombine to find equation in e only.\ne 3 A1\n5\n© UCLES 2023 Page 8 of 14\n\n4(a) Alternative method for question 4(a)\nLet speeds of A and B along line of centres after collision be V and V M1\nA B\nV V ucos30 (1) Allow sign errors, allow missing m.\nA B\nV V eucos30 (2) M1 Signs on LHS must be consistent with (1).\nA B\nSpeeds perpendicular to line of centres after collision are usin30 and 2u B1 SOI\nV V V 4V\nMoving in same direction, so A  B (3) B A\nusin30 2u\n1 1 M1 u 4u\nSolve (1) and (2): V  u1ecos30, V  u1ecos30 Note: V  3, V  3\nA 2 B 2 A 10 B 10\nSubstitute in (3) to find equation in e only .\ne 3 A1\n5\n5\n\n4(b) KE after = 1 2 m    V A 2     u 2    2    1 2 m (2u2 V B 2) B1 C af o te rr r e c c o t l e li x s p io re n s , s w io i n th f b o o r t K h E c o f m or p o o n n e e n o t f s . t he spheres,\n \n7 56 B1 119\nKE for A after = mu2 or KE for B after = mu2 Implied by total KE after = mu2.\n50 25 50\n9 6\nor KE loss for A = mu2 or KE gain for B = mu2\n25 25\n3 B1 1\nTotal loss in KE = mu2 Term m2u2 may be omitted from KE of B\n25 2\nbefore and after.\n3\n© UCLES 2023 Page 9 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2023-mj-33-q05",
      "question_id": "9231-2023-mj-33-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) Tcosmg B1\nmv2  27ag B1\nTsin  m\nr 4r\n \nr 12atan used M1\n27 3 M1 Finds value for tan OE.\nDivide: tan , so tan\n412tan 4 Reduces to equation in  or x, no k.\nr 9a, extension of string = 3a A1\nAlternative method for question 5(a)\nLet L be stretched length of string. B1 12a\nOr T mg\nTcosmg L\nmv2  27ag B1\nTsin  m\nr 4r\n \nr  Lsin used M1\n12a\n L2 144a20.5 M1\nUse cos and sin and eliminate T.\nL L\n[L2 144a2 81a2] L15a, extension of string = 3a A1\n5\n© UCLES 2023 Page 10 of 14\n\n5(b) kmgL12a B1\nHooke’s law: T \n12a\nkmgL12a mgL M1\nEliminate T: \n12a 12a\nL A1\nk  5\nL12a\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s23_ms_33-p11.png"
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    },
    {
      "id": "9231-2023-mj-33-q06",
      "question_id": "9231-2023-mj-33-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) dv B1 Mass must be seen at this point or earlier.\nm mg1kv\n[SUVAT does not apply.]\ndt\n1 M1 Separate variables and integrate to logarithm.\n ln1kvgt A\nk\nA1 Correct, with constant of integration.\nt0, v0 [A0] M1 Use initial condition to evaluate their constant.\nv\n1 1ekgt A1 Any correct form with v as subject.\nFinal A0 if numerical value of g present.\nk\n5\n© UCLES 2023 Page 11 of 14\n\n6(b)\nk 0.05 and so\ndx\n20\n 1e0.5t *M1 Attempt to integrate if expression contains a term of\ndt the form bect .\nIntegrate: x20  t2e0.5t B A1 1 1 \nx t ekgt B\nk gk \nt0, x0 B40 DM1 Use initial condition to evaluate their constant.\nWhen v12, from part (a), e0.5t 10.05120.4, t2ln0.4 M1 1.83…\nx40ln0.4400.440 12.7 A1 5\n40ln 24\n2\nAlternative method for question 6(b)\ndv  1  *M1 Separate variables and write in integrable form\nv g1kv leading to  1  dvkgdx\ndx  1kv\n1 DM1 A1 Dependent on previous M1. Attempt to integrate.\nv ln1kvkgxB\nk\nv0, x0 B0 and k 0.05, v12 1 220ln0.40.5x M1 Dependent on both previous M1s.\nUse initial condition to evaluate their constant and\nuse v12\nx12.7 A1 5\n40ln 24\n2\n5\n© UCLES 2023 Page 12 of 14",
      "source_pages": [
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        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s23_ms_33-p12.png"
      ]
    },
    {
      "id": "9231-2023-mj-33-q07",
      "question_id": "9231-2023-mj-33-q07",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) For aircraft, d 5T B1\n1 B1 To point of collision.\nFor ball, 4usin T  10T2\n2\nFor ball, ucos T d 85T 8 B1\nEliminate u: *M1\nu4 1 5\n 45T2\n55T 8\n4 T  10T2, u and u\n5 2 4T 3T\nDependent on LHS of second B1 being 4,\n3  45T2 45T 8 expression involving only T\n3T2 4T 40 DM1 Dependent on previous M1.\nObtain and solve 3-term quadratic.\nT 2 A1 Single correct answer.\n6 2\nNote d 8 used leads to T  B1B1B0M1M1A0\n3\n© UCLES 2023 Page 13 of 14\n\n7(b) usin10T M1 OE\ntan1\nAccept ‘tan = ……’\nucos\n8 A1 OE\ntan1\n9\nDirection is 41.6° below the horizontal A1 CAO\nNote: d 8 used leads to 20.9 above the\nhorizontal.\n3\n© UCLES 2023 Page 14 of 14",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s23_ms_33-p14.png"
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    },
    {
      "id": "9231-2023-mj-41-q01",
      "question_id": "9231-2023-mj-41-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 91.2 1  91.22    67  M1 Both.\nx  12 7.6 s x 2  11   695.8 12     0.2436  275    \n0.2436 M1 With a t-value.\nCI: 7.6t 0.975\n11 12\n2.201 seen B1\n7.60.3136=7.29, 7.91 A1\n4\n© UCLES 2023 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_41-p06.png"
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    },
    {
      "id": "9231-2023-mj-41-q02",
      "question_id": "9231-2023-mj-41-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2 99.6 21.5 43  M1 Both SOI.\nx  8.3 s 2  1.9545 or\n12 x 11   22  \nH :   H :   B1\n0 P Q 1 P Q\n92.16111.9545 M1 FT FT their 1.9545 only\nPooled estimate =\n10122\n2.047 (2.05) A1 May be implied, allow 2.04\n9.128.3 M1 FT FT their pooled estimate\nt\n1 1\n2.047 \n10 12\n1.339 (or 1.338) A1 Accept 1.34\n'1.339'1.725 M1 Compare with correct tabular value 1.725 and\nAccept H / not significant conclusion without context. Condone ‘reject H ’.\n0 1\nInsufficient evidence to suggest that the (mean) time taken at P is greater than A1 Correct conclusion in context, following correct\nthe (mean) time taken at Q. work, level of uncertainty in language (not ‘prove’,\nnot ‘there is no evidence’), no contradictions.\nA0 if H and H reversed.\n0 1\n8\n© UCLES 2023 Page 7 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_41-p07.png"
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    },
    {
      "id": "9231-2023-mj-41-q03",
      "question_id": "9231-2023-mj-41-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 2 1  4  1  4 x3 M1 Integration with correct powers and correct limits\n  x2  dx    seen.\n1.5 24x2  24  x 3 \na500.092014 = 4.6007 A1 AG\n53 1325\nor or 0.092014 or 4.60069 seen\n576 288\nb11.4211 B1 Or 11.421(0)\n3\n\n3(b) Combine first two columns: 6, 9.0278 M1 Must be seen, or implied by 1.0155.\nCalculate value of chi-squared: 1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885 M1 At least 2 correct values (at least 3sf) or\nexpressions seen.\nAllow columns not combined or three columns\ncombined.\n3.16 A1 CWO.\nDependent on M1M1 scored.\nSCB1 for 3.16 with no working\nH : f is a good fit for the data B1\n0\nH : f is not a good fit for the data\n1\nTabular value of chi-squared: 7.78 M1 Correct tabular value: allow correct FT value if\n‘3.16’ < 7.78 and accept H columns not combined (9.236) or three columns\n0\ncombined (6.251).\nInsufficient evidence to suggest that f is not a good fit for the data A1 Correct conclusion in context, following correct\nCondone: sufficient evidence to suggest that f is a good fit for the data. work, level of uncertainty in language (not ‘prove’,\nnot ‘there is no evidence’), no contradictions.\nA0 if hypotheses reversed\n6\n© UCLES 2023 Page 8 of 12",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2023-mj-41-q04",
      "question_id": "9231-2023-mj-41-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Differences (allow up to 3 errors)\n2 5 1 –6 42 11 –3 7 20 12 –8 –10 50\nA1 Correct rank order, ignore signs\n2 4 1 –5 12 9 –3 6 11 10 –7 –8 13\nP68, Q23 A1 CWO\nH : difference in population medians = 0 B1 ‘Population’ required. Accept use of m, not .\n0\nH 1 : population median in 2022 > population median in 2018 Do not accept ‘difference between population\nmedians > 0’ without 2022, 2018 OE specified\nCritical value 21 B1\n‘23’ > ‘21’ and accept H M1 Compare their 23 with their 21 and FT conclusion.\n0\nTheir 23 must be less than 46.\nInsufficient evidence to support researcher’s belief / insufficient evidence that A1 Correct conclusion in context, following correct\nthe (median) number of employees in 2022 is greater than the (median) number work, level of uncertainty in language (not ‘prove’,\nof employees in 2018 not ‘there is no evidence’), no contradictions.\nA0 if hypotheses reversed\n7\n\n4(b) Rank for G will now be +3, giving Q = 20 which is < 21 M1 Must include numbers, ft their 21 from part (a) and\nft their 20 (23 – 3)\nChange in conclusion A1 CWO\nCondone ‘reject H ’ OE.\n0\n2\n© UCLES 2023 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_41-p09.png"
      ]
    },
    {
      "id": "9231-2023-mj-41-q05",
      "question_id": "9231-2023-mj-41-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 9 26 24 B1 Two probabilities correct, seen anywhere.\nP(3H) = P(2H) = P(1H) = P(0H) =\n60 60 60 60\n24 26 9 1 M1 Cubic polynomial with 4 probabilities as\nG (t) =  t t2  t3\nX coefficients with the correct powers of t from their\n60 60 60 60\nworking. Equivalent forms are acceptable.\nA1 Correct, AEF.\n3\n\n5(b) 1 1 1 B1\nG (t) =  t t2\nY\n4 2 4\n24 26 9 1  1 1 1  M1 Attempt to multiply their two PGFs.\nG Z (t) =   t t2  t3   t t2 \n60 60 60 60  4 2 4 \n1  2474t85t2 45t311t4 t5 M1 Obtain quintic expression and collect terms\n240\nA1 Correct, AEF.\n4\n\n5(c) Differentiate: G' (t) = 1  74170t135t2 44t3 5t4 M1 Differentiate their G.\nZ\n240\n428 107 A1 Any correct form.\nG' (1) =  1.78\nZ\n240 60\n2\n© UCLES 2023 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_41-p10.png"
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    },
    {
      "id": "9231-2023-mj-41-q06",
      "question_id": "9231-2023-mj-41-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) 3  1 x  1 x  M1 + c not required.\nF(x) 2e2 8e 2 c\n28 \n \n3  1 x  1 x  9 A1\nF(x) 2e2 8e 2 \n28  14\n \nCorrect ranges including 0x2ln3 associated with their F(x) B1 No gaps in range.\nF(x) = 0 for x0 and 1 for x2ln3\n3\n\n6(b) 3  8  M1 y substituted into their F.\nG(y) 2y 6\n28 y \n1 y3 B1 Seen anywhere, condone 1yeln3.\n3  8  A1 Correct expression, not containing logs.\ng(y) 2 \n28 y2 \n3\n\n6(c) 3  8  3 M1 3\n2y 6 Their G(y) \n28 y  10 10\n5y2 8y200 M1 Obtain from G(y) and solve quadratic equation to\nfind y.\ny1.35 A1 Single correct answer.\n3\n© UCLES 2023 Page 11 of 12\n\n6(d) E(Y 4) =  1 3 2 3 8    2 y 8 2    y4dy  1 3 2 3 8  2y4 8y2 dy = 2 3 8   5 2 y5  8 3 y3    1 Integrate y4 their g(y). Limits must be 1 and 3.\n17.8 A1 89\n5\n2\n© UCLES 2023 Page 12 of 12",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_41-p11.png",
        "../answer-assets/9231_s23_ms_41-p12.png"
      ]
    },
    {
      "id": "9231-2023-mj-42-q01",
      "question_id": "9231-2023-mj-42-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 91.2 1  91.22    67  M1 Both.\nx  12 7.6 s x 2  11   695.8 12     0.2436  275    \n0.2436 M1 With a t-value.\nCI: 7.6t 0.975\n11 12\n2.201 seen B1\n7.60.3136=7.29, 7.91 A1\n4\n© UCLES 2023 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p06.png"
      ]
    },
    {
      "id": "9231-2023-mj-42-q02",
      "question_id": "9231-2023-mj-42-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2 99.6 21.5 43  M1 Both SOI.\nx  8.3 s 2  1.9545 or\n12 x 11   22  \nH :   H :   B1\n0 P Q 1 P Q\n92.16111.9545 M1 FT FT their 1.9545 only\nPooled estimate =\n10122\n2.047 (2.05) A1 May be implied, allow 2.04\n9.128.3 M1 FT FT their pooled estimate\nt\n1 1\n2.047 \n10 12\n1.339 (or 1.338) A1 Accept 1.34\n'1.339'1.725 M1 Compare with correct tabular value 1.725 and\nAccept H / not significant conclusion without context. Condone ‘reject H ’.\n0 1\nInsufficient evidence to suggest that the (mean) time taken at P is greater than A1 Correct conclusion in context, following correct\nthe (mean) time taken at Q. work, level of uncertainty in language (not ‘prove’,\nnot ‘there is no evidence’), no contradictions.\nA0 if H and H reversed.\n0 1\n8\n© UCLES 2023 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p07.png"
      ]
    },
    {
      "id": "9231-2023-mj-42-q03",
      "question_id": "9231-2023-mj-42-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 2 1  4  1  4 x3 M1 Integration with correct powers and correct limits\n  x2  dx    seen.\n1.5 24x2  24  x 3 \na500.092014 = 4.6007 A1 AG\n53 1325\nor or 0.092014 or 4.60069 seen\n576 288\nb11.4211 B1 Or 11.421(0)\n3\n\n3(b) Combine first two columns: 6, 9.0278 M1 Must be seen, or implied by 1.0155.\nCalculate value of chi-squared: 1.0155 + 0.5715 + 0.7661 + 0.2183 + 0.5885 M1 At least 2 correct values (at least 3sf) or\nexpressions seen.\nAllow columns not combined or three columns\ncombined.\n3.16 A1 CWO.\nDependent on M1M1 scored.\nSCB1 for 3.16 with no working\nH : f is a good fit for the data B1\n0\nH : f is not a good fit for the data\n1\nTabular value of chi-squared: 7.78 M1 Correct tabular value: allow correct FT value if\n‘3.16’ < 7.78 and accept H columns not combined (9.236) or three columns\n0\ncombined (6.251).\nInsufficient evidence to suggest that f is not a good fit for the data A1 Correct conclusion in context, following correct\nCondone: sufficient evidence to suggest that f is a good fit for the data. work, level of uncertainty in language (not ‘prove’,\nnot ‘there is no evidence’), no contradictions.\nA0 if hypotheses reversed\n6\n© UCLES 2023 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p08.png"
      ]
    },
    {
      "id": "9231-2023-mj-42-q04",
      "question_id": "9231-2023-mj-42-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Differences (allow up to 3 errors)\n2 5 1 –6 42 11 –3 7 20 12 –8 –10 50\nA1 Correct rank order, ignore signs\n2 4 1 –5 12 9 –3 6 11 10 –7 –8 13\nP68, Q23 A1 CWO\nH : difference in population medians = 0 B1 ‘Population’ required. Accept use of m, not .\n0\nH 1 : population median in 2022 > population median in 2018 Do not accept ‘difference between population\nmedians > 0’ without 2022, 2018 OE specified\nCritical value 21 B1\n‘23’ > ‘21’ and accept H M1 Compare their 23 with their 21 and FT conclusion.\n0\nTheir 23 must be less than 46.\nInsufficient evidence to support researcher’s belief / insufficient evidence that A1 Correct conclusion in context, following correct\nthe (median) number of employees in 2022 is greater than the (median) number work, level of uncertainty in language (not ‘prove’,\nof employees in 2018 not ‘there is no evidence’), no contradictions.\nA0 if hypotheses reversed\n7\n\n4(b) Rank for G will now be +3, giving Q = 20 which is < 21 M1 Must include numbers, ft their 21 from part (a) and\nft their 20 (23 – 3)\nChange in conclusion A1 CWO\nCondone ‘reject H ’ OE.\n0\n2\n© UCLES 2023 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p09.png"
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    },
    {
      "id": "9231-2023-mj-42-q05",
      "question_id": "9231-2023-mj-42-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 9 26 24 B1 Two probabilities correct, seen anywhere.\nP(3H) = P(2H) = P(1H) = P(0H) =\n60 60 60 60\n24 26 9 1 M1 Cubic polynomial with 4 probabilities as\nG (t) =  t t2  t3\nX coefficients with the correct powers of t from their\n60 60 60 60\nworking. Equivalent forms are acceptable.\nA1 Correct, AEF.\n3\n\n5(b) 1 1 1 B1\nG (t) =  t t2\nY\n4 2 4\n24 26 9 1  1 1 1  M1 Attempt to multiply their two PGFs.\nG Z (t) =   t t2  t3   t t2 \n60 60 60 60  4 2 4 \n1  2474t85t2 45t311t4 t5 M1 Obtain quintic expression and collect terms\n240\nA1 Correct, AEF.\n4\n\n5(c) Differentiate: G' (t) = 1  74170t135t2 44t3 5t4 M1 Differentiate their G.\nZ\n240\n428 107 A1 Any correct form.\nG' (1) =  1.78\nZ\n240 60\n2\n© UCLES 2023 Page 10 of 12",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p10.png"
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    },
    {
      "id": "9231-2023-mj-42-q06",
      "question_id": "9231-2023-mj-42-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) 3  1 x  1 x  M1 + c not required.\nF(x) 2e2 8e 2 c\n28 \n \n3  1 x  1 x  9 A1\nF(x) 2e2 8e 2 \n28  14\n \nCorrect ranges including 0x2ln3 associated with their F(x) B1 No gaps in range.\nF(x) = 0 for x0 and 1 for x2ln3\n3\n\n6(b) 3  8  M1 y substituted into their F.\nG(y) 2y 6\n28 y \n1 y3 B1 Seen anywhere, condone 1yeln3.\n3  8  A1 Correct expression, not containing logs.\ng(y) 2 \n28 y2 \n3\n\n6(c) 3  8  3 M1 3\n2y 6 Their G(y) \n28 y  10 10\n5y2 8y200 M1 Obtain from G(y) and solve quadratic equation to\nfind y.\ny1.35 A1 Single correct answer.\n3\n© UCLES 2023 Page 11 of 12\n\n6(d) E(Y 4) =  1 3 2 3 8    2 y 8 2    y4dy  1 3 2 3 8  2y4 8y2 dy = 2 3 8   5 2 y5  8 3 y3    1 Integrate y4 their g(y). Limits must be 1 and 3.\n17.8 A1 89\n5\n2\n© UCLES 2023 Page 12 of 12",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
        "../answer-assets/9231_s23_ms_42-p11.png",
        "../answer-assets/9231_s23_ms_42-p12.png"
      ]
    },
    {
      "id": "9231-2023-mj-43-q01",
      "question_id": "9231-2023-mj-43-q01",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) 13 2 1 M1 Integrate f(x), + c not required, at least one power correct.\nF(x)  x3 3x3c\n6 2\n \n1 2 1 1 1 A1 AEF\nx3  x3 \n4 2 4\nCorrect ranges including 1x27 associated with their F(x) A1 No gaps.\nF(x) = 0 for x 1 and F(x) = 1 for x 27\n3\n\n1(b) 1 M1 1\nG(y)  y2 2y1  Using y  x3 in their F(x).\n4\n1 A1 Correct, AEF, 0 otherwise not required.\ng(y) y1\n2\nFor 1 y3 B1 Seen anywhere, correct variable.\n3\n\n1(c) 1 1 M1 Equate their G(y) to ½ and solve to find y.\n y2 2y1   , y12 2 OE\n4 2\ny  1  2 A1 Only.\n2\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-43-q02",
      "question_id": "9231-2023-mj-43-q02",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 M1 SOI\ns2 s2\nx 1.771 1.14 or x 1.771 1.11 Allow incorrect t-value, not z-value.\n14 14\n1 B1 Does not depend on use of a t-value.\n[Add: x  1.141.11 1.125] x15.75\n2\ns2 11.141.11 M1 Allow incorrect t-value, but not a z-value.\nSubtract or substitute:   \n14 2 1.771 \ns2 0.001004 or s = 0.0316[9] A1 450\n, implied by correct final answer.\n448063\nx2 M1 OE\nx2 13s2 \n14\nx2 17.7(3) A1 CWO\n6\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2023-mj-43-q03",
      "question_id": "9231-2023-mj-43-q03",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) M1 Differences, allow at most 2 errors.\n5 7 -6 1 –4 2 8 26 –10\n4 6 -5 1 –3 2 7 9 –8\nA1 Correct rank order, ignore signs.\n[P = 29] Q = 16 A1 Condone P not excluded.\nH : population medians equal or m m B1 ‘Population’ required. Accept use of m, not .\n0 1 2\nDo not accept ‘difference between population medians > 0’\nH : population median X > population median Y or m m\n1 1 2 without X, Y OE specified.\nCritical value = 8 B1\n16 > 8 and accept H / not significant M1 Compare their ‘16’ with their ‘8’ and conclusion.\no\nTheir ‘16’ must be less than 23. Ignore their hypotheses.\nCondone ‘reject H ’.\n1\nInsufficient evidence to support teacher’s belief A1 Correct conclusion in context, following correct work, level of\nor uncertainty in language (not ‘prove’, not ‘there is no evidence’),\ninsufficient evidence that the marks/median in Paper 1 are/is no contradictions.\nhigher than the marks/median in Paper 2 e.g. Proves that the teacher is incorrect scores A0. A0 if\nhypotheses wrong way round.\n7\n\n3(b) The population differences are symmetrical (about the median B1 Words in bold, or their equivalent, are required.\ndifference)\n1\n© UCLES 2023 Page 8 of 13",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-mj-43-q04",
      "question_id": "9231-2023-mj-43-q04",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) H :   H :   B1\n0 x y 1 x y\n1  1642  B1\ns 2  338.1  0.02405 and\nx 79 80\n  19 379\nBoth. Implied by , or 3sf.\n1  124.82  790 14750\ns 2  261.1  0.02569\ny 59 60\n \n0.02405 0.02569 M1\ns2  \n80 60\ns2 0.0007289 or s 0.026998 or 0.0270 A1 Implied by z  1.11\n164.0 124.8 M1 FT their value for s.\n\n80 60\nz \ns\n1.11 A1\n‘1.11’ < 1.282 M1 Compare with correct z-value 1.282 and consistent signs. Condone\nAccept H / not significant ‘reject H ’.\n0 1\nUsing probabilities, P(Z >1.11) = 0.1333 > 0.1.\nInsufficient evidence to support the inspector’s suspicion/ A1 Correct conclusion in context, following correct work, level of\nInsufficient evidence that the (mean) lengths of rods from uncertainty in language (not ‘prove’, not ‘there is no evidence’),\nmachine X are shorter than the (mean) lengths of rods from no contradictions.\nmachine Y e.g. proves that the inspector is incorrect scores A0\nA0 if hypotheses wrong way round.\n8\n© UCLES 2023 Page 9 of 13\n\n4(b) Large sample sizes OR central limit theorem applies. B1\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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        "../answer-assets/9231_s23_ms_43-p10.png"
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    },
    {
      "id": "9231-2023-mj-43-q05",
      "question_id": "9231-2023-mj-43-q05",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 B1 Working required.\nG (t)k  13t4t2 k1341, k \n X  8\nG' (t) = k38t, M1 1 3 4 11\nX Or  px 0 1 2 \n \n8 8 8  8 \n11 A1 AG\nE(X) = G' (1) =\nX Evidence of using t 1 or  px required.\n8\nCWO\n3\n\n5(b) 1 1 M1 Multiply the two PGFs to obtain a single polynomial of degree 5.\nG (t) =  13t4t2  t212t\nZ\n8 3\n1  t2 5t310t4 8t5 A1 May have t2 as a factor.\n24\n2\n© UCLES 2023 Page 10 of 13\n\n5(c) 1 M1 Differentiate twice, allow one slip.\nG' (t) =  2t15t2 40t340t4\nZ\n24\n1\nG'' (t) =  230t120t2 160t3\nZ\n24\n2 M1 Use correct formula.\n1 97 97\nVar(X) = G'' Z (1) + G' Z (1) – (G' Z (1))2 = 312   \n24 24 24\n0.707 A1 407\n576\n3\n\n5(d) Z  4 B1 FT FT their final polynomial in part (b)\n1\n© UCLES 2023 Page 11 of 13",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2023-mj-43-q06",
      "question_id": "9231-2023-mj-43-q06",
      "subject": "9231",
      "year": 2023,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) H : grade is independent of age B1 Condone ‘Ability to remember’ instead of ‘grade’.\n0\nH : grade is not independent of age\n1\nCalculate expected values, shown in table in bold M1 At least 2 correct values or expressions.\n25 21.2 16 24.4 19 14.4 60\nA1 6 correct values or expressions.\n28 31.8 45 36.6 17 21.6 90\n53 61 36 150\nCalculate chi-squared values: M1 At least 2 correct values (at least 3sf) or expressions seen.\n0.6811 + 2.8918 + 1.4694 + 0.4541 + 1.9279 + 0.9796\n8.40[4] A1\nTabular value, 2 degrees of freedom = 7.378 M1 Compare their value with 7.378 and conclusion without context.\n‘8.404’ > 7.378 and reject H /significant Condone ‘accept H ’.\n0 1\nSufficient evidence to suggest that grade is not independent of A1 Correct conclusion in context, following correct work, level of\nage. uncertainty in language (not ‘prove’, not ‘there is no evidence’),\nno contradictions.\nA0 if hypotheses wrong way round.\n7\n\n6(b) Value, 4 degrees of freedom = 11.14 M1 Condone ‘reject H ’.\n1\n10.91 < 11.14 Accept H /not significant\n0\nInsufficient evidence to suggest that grade is not independent of A1 CAO\nage.\n2\n© UCLES 2023 Page 12 of 13\n\n6(c) For example, result in part (b) because the table contains more B1 Any appropriate comment to support part (a) or part (b).\ninformation Allow ‘more specific’. Not ‘more data’ or ‘more accurate’ on its\ne.g. More degrees of freedom, more groups, more detail own.\n1\n© UCLES 2023 Page 13 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-May-June/ms/9231_s23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-may-june/9231_s23_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s23_ms_43-p13.png"
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    },
    {
      "id": "9231-2023-on-11-q01",
      "question_id": "9231-2023-on-11-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) r2 +2r+1−r2 =2r+1 B1 Expands\nn M1 A1 Uses method of differences and sums both sides.\n2r+n=(n+1)2 −12\nr=1\nn A1 AG.\n2r=n2 +n=n(n+1)\nr=1\n4\n\n1(b) n n M1 Relates with r.\n(r+a)=r+an\nr=1 r=1\n1n(n+1)+an=n M1 n\n2 Applies r= 1n(n+1).\n2\nr=1\na= 1(1−n) A1\n2\n3\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_11-p06.png"
      ]
    },
    {
      "id": "9231-2023-on-11-q02",
      "question_id": "9231-2023-on-11-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 1−2x+x2 (1−x)2 B1 Checks base case.\n1= = so H is true.\n(1−x)2 (1−x)2 1\nk 1−(k+1)xk +kxk+1 B1 States inductive hypothesis.\nAssume that rxr−1 = .\n(1−x)2\nr=1\nk+1 1−(k+1)xk +kxk+1 M1 Considers sum to k+1.\nrxr−1= +(k+1)xk\n(1−x)2\nr=1\n1−(k+1)xk +kxk+1+(k+1)xk( 1−2x+x2) M1 Puts over a common denominator.\n(1−x)2\n1+kxk+1+(k+1)xk( −2x+x2)\n1−(k+2)xk+1+(k+1)xk+2\nA1\n=\n(1−x)2 (1−x)2\nSo H is true. By induction, H is true for all positive integers n. A1 States conclusion.\nk+1 n\n6\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_11-p07.png"
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    },
    {
      "id": "9231-2023-on-11-q03",
      "question_id": "9231-2023-on-11-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) b=−(+++)=−3 B1\n5=(−3)2\n−2(+++++)\nM1 A1 Uses formula for sum of squares.\nc=2 A1\n+++ −d M1 Uses\n6= =\n+++\n −2 −1+−1+−1+−1 = .\n\nd =12 A1 Equation is x4 −3x3+2x2 +12x−2=0.\n6\n\n3(b) 4+4+4+4 =3(−27)−2(5)−12(3)+2(4) M1 Uses their quartic equation derived in (a).\n–119 A1\n2\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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        "../answer-assets/9231_w23_ms_11-p08.png"
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    },
    {
      "id": "9231-2023-on-11-q04",
      "question_id": "9231-2023-on-11-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)  2  −2 4 B1 Finds direction of one line to another.\n     \n−2 − −3 = 1\n     \n     \n 3  −5 8\ni j k 18 9 M1 A1 Find common perpendicular.\n   \n−4 3 5 = 14 ~ 7\n   \n   \n2 −3 1  6  3\n4 9 M1 A1 Uses formula for shortest distance.\n1     67\n1 7 = (=5.68)\n   \n139     139\n8 3\n5\n\n4(b) i j k  3   1  M1 A1 Finds vector perpendicular to the plane.\n   \n1 0 1 = 9 ~ 3\n   \n   \n−4 3 5 −3 −1\n1(−1)+3(−3)−1(−4)=−6 x+3y−z=−6 M1 A1 Uses point in the plane.\n4\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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        "../answer-assets/9231_w23_ms_11-p09.png"
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    },
    {
      "id": "9231-2023-on-11-q05",
      "question_id": "9231-2023-on-11-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 3 2 3 2 1 M1 A1 Sets determinant of A equal to zero.\n−k +3 =0−1−k+3=0k=2\n2 5 3 5 3 2\n1 2 3 0 −2 −2 4  M1 Multiplying two matrices correctly, correct\n−2 −1 1   −2 −1 1  dimensions.\n  2 1 3  −1 3  =  −1 −1 \n 1 1 3    1 1 3 \n3 2 5 0 0  −2 0 \n 3 −7 M1 A1 Completing matrix multiplication, AG.\n \n−9 3 \n5\n\n5(b)  3 −7x 3x−7y  B1 x X\n  =  Transforms   to  .\n−9 3 y −9x+3y y Y \n−9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX .\n−9+3m=3m−7m2 7m2 =9 A1\ny= 3 x and y=− 3 x A1\n7 7\n5\n© UCLES 2023 Page 10 of 15\n\n5(c)  0 B1\nD= \n0 \n 0 B1\nE= \n0 1\n0 1 B1\nF= \n1 0\n 3 −7  0 0  M1 Setting up simultaneous equations using their D and\n = −9  E.\n−9 3  0  1 0\n3 0 7 0 A1 Condone =3,= 7 if it is clear that they refer to\nD=  E= 9  9\n0 3 0 1 the correct matrices.\n5\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2023-on-11-q06",
      "question_id": "9231-2023-on-11-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2−x+1+y2 =1r2−rcos+1 =1 B1 Uses x2 + y2 =r2 and x=rcos.\n4 4 4 4\nr(r−cos)=0 M1 Factorises.\nr 0r =cos A1 AG.\n3\n© UCLES 2023 Page 11 of 15\n\n6(b) sin2=cos2sincos=cos M1 Sets r values equal and uses sin2=2sincos.\ncos0sin= 1 A1 cos0 must be recognised.\n2\n( 1 3,1π ) A1\n2 6\n3\n\n6(c) B1 Initial line drawn and one curve correct.\nB1 Other curve correct.\nB1 Intersection marked in correct position and both\ncurves labelled.\nP\no\n=0\n3\n© UCLES 2023 Page 12 of 15\n\n6(d) 1 1 6  sin22d+1 1 2  cos2d M1 Uses 1\n2\nr2d with correct limits.\n2 0 2 1\n6\n1 1 M1 Integratessin22 using identity.\n16 sin22d=16 1−cos4d\n2 4\n0 0\n1 A1\n= 1−1sin46\n \n4 4 0\n1 1 M1 Integrates cos2 using identity.\n12 cos2d= 12 1+cos2d\n2 1 4 1\n6 6\n1 A1\n= 1+ 1sin22\n \n4 2 1\n6\n1 ( 1−1 3 ) +1 ( 1−1−1 3 ) =1 ( −3 3 ) A1\n4 6 8 4 2 6 4 8 4\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12,
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_11-p13.png"
      ]
    },
    {
      "id": "9231-2023-on-11-q07",
      "question_id": "9231-2023-on-11-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1, x=2 B1 Vertical asymptotes.\ny=1 B1 Horizontal asymptote.\n2\n© UCLES 2023 Page 13 of 15\n\n7(b) dy (x2 −x−2)(2x)−(x2 +2)(2x−1) M1* dy\n= Finds .\ndx ( x2 −x−2 )2 dx\nx2 +8x−2=0 DM1 Sets equal to 0 and forms equation.\n(−8.2,0.9), (0.2,−0.9). A1 A1 Condone\n( ) ( )\n−4−3 2,2 2 , −4+3 2,−2 2 .\n3 3\n4\ny\n\n7(c) B1 Axes and all three asymptotes.\nB1 Correct shape and position, crossing horizontal\nasymptote.\nB1 States (0,−1)coordinates of intersection with axes,\nmay be seen on diagram.\nx\n(0,−1)\n3\n© UCLES 2023 Page 14 of 15\n\n7(d) B1 FT FT from sketch in (c)\nB1 All correct.\nx\n2\n\n7(e) x2 +2 x2 +2 M2 Finds critical points, award M1 for each case.\n=1 or =−1\nx2 −x−2 x2 −x−2\nx+4=0 or 2x2 −x=0\nx=−4 or x=0, x= 1 A1\n2\n−4x−1, 0 x 1, x2 B1 Must have three distinct regions. Condone −1\n2\nand 2.\n4\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
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        14,
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_11-p15.png"
      ]
    },
    {
      "id": "9231-2023-on-12-q01",
      "question_id": "9231-2023-on-12-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "1(a) 1n(n+1)(2n+1)+ 3n(n+1)+n M1 A1 Substitutes correct formulae from\n2 2\nMF19.\nn3+3n2+3n A1 Simplifies\n3\n\n1(b) 1 1 (r+1)3−r3 r3+3r2 +3r+1−r3 3r2 +3r+1 M1 A1 Puts over a common denominator and\n− = = = expands, AG.\nr3 (r+1)3 r3(r+1)3 r3(r+1)3 r3(r+1)3\nn 3r2 +3r+1 n  1 1  M1 A1 Shows three complete terms, including\n =  −  last.\nr3(r+1)3 r3 (r+1)3\n\nr=1 r=1\n1 1 1 1 1\n=1− + − + + −\n23 23 33 n3 (n+1)3\n1 A1\n1−\n(n+1)3\n5\n\n1(c) 1 B1FT FT from their answer to part (b).\n1\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_12-p06.png"
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    },
    {
      "id": "9231-2023-on-12-q02",
      "question_id": "9231-2023-on-12-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 d ( x2ex) =x2ex +2xex = ( x2 +2x ) ex so true when n=1. M1 A1 Differentiates once using the product\nrule.\ndx\nAssume that\ndk\n( x2ex) = ( x2 +2kx+k(k−1) ) ex\nB1 States inductive hypothesis.\ndxk\n[for some value of k].\ndk+1\n( x2ex) = ( x2 +2kx+k(k−1) ) ex +ex(2x+2k)\nM1 Differentiates kth derivative.\ndxk+1\n( x2+2(k+1)x+k(k+1) ) ex A1\nSo true when n=k+1. By induction, true for all positive integers n. A1 States conclusion.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_12-p07.png"
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    },
    {
      "id": "9231-2023-on-12-q03",
      "question_id": "9231-2023-on-12-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Shear followed by a stretch. B2 Award B1 if given in the wrong order.\n2\n\n3(b) OPQR = detM = k B1\n1 1 0 M1 A1\nM−1 =  \nk−1 k\n3\n© UCLES 2023 Page 7 of 15\n\n3(c) k 0 x  B1 Sets y= 1 x.\nk−1\n  \n1 1 1 x\n\nk−1\nk 0 x   kx   kx  M1 Shows that Y = 1 X.\n  = =  k−1\n1 1 1 x\n \nx+ 1 x\n \nk x\n\nk−1 k−1 k−1\n x  A1\nk \n1 x\n \nk−1\nAlternative method for 3(c)\nk 0x  kx  B1 x X\n  =  Transforms   to  \n1 1y x+ y y Y \nX =kx and mX =x+ y M1 Uses y = mx and Y = mX\nmkx=x+mx\n1 A1 AG\nm=\nk−1\n1\ny= x\nk−1\n3\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_12-p08.png"
      ]
    },
    {
      "id": "9231-2023-on-12-q04",
      "question_id": "9231-2023-on-12-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) y=3x+1x= 1(y−1) B1 Substitutes.\n3\n3 2\n y−1  y−1  y−1\n27 +18 +6 −1=0\n     \n 3   3   3 \n(y−1)3 +2(y−1)2\n+2(y−1)−1=0\nM1\n y3 −3y2 +3y−1+ 2y2 −4y+2+2y−2−1=0 Expands.\ny3 − y2 + y−2=0 A1 AG.\n3\n\n4(b) S =12 −2(1)=−1 M1 A1 Uses formula for sum of squares, AG.\n2\nS =(3+1)3+(3+1)3+(3+1)3 =−1−(1)+6 M1 Uses y3 = y2 − y+2 or expands and\n3\nuses original equation.\n4 A1\n4\n\n4(c) (3+1)(3+1)+(3+1)(3+1)+(3+1)(3+1) 1 B1\nS = =\n−1 (3+1)(3+1)(3+1) 2\n2S =S −3+S =1−3+ 1 M1 Uses 2y−2 = y−1+ y−1.\n−2 1 −1 2\nS =−3 A1 CAO\n−2 4\n3\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_12-p09.png"
      ]
    },
    {
      "id": "9231-2023-on-12-q05",
      "question_id": "9231-2023-on-12-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k  2   1  M1 A1 Finds perpendicular to .\n1\n   \n1 −2 −3 = −8 ~ −4\n   \n   \n3 0 −1  6   3 \n1(1)−4(−1)+3(−2)=−1 M1 Uses point on .\n1\nx−4y+3z=−1 A1\n4\n\n5(b)  1 1 M1 A1 Shows dot product with direction of\n   line is 0.\n−4 1 =1−4+3=0\n  \n  \n 3 1\n2\n\n5(c) −4  1  −3  1   M1 A1 Uses correct formula for distance from\n1     1      point on l to .\n1 −4 or 0 −4 +1 1\n        \n12 +42 +32   3     3   12 +42 +32     1     3     1 (−3.1+ 0.−4+1.3+1)\n12 +42 +32\n1 A1\n(=0.196)\n26\n3\n© UCLES 2023 Page 10 of 15\n\n5(d)  1  B1  \n 15   5   0 \nStates point common to both planes e.g.  4 .  7   \n15   5\n   \n0  0  or or alternative.\n  17\n \n−4  \n  1\n \n 7   \n17\ni j k −17 M1 A1 Finds direction of line.\n \n1 −4 3 = 7\n \n \n3 3 2  15 \n 5  −17 A1 OE.\n7\n   \nr= 0 +\n\n7\n\n   \n − 7 4   15 \n4\n© UCLES 2023 Page 11 of 15",
      "source_pages": [
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        11
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_12-p11.png"
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    },
    {
      "id": "9231-2023-on-12-q06",
      "question_id": "9231-2023-on-12-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) =0 B1 Initial line drawn. Correct shape, r\nstrictly decreasing.\nB1 Correct shape at extremities.\n1−e−1\n2\nπ B1 May be seen on their diagram.\n3\n\n6(b) 1 1 2 π( e−−e−1 2 π )2 d M1 Uses correct formula with correct\n2 limits.\n0\n1 1 2 π e−2−2e−−1 2 π +e−πd A1\n2\n0\n1π M1 A1 Integrates.\n1−1e−2+2e−−1\n2\nπ +e−π2\n2 2 \n0\n1 ( −1e−π +2e−π + 1πe−π + 1−2e−1 2 π ) = 3e−π + 1πe−π −e−1 2 π + 1 A1\n2 2 2 2 4 4 4\n5\n© UCLES 2023 Page 12 of 15\n\n6(c) y= ( e−−e−1 2 π ) sin B1 Uses y=rsin\ndy = ( e−−e−1 2 π ) cos+sin ( −e−) =0 M1 A1 Sets derivative equal to zero.\nd\n −e−  A1 AG.\n   1  1+ tan=01−e−1 2 π −tan=0\n2 e−−e−1\n2\nπ\n\n1−e0.56−1 2 π −tan0.56=0.00912 and1−e0.57−1 2 π −tan0.57=−0.00856 B1 Shows sign change.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_12-p12.png",
        "../answer-assets/9231_w23_ms_12-p13.png"
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    },
    {
      "id": "9231-2023-on-12-q07",
      "question_id": "9231-2023-on-12-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1 B1 Vertical asymptote.\n(x+1)(x−1)+1 M1 Oblique asymptote.\ny=\nx+1\ny=x−1 A1\n3\n\n7(b) dy x2 +2x M1 dy\n= =0 Sets =0.\ndx (x+1)2 dx\n(0,0), (−2,−4) A1\n2\n© UCLES 2023 Page 13 of 15\n\n7(c) y B1 Axes and asymptotes.\nB1 Left branch correct.\nB1 Right branch correct.\nx\n3\n\n7(d) ( −2,−1) B1 B1 B1 for each correct coordinate. SC B1\n4 for ( −2,−1) and (0,0).\n4\n2\n© UCLES 2023 Page 14 of 15\n\n7(e) y B1 Left branch correct.\nB1 Right branch correct.\nx\nx2 x2 M2 Finds critical points, award M1 for\n=1 or =−1 each case.\nx+1 x+1\nx2 −x−1=0\nx=1−1 5 or x=1+1 5 A1\n2 2 2 2\nx−1, 1−1 5x1+1 5, x0 B1 Condone missing x0.\n2 2 2 2\n6\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
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        14,
        15
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2023-on-13-q01",
      "question_id": "9231-2023-on-13-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) r2 +2r+1−r2 =2r+1 B1 Expands\nn M1 A1 Uses method of differences and sums both sides.\n2r+n=(n+1)2 −12\nr=1\nn A1 AG.\n2r=n2 +n=n(n+1)\nr=1\n4\n\n1(b) n n M1 Relates with r.\n(r+a)=r+an\nr=1 r=1\n1n(n+1)+an=n M1 n\n2 Applies r= 1n(n+1).\n2\nr=1\na= 1(1−n) A1\n2\n3\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p06.png"
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    },
    {
      "id": "9231-2023-on-13-q02",
      "question_id": "9231-2023-on-13-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 1−2x+x2 (1−x)2 B1 Checks base case.\n1= = so H is true.\n(1−x)2 (1−x)2 1\nk 1−(k+1)xk +kxk+1 B1 States inductive hypothesis.\nAssume that rxr−1 = .\n(1−x)2\nr=1\nk+1 1−(k+1)xk +kxk+1 M1 Considers sum to k+1.\nrxr−1= +(k+1)xk\n(1−x)2\nr=1\n1−(k+1)xk +kxk+1+(k+1)xk( 1−2x+x2) M1 Puts over a common denominator.\n(1−x)2\n1+kxk+1+(k+1)xk( −2x+x2)\n1−(k+2)xk+1+(k+1)xk+2\nA1\n=\n(1−x)2 (1−x)2\nSo H is true. By induction, H is true for all positive integers n. A1 States conclusion.\nk+1 n\n6\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p07.png"
      ]
    },
    {
      "id": "9231-2023-on-13-q03",
      "question_id": "9231-2023-on-13-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) b=−(+++)=−3 B1\n5=(−3)2\n−2(+++++)\nM1 A1 Uses formula for sum of squares.\nc=2 A1\n+++ −d M1 Uses\n6= =\n+++\n −2 −1+−1+−1+−1 = .\n\nd =12 A1 Equation is x4 −3x3+2x2 +12x−2=0.\n6\n\n3(b) 4+4+4+4 =3(−27)−2(5)−12(3)+2(4) M1 Uses their quartic equation derived in (a).\n–119 A1\n2\n© UCLES 2023 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p08.png"
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    },
    {
      "id": "9231-2023-on-13-q04",
      "question_id": "9231-2023-on-13-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)  2  −2 4 B1 Finds direction of one line to another.\n     \n−2 − −3 = 1\n     \n     \n 3  −5 8\ni j k 18 9 M1 A1 Find common perpendicular.\n   \n−4 3 5 = 14 ~ 7\n   \n   \n2 −3 1  6  3\n4 9 M1 A1 Uses formula for shortest distance.\n1     67\n1 7 = (=5.68)\n   \n139     139\n8 3\n5\n\n4(b) i j k  3   1  M1 A1 Finds vector perpendicular to the plane.\n   \n1 0 1 = 9 ~ 3\n   \n   \n−4 3 5 −3 −1\n1(−1)+3(−3)−1(−4)=−6 x+3y−z=−6 M1 A1 Uses point in the plane.\n4\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p09.png"
      ]
    },
    {
      "id": "9231-2023-on-13-q05",
      "question_id": "9231-2023-on-13-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 3 2 3 2 1 M1 A1 Sets determinant of A equal to zero.\n−k +3 =0−1−k+3=0k=2\n2 5 3 5 3 2\n1 2 3 0 −2 −2 4  M1 Multiplying two matrices correctly, correct\n−2 −1 1   −2 −1 1  dimensions.\n  2 1 3  −1 3  =  −1 −1 \n 1 1 3    1 1 3 \n3 2 5 0 0  −2 0 \n 3 −7 M1 A1 Completing matrix multiplication, AG.\n \n−9 3 \n5\n\n5(b)  3 −7x 3x−7y  B1 x X\n  =  Transforms   to  .\n−9 3 y −9x+3y y Y \n−9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX .\n−9+3m=3m−7m2 7m2 =9 A1\ny= 3 x and y=− 3 x A1\n7 7\n5\n© UCLES 2023 Page 10 of 15\n\n5(c)  0 B1\nD= \n0 \n 0 B1\nE= \n0 1\n0 1 B1\nF= \n1 0\n 3 −7  0 0  M1 Setting up simultaneous equations using their D and\n = −9  E.\n−9 3  0  1 0\n3 0 7 0 A1 Condone =3,= 7 if it is clear that they refer to\nD=  E= 9  9\n0 3 0 1 the correct matrices.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p10.png",
        "../answer-assets/9231_w23_ms_13-p11.png"
      ]
    },
    {
      "id": "9231-2023-on-13-q06",
      "question_id": "9231-2023-on-13-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2−x+1+y2 =1r2−rcos+1 =1 B1 Uses x2 + y2 =r2 and x=rcos.\n4 4 4 4\nr(r−cos)=0 M1 Factorises.\nr 0r =cos A1 AG.\n3\n© UCLES 2023 Page 11 of 15\n\n6(b) sin2=cos2sincos=cos M1 Sets r values equal and uses sin2=2sincos.\ncos0sin= 1 A1 cos0 must be recognised.\n2\n( 1 3,1π ) A1\n2 6\n3\n\n6(c) B1 Initial line drawn and one curve correct.\nB1 Other curve correct.\nB1 Intersection marked in correct position and both\ncurves labelled.\nP\no\n=0\n3\n© UCLES 2023 Page 12 of 15\n\n6(d) 1 1 6  sin22d+1 1 2  cos2d M1 Uses 1\n2\nr2d with correct limits.\n2 0 2 1\n6\n1 1 M1 Integratessin22 using identity.\n16 sin22d=16 1−cos4d\n2 4\n0 0\n1 A1\n= 1−1sin46\n \n4 4 0\n1 1 M1 Integrates cos2 using identity.\n12 cos2d= 12 1+cos2d\n2 1 4 1\n6 6\n1 A1\n= 1+ 1sin22\n \n4 2 1\n6\n1 ( 1−1 3 ) +1 ( 1−1−1 3 ) =1 ( −3 3 ) A1\n4 6 8 4 2 6 4 8 4\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p11.png",
        "../answer-assets/9231_w23_ms_13-p12.png",
        "../answer-assets/9231_w23_ms_13-p13.png"
      ]
    },
    {
      "id": "9231-2023-on-13-q07",
      "question_id": "9231-2023-on-13-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1, x=2 B1 Vertical asymptotes.\ny=1 B1 Horizontal asymptote.\n2\n© UCLES 2023 Page 13 of 15\n\n7(b) dy (x2 −x−2)(2x)−(x2 +2)(2x−1) M1* dy\n= Finds .\ndx ( x2 −x−2 )2 dx\nx2 +8x−2=0 DM1 Sets equal to 0 and forms equation.\n(−8.2,0.9), (0.2,−0.9). A1 A1 Condone\n( ) ( )\n−4−3 2,2 2 , −4+3 2,−2 2 .\n3 3\n4\ny\n\n7(c) B1 Axes and all three asymptotes.\nB1 Correct shape and position, crossing horizontal\nasymptote.\nB1 States (0,−1)coordinates of intersection with axes,\nmay be seen on diagram.\nx\n(0,−1)\n3\n© UCLES 2023 Page 14 of 15\n\n7(d) B1 FT FT from sketch in (c)\nB1 All correct.\nx\n2\n\n7(e) x2 +2 x2 +2 M2 Finds critical points, award M1 for each case.\n=1 or =−1\nx2 −x−2 x2 −x−2\nx+4=0 or 2x2 −x=0\nx=−4 or x=0, x= 1 A1\n2\n−4x−1, 0 x 1, x2 B1 Must have three distinct regions. Condone −1\n2\nand 2.\n4\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
        13,
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_13-p13.png",
        "../answer-assets/9231_w23_ms_13-p14.png",
        "../answer-assets/9231_w23_ms_13-p15.png"
      ]
    },
    {
      "id": "9231-2023-on-21-q01",
      "question_id": "9231-2023-on-21-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 14 −4 6 M1 A1 Evaluates determinant. Can expand along any row e.g.\n1 k 1 k 1 1\n−(36−36)+(−126+126)+k(84−84).\n1 1 k =14 +4 +6\n6 −9 −21 −9 −21 6\n−21 6 −9\nIf using row operations, they must show an inconsistent\n=14(−9−6k)+4(−9+21k)+6(6+21)=0\nsystem for M1. All their row operations must be correct\nfor A1.\nTwo parallel planes, not identical. B1\nOther plane not parallel. B1\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2023-on-21-q02",
      "question_id": "9231-2023-on-21-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1 π B1 Finds modulus and argument of 4+4i 3.\n(z+5i)3 =4+4i 3=8e3\n 1 1  1  1  M1 A1 Finds one root.\nz =2cos π+isin π−5i = 2cos π+i2sin π−5\n1  9 9  9  9 \n7  7  13  13  A1 FT Finds other two roots. FT on their modulus.\nz =2cos π+i  2sin π−5 , z =2cos π+i  2sin π−5 \n2 9  9  3 9  9 \nA1 FT\n5\n© UCLES 2023 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2023-on-21-q03",
      "question_id": "9231-2023-on-21-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 1 B1\nex\ndy 2\n=\ndx 1\n1− e2x\n4\n 1 1  1  1  B1\n1− e2x  ex − ex − e2x \nd2y  4 2  2  2 \n=\ndx2\n 1 \n2\n1− e2x \n 4 \n2 10 M1 Evaluates derivatives at x=0.\nf '(0)= f ''(0)=\n3 9\n1 1 3  M1 Uses logarithmic form of tanh−1.\nf(0)=tanh−1\n \n= ln\n\n2\n\n2 2 2 \n1 2 5 M1 A1 1\nln3+ x+ x2 Applies f(x)= f(0)+ f '(0)x+ f ''(0)x2\n2 3 9 2!\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_21-p07.png"
      ]
    },
    {
      "id": "9231-2023-on-21-q04",
      "question_id": "9231-2023-on-21-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4 m2 +2m+3=0 M1 Auxiliary equation.\n[y=]e−x(\nAcos 2x+Bsin 2x\n) A1 Complementary function. Allow with “y=” missing.\ny= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives.\n3p=27 3q+4p=0 2p+2q+3r=0 M1 Substitutes and equates coefficients.\n© UCLES 2023 Page 7 of 14\n\n4 p=9 q=−12 r=2 A1\ny=e−x( Acos 2x+Bsin 2x ) +9x2 −12x+2 A1 General solution. Must have “y=”.\ny'=e−x(\n− 2Asin 2x+ 2Bcos 2x\n) −e−x(\nAcos 2x+Bsin 2x\n)\n+18x−12\nM1* Differentiates. Must use product rule.\nA+2=2 2B−A−12=−8 DM1 Uses initial conditions\nA1\n A=0,B=2 2\ny=2 2e−xsin 2x+9x2 −12x+2 A1 Must have “y=”.\n10\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_21-p07.png",
        "../answer-assets/9231_w23_ms_21-p08.png"
      ]
    },
    {
      "id": "9231-2023-on-21-q05",
      "question_id": "9231-2023-on-21-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 1 B1 Differentiates x and y with respect to t.\n−\nx=t2 −t 2, y=2\n 1 1  2  1 1 2 M1 A1 Factorises x2 + y2.\n− −\nx2 + y2 =t2 −t 2 +4=t+2+t−1 =t2 +t 2\n   \n   \n3 M1 A1 Applies correct formula for arc length.\n3\n1\n−\n1 2 3 1\n\n0\nt2 +t 2dt =\n 3\nt2 +2t2\n\n=4 3\nM1 for their \n3\nx2 + y2 dt.\n0 0\nAnswer must be simplified to 4 3 for A1.\n5\n© UCLES 2023 Page 8 of 14\n\n5(b) dy y 2 B1 Finds first derivative.\n= =\ndx x 1 − 1\nt2 −t 2\n1 − 1 1 − 3 B1 Differentiates dy with respect to t.\n−2 t 2 + t 2\nd   2     2 2   dx\n=\ndt 1\n−\n1 \n 1 1 \n2\nt2 −t 2  −\nt2 −t 2\n \n \n1 − 1 1 − 3 M1 A1 Applies chain rule. OE. Does not have to be simplified\n−2 t 2 + t 2 for A1.\nd2y d   2   dt   2 2   t+1\n=  = =−\ndx2 dt 1 − 1  dx  1 1  3 (t−1)3\nt2 −t 2  −\nt2 −t 2\n \n \n0t1 A1 Accept −1t1. CWO.\n5\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2023-on-21-q06",
      "question_id": "9231-2023-on-21-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) coshx= 1( ex +e−x) sinhx= 1( ex −e−x) B1\n2 2\n1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x M1 A1 Expands, AG.\n2 2\n3\n© UCLES 2023 Page 9 of 14\n\n6(b) u=sinhxdu=coshxdx B1\nsinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) du M1 Applies identities to find integral in terms of u.\n=4u2( u2 +1 ) du A1\n1 1  1 1  A1\n=4\n\nu5 + u3\n\n(+C)=4\n\nsinh5x+ sinh3x\n\n(+C)\n5 3  5 3 \n4\n\n6(c)\ne\ntanhxdx\n=elncoshx =coshx\nM1 A1 Finds integrating factor.\nd M1 Correct form on LHS and attempt to integrate RHS.\n(ycoshx)=sinh22xcoshx\ndx\n1 1  M1 A1 Integrates RHS using their part (b).\nycoshx=4\n\nsinh5x+ sinh3x\n\n+C\n5 3 \n4=C M1 Substitutes initial conditions.\n1 1  A1\ny=4sechx\n\nsinh5x+ sinh3x+1\n\n5 3 \n7\n© UCLES 2023 Page 10 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2023-on-21-q07",
      "question_id": "9231-2023-on-21-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) =−6, =−2, =8 B1\ni j k 8 1 M1 A1 Uses vector product (or equations) to find corresponding\n    eigenvectors.\n=−6: 0 4 5 = 0 0\n   \n   \n0 0 2 0 0\ni j k 10 1 A1\n   \n=−2: −4 2 13 = 20 2\n   \n   \n0 0 5  0  0\ni j k 140 2 A1\n   \n=8: −14 2 13 = 70 1\n   \n   \n0 −10 5 140 2\n1 1 2 −1 0 0 M1 A1 Or correctly matched permutations of columns.\n6\n    M1 for their (non-zero) eigenvectors matched to their\nThus P= 0 2 1 and D= 0 −1 0\n   2  eigenvalues.\n 0 0 2     0 0 1 8  \n7\n\n7(b) A3−52A−96I=0 M1 Substitutes A into characteristic equation.\n96A−1 =A2 −52I M1 Multiples through by A−1.\n36 −16 36 B1\n \nA2 = 0 4 30\n \n \n 0 0 64\n© UCLES 2023 Page 11 of 14\n\n7(b) −1 −1 3  A1\n6 6 8\n \nA−1 = 0 −1 5\n 2 16\n 0 0 1 \n 8 \n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2023-on-21-q08",
      "question_id": "9231-2023-on-21-q08",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "8(a) zn −1 B1\nz−1\n1\n\n8(b) zn −1 cosn−1+isinn B1\n=\nz−1 cos−1+isin\n(cosn−1+isinn)(cos−1−isin) M1 Multiplies numerator and denominator by complex\nconjugate.\n(cos−1+isin)(cos−1−isin)\nzn −1 cosncos+sinnsin−cosn−cos+1 M1 Takes real part.\nRe = cos(n−1)=cosncos+sinnsin\n\nz−1\n\n(cos−1)2 +sin2\ncosncos+sinnsin−cosn 1 A1\n= +\n2(1−cos) 2\ncosn(cos−1)+sinnsin 1 M1 Factorises.\n= +\n2(1−cos) 2\n© UCLES 2023 Page 12 of 14\n\n8(b) 1 sinnsin M1 A1 Divides through by denominator. AG.\n=\n\n1−cosn+\n\n2 1−cos \nAlternative method for question 8(b)\nzn −1 ein−1 B1\n=\nz−1 ei−1\ne i(n−1 2 ) −e−i1 2  cos(n−1)+isin(n−1)−cos1+isin(1) M1\n= 2 2 2 2\nei1\n2\n−e−i1\n2\n 2isin1\n2\nzn −1 sin(n−1)+sin1 M1 Takes real part\nRe = 2 2\nz−1 2sin1\n \n2\nsin(n−1)\n1\nA1\n= 2 +\n2sin1 2\n2\nsinncos1−cosnsin1 1 M1 Uses compound angle identity\n= 2 2 +\n2sin1 2\n2\nsinncos1 1 1 M1 Divides through by denominator.\n= 2 − cosn+\n2sin1 2 2\n2\nsinnsin 1 1 sinnsin 1 1 A1 AG. sin=2sin1cos1 and\n− cosn+ = − cosn+ 2 2\n4sin2 1\n2\n 2 2 2(1−cos) 2 2 2sin2 1=1−cos.\n2\n7\n© UCLES 2023 Page 13 of 14\n\n8(c) 1 1 1 1 1 2 1 n−1 M1 A1 Forms sum of areas of rectangles given in the diagram.\n cosxdx + cos + cos + cos\nA0 if comparison with integral missing or unclear.\n0 n n n n n n n\n n 1 M1 A1 1\nsin sin Applies result from part (b) with = . AG.\n1 1 2 n−1 1  n n n  n\n=  1+cos +cos ++cos  = 1−cos + \nn n n n  2n  n 1−cos 1 \n n \n4\n\n8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison\n cosxdx cos + cos + cos\nwith integral missing or unclear.\n0 n n n n n n\n 1  1 A1\nsin1sin sin1sin\n1  n  1 1 1  n \n= 1−cos1+ + cos1− = cos1−1+ \n2n 1 n n 2n 1\n 1−cos   1−cos \n n   n \n3\n© UCLES 2023 Page 14 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2023-on-22-q01",
      "question_id": "9231-2023-on-22-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 EITHER Solution 1 M1 Changes ln(x+2) or ln ( x2 +5 )\nln(x+2)=ln ( 2(1+1x)) =ln2+ln(1+1x)\n2 2 so that the formula given in the list\nof formulae (MF19) can be applied.\nln(x+2)=ln2+1x−1x2+ A1\n2 8\nln ( x2 +5 ) =ln ( 5 ( 1+1x2)) =ln5+1x2 + A1\n5 5\n( ln2+1x−1x2 + ) + ( ln5+1x2+ ) M1 Sums power series\n2 8 5\nOR Solution 2 (M1 A1) Finds first derivative.\nf '(x)=(x+2)−1 +2x ( x2 +5 )−1\nf ''(x)=−(x+2)−2 −(2x)2( x2 +5 )−2 +2 ( x2 +5 )−1 (A1) Finds second derivative.\nf(0)=ln10 f '(0)= 1 f ''(0)= 3 (M1) Evaluates derivatives at zero.\n2 20\nln(x+2)+ln ( x2 +5 ) =ln10+1x+ 3 x2 A1 Accept ln10 written as ln2+ln5\n2 40 but do not accept decimals. WWW.\n5\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-22-q02",
      "question_id": "9231-2023-on-22-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) dy dx B1\n=et(t+1), =−t−2\ndt dt\ndy\n=\ndy\n\ndt =−t2et(t+1)=−et( t3+t2) M1 A1 Uses chain rule, AG.\ndx dt dx\n3\n\n2(b) d \n\ndy\n\n=−et( 3t2 +2t ) −et( t3+t2) =−et( t3+4t2 +2t ) M1 A1 Applies product rule.\ndtdx\nd2y\n=t3et( t2 +4t+2 )\nM1 A1 Uses chain rule, multiplies by their\ndt\ndx2\nfor M1.\ndx\nAlternative method for question 2(b)\nd2y =−et  3t2 dt +2t dt  −et( t3 +t2)dt (M1 A1) Differentiates −et( t3+t2)\ndx2  dx dx dx implicitly.\nd2y =et( −4t2 −2t−t3)dt =t3et( t2 +4t+2 ) (M1 A1) Substitutes dt =−t2.\ndx2 dx dx\n4\n© UCLES 2023 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
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    },
    {
      "id": "9231-2023-on-22-q03",
      "question_id": "9231-2023-on-22-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) cos5=Re(cos+isin)5 =cos5−10sin2cos3+5sin4cos M1 A1 Expands and takes real part. Accept\n1\nRHS to LHS using 2cos=z+ .\nz\n=cos5−10cos3 ( 1−cos2 ) +5cos ( 1−2cos2+cos4 ) M1 Applies sin2=1−cos2.\n=16cos5−20cos3+5cos A1 AG\n4\n\n3(b) x=cos, cos5=1 2 M1 Applies identify given in (b).\n2\n5=1+2k M1 Solves cos5=1 2\n4 2\ncos ( 1 π ) A1 Gives one correct solution. Accept\n20\nq= 1 .\n20\ncos(9 π),cos(17π),cos(25π),cos(33π) A1 Gives other solutions. OE. A0 for\n20 20 20 20\nrepeated roots.\n4\n© UCLES 2023 Page 8 of 15",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2023-on-22-q04",
      "question_id": "9231-2023-on-22-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4 e3dx =e3x M1 A1 Finds integrating factor.\nd ( ye3x) =e3xsinx M1 Correct form on LHS and attempt\nto integrate RHS.\ndx\nEITHER M1 A1 Integrates by parts once or uses\ne3xsinxdx=−e3xcosx+3e3xcosxdx eix −e-ix\nsinx= .\n2i\nOR\ne3xsinxdx=1e3xsinx−1e3xcosxdx\n3 3\nEITHER M1 Integrates by parts again or\ne3xsinxdx=−e3xcosx+3 ( e3xsinx−3e3xsinxdx ) eix −e-ix\nsubstitutes =sinx and\n2i\nOR\n( )\neix +e-ix\ne3xsinxdx= 1e3xsinx−1 1e3xcosx+1e3xsinxdx =cosx.\n3 3 3 3 2\nye3x = 1 e3x(3sinx−cosx)+C A1 Must not see i.\n10\n1=− 1 +C M1 Finds C. Substitutes into their\n10\nexpression (must be integrated).\ny= 3sinx− 1 cosx+11e−3x A1 Divides through by coefficient of y.\n10 10 10\n9\n© UCLES 2023 Page 9 of 15",
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-22-q05",
      "question_id": "9231-2023-on-22-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) dy M1 A1 Differentiating using product rule\n=−2xsech2xtanhx+sech2x\nand chain rule for M1.\ndx\n=1−tanh2x−2xtanhx+2xtanh3x\n \nsech2x02xtanhx−1=0 A1 AG.\n2(0.7)tanh(0.7)−1=−0.1540 B1 Shows sign change. Must write\n2(0.8)tanh(0.8)−1=0.0620 down values correct to at least 1dp\nfor B1.\n4\n\n5(b) n M1 A1 Compares sum with integral.\nn\nrsech2r xsech2xdx\nConsistent limits for M1.\n1\nr=2\n n xsech2xdx=xtanhxn − n tanhxdx M1 A1 Integrates by parts.\n1 1 1\n=xtanhx+lnsechxn A1\n1\n=ntanhn+lnsechn−(tanh1+lnsech1) A1 AG. Must have gained all previous\nmarks.\n6\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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    },
    {
      "id": "9231-2023-on-22-q06",
      "question_id": "9231-2023-on-22-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) 1, 2, −1 B1\n1\n© UCLES 2023 Page 10 of 15\n\n6(b) P3−2P2 −P+2I=0 B1 States that P satisfies its\ncharacteristic equation.\n2P−1 =I+2P−P2 M1 Multiplies through by P−1.\n1 −3 −1 1 1 3  M1 A1\n2 2\n   \nP2 = 0 4 1 P−1 = 0 1 1\n   2 2 \n   \n0 0 1 0 0 −1\n   \n4\n\n6(c) 1 0 0 B1\na\n \nD= 0 2 0\n \n \n0 0 1\n \n2\n1 0 0 M1 Applies A−1 =PDP−1.\na\n \nA−1 =P 0 2 0 P−1\n \n \n0 0 1\n \n2\n1 −2 1 1 1 3  M1 A1 Multiplies two adjacent matrices.\na 2 2 2\n  \n= 0 4 1 0 1 1\n 2  2 2 \n  \n0 0 −1 0 0 −1\n  \n2\n1 1−2a 3−3a A1\na 2a 2a\n \n0 2 3\n 2 \n \n0 0 1\n \n2\n5\n© UCLES 2023 Page 11 of 15",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_22-p11.png"
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    },
    {
      "id": "9231-2023-on-22-q07",
      "question_id": "9231-2023-on-22-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a)\ncosh2A= 1\n( e2A+e−2A)\nsinhA= 1\n( eA−e−A) B1\n2 2\n2sinh2 A= 1 ( eA −e−A)2 = 1 ( e2A −2+e−2A) =cosh2A−1 M1 A1 Expands, AG. A0 for mixing\n2 2 variables e.g. sinhA=1 ( ex −e−x) .\n2\n3\n© UCLES 2023 Page 12 of 15\n\n7(b) 2 M1 A1 Correct formula with correct limits.\n3 dy 2 2\nS = 2y 1+ dx=23x2 1+4x2dx Correct limits for M1. (Limits may\n  \n dx 0 be recovered.)\n0\nsinh−14 M1 Applies given substitution to their\nS =π 31sinh2u 1+sinh2ucoshudu\n4 expression with correct limits.\n0\n(Limits may be recovered.)\nsinh−14 A1 Must be simplified.\n=1π 3sinh2ucosh2udu\n4\n0\nsinh−14 M1 Applies sinh2u=2sinhucoshu.\n= 1 π 3sinh22udu\n16 May use double angle formulae for\n0\ncosh and sinh instead.\nsinh−14 M1 Applies sinh2 A=1(cosh2A−1). S\n= 1 π 3cosh4u−1du 2\n32\n0 must have the form\nasinh2ucosh2udu.\nsinh−14 A1\n= 1 π1sinh4u−u 3\n \n32 4 0\nsinh−14 =ln3 B1\n3\nS = 1 π ( 1 ( eln81−e−ln81) −ln3 ) = 1 π ( 1( 81− 1 ) −ln3 ) = 1 π (820−ln3 ) A1 AG.\n32 8 32 8 81 32 81\n9\n© UCLES 2023 Page 13 of 15",
      "source_pages": [
        12,
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_22-p12.png",
        "../answer-assets/9231_w23_ms_22-p13.png"
      ]
    },
    {
      "id": "9231-2023-on-22-q08",
      "question_id": "9231-2023-on-22-q08",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) dv dy B1 v= y4\n=4y3\ndx dx\nd2v d2y dy 2 B1\n=4y3 +12y2\n \ndx2 dx2 dx\nd2v dv d2y dy 2 dy M1 Uses substitution to find v-x\n+ +4v=4y3 +12y2   +4y3 +4y4 equation, AG.\ndx2 dx dx2 dx dx\n d2y dy 2 dy  A1 AG.\n=4y3 +3y2   + y3 + y4=4e−2x\n dx2 dx dx \n \nAlternative method for question 8(a)\ndy = 1 dv = 1 dv (B1) y=v 1 4\ndx 4y3 dx 4v 3 4 dx\nd2y 1 d2v 3 dv 2 (B1)\n= −  \ndx2 4v 3 4 dx2 16v 7 4 dx\nv 3 4   1 d2v − 3   dv  2 +3v 1 2   1 dv  2 +v4 3   1 dv +v=e−2x (M1) U eq s u e a s t i s o u n b , s A tit G ut . i on to find v-x\n 4v 3 4 dx2 16v 7 4 dx   4v 3 4 dx 4v 3 4 dx\n1d2v 3 dv 2 3 dv 2 1dv (A1)\n− + + +v=e−2x\n   \n4dx2 16vdx 16vdx 4dx\n4\n© UCLES 2023 Page 14 of 15\n\n8(b) m2 +m+4=0 M1 Auxiliary equation.\nv=e−1 2 x( Acos 15 x+Bsin 15 x ) A1 Complimentary function. Allow\n2 2 ‘v =’ missing.\nv=ke−2x v'=−2ke−2x v''=4ke−2x B1 Particular integral and its\nderivatives.\n4ke−2x −2ke−2x +4ke−2x =4e−2x M1 Substitutes and equates\ncoefficients.\nk = 2 A1\n3\nv= y4 =e−1 2 x( Acos 15 x+Bsin 15 x ) + 2e−2x A1 Substitutes for y and find v in terms\n2 2 3 of x. Must not see i.\nv'=4y3 dy =e−1 2 x( − 15 Asin 15 x+ 15Bcos 15x ) −1e−1 2 x( Acos 15x+Bsin 15x ) −4e−2x B1\ndx 2 2 2 2 2 2 2 3\n1= A+ 2 −3 = 15B−1 A−4  A= 1, B=0 M1 A1 Substitutes initial conditions and\n3 2 2 2 3 3 forms simultaneous equations.\nMust have used the product rule\nwhen differentiating their v for M1.\ny= ( 1e−1 2 xcos 15 x+2e−2x )1 4 A1 Accept\n3 2 3 y= ( 1e−1 2 xcos 15 x+2e−2x )1 4\n3 2 3\n10\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_22-p15.png"
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    },
    {
      "id": "9231-2023-on-23-q01",
      "question_id": "9231-2023-on-23-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 14 −4 6 M1 A1 Evaluates determinant. Can expand along any row e.g.\n1 k 1 k 1 1\n−(36−36)+(−126+126)+k(84−84).\n1 1 k =14 +4 +6\n6 −9 −21 −9 −21 6\n−21 6 −9\nIf using row operations, they must show an inconsistent\n=14(−9−6k)+4(−9+21k)+6(6+21)=0\nsystem for M1. All their row operations must be correct\nfor A1.\nTwo parallel planes, not identical. B1\nOther plane not parallel. B1\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2023-on-23-q02",
      "question_id": "9231-2023-on-23-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1 π B1 Finds modulus and argument of 4+4i 3.\n(z+5i)3 =4+4i 3=8e3\n 1 1  1  1  M1 A1 Finds one root.\nz =2cos π+isin π−5i = 2cos π+i2sin π−5\n1  9 9  9  9 \n7  7  13  13  A1 FT Finds other two roots. FT on their modulus.\nz =2cos π+i  2sin π−5 , z =2cos π+i  2sin π−5 \n2 9  9  3 9  9 \nA1 FT\n5\n© UCLES 2023 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2023-on-23-q03",
      "question_id": "9231-2023-on-23-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 1 B1\nex\ndy 2\n=\ndx 1\n1− e2x\n4\n 1 1  1  1  B1\n1− e2x  ex − ex − e2x \nd2y  4 2  2  2 \n=\ndx2\n 1 \n2\n1− e2x \n 4 \n2 10 M1 Evaluates derivatives at x=0.\nf '(0)= f ''(0)=\n3 9\n1 1 3  M1 Uses logarithmic form of tanh−1.\nf(0)=tanh−1\n \n= ln\n\n2\n\n2 2 2 \n1 2 5 M1 A1 1\nln3+ x+ x2 Applies f(x)= f(0)+ f '(0)x+ f ''(0)x2\n2 3 9 2!\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p07.png"
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    },
    {
      "id": "9231-2023-on-23-q04",
      "question_id": "9231-2023-on-23-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4 m2 +2m+3=0 M1 Auxiliary equation.\n[y=]e−x(\nAcos 2x+Bsin 2x\n) A1 Complementary function. Allow with “y=” missing.\ny= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives.\n3p=27 3q+4p=0 2p+2q+3r=0 M1 Substitutes and equates coefficients.\n© UCLES 2023 Page 7 of 14\n\n4 p=9 q=−12 r=2 A1\ny=e−x( Acos 2x+Bsin 2x ) +9x2 −12x+2 A1 General solution. Must have “y=”.\ny'=e−x(\n− 2Asin 2x+ 2Bcos 2x\n) −e−x(\nAcos 2x+Bsin 2x\n)\n+18x−12\nM1* Differentiates. Must use product rule.\nA+2=2 2B−A−12=−8 DM1 Uses initial conditions\nA1\n A=0,B=2 2\ny=2 2e−xsin 2x+9x2 −12x+2 A1 Must have “y=”.\n10\nQuestion Answer Marks Guidance",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p07.png",
        "../answer-assets/9231_w23_ms_23-p08.png"
      ]
    },
    {
      "id": "9231-2023-on-23-q05",
      "question_id": "9231-2023-on-23-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 1 1 B1 Differentiates x and y with respect to t.\n−\nx=t2 −t 2, y=2\n 1 1  2  1 1 2 M1 A1 Factorises x2 + y2.\n− −\nx2 + y2 =t2 −t 2 +4=t+2+t−1 =t2 +t 2\n   \n   \n3 M1 A1 Applies correct formula for arc length.\n3\n1\n−\n1 2 3 1\n\n0\nt2 +t 2dt =\n 3\nt2 +2t2\n\n=4 3\nM1 for their \n3\nx2 + y2 dt.\n0 0\nAnswer must be simplified to 4 3 for A1.\n5\n© UCLES 2023 Page 8 of 14\n\n5(b) dy y 2 B1 Finds first derivative.\n= =\ndx x 1 − 1\nt2 −t 2\n1 − 1 1 − 3 B1 Differentiates dy with respect to t.\n−2 t 2 + t 2\nd   2     2 2   dx\n=\ndt 1\n−\n1 \n 1 1 \n2\nt2 −t 2  −\nt2 −t 2\n \n \n1 − 1 1 − 3 M1 A1 Applies chain rule. OE. Does not have to be simplified\n−2 t 2 + t 2 for A1.\nd2y d   2   dt   2 2   t+1\n=  = =−\ndx2 dt 1 − 1  dx  1 1  3 (t−1)3\nt2 −t 2  −\nt2 −t 2\n \n \n0t1 A1 Accept −1t1. CWO.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p08.png",
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    },
    {
      "id": "9231-2023-on-23-q06",
      "question_id": "9231-2023-on-23-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) coshx= 1( ex +e−x) sinhx= 1( ex −e−x) B1\n2 2\n1( ex −e−x)( ex+e−x) = 1( e2x −e−2x) =sinh2x M1 A1 Expands, AG.\n2 2\n3\n© UCLES 2023 Page 9 of 14\n\n6(b) u=sinhxdu=coshxdx B1\nsinh22xcoshxdx=4sinh2xcosh2xdu=4sinh2x ( sinh2x+1 ) du M1 Applies identities to find integral in terms of u.\n=4u2( u2 +1 ) du A1\n1 1  1 1  A1\n=4\n\nu5 + u3\n\n(+C)=4\n\nsinh5x+ sinh3x\n\n(+C)\n5 3  5 3 \n4\n\n6(c)\ne\ntanhxdx\n=elncoshx =coshx\nM1 A1 Finds integrating factor.\nd M1 Correct form on LHS and attempt to integrate RHS.\n(ycoshx)=sinh22xcoshx\ndx\n1 1  M1 A1 Integrates RHS using their part (b).\nycoshx=4\n\nsinh5x+ sinh3x\n\n+C\n5 3 \n4=C M1 Substitutes initial conditions.\n1 1  A1\ny=4sechx\n\nsinh5x+ sinh3x+1\n\n5 3 \n7\n© UCLES 2023 Page 10 of 14",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_23-p09.png",
        "../answer-assets/9231_w23_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2023-on-23-q07",
      "question_id": "9231-2023-on-23-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) =−6, =−2, =8 B1\ni j k 8 1 M1 A1 Uses vector product (or equations) to find corresponding\n    eigenvectors.\n=−6: 0 4 5 = 0 0\n   \n   \n0 0 2 0 0\ni j k 10 1 A1\n   \n=−2: −4 2 13 = 20 2\n   \n   \n0 0 5  0  0\ni j k 140 2 A1\n   \n=8: −14 2 13 = 70 1\n   \n   \n0 −10 5 140 2\n1 1 2 −1 0 0 M1 A1 Or correctly matched permutations of columns.\n6\n    M1 for their (non-zero) eigenvectors matched to their\nThus P= 0 2 1 and D= 0 −1 0\n   2  eigenvalues.\n 0 0 2     0 0 1 8  \n7\n\n7(b) A3−52A−96I=0 M1 Substitutes A into characteristic equation.\n96A−1 =A2 −52I M1 Multiples through by A−1.\n36 −16 36 B1\n \nA2 = 0 4 30\n \n \n 0 0 64\n© UCLES 2023 Page 11 of 14\n\n7(b) −1 −1 3  A1\n6 6 8\n \nA−1 = 0 −1 5\n 2 16\n 0 0 1 \n 8 \n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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      ]
    },
    {
      "id": "9231-2023-on-23-q08",
      "question_id": "9231-2023-on-23-q08",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "8(a) zn −1 B1\nz−1\n1\n\n8(b) zn −1 cosn−1+isinn B1\n=\nz−1 cos−1+isin\n(cosn−1+isinn)(cos−1−isin) M1 Multiplies numerator and denominator by complex\nconjugate.\n(cos−1+isin)(cos−1−isin)\nzn −1 cosncos+sinnsin−cosn−cos+1 M1 Takes real part.\nRe = cos(n−1)=cosncos+sinnsin\n\nz−1\n\n(cos−1)2 +sin2\ncosncos+sinnsin−cosn 1 A1\n= +\n2(1−cos) 2\ncosn(cos−1)+sinnsin 1 M1 Factorises.\n= +\n2(1−cos) 2\n© UCLES 2023 Page 12 of 14\n\n8(b) 1 sinnsin M1 A1 Divides through by denominator. AG.\n=\n\n1−cosn+\n\n2 1−cos \nAlternative method for question 8(b)\nzn −1 ein−1 B1\n=\nz−1 ei−1\ne i(n−1 2 ) −e−i1 2  cos(n−1)+isin(n−1)−cos1+isin(1) M1\n= 2 2 2 2\nei1\n2\n−e−i1\n2\n 2isin1\n2\nzn −1 sin(n−1)+sin1 M1 Takes real part\nRe = 2 2\nz−1 2sin1\n \n2\nsin(n−1)\n1\nA1\n= 2 +\n2sin1 2\n2\nsinncos1−cosnsin1 1 M1 Uses compound angle identity\n= 2 2 +\n2sin1 2\n2\nsinncos1 1 1 M1 Divides through by denominator.\n= 2 − cosn+\n2sin1 2 2\n2\nsinnsin 1 1 sinnsin 1 1 A1 AG. sin=2sin1cos1 and\n− cosn+ = − cosn+ 2 2\n4sin2 1\n2\n 2 2 2(1−cos) 2 2 2sin2 1=1−cos.\n2\n7\n© UCLES 2023 Page 13 of 14\n\n8(c) 1 1 1 1 1 2 1 n−1 M1 A1 Forms sum of areas of rectangles given in the diagram.\n cosxdx + cos + cos + cos\nA0 if comparison with integral missing or unclear.\n0 n n n n n n n\n n 1 M1 A1 1\nsin sin Applies result from part (b) with = . AG.\n1 1 2 n−1 1  n n n  n\n=  1+cos +cos ++cos  = 1−cos + \nn n n n  2n  n 1−cos 1 \n n \n4\n\n8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison\n cosxdx cos + cos + cos\nwith integral missing or unclear.\n0 n n n n n n\n 1  1 A1\nsin1sin sin1sin\n1  n  1 1 1  n \n= 1−cos1+ + cos1− = cos1−1+ \n2n 1 n n 2n 1\n 1−cos   1−cos \n n   n \n3\n© UCLES 2023 Page 14 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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        "../answer-assets/9231_w23_ms_23-p14.png"
      ]
    },
    {
      "id": "9231-2023-on-31-q01",
      "question_id": "9231-2023-on-31-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) PCLM: mv=−mucos60+2mucos B1 First line must be seen.\n1 1\nv=− u+2ucos=v= u(4cos−1)\nAG\n2 2\n1\n\n1(b)\nKE of A =\n1\nm\n(\nv2\n+(usin60)2) B1\n2\nKE of A = 2  KE of B, so M1\n1 m ( v2 +(usin60)2) =2 1 2m(usin)2\n2 2\n2 M1 Use result of (a) and rearrange.\n  1 u(4cos−1)   + 3 u2 =4u2(sin)2 Obtain 3-term quadratic.\n2  4\n8cos2−2cos−3=0\n3 1 3 A1\ncos= , − but angle is acute, so cos=\n4 2 4\n4\n\n1(c) NEL: v=e(ucos60+ucos) M1\n1 3  A1\nv=u, u=e\n\nu+ u\n\n2 4 \n4\ne=\n5\n2\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-31-q02",
      "question_id": "9231-2023-on-31-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) dv B1\n2 =2g−0.2v2\ndt\ndv M1 Integrate to a ln term of the correct form.\nSeparate variables and attempt to integrate =dt\n0.1\n( 100−v2)\n1 10+v A1\nln\n \n=0.1t+c\n20 10−v\n1 M1 Use initial condition.\nt=0, v=5, c= ln3\n20\n10+v 10+v M1 Rearrange, removing ln.\n2t =ln , e2t =\n3(10−v) 3(10−v)\n30−10e−2t A1 AEF\nv=\n3+e−2t\n6\n\n2(b) v→10 B1FT FT from expression of correct form.\n1\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_31-p07.png"
      ]
    },
    {
      "id": "9231-2023-on-31-q03",
      "question_id": "9231-2023-on-31-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3 Let N be normal reaction at B and F the frictional force acting downwards B1\nTcos=F+W\n→Tsin=N B1\nMoments about B: Ta=Wsina+Wcosa M1A1 A moments equation with all relevant forces.\n1 M1\nF = N used\n2\n 1  M1 Combine to obtain equation in .\n(cos+sin)  cos− sin  =1 oe Equation in trigonometric functions only.\n 2 \n1 3 M1 Solve trigonometric equation.\ncossin=\n(sin)2\n2 2\nsin(cos−3sin)=0\n1 A1\ntan=\n3\n8\n© UCLES 2023 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_31-p08.png"
      ]
    },
    {
      "id": "9231-2023-on-31-q04",
      "question_id": "9231-2023-on-31-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) Energy equation: gain in KE = loss in EPE B1 At least one correct EPE.\nM1 3-term energy equation.\n1\nm80ag=\n1\n\n5mg( (2L−8a)2 −(4a)2) A1 All correct.\n2 2 8a\n4L2 −8aL−20a2 =0 M1 Simplify to quadratic in L.\nL=10a A1 Correct single answer.\n5\n\n4(b) At P: 2Tcos=macceleration M1\n5mg 15 M1\nHooke’s law: T = 12a= mg\n8a 2\nSolve, acceleration = 12g A1\n3\n© UCLES 2023 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_31-p09.png"
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    },
    {
      "id": "9231-2023-on-31-q05",
      "question_id": "9231-2023-on-31-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) gx2 M1\nUse equation of trajectory: y=xtan− sec2:\n2u2\n3 1\n(3a)2\n 1\na=3a −g 1+\n \n8 3 2u2  9\n5 5ga A1 At least one step of working.\na= , u2 =8ga\n8 u2 AG\n2\n© UCLES 2023 Page 10 of 13\n\n5(b) 2usin B1\nFor P, time of flight T =\ng\nFor P, range = Tucos B1\n1 1 B1\nFor Q, time of flight = T, so range = TVcos\n2 2\nEquate: Vcos=2ucos (1)\n1 M1\ngT2\n1 2 4Vsin\nFor Q vertically: 0=Vsin T − , so T =\n2 4 g\nEquate with result for P: A1\n4Vsin 2usin 1\n= so Vsin= usin (2)\ng g 2\nFrom (1) and (2): V2 =u2   4(cos)2 + 1 (sin)2  M1\n 4 \n 9 1 1  145 A1\nV2 =u2\n\n4 + \n\n= u2\n 10 4 10 40\nV = 29ag\n7\n© UCLES 2023 Page 11 of 13",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_31-p11.png"
      ]
    },
    {
      "id": "9231-2023-on-31-q06",
      "question_id": "9231-2023-on-31-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Angular speeds of P and Q are equal, M1\nv v\nso Q = P\nx 3a−x\nA1 Shown convincingly: angular speeds equal stated.\n2x ag\nv = AG\nQ 3a−x\n2\n\n6(b) m4ag B1\nFor P: T +mgcos60=\n3a−x\nmv2 B1\nFor Q: T −mgcos60= Q\nx\nm.4ag\nmv2 M1\nEliminate T: −mgcos60+ =mgcos60+ Q\n3a−x x\nm4ag mx4ag M1 Solve to find x.\n=1+\n3a−x (3a−x)2 Obtain 3-term quadratic equation.\n4a(3a−x)=(3a−x)2 +4ax, x2 +2ax−3a2 =0\n(x−a)(x+3a)=0, x=a A1\n5\n© UCLES 2023 Page 12 of 13\n\n6(c) Energy changes from initial position: B1 KEs correct.\nGain in KE of P:\n1\nm\n( 4ag−u2)\nB1FT GPEs correct.\n2\n1  u 2 \nLoss in KE of Q: 2 m   2   −v Q 2  \n \nLoss in GPE of P = mg(3a−x)(1−cos60) (=mga)\n 1 \nGain in GPE of Q = mgx(1−cos60)  = mga \n 2 \n1 2 m ( 4ag−u2) − 1 2 m       u 2    2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60) M1 Energy equation.\n \n5 A1 AEF\nSimplify: 4ag− u2 +ag =ag\n4\n16 4\nu2 = ag, u= 5ag\n5 5\n4\n© UCLES 2023 Page 13 of 13",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2023-on-32-q01",
      "question_id": "9231-2023-on-32-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 ↑Tcosθ=mg B1\nmv2 B1\n→ Tsinθ=\nasinθ\nEliminate T and substitute for 𝜃 M1\nA1\nv= 5ag\n6\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_32-p06.png"
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    },
    {
      "id": "9231-2023-on-32-q02",
      "question_id": "9231-2023-on-32-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 dv 150 450 M1 Allow sign errors.\n0.5v = −\ndx (x+1)2 (x+1)3\n300 450 M1A1 Correct powers, allow sign errors.\nIntegrate: 0.5v2 =− + + A\nx+1 (x+1)2\nx=0, v=20; A=50 M1 Use initial condition.\n100(x2 −4x+4) A1 AEF\nRearrange: v2 =\n(x+1)2\n100(x−2)2 10(x−2) A1\nv2 = so v=\n(x+1)2 (x+1)\n20−10x\nFrom initial condition, sign must be negative, v= Signs dealt with convincingly.\nx+1\n6\n© UCLES 2023 Page 6 of 15",
      "source_pages": [
        6
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-32-q03",
      "question_id": "9231-2023-on-32-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) Mass is proportional to area B1 At least two areas correct, at least one distance correct\nCentre of mass\nArea\nfrom AC\n1 1\nABC 2aa a\n2 3\n1 a 2 2a\nBDE 2  a−  \n2 k 3k\n2\n1 a\nADEC − 2  +a2 x\n2 k\n 1 a 2  1  2a a 2 M1A1 Moments equation, dimensionally correct, correct number\nx− 2  +a2= aa2 −a−    of terms.\n  2 k   3  3k k\na ( k2 +k−2 ) A1 a ( k3−3k+2 )\nx = Allow unsimplified single fraction .\n3k(k+1) 3k ( k2 −1 )\n4\n© UCLES 2023 Page 7 of 15\n\n3(b) x  5  B1 With their x from part (a).\ntan= =\n \na  18\nSo 18 ( k2 +k−2 ) =5 ( 3k2 +3k ) M1 Obtain a polynomial in k only, e.g. k3−13k+12=0,\nmay be implied.\nk2 +k−12=0\n(k+4)(k−3)=0, A1 CWO\nk=3 only\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_w23_ms_32-p08.png"
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    },
    {
      "id": "9231-2023-on-32-q04",
      "question_id": "9231-2023-on-32-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) After collision, A has velocity v towards wall on right and B has component B1\nA\nof velocity towards lower wall of u sin . Same distance and time, so\nv = u sin .\nA\nAlong line of centres: M1 Both, consistent signs, must be cos.\nPCLM: mv +mv =mucos\nA B\nNEL: v −v =eucos\nA B\n2sin=(1+e)cos M1 Eliminating v and v to find an equation in .\nA B\nCondone common factor of u.\n3 1\nNote: v = ucos and v = ucos.\nA 4 B 4\ntan= 3 A1\n4\n4\n© UCLES 2023 Page 8 of 15\n\n4(b) Final KE = 1 m ( v2 +(usin)2 +v2 ) B1 Both components of velocity of B needed.\n2 A B\nLoss = 1 2 mu2 − 1 2 m ( v A 2 +(usin)2 +v B 2 ) = 1 2 mu2    1− 2 9 5 − 2 9 5 − 2 1 5    M1 v co A , m v p B , o  n e s n u t b s s o ti f t u v t e e l d o , c f i t t y t h o e f i r B f i i n n a c l l u K d E ed w . ith both\n3 A1\nLoss = mu2\n25\nPercentage loss = 24%\n3\n© UCLES 2023 Page 9 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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        "../answer-assets/9231_w23_ms_32-p09.png"
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    },
    {
      "id": "9231-2023-on-32-q05",
      "question_id": "9231-2023-on-32-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) mv2 B1\nAt A: R −mgcos= A\nA a\nmv2 B1\nAt B: −R +mgcos= B\nB a\n1 1 M1 All terms present, allow sign errors, cos/sin mix.\nEnergy: mv2 − mv2 =mga(cos+cos)\n2 A 2 B\nA1 AEF\n1 M1 Any equivalent working, leading to a dimensionally\nEliminate velocities, use R = R and substitute angle values\nB 6 A correct equation in R and mg.\nB\n1 1\na( R −mgcos)− a(−R +mgcos)=mga(cos+cos)\n2 A 2 B\n3 4 7\n6R − mg+R − mg=2mg\nB 5 B 5 5\n3 A1\nR = mg\nB 5\n6\n\n5(b) 3 3 mv2 M1\nFrom first equation in (a), 6 mg− mg = A\n5 5 a\nv2 =3ag, so A1\nA\nCAO\nk=3\n2\n© UCLES 2023 Page 10 of 15",
      "source_pages": [
        10
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-32-q06",
      "question_id": "9231-2023-on-32-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) → x=ucos t\n1 B1 Both correct.\n y=usin t− gt2\n2\n2 M1 Eliminate.\nx 1  x \nEliminate t: y=usin − g \nucos 2 ucosα\ngx2 A1\ny=xtan− sec2 AG\n2u2\n3\n© UCLES 2023 Page 11 of 15\n\n6(b) 4 g322 16 B1\nh=32tan− sec2 [= 32tan− sec2]\n5 2402 5\ng322 8 B1\nh=32tan− sec2 [= 32tan− sec2]\n( )2 5\n2 40 2\n8 5 16 M1 Equate expressions for h and obtain a 3-term quadratic in\n32tan− sec2= (32tan− sec2)\ntan.\n5 4 5\n32t−\n8( 1+t2)\n=40t−4\n( 1+t2)\n5\n3t2 −10t+3=0\n1 A1 Both correct\nt=3,\n3\nh=80 M1 For using their value of t to work out one value of h.\n80\nh=\nA1 Both correct\n9\n6\n© UCLES 2023 Page 12 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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        "../answer-assets/9231_w23_ms_32-p12.png"
      ]
    },
    {
      "id": "9231-2023-on-32-q07",
      "question_id": "9231-2023-on-32-q07",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) 2 B1\n1 6 \nGain in KE = m 2ag and Gain in GPE = mg3asin\n \n2 5 \n1 B1\nkmg\nLoss in EPE = 2 ( (3a)2 −a2 )\na\n1 M1 KE, GPE and at least one EPE present, allow sign errors,\n2 kmg\nEnergy equation: 1 m   6 2ag   +mg3asin= 2 ( (3a)2 −a2 ) dimensionally correct.\n2 5  a\nA1 All correct.\n36 12 A1\nmg+ mg =4kmg\n25 5\n24\nk=\n25\n5\n© UCLES 2023 Page 13 of 15\n\n7(b) 3a 72 M1 Hooke’s law.\nIn lower position, tension T in spring = kmg =3kmg = mg\na 25\nPerpendicular to rod, T =(F +mg)cos\n72 A1\nSo, (F+mg)cos= mg,\n25\n19\nF = mg\n5\nAlternative method for question 7(b)\n3a 72 M1\nIn lower position, tension T in spring = kmg =3kmg = mg\nHooke’s law.\na 25\nEliminate T.\nTcos=Tsin\nTsin+Tcos =F+mg\n19 A1\nF = mg\n5\n2\n© UCLES 2023 Page 14 of 15\n\n7(c) Let tension in rod = T' M1\nParallel to rod, T=(F +mg)sin\n96 A1\nT= mg\n25\nAlternative method for question 7(c)\nTcos=Tsin M1 At least one equation seen with their T and/or F.\nTsin+Tcos =F+mg\n96 A1\nT= mg\n25\n2\n© UCLES 2023 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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        "../answer-assets/9231_w23_ms_32-p14.png",
        "../answer-assets/9231_w23_ms_32-p15.png"
      ]
    },
    {
      "id": "9231-2023-on-33-q01",
      "question_id": "9231-2023-on-33-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) PCLM: mv=−mucos60+2mucos B1 First line must be seen.\n1 1\nv=− u+2ucos=v= u(4cos−1)\nAG\n2 2\n1\n\n1(b)\nKE of A =\n1\nm\n(\nv2\n+(usin60)2) B1\n2\nKE of A = 2  KE of B, so M1\n1 m ( v2 +(usin60)2) =2 1 2m(usin)2\n2 2\n2 M1 Use result of (a) and rearrange.\n  1 u(4cos−1)   + 3 u2 =4u2(sin)2 Obtain 3-term quadratic.\n2  4\n8cos2−2cos−3=0\n3 1 3 A1\ncos= , − but angle is acute, so cos=\n4 2 4\n4\n\n1(c) NEL: v=e(ucos60+ucos) M1\n1 3  A1\nv=u, u=e\n\nu+ u\n\n2 4 \n4\ne=\n5\n2\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_33-p06.png"
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    },
    {
      "id": "9231-2023-on-33-q02",
      "question_id": "9231-2023-on-33-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) dv B1\n2 =2g−0.2v2\ndt\ndv M1 Integrate to a ln term of the correct form.\nSeparate variables and attempt to integrate =dt\n0.1\n( 100−v2)\n1 10+v A1\nln\n \n=0.1t+c\n20 10−v\n1 M1 Use initial condition.\nt=0, v=5, c= ln3\n20\n10+v 10+v M1 Rearrange, removing ln.\n2t =ln , e2t =\n3(10−v) 3(10−v)\n30−10e−2t A1 AEF\nv=\n3+e−2t\n6\n\n2(b) v→10 B1FT FT from expression of correct form.\n1\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_33-p07.png"
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    },
    {
      "id": "9231-2023-on-33-q03",
      "question_id": "9231-2023-on-33-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3 Let N be normal reaction at B and F the frictional force acting downwards B1\nTcos=F+W\n→Tsin=N B1\nMoments about B: Ta=Wsina+Wcosa M1A1 A moments equation with all relevant forces.\n1 M1\nF = N used\n2\n 1  M1 Combine to obtain equation in .\n(cos+sin)  cos− sin  =1 oe Equation in trigonometric functions only.\n 2 \n1 3 M1 Solve trigonometric equation.\ncossin=\n(sin)2\n2 2\nsin(cos−3sin)=0\n1 A1\ntan=\n3\n8\n© UCLES 2023 Page 8 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_33-p08.png"
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    },
    {
      "id": "9231-2023-on-33-q04",
      "question_id": "9231-2023-on-33-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) Energy equation: gain in KE = loss in EPE B1 At least one correct EPE.\nM1 3-term energy equation.\n1\nm80ag=\n1\n\n5mg( (2L−8a)2 −(4a)2) A1 All correct.\n2 2 8a\n4L2 −8aL−20a2 =0 M1 Simplify to quadratic in L.\nL=10a A1 Correct single answer.\n5\n\n4(b) At P: 2Tcos=macceleration M1\n5mg 15 M1\nHooke’s law: T = 12a= mg\n8a 2\nSolve, acceleration = 12g A1\n3\n© UCLES 2023 Page 9 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w23_ms_33-p09.png"
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    },
    {
      "id": "9231-2023-on-33-q05",
      "question_id": "9231-2023-on-33-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) gx2 M1\nUse equation of trajectory: y=xtan− sec2:\n2u2\n3 1\n(3a)2\n 1\na=3a −g 1+\n \n8 3 2u2  9\n5 5ga A1 At least one step of working.\na= , u2 =8ga\n8 u2 AG\n2\n© UCLES 2023 Page 10 of 13\n\n5(b) 2usin B1\nFor P, time of flight T =\ng\nFor P, range = Tucos B1\n1 1 B1\nFor Q, time of flight = T, so range = TVcos\n2 2\nEquate: Vcos=2ucos (1)\n1 M1\ngT2\n1 2 4Vsin\nFor Q vertically: 0=Vsin T − , so T =\n2 4 g\nEquate with result for P: A1\n4Vsin 2usin 1\n= so Vsin= usin (2)\ng g 2\nFrom (1) and (2): V2 =u2   4(cos)2 + 1 (sin)2  M1\n 4 \n 9 1 1  145 A1\nV2 =u2\n\n4 + \n\n= u2\n 10 4 10 40\nV = 29ag\n7\n© UCLES 2023 Page 11 of 13",
      "source_pages": [
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        11
      ],
      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w23_ms_33-p11.png"
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    },
    {
      "id": "9231-2023-on-33-q06",
      "question_id": "9231-2023-on-33-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Angular speeds of P and Q are equal, M1\nv v\nso Q = P\nx 3a−x\nA1 Shown convincingly: angular speeds equal stated.\n2x ag\nv = AG\nQ 3a−x\n2\n\n6(b) m4ag B1\nFor P: T +mgcos60=\n3a−x\nmv2 B1\nFor Q: T −mgcos60= Q\nx\nm.4ag\nmv2 M1\nEliminate T: −mgcos60+ =mgcos60+ Q\n3a−x x\nm4ag mx4ag M1 Solve to find x.\n=1+\n3a−x (3a−x)2 Obtain 3-term quadratic equation.\n4a(3a−x)=(3a−x)2 +4ax, x2 +2ax−3a2 =0\n(x−a)(x+3a)=0, x=a A1\n5\n© UCLES 2023 Page 12 of 13\n\n6(c) Energy changes from initial position: B1 KEs correct.\nGain in KE of P:\n1\nm\n( 4ag−u2)\nB1FT GPEs correct.\n2\n1  u 2 \nLoss in KE of Q: 2 m   2   −v Q 2  \n \nLoss in GPE of P = mg(3a−x)(1−cos60) (=mga)\n 1 \nGain in GPE of Q = mgx(1−cos60)  = mga \n 2 \n1 2 m ( 4ag−u2) − 1 2 m       u 2    2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60) M1 Energy equation.\n \n5 A1 AEF\nSimplify: 4ag− u2 +ag =ag\n4\n16 4\nu2 = ag, u= 5ag\n5 5\n4\n© UCLES 2023 Page 13 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2023-on-41-q01",
      "question_id": "9231-2023-on-41-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 H : =4.22 H : 4.22 B1\n0 1\n42.05 1 42.052  13  M1\nx = =4.205, s2 = 176.83−  = = 0.0010833 \n10 9  10  12000 \n4.205−4.22 M1 Their 4.205, their s2.\nt =\ns2\n10\nt=−1.44 A1 Condone sign.\nTabular value = 1.833: ‘1.44’ < 1.833, accept H M1 Compare their value with correct tabular value, signs\n0\nconsistent, allow ‘not significant’.\nThere is insufficient evidence to support Maya’s belief. A1 Correct conclusion in context, following correct work,\nlevel of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n6\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2023-on-41-q02",
      "question_id": "9231-2023-on-41-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 H : opinion is independent of street H : opinion is not independent of street B1\n0 1\nM1 At least 2 correct values or expressions.\n33 32.24 54 57.66 42 43.4 26 21.7 155\nA1 6 correct values or expressions.\n19 19.76 39 35.34 28 26.6 9 13.3 95\n52 93 70 35 250\nCalculate chi-squared values: M1 Correct to 3 decimal places.\n0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902 At least 2 correct values or expressions.\n3.02 A1 SC B1 3.02 following M1M0\nSC B2 3.02 following M0M0\nTabular value, 3 degrees of freedom 7.815: ‘3.02’ < 7.815, accept H M1 Allow ‘not significant’.\n0\nInsufficient evidence to suggest that opinion depends on street. A1 CWO. Correct conclusion in context, following correct\nwork, level of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n7\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-41-q03",
      "question_id": "9231-2023-on-41-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) H :  − =2 H :  − 2 B1 May use  , but must be consistent with direction of\n0 B E 1 B E d\ndifferences found.\nM1 Differences, at most 2 errors.\n8 8 15 –2 –3 20 –3 0 5 4\n1 522  2728  M1 Their values but must see 9 and 10 used correctly.\nd =52, d2 =816, s2 = 816−  = =60.62 \n9  10  45 \n52 M1\n−2\n10\nt =\ns2\n10\nt=1.29967, 1 .30 A1\nTabular value = 2.262. ‘1.30’ < 2.262, accept H M1 Allow ‘not significant’.\n0\nInsufficient evidence to suggest that cholesterol level has reduced by more A1 CWO. Correct conclusion in context, following correct\nthan 2. work, level of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n7\n\n3(b) Population differences are normally distributed B1\n1\n© UCLES 2023 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2023-on-41-q04",
      "question_id": "9231-2023-on-41-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 2 B1 AG\n1 mx2 1\nP(X ⩽ 2) = , so   = Reasoning required.\n3  2  3\n0\n4 1 1\nm\n \n= , m=\n2 3 6\n6 6 M1 Attempt to integrate and correct use of correct limits to\n k  2  k  1 2\n  x2 +c  dx= 3 so   − x +cx   =4c+ 3 k = 3 or k+12c=2 form linear equation in k and c.\n2 2\nk 1 k M1 Matching at x=2.\nAlso, 2m= +c or = +c or 3k+12c=4\n4 3 4\n1 A1\nSolve, k =1, c=\n12\n4\n© UCLES 2023 Page 9 of 13\n\n4(b) L B1\n1 1\nLQ:  xdx= , L= 3\n6 4\n0\nU M1 Or find and use CDF. May be in terms of m.\n1  k  3\nUQ: +  +cdx=\n3 x2  4\n2\nk k 5 M1 Attempt at integral and correct use of correct limits.\n− +Uc+ −2c=\nSimplify to quadratic equation, may be in terms of k and\nU 2 12\nc.\nU2 −U −12=0\nU =4 A1\nIQR = 4− 3 A1 FT FT their UQ – their LQ, exact values only.\n5\n© UCLES 2023 Page 10 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2023-on-41-q05",
      "question_id": "9231-2023-on-41-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) P(X = r) = p(1− p)r−1 = pqr−1 B1 Implied by G(t) = pqr−1 tr .\nG(t) =\npt(qt)r−1 M1 (Infinite summation) with common ratio (qt) and first\nterm or common factor pt indicated or GP identified.\n 1  pt A1 AG\nG(t) = pt = Convincingly obtained.\n1−qt 1−qt\n3\n\n5(b)\nG'(t) =\npqt(1−qt)−2\n+\np(1−qt)−1\n[=\np(1−qt)−2\n]\nM1 Attempt to differentiate as a product/quotient.\nG''(t) =\n2pq2t(1−qt)−3 +2pq(1−qt)−2 [=2qp(1−qt)−3\n]\nM1 Attempt to differentiate their G'(t).\n1 M1 Substitute t=1 into their G'(t) and G''(t) and use formula\nG'(1) =\np(1−q)−2\n= , G''(1) =\n2𝑞\np 𝑃2 for Var(X) (at some point).\n2𝑞 1 1 2 M1 (1−q) replaced by p in denominator throughout, and\nVar(X) = + −( )\n𝑃2 𝑝 𝑝\nsome cancellation seen (at some point).\nq A1 AG\np2 Convincingly obtained.\n5\n© UCLES 2023 Page 11 of 13\n\n5(c) 1t\n t \nB1 Implied by correct final answer.\nG (t) = 6 =\nX 1−5t   6−5t  \n6\n2 B1 2\n t   1t \nG (t) = (G (t))2 =\nZ X   Allow  6  .\n6−5t 1−5t\n \n6\n2\n© UCLES 2023 Page 12 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2023-on-41-q06",
      "question_id": "9231-2023-on-41-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 H : population medians are equal B1 Do not accept ‘differences between population medians <\n0\nH : population median after > population median before 0’ unless difference defined.\n1\nAllow H :m=22, H :m22.0\n0 1\nM1 Signed differences, allow at most 4 errors\n–0.8 -7 1.5 12 0.9 8 –3.4 –19 –2.6 –18\n0.1 1 4.5 21 –1.8 –14 3.7 20 –1.4 –11 M1 Attempt at ranks (ignore signs).\n0.3 3 –4.6 –22 0.2 2 5 23 1.9 15\n6.2 25 0.6 5 5.2 24 1 9 1.7 13\n–2.2 -16 0.7 6 1.3 10 –0.5 –4 2.3 17\n(W = 214) W = 111 A1 Cao identified, or used, as test statistic.\n+ −\n1 1 B1\nNormal: mean = n(n+1)= 2526=162.5\n4 4\n1 1 B1\nVariance = n(n+1)(2n+1)= 252651=1381.25\n24 24\n111.5−162.5 M1 Allow missing or incorrect continuity correction.\nz- value:\nTheir 111 must come from ranks.\n1381.25\n−1.37 A1 CAO\nTabular value is −1.645: ‘−1.37'−1.645, or 0.915<0.95 oe, accept H , M1 Consistent signs. Allow ‘not significant’.\n0\nInsufficient evidence to support the teacher’s claim A1 All correct. Correct conclusion in context, following\nInsufficient evidence to suggest that the distances thrown are further when it correct work, level of uncertainty in language.\nis hot\nA0 if hypotheses wrong way round or missing\n10\n© UCLES 2023 Page 13 of 13",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2023-on-42-q01",
      "question_id": "9231-2023-on-42-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 243.8 1219  384.9 3849  B1 Implied by correct sor pooled estimate\ns2 = = =4.132 , s2 = = =5.201\nx 59   295   y 74   740   243.8+384.9 =4.727\n60+75−2\n4.132 5.201 M1 Using their sample variances.\ns2 = + =0.1382 or s=0.3718\nPooled estimate M0.\n60 75\nA1\n6345 7614 M1 With a z-value\nCI: − 1.96'0.3718'\n60 75\nA1 With 1.96 (with their s)\n6345 7614 1 1\nPooled − 1.962.174 +\n60 75 60 75\n3.50, 4.96 or (3.5, 4.96) A1 4.230.729 is A0, condone 4.96, 3.5 etc.\n \n6\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2023-on-42-q02",
      "question_id": "9231-2023-on-42-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) B1 Each.\n0 5 13 17 21 16 9 5 4\nB1\n2.718 9.512 16.646 19.421 16.993 11.895 6.939 3.469 2.407\n2\n\n2(b) H : Po(3.5) is a good fit to the data B1\n0\nH : Po(3.5) is not a good fit to the data\n1\nCombine first 2 columns: 5, 12.23 M1 Both.\nAnd last 2 columns: 9, 5.876\nChi-squared values: M1 Allow if no or incorrect number of columns\n4.274 + 0.799 + 0.302 + 0.945 + 1.417 + 0.612 + 1.661 added.\nAt least two ‘correct’ values (3 sf) or\nexpressions seen from their grouping (or lack\nof).\n2.718 + 2.140 + … + 0.6757 + 1.054\n10.0 A1 AWRT 10.0\nIf M0 awarded then SC B1 for 10.0.\nTabular value: 10.64 ‘10.0’ < 10.64, accept H /not significant M1 Allow equivalent to 10.64 if columns not\n0\ncombined or only one pair combined\n(12.02 one pair combined, 13.36 none\ncombined).\nInsufficient evidence to suggest that Po(3.5) is not a good fit to the data A1 Correct conclusion in context, following correct\nwork, level of uncertainty in language.\nA0 if hypotheses the wrong way round or\nmissing.\n6\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2023-on-42-q03",
      "question_id": "9231-2023-on-42-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) 120 270 108 6 B1 2 correct probabilities.\nP(3R) = P(2R) = P(1R) = P(0R) =\n504 504 504 504\nG (t) = 1 ( 6+108t+270t2 +120t3) = 1 ( 1+18t+45t2 +20t3) M1 Cubic polynomial with their probabilities.\nX\n504 84\n1 3 15 5 A1 Correct.\n+ t+ t2 + t3\n84 14 28 21\n3\n\n3(b) G (t) = 1 ( 1+18t+45t2 +20t3)  1 ( 1+6t+5t2) M1 Attempt to multiply out the brackets.\nZ\n84 12\n1 ( 1+24t+158t2 +380t3+345t4 +100t5) M1 Obtain quintic polynomial (may not be\nsimplified).\n1008\n1 1 79 95 115 25 A1 Correct.\n+ t+ t2 + t3+ t4 + t5\n1008 42 504 252 336 252\n3\n© UCLES 2023 Page 8 of 13\n\n3(c) G' = 1 ( 24+316t+1140t2 +1380t3+500t4) M1 Differentiate twice. May not see derivatives in\nterms of t.\n1008\nG'' = 1 ( 316+2280t+4140t2 +2000t3)\n1008\n 3360 10 B1 Seen or implied, NFWW (e.g. sampling with\n\nE(Z)=\n replacement).\n 1008 3\n8736 10 10 2  26 10 100 M1 Use formula with their G(1) and G'(1).\nVar(Z) = + − = − −\n   \n1008 3  3   3 3 9 \n8 A1 Or 0.889\n9\n4\n© UCLES 2023 Page 9 of 13",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_42.pdf?download=true",
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    {
      "id": "9231-2023-on-42-q04",
      "question_id": "9231-2023-on-42-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) 6 B1 AG.\n1 1 1\nUQ = 4 so cdx= : c6−4= , c= Some indication of reasoning required.\n4 4 8\n4\n4 M1 Integrate and use correct limits to form equation\n 1 ( 4ax−bx3) dx= 3 , 1   2ax2 − b x4   = 3 , 32a−64b=96 in a and b.\n128 4 128 4  4\n0\nEquate at x=4: a−4b=1 M1\nSolve: a=5, b=1 A1\n4\n\n4(b) m M1* Attempt at integral and use of correct limits.\n\n1 ( 20x−x3)\ndx=\n1\nCould be in terms of a, b, c\n128 2\n0 Or attempt at relevant part of CDF and set\nOR\n1\nF(m)= .\n1  b  5 1\nF(x)=\n\n2ax2 − x4\n\n= x2 − x4 2\n128 4  64 512 Could be in terms of a, b, c.\nm4 −40m2 +256=0 M1 Simplify to quartic. Could be in terms of a, b, c.\nm=2 2 A1 Positive answer only.\n3\n© UCLES 2023 Page 10 of 13\n\n4(c) ( ) 1 4 3 7 6 1 1 M1 Their PDF multiplied by x with correct limits.\nE X = 20x2 −x2 dx+ x2dx\n128 8 Could be in terms of a, b, c.\n0 4\n4 6 M1 Correct use of correct limits in their integral.\n1  5 2 9 1  3\n8x2 − x2 + x2 Could be in terms of a, b, c.\n128 9  12  May be implied by correct final answer.\n0 4\n1  1  1 ( ) 4 1 A1 CAO\n\n28 − 210\n\n+ 6 6−8 = + 6 =1.67\n1.67 following (first) M0 SC B2\n128 9  12 9 2\n3\n© UCLES 2023 Page 11 of 13",
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    },
    {
      "id": "9231-2023-on-42-q05",
      "question_id": "9231-2023-on-42-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "5(a) H : population medians are equal or m = m B1 Do not accept ‘difference between population\n0 x y\nH : population median for X < population median for Y or m < m medians < 0’ without X or Y oe specified.\n1 x y\nM1 Rankings, allow at most 3 errors.\nX Y\n4.0 1 4.5 2\n4.6 3 4.9 6\n4.7 4 5.1 8\n4.8 5 5.3 10\n5.0 7 5.4 11\n5.2 9 5.7 13\n5.6 12 5.9 15\n5.8 14 6.3 16\n6.4 17\nSum: 55 98\nTest statistic = 55 A1\nTabular value for m=8, n=9 is 54 B1\n‘55’ > ‘54’, accept H /not significant M1 Ft their ‘55’ Must come from ranks.\n0\nFt their ‘54’, must come from table.\nInsufficient evidence to support manager’s claim. A1 Correct conclusion in context, following correct\nInsufficient evidence to suggest that the median time of machine X is less than the work, level of uncertainty in language.\nmedian time of machine Y. A0 if hypotheses the wrong way round or\nmissing.\n6\n© UCLES 2023 Page 12 of 13\n\n5(b) H : 𝜇 = 𝜇 H : 𝜇 < 𝜇 B1\n0 𝑥 𝑦 1 𝑥 𝑦\nx=39.7 x2 =199.33 y=49.5 y2 =275.47 B1\n1 39.72 \ns\nx\n2 =\n7  \n199.33−\n8  \n=0.33125\n1 49.52  B1\ns2\ny\n=\n8  \n275.47−\n9  \n=0.4025\n70.33125+80.4025 M1\nPooled variance s2 =\n8+9−2\n0.36925 A1\n39.7 −49.5 M1\nt = 8 9\ns 1+1\n8 9\nt=−1.82 A1\nTabular value = 1.753: 1.82 > 1.753 M1\nReject H , sufficient evidence that mean for machine X is less than mean for machine A1 CWO\n0\nY. Correct conclusion in context, following correct\nwork, level of uncertainty in language.\n9\n\n5(c) t-test is assuming a normal distribution, and with equal variances. This may not be B1 Not specific to data in question.\ntrue. So, no reason to expect results to be the same. Mention of normal distribution is not enough.\nOutliers affect part (b) but not part (a).\n1\n© UCLES 2023 Page 13 of 13",
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_42.pdf",
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    },
    {
      "id": "9231-2023-on-43-q01",
      "question_id": "9231-2023-on-43-q01",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 H : =4.22 H : 4.22 B1\n0 1\n42.05 1 42.052  13  M1\nx = =4.205, s2 = 176.83−  = = 0.0010833 \n10 9  10  12000 \n4.205−4.22 M1 Their 4.205, their s2.\nt =\ns2\n10\nt=−1.44 A1 Condone sign.\nTabular value = 1.833: ‘1.44’ < 1.833, accept H M1 Compare their value with correct tabular value, signs\n0\nconsistent, allow ‘not significant’.\nThere is insufficient evidence to support Maya’s belief. A1 Correct conclusion in context, following correct work,\nlevel of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n6\n© UCLES 2023 Page 6 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2023-on-43-q02",
      "question_id": "9231-2023-on-43-q02",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 H : opinion is independent of street H : opinion is not independent of street B1\n0 1\nM1 At least 2 correct values or expressions.\n33 32.24 54 57.66 42 43.4 26 21.7 155\nA1 6 correct values or expressions.\n19 19.76 39 35.34 28 26.6 9 13.3 95\n52 93 70 35 250\nCalculate chi-squared values: M1 Correct to 3 decimal places.\n0.0179 + 0.2323 + 0.0452 + 0.8521 + 0.0292 + 0.3790 + 0.0737 + 1.3902 At least 2 correct values or expressions.\n3.02 A1 SC B1 3.02 following M1M0\nSC B2 3.02 following M0M0\nTabular value, 3 degrees of freedom 7.815: ‘3.02’ < 7.815, accept H M1 Allow ‘not significant’.\n0\nInsufficient evidence to suggest that opinion depends on street. A1 CWO. Correct conclusion in context, following correct\nwork, level of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n7\n© UCLES 2023 Page 7 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2023-on-43-q03",
      "question_id": "9231-2023-on-43-q03",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) H :  − =2 H :  − 2 B1 May use  , but must be consistent with direction of\n0 B E 1 B E d\ndifferences found.\nM1 Differences, at most 2 errors.\n8 8 15 –2 –3 20 –3 0 5 4\n1 522  2728  M1 Their values but must see 9 and 10 used correctly.\nd =52, d2 =816, s2 = 816−  = =60.62 \n9  10  45 \n52 M1\n−2\n10\nt =\ns2\n10\nt=1.29967, 1 .30 A1\nTabular value = 2.262. ‘1.30’ < 2.262, accept H M1 Allow ‘not significant’.\n0\nInsufficient evidence to suggest that cholesterol level has reduced by more A1 CWO. Correct conclusion in context, following correct\nthan 2. work, level of uncertainty in language.\nA0 if hypotheses wrong way round or missing.\n7\n\n3(b) Population differences are normally distributed B1\n1\n© UCLES 2023 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
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    },
    {
      "id": "9231-2023-on-43-q04",
      "question_id": "9231-2023-on-43-q04",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 2 B1 AG\n1 mx2 1\nP(X ⩽ 2) = , so   = Reasoning required.\n3  2  3\n0\n4 1 1\nm\n \n= , m=\n2 3 6\n6 6 M1 Attempt to integrate and correct use of correct limits to\n k  2  k  1 2\n  x2 +c  dx= 3 so   − x +cx   =4c+ 3 k = 3 or k+12c=2 form linear equation in k and c.\n2 2\nk 1 k M1 Matching at x=2.\nAlso, 2m= +c or = +c or 3k+12c=4\n4 3 4\n1 A1\nSolve, k =1, c=\n12\n4\n© UCLES 2023 Page 9 of 13\n\n4(b) L B1\n1 1\nLQ:  xdx= , L= 3\n6 4\n0\nU M1 Or find and use CDF. May be in terms of m.\n1  k  3\nUQ: +  +cdx=\n3 x2  4\n2\nk k 5 M1 Attempt at integral and correct use of correct limits.\n− +Uc+ −2c=\nSimplify to quadratic equation, may be in terms of k and\nU 2 12\nc.\nU2 −U −12=0\nU =4 A1\nIQR = 4− 3 A1 FT FT their UQ – their LQ, exact values only.\n5\n© UCLES 2023 Page 10 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2023-on-43-q05",
      "question_id": "9231-2023-on-43-q05",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) P(X = r) = p(1− p)r−1 = pqr−1 B1 Implied by G(t) = pqr−1 tr .\nG(t) =\npt(qt)r−1 M1 (Infinite summation) with common ratio (qt) and first\nterm or common factor pt indicated or GP identified.\n 1  pt A1 AG\nG(t) = pt = Convincingly obtained.\n1−qt 1−qt\n3\n\n5(b)\nG'(t) =\npqt(1−qt)−2\n+\np(1−qt)−1\n[=\np(1−qt)−2\n]\nM1 Attempt to differentiate as a product/quotient.\nG''(t) =\n2pq2t(1−qt)−3 +2pq(1−qt)−2 [=2qp(1−qt)−3\n]\nM1 Attempt to differentiate their G'(t).\n1 M1 Substitute t=1 into their G'(t) and G''(t) and use formula\nG'(1) =\np(1−q)−2\n= , G''(1) =\n2𝑞\np 𝑃2 for Var(X) (at some point).\n2𝑞 1 1 2 M1 (1−q) replaced by p in denominator throughout, and\nVar(X) = + −( )\n𝑃2 𝑝 𝑝\nsome cancellation seen (at some point).\nq A1 AG\np2 Convincingly obtained.\n5\n© UCLES 2023 Page 11 of 13\n\n5(c) 1t\n t \nB1 Implied by correct final answer.\nG (t) = 6 =\nX 1−5t   6−5t  \n6\n2 B1 2\n t   1t \nG (t) = (G (t))2 =\nZ X   Allow  6  .\n6−5t 1−5t\n \n6\n2\n© UCLES 2023 Page 12 of 13",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2023-on-43-q06",
      "question_id": "9231-2023-on-43-q06",
      "subject": "9231",
      "year": 2023,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 H : population medians are equal B1 Do not accept ‘differences between population medians <\n0\nH : population median after > population median before 0’ unless difference defined.\n1\nAllow H :m=22, H :m22.0\n0 1\nM1 Signed differences, allow at most 4 errors\n–0.8 -7 1.5 12 0.9 8 –3.4 –19 –2.6 –18\n0.1 1 4.5 21 –1.8 –14 3.7 20 –1.4 –11 M1 Attempt at ranks (ignore signs).\n0.3 3 –4.6 –22 0.2 2 5 23 1.9 15\n6.2 25 0.6 5 5.2 24 1 9 1.7 13\n–2.2 -16 0.7 6 1.3 10 –0.5 –4 2.3 17\n(W = 214) W = 111 A1 Cao identified, or used, as test statistic.\n+ −\n1 1 B1\nNormal: mean = n(n+1)= 2526=162.5\n4 4\n1 1 B1\nVariance = n(n+1)(2n+1)= 252651=1381.25\n24 24\n111.5−162.5 M1 Allow missing or incorrect continuity correction.\nz- value:\nTheir 111 must come from ranks.\n1381.25\n−1.37 A1 CAO\nTabular value is −1.645: ‘−1.37'−1.645, or 0.915<0.95 oe, accept H , M1 Consistent signs. Allow ‘not significant’.\n0\nInsufficient evidence to support the teacher’s claim A1 All correct. Correct conclusion in context, following\nInsufficient evidence to suggest that the distances thrown are further when it correct work, level of uncertainty in language.\nis hot\nA0 if hypotheses wrong way round or missing\n10\n© UCLES 2023 Page 13 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9231_w23_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2023-oct-nov/9231_w23_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
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    },
    {
      "id": "9231-2024-mj-11-q01",
      "question_id": "9231-2024-mj-11-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 p B1\n2\n1\n\n1(b) 222  51 M1 Factorises.\n2 2\n5 A1\n4\n2\n\n1(c) 4z3 2pz2 5z250 B1 FT OE\nFT on their value of  (in terms of p).\n1\n\n1(d) 1  1  p M1 Uses 2 2 2 2 2()\n3 4\np 1 A1 CAO\n12\n2\n© Cambridge University Press & Assessment 2024 Page 6 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2024-mj-11-q02",
      "question_id": "9231-2024-mj-11-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 64 3821332 is divisible by 74. B1 Checks base case.\nAssume that 64k 38k 2 is divisible by 74 for some positive integer k. B1 States inductive hypothesis.\nThen 64k4 38k12(12951)64k (371)38k 2 M1 A1 Separates 64k 38k 2 or considers difference.\nIs divisible by 74 because 129564k 3738k is divisible by 74. A1\nHence, by induction, 64k 38k 2 is divisible by 74, is true for every positive integer A1\nn.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-11-q03",
      "question_id": "9231-2024-mj-11-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) N N N N B1 Expands.\nr(r1)3r43r3 7r2 4r\nr1 r1 r1 r1\n3N2N 12  7 N(N 1)(2N 1)2NN 1 M1 Substitutes formulae from MF19.\n4 6\n1 N  9N  N2 2N 1  14  2N2 3N 1  24N 24  A1\n12\n1 N  9N346N2 75N 38   1 N(N 1)(N 2)(9N 19) AG\n12 12\n3\n© Cambridge University Press & Assessment 2024 Page 7 of 16\n\n3(b) 3r4 4 1 M1 A1 Finds partial fractions.\n \nr(r1) r r1\nN 3r4 1 r1  1 4 1  1 4 1  1  4 1  M1 Writes at least three terms, including last.\n\n \n\n \n\n\n\n \n\n\n\n \n\n\nr(r1)4 42 1 2 43 2 3 4N1N N 1\nr1\n1 1 A1\n\n4 4N1(N 1)\n4\n\n3(c) 1 B1 CAO\n4\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2024-mj-11-q04",
      "question_id": "9231-2024-mj-11-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "4(a) [One way] stretch, rotation B1\nStretch followed by rotation B1 Correct order.\nStretch parallel to the x-axis, scale factor 14 B1\nRotation, 1π [anticlockwise] about the origin. B1\n3\n4\n© Cambridge University Press & Assessment 2024 Page 8 of 16\n\n4(b)  1 1 3 1  1 1 3 14 0 1 1 1 0  B1 Both inverses correct.\n 2 2   2 2  ,     \n1 3 1   1 3 1   0 1 140 14\n   \n2 2 2 2\n 1 0 1 1 3 B1 Correct order.\nM1  14  2 2 \n0 1 \n\n1 3 1 \n\n2 2\n2\n\n4(c)  7 1 3 B1\nM 2 \n 7 3 1 \n \n2\n 7 1 3x 7x1 3y B1 x X\n   7 3 2 1 2     y      7 3x 2  1 2 y    Transforms  y   to  Y  \n7 3x 1mxm  7x1 3mx  M1 Uses ymx and Y mX .\n2 2\n7 3 1m7m1 3m2  1 3m2 13m7 30 A1\n2 2 2 2\ny2 3x and y 7 3x A1\n3\n5\n\n4(d) 2814 ABC M1 Uses DEF  detM ABC .\n2cm2 A1 Allow with units missing.\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2024-mj-11-q05",
      "question_id": "9231-2024-mj-11-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) AB2j5k AC 5i5j BC 5i7j5k B1 Finds direction vectors of two lines in the plane.\ni j k 25 5 M1 A1 Finds normal to the plane ABC.\n   \n0 2 5  25 ~ 5\n   \n   \n5 5 0  10   2 \n5(2)5(2)2(4)85x5y2z8 M1 A1 Substitutes point. AEF for final answer.\n5\n\n5(b) 82ij3k5i5j2k 7 M1 A1 Correct formula for distance from point to plane.\n 0.953\n52 52 22 54\n2\n\n5(c) 2 3  5  B1\n\n     \nCD 1  3  4\n     \n     \n3  4  1\ni j k  18  M1 A1 Find common perpendicular.\n \n0 2 5  25\n \n \n5 4 1 10\n5  18  M1 A1 Uses formula for shortest distance.\n1     35\n5  25  1.08\n   \n1049 1049\n   \n0 10\n   \n5\n© Cambridge University Press & Assessment 2024 Page 10 of 16",
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-11-q06",
      "question_id": "9231-2024-mj-11-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 B1\nx2(xa2)2a5 52a M1\ny xa2\nx2 x2\nyxa2 A1\n3\n\n6(b) dy x22xa  x2 ax1  M1 Differentiates.\n\ndx x22\n  A1 dy\nx2 4x2a10 or dy 1 2a5  Forms quadratic equation or simplifies .\ndx\n dx x22 \n  Not from wrong working.\n1642a1208a0 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with\ncorrect conclusion.\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 16\n\n6(c) B1 Axes and asymptotes labelled.\ny  xa2\nB1 Branches correct. (Asymptotes may cross above, on\nor below the x-axis.)\nx\nx2\n0, 1 B1 May be seen on their diagram.\n2\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 16\n\n6(d)(i) B1 FT from their attempt in part 6(c).\nB1 Everything correct (approach to vertical asymptote\nand cusps correct).\nx\n2\n\n6(d)(ii) y B1\nya\nx\n1\n© Cambridge University Press & Assessment 2024 Page 13 of 16\n\n6(d)(iii) x2 ax1 x2 ax1 M1 Or direct use of x3\na or a\nx2 x2\nx2 12a0 or x2 2ax12a0\na a2 2a1 x 2a1,a a2 2a1 x 2a1\na5 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_11-p14.png"
      ]
    },
    {
      "id": "9231-2024-mj-11-q07",
      "question_id": "9231-2024-mj-11-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) B1 Correct shape.\nB1 Section 1ππ correct.\n2\n0\n  B1 May be seen on their diagram.\nπtan1π, 0 (1.99, 0)\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 16\n\n7(b) π 0 M1 Uses correct formula.\nA 1 πtan1πd or A1 utan1udu\n2 2 \n0 π or A 1 utan1udu\n2\n0\n11u2 tan1u π 1 π u2 du M1 A1 Integrates by parts.\n22  0 4 0 1u2\nπ u2 π1u2 1 π 1 M1 Rearranges integral into form which can be\n du du  1 du integrated.\n0 1u2 0 1u2 0 1u2\nπ A1\nutan1u\n \n0\n. A 11u2 tan1u π 1utan1u π A1\n22 \n0\n4 \n0\nA 1π2tan1π1  πtan1π   1  π2 1  tan1π1π2.65 A1\n4 4 4 4\n7\n© Cambridge University Press & Assessment 2024 Page 15 of 16\n\n7(c) y πtan1πsin B1 Uses yrsin\nπ\n 1π21\ntan1π\nM1 A1 Finds derivative.\ndy\n πtan1πcos sin\nd 2 πtan1π\nπ A1 dy\n0,π2πtan1πcot tan1π0 Puts 0 and forms equation. AG.\n1π2 d\nπ1.2 B1 Shows sign change.\n2π1.2tan1π1.2cot1.2 tan1π1.20.151\n1π1.22\nπ1.3\nand 2π1.3tan1π1.3cot1.3 tan1π1.30.395\n1π1.32\n5\n© Cambridge University Press & Assessment 2024 Page 16 of 16",
      "source_pages": [
        14,
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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      ]
    },
    {
      "id": "9231-2024-mj-12-q01",
      "question_id": "9231-2024-mj-12-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 p B1\n2\n1\n\n1(b) 222  51 M1 Factorises.\n2 2\n5 A1\n4\n2\n\n1(c) 4z32pz2 5z250 B1 FT OE\nFT on their value of  (in terms of p).\n1\n\n1(d) 1  1  p M1 Uses 2 2 2 2 2()\n3 4\np 1 A1 CAO\n12\n2\n© Cambridge University Press & Assessment 2024 Page 6 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2024-mj-12-q02",
      "question_id": "9231-2024-mj-12-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 64 3821332 is divisible by 74. B1 Checks base case.\nAssume that 64k 38k 2 is divisible by 74 for some positive integer k. B1 States inductive hypothesis.\nThen 64k4 38k12(12951)64k (371)38k 2 M1 A1 Separates 64k 38k 2 or considers difference.\nIs divisible by 74 because 129564k 3738k is divisible by 74. A1\nHence, by induction, 64k 38k 2 is divisible by 74, is true for every positive integer A1\nn.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-12-q03",
      "question_id": "9231-2024-mj-12-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) N N N N B1 Expands.\nr(r1)3r43r3 7r2 4r\nr1 r1 r1 r1\n3N2N 12  7N(N 1)(2N 1)2NN 1 M1 Substitutes formulae from MF19.\n4 6\n1 N  9N  N2 2N 1  14  2N2 3N 1  24N 24  A1\n12\n1 N  9N3 46N2 75N 38   1 N(N 1)(N 2)(9N 19) AG\n12 12\n3\n© Cambridge University Press & Assessment 2024 Page 7 of 16\n\n3(b) 3r4 4 1 M1 A1 Finds partial fractions.\n \nr(r1) r r1\nN 3r4 1 r1  1 4 1  1 4 1  1  4 1  M1 Writes at least three terms, including last.\n\n \n\n \n\n\n\n \n\n\n\n \n\n\nr(r1)4 42 1 2 43 2 3 4N1N N 1\nr1\n1 1 A1\n\n4 4N1(N 1)\n4\n\n3(c) 1 B1 CAO\n4\n1\nQuestion Answer Marks Guidance",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2024-mj-12-q04",
      "question_id": "9231-2024-mj-12-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "4(a) [One way] stretch, rotation B1\nStretch followed by rotation B1 Correct order.\nStretch parallel to the x-axis, scale factor 14 B1\nRotation, 1π [anticlockwise] about the origin. B1\n3\n4\n© Cambridge University Press & Assessment 2024 Page 8 of 16\n\n4(b)  1 1 3 1  1 1 3 14 0 1 1 1 0  B1 Both inverses correct.\n 2 2   2 2  ,     \n1 3 1   1 3 1   0 1 140 14\n   \n2 2 2 2\n 1 0 1 1 3 B1 Correct order.\nM1  14  2 2 \n0 1 \n\n1 3 1 \n\n2 2\n2\n\n4(c)  7 1 3 B1\nM 2 \n 7 3 1 \n \n2\n 7 1 3x 7x1 3y B1 x X\n   7 3 2 1 2      y     7 3x 2  1 2 y    Transforms  y   to  Y  \n7 3x1mxm  7x1 3mx  M1 Uses ymx and Y mX .\n2 2\n7 31m7m1 3m2  1 3m2 13m7 30 A1\n2 2 2 2\ny2 3x and y7 3x A1\n3\n5\n\n4(d) 2814 ABC M1 Uses DEF  detM ABC .\n2cm2 A1 Allow with units missing.\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 16",
      "source_pages": [
        8,
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      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_12-p08.png",
        "../answer-assets/9231_s24_ms_12-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-12-q05",
      "question_id": "9231-2024-mj-12-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) AB2j5k AC5i5j BC5i7j5k B1 Finds direction vectors of two lines in the plane.\ni j k 25 5 M1 A1 Finds normal to the plane ABC.\n   \n0 2 5  25 ~ 5\n   \n   \n5 5 0  10   2 \n5(2)5(2)2(4)85x5y2z8 M1 A1 Substitutes point. AEF for final answer.\n5\n\n5(b) 82ij3k5i5j2k 7 M1 A1 Correct formula for distance from point to plane.\n 0.953\n52 52 22 54\n2\n\n5(c) 2 3  5  B1\n\n     \nCD 1  3  4\n     \n     \n3  4  1\ni j k  18  M1 A1 Find common perpendicular.\n \n0 2 5  25\n \n \n5 4 1 10\n5  18  M1 A1 Uses formula for shortest distance.\n1     35\n5  25  1.08\n   \n1049 1049\n   \n 0  10\n5\n© Cambridge University Press & Assessment 2024 Page 10 of 16",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_12-p10.png"
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    },
    {
      "id": "9231-2024-mj-12-q06",
      "question_id": "9231-2024-mj-12-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 B1\nx2(xa2)2a5 52a M1\ny xa2\nx2 x2\nyxa2 A1\n3\n\n6(b)\ndy\nx22xa  x2 ax1  M1 Differentiates.\n\ndx x22\n  A1 dy\nx2 4x2a10 or dy 1 2a5  Forms quadratic equation or simplifies .\ndx\n dx x22 \n  Not from wrong working.\n1642a1208a0 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with\ncorrect conclusion.\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 16\n\n6(c) B1 Axes and asymptotes labelled.\ny  xa2\nB1 Branches correct. (Asymptotes may cross above, on\nor below the x-axis.)\nx\nx2\n0, 1 B1 May be seen on their diagram.\n2\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 16\n\n6(d)(i) B1 FT from their attempt in part 6(c).\nB1 Everything correct (approach to vertical asymptote\nand cusps correct).\nx\n2\n\n6(d)(ii) y B1\nya\nx\n1\n© Cambridge University Press & Assessment 2024 Page 13 of 16\n\n6(d)(iii) x2 ax1 x2 ax1 M1 Or direct use of x3\na or a\nx2 x2\nx2 12a0 or x2 2ax12a0\na a2 2a1x 2a1,a a2 2a1x 2a1\na5 A1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12,
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_12-p11.png",
        "../answer-assets/9231_s24_ms_12-p12.png",
        "../answer-assets/9231_s24_ms_12-p13.png",
        "../answer-assets/9231_s24_ms_12-p14.png"
      ]
    },
    {
      "id": "9231-2024-mj-12-q07",
      "question_id": "9231-2024-mj-12-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) B1 Correct shape.\nB1 Section 1ππ correct.\n2\n0\n  B1 May be seen on their diagram.\nπtan1π, 0 (1.99, 0)\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 16\n\n7(b) π 0 M1 Uses correct formula.\nA 1 πtan1πd or A1 utan1udu\n2 2 \n0 π or A 1 utan1udu\n2\n0\n11u2 tan1u π 1 π u2 du M1 A1 Integrates by parts.\n22  0 4 0 1u2\nπ u2 π1u2 1 π 1 M1 Rearranges integral into form which can be\n du du  1 du\nintegrated.\n0 1u2 0 1u2 0 1u2\nπ A1\nutan1u\n \n0\n. A 11u2 tan1u π 1utan1u π A1\n22 \n0\n4 \n0\nA 1π2tan1π1  πtan1π   1  π2 1  tan1π1π2.65 A1\n4 4 4 4\n7\n© Cambridge University Press & Assessment 2024 Page 15 of 16\n\n7(c) y πtan1πsin B1 Uses yrsin\nπ\n 1π21\ntan1π\nM1 A1 Finds derivative.\ndy\n πtan1πcos sin\nd 2 πtan1π\nπ A1 dy\n0,π2πtan1πcot tan1π0 Puts 0 and forms equation. AG.\n1π2 d\nπ1.2 B1 Shows sign change.\n2π1.2tan1π1.2cot1.2 tan1π1.20.151\n1π1.22\nπ1.3\nand 2π1.3tan1π1.3cot1.3 tan1π1.30.395\n1π1.32\n5\n© Cambridge University Press & Assessment 2024 Page 16 of 16",
      "source_pages": [
        14,
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_12-p14.png",
        "../answer-assets/9231_s24_ms_12-p15.png",
        "../answer-assets/9231_s24_ms_12-p16.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q01",
      "question_id": "9231-2024-mj-13-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 5 2 6 2 M1 A1 Evaluates determinant, forms quadratic\nk  k(5k6)(6k2)5k2 2 expression.\n3 k 1 k\n(Allow 1 slip in the calculation of the\ndeterminant for M1 only)\nNo real value of k  Non-singular A1 Convincing conclusion using the\ndiscriminant or determinant.\n3\n\n1(b) 3k11 or  1  1 M1 Uses AA1 Iordet(A1)(detA)1to\n2 5k2 2\nfind equation in k. Could also use a minor\ndeterminant e.g. k1300\nk 0 A1 Only k = 0, A0 for any additional solutions.\n2\n© Cambridge University Press & Assessment 2024 Page 6 of 16",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_13-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q02",
      "question_id": "9231-2024-mj-13-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) yx2 1 B1 Uses correct substitution. May use a\ndifferent letter.\n(y1) 3 2 2(y1)3y11 2 10y11 2 y132y11 M1 Substitutes and obtains an equation not\ninvolving radicals.\ny1y22 2y12 y1 y2 4y4  4y2 4y1\nOR\n(x3 3x)2 (2x2 1)2 x6 2x4 5x2 10\ny13 2y12\n5(y1)10\ny3y2 4y50 A1 OE\n3\n\n2(b)  2 1 2   2 1 2   2 1 2 12(4) M1 Uses\n2 2 2 ()2\n2().\non their answer for part 2(a).\n7 A1 CAO\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 16\n\n2(c)  2 1 3   2 1 3   2 1 3 7453 M1 Uses their equation from part 2(a).\n4 A1 CAO\nAlternative method for question 2(c)\nUses M1\n 2 1 3 \n 2 1 3 3  2 1  2 1  2 1  3  2 1  2 1  2 1 \n13 31435 Uses their equation from part 2(a) in a valid\nformula.\n4 A1 CAO\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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        "../answer-assets/9231_s24_ms_13-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q03",
      "question_id": "9231-2024-mj-13-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "3(a) [One-way] stretch, shear B1 Both types.\nStretch followed by shear B1 Correct order.\nStretch parallel to the x-axis, scale factor 7 B1\nShear, x-axis fixed, with (0,1) mapped to (2,1). B1\n4\n© Cambridge University Press & Assessment 2024 Page 8 of 16\n\n3(b) 7 2 B1\nM \n0 1\n7 2x 7x2y B1 FT x X\n    Transforms   to  .\n0 1y  y  y Y \nmxm7x2mx M1 Uses ymx and Y mX .\n2m2 6m0 A1\ny0 and y3x A1 SCB1 if M0 and both straight lines correct\n5\n\n3(c) Area of PQR  |7|Area of DEF M1\nArea of DEF5cm2 A1 SCB1 for an answer of 245 (35  7)\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 16",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_13-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q04",
      "question_id": "9231-2024-mj-13-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "4(a) 12  1(1)(2)(3) so H is true. B1 Checks base case.\n6 1\nk B1 States inductive hypothesis.\nAssume that r2  1k(k1)(2k1).\n6\nr1\nk1 M1 Considers sum to k1.\nr2  1k(k1)(2k1)k12\n6\nr1\n1(k1)  2k2 k6k6   1(k1)  2k2 7k6   1(k1)k22k3 A1\n6 6 6\nSo H is true. By induction, H is true for all positive integers n. A1 States conclusion.\nk1 n\n5\n\n4(b) 35 15 55 35 75 552n15 (2n1)5 M1 Applies method of differences to LHS,\nwrites complete terms for at least three\nvalues of r including the first and last.\n2n15\n1\nA1 SCB1 for insufficient complete terms\nshown.\n160S  40n(n1)(2n1)2n M1 A1 Sums RHS and applies standard formulae.\nn 3\n160S 2n15  40n(n1)(2n1)(2n1) M1 Makes S n the subject and factorises.\nn 3\n160S (2n1)  2n14  40n(n1)1 \nn 3\n2n1 16n4 32n3  32n2 16n  16n2n1 3n3 6n2 2n1  A1 AG\n3 3 3\nS  1 n(n1)(2n1)(3n2 3n1)\nn 30\n6\n© Cambridge University Press & Assessment 2024 Page 10 of 16\n\n4(c) n5S  1 1 12 1 3 3  1  M1 Divides by n5.\nn 30 n n n n2\n1 A1\n5\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_13-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q05",
      "question_id": "9231-2024-mj-13-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 3  1  4 B1 OE.\n      Finds direction of one line to another.\n4  4  0\n     \n     \n0 1 1\n     \ni j k  5  M1 A1 Find common perpendicular.\n \n0 1 2  2\n \n \n1 2 1 1\n4  5  M1 A1 Uses formula for shortest distance.\n1     21\n0  2  (3.83)\n   \n30 30\n   \n 1  1\n5\n\n5(b) 5(1)2(4)1(1)2 5x2yz2 M1 Uses point in the plane.\nA1 FT Follow through their normal in part 5(a)\n2\n© Cambridge University Press & Assessment 2024 Page 11 of 16\n\n5(c) ijk B1 Finding direction between the two given\npoints.\ni j k 1 M1 Find common perpendicular. Allow use of\n  their normal from part (b).\n1 1 1  2\n \n \n1 4 3 3\nx2y3z6 A1 OE\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_13-p11.png",
        "../answer-assets/9231_s24_ms_13-p12.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q06",
      "question_id": "9231-2024-mj-13-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 30 has no real roots. B1 OE\ny0 B1 Horizontal asymptote.\n2\n\n6(b) dy (x2 3)(x1)(2x) M1 dy\n Finds .\ndx  x2 3 2 dx\nx2 2x30 M1 Sets equal to 0 and simplifies to quadratic\nequation.\n 3, 1 , 1, 1 A1 A1\n6 2\n4\n© Cambridge University Press & Assessment 2024 Page 12 of 16\n\n6(c) y B1 Axes and approach to asymptote.\nB1 Correct smooth shape and position.\nx\n(1, 0),  0, 1 B1 States coordinates of intersections with\n3\naxes, may be shown on their graph.\n3\n\n6(d) y B1 Sections for x0 and x1 correct\nB1 1x0 correct.\nx\n  B1 States coordinates of intersections with\n(1, 0), 0,  1\n3 axes, may be shown on their graph.\n  B1 May be shown on their graph.\n1,  1\n2\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 16",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_13-p12.png",
        "../answer-assets/9231_s24_ms_13-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-13-q07",
      "question_id": "9231-2024-mj-13-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) B1 One loop correct with initial line.\nB1 Second loop correct with correct form at the\npole.\n0\n1π B1 May be seen on their diagram.\n2\n3\n\n7(b) r5 2(rsin)(rcos)2 M1 Use of sin22sincos, and xrcos\nor yrsin.\nr5 2yx2 A1\n x2  y25 2 2x2y A1 AEF\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 16\n\n7(c) 1 π sin(2)cosd M1 Applies 1r2 d.\n2\n2\n0\n1 M1 Attempt to integrate in a valid way.\nsin2cos d\nMay apply sin(2)2sincos .\n2\n1 A1 Correct answer (there are alternative forms).\n sincos2d cos3  [c]\n3\nπ A1 CAO\n1cos3  2\n 3  0 3\n4\n\n7(d) dr  1sin2cos1 22cos2cossin2sin *M1 A1 Differentiates with respect to.\nd 2\n2cos2cossin2sin6cos34cos dM1 Correct use of relevant identities to express\nin terms of a single trig function.\n6cos34cos0cos2 2 dM1 A1 Sets derivative equal to 0 and solves to find\n3\n1 2\nsin2 , cos2 ,\n3 3\none of\n1\ntan2 = or =0.6155\n2\nA1\n4 4 3\nr   0.877\n3 3 9\n© Cambridge University Press & Assessment 2024 Page 15 of 16\n\n7(d) Alternative method for question 7(d)\n2r dr 2  2sin2coscos3  *M1 A1 Differentiates [RHS] with respect to.\nd\n2  2  1cos2  coscos3  6cos34cos dM1 Correct use of relevant identities in terms of\na single trig function\n6cos34cos0cos2 2 dM1 A1 Sets derivative equal to 0 and solves to find\n3\n1 2\nsin2 , cos2 ,\n3 3\none of\n1\ntan2 = or =0.6155\n2\nA1\n4 4 3\nr   0.877\n3 3 9\n6\n© Cambridge University Press & Assessment 2024 Page 16 of 16",
      "source_pages": [
        14,
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_13-p14.png",
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        "../answer-assets/9231_s24_ms_13-p16.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q01",
      "question_id": "9231-2024-mj-21-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 z3 216cos5π i216sin5π B1 Finds modulus and argument of 108 3108i\n6 6\nz 6cos5π i6sin5π M1 A1FT Finds one root. FT on their modulus. For M1\n1 18 18\nmust have divided their argument of\n108 3108i by 3.\nz 6cos17π i6sin17π,z 6cos29π i6sin29π M1 A1 Finds other two roots.\n2 18 18 3 18 18\n5\n© Cambridge University Press & Assessment 2024 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q02",
      "question_id": "9231-2024-mj-21-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 e1x2 e  ex2 e  1x2   B1 Changes e1x2 or e1x so that the formula given\nin the list of formulae (MF19) can be applied.\ne1x e  1x 1x2   B1\n2\ne  1x2   e  1x 1x2   M1 Sums power series.\n2\ne1x2 e1x 2eex 3ex2  A1\n2\nAlternative method for question 2\nf'(x)2xe1x2 e1x B1 Finds first derivative.\nf''(x)4x2e1x2 2e1x2 e1x B1 Finds second derivative.\nf(0)2e f'(0)e f''(0)3e M1 Evaluates their derivatives at zero.\ne1x2 e1x 2eex 3ex2  A1\n2\n4\n© Cambridge University Press & Assessment 2024 Page 7 of 14",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q03",
      "question_id": "9231-2024-mj-21-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) dy t dx 1 B1\ncos1t , \ndt 1t2 dt 1t2\ndy dy dt M1 A1 Uses chain rule, AG.\n  t 1t2 cos1t\ndx dt dx\n3\n\n3(b) d dy 1t2 tcos1t tcos1t M1 A1 Applies product rule.\n 1  2\ndtdx 1t2 1t2 1t2\nd2y M1 A1 Uses chain rule.\n2 1t2 tcos1t\ndx2\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q04",
      "question_id": "9231-2024-mj-21-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) ln3 B1 Separates into correct structure.\nI  sechn2 xsech2 xdx\nn\n0\nln3 ln3 ln3 *M1 Uses integration by parts correctly.\nI  sechn2xsech2 xdxsechn2xtanhx (n2) sechn2xtanh2xdx\nn  \n0 0 0\n3n24(n2)I I  DM1 A1 Uses 1sech2 xtanh2 x.\n5 5 n2 n\n1n2I 3n24(n2)I (n1)I 3n24(n2)I A1 AG\nn 5 5 n2 n 5 5 n2\n5\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n4(b) I  4 B1\n2 5\n3I \n324\n2I I \n236\n0.629\nM1 A1 Applies reduction formula with n4.\n4 5 5 2 4 375\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p08.png",
        "../answer-assets/9231_s24_ms_21-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q05",
      "question_id": "9231-2024-mj-21-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a)   1 2xx2dx 1  21  12  1  22  22  1  2n  n2 M1 A1 Forms the sum of the areas of the rectangles,\n  0   n n n n n n n n n may be in summation form. M1 is for correct\nnumber of rectangles.\nn n M1 A1 Applies formulae from MF19. Summations must\n 2 r 1 r2  1 n(n1) 1 n(n1)(2n1) have (correct) limits for A1. Substituted in\nn2 n3 n2 6n3\nr1 r1 correctly once for M1.\n1 1 11 12 1  1 12 1   2 1  1 A1 AG. This A1 requires the previous A1.\nn 6 n n n 3 6n 3 2n 6n2\n5\n\n5(b)  1 2xx2dx1  2112 1  2222 1  2n1n12 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n n n n rectangles. M1 is for seeing correct height of last\n0\nrectangle.\nn1 n1 M1 Applies formulae from MF19, substitutes correct\n2 r 1 r2  1 (n1)(n) 1 (n1)(n)(2n1) limit. Substituted in correctly once for M1.\nn2 n3 n2 6n3\nr1 r1\n112 1   2 1  1 A1  112 1  1 or fully expanded or fully\nn 3 6n 3 2n 6n2 n 3 6n n\nfactorised.\n4\n© Cambridge University Press & Assessment 2024 Page 9 of 14\n\n5(c) U L   1 12 1    112 1  ( 1) M1 Expresses their U L in terms of 1.\nn n n 3 6n n 3 6n n n n n\nFor, M1 they must taking the limit of an\nexpression that tends to a constant.\n(Must be fully expanded if they do not take the\nlimit at any stage.)\n2 2 A1 AG, CWO.\n1 0 as n or  0\nn 3 3 Their L n must be correct.\n2\nQuestion Answer Marks Guidance",
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      ],
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p09.png",
        "../answer-assets/9231_s24_ms_21-p10.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q06",
      "question_id": "9231-2024-mj-21-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) coshxsinhx1 2   1  ex ex  1  ex ex1 2  M1 Substitutes sinh and cosh in terms of\n2 2 exponentials. \n ex1 2 e 1 2 x A1 AG\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 14\n\n6(b) m2 m30m1  1i 11 M1 Auxiliary equation.\n2 2\nye1 2 x Acos 11xBsin 11x  A1 Complimentary function.\n2 2 Allow ‘y’ missing.\nyke 1 2 x  y' 1ke 1 2 x  y'' 1ke 1 2 x B1 Particular integral and its derivatives.\n2 4\n1ke 1 2 x 1ke 1 2 x 3ke 1 2 x 5e 1 2 x  1k1k3k 5 M1 Substitutes and equates coefficients.\n4 2 4 2 PI must be correct form (aebx).\nk  4 A1\n3\nye1 2 x Acos 11xBsin 11x   4e 1 2 x A1 FT Must have ‘y’. FT on their CF.\n2 2 3\ndy e1 2 x  11 Asin 11x 11Bcos 11x  1e1 2 x Acos 11xBsin 11x   2e 1 2 x B1\ndx 2 2 2 2 2 2 2 3\n1 A 4 4  11B1 A 2  A1,B 1 M1 A1 Substitutes initial conditions and forms\n3 3 2 2 3 3 11 simultaneous equations. For M1, CF must be\ncorrect form\nye1 2 x 1cos 11x 1 sin 11x   4e 1 2 x A1 Must have ‘y’.\n3 2 11 2 3\n10\n© Cambridge University Press & Assessment 2024 Page 11 of 14",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p10.png",
        "../answer-assets/9231_s24_ms_21-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q07",
      "question_id": "9231-2024-mj-21-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a)  x  1 M1 A1 Applies substitution.\n dx 1 du u C 1x2 C\n 1x2 2 u\n2\n\n7(b) dy y B1 Divides through by x.\n xsinh1x\ndx x\ne\nx1dx\nelnx x1\nM1 A1 Finds integrating factor.\nd  x1y  sinh1x M1 Correct form on LHS. d Iy for their\ndx dx\nintegrating factor I.\n x M1 A1 Integrates RHS. RHS must be of the form\nx1y xsinh1x dx\n 1x2\ncsinh1x.\nx1yxsinh1x 1x2 C A1 sinh1xln  x x2 1 \n  *M1 Substitutes initial conditions.\n1ln 1 2  2C\nyx2sinh1xx 1x2 xln  1 2  x  1 2  DM1 A1 Divides through by their integrating factor.\nAccept\nyx2sinh1xx 1x2 xsinh11x  1 2  .\n10\n© Cambridge University Press & Assessment 2024 Page 12 of 14",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p12.png"
      ]
    },
    {
      "id": "9231-2024-mj-21-q08",
      "question_id": "9231-2024-mj-21-q08",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "8(a) 6 a 0 M1 Finds determinant or solves equations.\n2 1 0 8a24\n1 5 4\na3 A1\n2\n\n8(b) If a3 then system has unique solution (so consistent). B1 FT\n6x 3y  3, M1 Eliminates variable using all 3 equations. Or\nstates there are two distinct equations and three\nIf a3 then 2x y  1, 11x4z7\nunknowns. Or gives accurate geometrical\nx 5y 4z 2,\ndescription with a3.\nSo the system has infinitely many solutions (so consistent). A1\nAlternative method for question 8(b)\nSetsy0. B1\n1,0,3 is a solution so system is consistent for all values of a. M1 A1 Finds a solution with y0 and states\n2 8\nconclusion.\n3\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(c) Eigenvalues of A are 6, 1 and 4. B1 Lower diagonal matrix or characteristic\nequation.\ni j k 14 M1 A1 Uses vector product (or equations) to find\n  corresponding eigenvectors.\n6: 2 7 0  4\n \n \n1 5 2 17\ni j k  0   0  A1\n   \n1: 2 0 0  10  1\n   \n   \n1 5 5  10   1 \ni j k  0  0 A1\n   \n4: 2 5 0  0  0\n   \n   \n1 5 0 15 1\n14 0 0 11664 0 0  M1 A1 Or correctly matched permutations of columns.\n    Column of zeros in P gets M0.\nThus P 4 1 0 and D 0 100 0\n   \n   \n17 1 1  0 0 6400\n7\n\n8(d) 6143 92 14240 B1 Characteristic equation.\n14A24IA3 9A2 M1 Substitutes for A and makes 14A24I the\nsubject.\n14A24I2   A3 9A22 A4A9I2 M1 A1 Squares and factorises. CWO.\n4\n© Cambridge University Press & Assessment 2024 Page 14 of 14",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_21-p13.png",
        "../answer-assets/9231_s24_ms_21-p14.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q01",
      "question_id": "9231-2024-mj-22-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 z3 216cos5π i216sin5π B1 Finds modulus and argument of 108 3108i\n6 6\nz 6cos5π i6sin5π M1 A1FT Finds one root. FT on their modulus. For M1\n1 18 18\nmust have divided their argument of\n108 3108i by 3.\nz 6cos17π i6sin17π,z 6cos29π i6sin29π M1 A1 Finds other two roots.\n2 18 18 3 18 18\n5\n© Cambridge University Press & Assessment 2024 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q02",
      "question_id": "9231-2024-mj-22-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 e1x2 e  ex2 e  1x2   B1 Changes e1x2 or e1x so that the formula given\nin the list of formulae (MF19) can be applied.\ne1x e  1x 1x2   B1\n2\ne  1x2   e  1x 1x2   M1 Sums power series.\n2\ne1x2 e1x 2eex 3ex2  A1\n2\nAlternative method for question 2\nf'(x)2xe1x2 e1x B1 Finds first derivative.\nf''(x)4x2e1x2 2e1x2 e1x B1 Finds second derivative.\nf(0)2e f'(0)e f''(0)3e M1 Evaluates their derivatives at zero.\ne1x2 e1x 2eex 3ex2  A1\n2\n4\n© Cambridge University Press & Assessment 2024 Page 7 of 14",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q03",
      "question_id": "9231-2024-mj-22-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) dy t dx 1 B1\ncos1t , \ndt 1t2 dt 1t2\ndy dy dt M1 A1 Uses chain rule, AG.\n  t 1t2 cos1t\ndx dt dx\n3\n\n3(b) d dy 1t2 tcos1t tcos1t M1 A1 Applies product rule.\n 1  2\ndtdx 1t2 1t2 1t2\nd2y M1 A1 Uses chain rule.\n2 1t2 tcos1t\ndx2\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q04",
      "question_id": "9231-2024-mj-22-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) ln3 B1 Separates into correct structure.\nI  sechn2 xsech2 xdx\nn\n0\nln3 ln3 ln3 *M1 Uses integration by parts correctly.\nI  sechn2xsech2 xdxsechn2xtanhx (n2) sechn2xtanh2xdx\nn  \n0 0 0\n3n24(n2)I I  DM1 A1 Uses 1sech2 xtanh2 x.\n5 5 n2 n\n1n2I 3n24(n2)I (n1)I 3n24(n2)I A1 AG\nn 5 5 n2 n 5 5 n2\n5\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n4(b) I  4 B1\n2 5\n3I \n324\n2I I \n236\n0.629\nM1 A1 Applies reduction formula with n4.\n4 5 5 2 4 375\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p08.png",
        "../answer-assets/9231_s24_ms_22-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q05",
      "question_id": "9231-2024-mj-22-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a)   1 2xx2dx 1  21  12  1  22  22  1  2n  n2 M1 A1 Forms the sum of the areas of the rectangles,\n  0   n n n n n n n n n may be in summation form. M1 is for correct\nnumber of rectangles.\nn n M1 A1 Applies formulae from MF19. Summations must\n 2 r 1 r2  1 n(n1) 1 n(n1)(2n1) have (correct) limits for A1. Substituted in\nn2 n3 n2 6n3\nr1 r1 correctly once for M1.\n1 1 11 12 1  1 12 1   2 1  1 A1 AG. This A1 requires the previous A1.\nn 6 n n n 3 6n 3 2n 6n2\n5\n\n5(b)  1 2xx2dx1  2112 1  2222 1  2n1n12 M1 A1 Forms the sum of the areas of appropriate\nn n n n n n n n n rectangles. M1 is for seeing correct height of last\n0\nrectangle.\nn1 n1 M1 Applies formulae from MF19, substitutes correct\n2 r 1 r2  1 (n1)(n) 1 (n1)(n)(2n1) limit. Substituted in correctly once for M1.\nn2 n3 n2 6n3\nr1 r1\n112 1   2 1  1 A1  112 1  1 or fully expanded or fully\nn 3 6n 3 2n 6n2 n 3 6n n\nfactorised.\n4\n© Cambridge University Press & Assessment 2024 Page 9 of 14\n\n5(c) U L   1 12 1    112 1  ( 1) M1 Expresses their U L in terms of 1.\nn n n 3 6n n 3 6n n n n n\nFor, M1 they must taking the limit of an\nexpression that tends to a constant.\n(Must be fully expanded if they do not take the\nlimit at any stage.)\n2 2 A1 AG, CWO.\n1 0 as n or  0\nn 3 3 Their L n must be correct.\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q06",
      "question_id": "9231-2024-mj-22-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) coshxsinhx1 2   1  ex ex  1  ex ex1 2  M1 Substitutes sinh and cosh in terms of\n2 2 exponentials. \n ex1 2 e 1 2 x A1 AG\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 14\n\n6(b) m2 m30m1  1i 11 M1 Auxiliary equation.\n2 2\nye1 2 x Acos 11xBsin 11x  A1 Complimentary function.\n2 2 Allow ‘y’ missing.\nyke 1 2 x  y' 1ke 1 2 x  y'' 1ke 1 2 x B1 Particular integral and its derivatives.\n2 4\n1ke 1 2 x 1ke 1 2 x 3ke 1 2 x 5e 1 2 x  1k1k3k 5 M1 Substitutes and equates coefficients.\n4 2 4 2 PI must be correct form (aebx).\nk  4 A1\n3\nye1 2 x Acos 11xBsin 11x   4e 1 2 x A1 FT Must have ‘y’. FT on their CF.\n2 2 3\ndy e1 2 x  11 Asin 11x 11Bcos 11x  1e1 2 x Acos 11xBsin 11x   2e 1 2 x B1\ndx 2 2 2 2 2 2 2 3\n1 A 4 4  11B1 A 2  A1,B 1 M1 A1 Substitutes initial conditions and forms\n3 3 2 2 3 3 11 simultaneous equations. For M1, CF must be\ncorrect form\nye1 2 x 1cos 11x 1 sin 11x   4e 1 2 x A1 Must have ‘y’.\n3 2 11 2 3\n10\n© Cambridge University Press & Assessment 2024 Page 11 of 14",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p10.png",
        "../answer-assets/9231_s24_ms_22-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q07",
      "question_id": "9231-2024-mj-22-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a)  x  1 M1 A1 Applies substitution.\n dx 1 du u C 1x2 C\n 1x2 2 u\n2\n\n7(b) dy y B1 Divides through by x.\n xsinh1x\ndx x\ne\nx1dx\nelnx x1\nM1 A1 Finds integrating factor.\nd  x1y  sinh1x M1 Correct form on LHS. d Iy for their\ndx dx\nintegrating factor I.\n x M1 A1 Integrates RHS. RHS must be of the form\nx1y xsinh1x dx\n 1x2\ncsinh1x.\nx1yxsinh1x 1x2 C A1 sinh1xln  x x2 1 \n  *M1 Substitutes initial conditions.\n1ln 1 2  2C\nyx2sinh1xx 1x2 xln  1 2  x  1 2  DM1 A1 Divides through by their integrating factor.\nAccept\nyx2sinh1xx 1x2 xsinh11x  1 2  .\n10\n© Cambridge University Press & Assessment 2024 Page 12 of 14",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p12.png"
      ]
    },
    {
      "id": "9231-2024-mj-22-q08",
      "question_id": "9231-2024-mj-22-q08",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "8(a) 6 a 0 M1 Finds determinant or solves equations.\n2 1 0 8a24\n1 5 4\na3 A1\n2\n\n8(b) If a3 then system has unique solution (so consistent). B1 FT\n6x 3y  3, M1 Eliminates variable using all 3 equations. Or\nstates there are two distinct equations and three\nIf a3 then 2x y  1, 11x4z7\nunknowns. Or gives accurate geometrical\nx 5y 4z 2,\ndescription with a3.\nSo the system has infinitely many solutions (so consistent). A1\nAlternative method for question 8(b)\nSetsy0. B1\n1,0,3 is a solution so system is consistent for all values of a. M1 A1 Finds a solution with y0 and states\n2 8\nconclusion.\n3\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(c) Eigenvalues of A are 6, 1 and 4. B1 Lower diagonal matrix or characteristic\nequation.\ni j k 14 M1 A1 Uses vector product (or equations) to find\n  corresponding eigenvectors.\n6: 2 7 0  4\n \n \n1 5 2 17\ni j k  0   0  A1\n   \n1: 2 0 0  10  1\n   \n   \n1 5 5  10   1 \ni j k  0  0 A1\n   \n4: 2 5 0  0  0\n   \n   \n1 5 0 15 1\n14 0 0 11664 0 0  M1 A1 Or correctly matched permutations of columns.\n    Column of zeros in P gets M0.\nThus P 4 1 0 and D 0 100 0\n   \n   \n17 1 1  0 0 6400\n7\n\n8(d) 6143 92 14240 B1 Characteristic equation.\n14A24IA3 9A2 M1 Substitutes for A and makes 14A24I the\nsubject.\n14A24I2   A3 9A22 A4A9I2 M1 A1 Squares and factorises. CWO.\n4\n© Cambridge University Press & Assessment 2024 Page 14 of 14",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_22-p13.png",
        "../answer-assets/9231_s24_ms_22-p14.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q01",
      "question_id": "9231-2024-mj-23-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 x2 4x13(x2)2 B1 Completes the square.\n7 7 M1 A1 Applies formula.\n2 1  x22\n dxsin1    x2 7 2  x2 7 2\n\n2\n3(x2)2   3 \n2\nsin1\n \ncos1\n \n  3    3 \n2 2\nsin1 1 3  sin10 1π M1 A1 Uses limits. Answer must be in radians.\n2 3\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q02",
      "question_id": "9231-2024-mj-23-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) xsinht y cosht B1 Differentiates x and y with respect to t.\nx2  y2 sinh2tcosh2t cosh2t M1 A1 Applies cosh2tsinh2tcosh2t. May use\nexponential form or\n dy 2 1 2  cosh2t 1 2\n1   dx1  sinhtdt.\n\n\ndx \n \nsinh2t\n\n3 A1 AG\ns5 cosh2tdt\n0\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 15\n\n2(b) d \ncosh2t\n\nsinh2t sech2t\nsinh2t(cosh2t)1\n2\nB1 Finds first derivative.\ndt\nd2\n cosh2t  2 cosh2t tanh22t cosh2t 2 cosh2t sinh22tcosh2t3 2\nB1 Finds second derivative\ndt2\n3 3 M1 A1 Applies Maclaurin’s series.\ns51t2dt t1t35\n0  3  0\n 84 0.672 A1 CWO.\n125\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p06.png",
        "../answer-assets/9231_s24_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q03",
      "question_id": "9231-2024-mj-23-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) d  8y3 24y2y' B1 Differentiates y3 correctly.\ndx\nd B1 Differentiates xy correctly.\n2xy2xy' y\ndx\n3(2)2 2(2)y'2(1)24(1)2y'01020y'0 B1 Substitutes (2,1) into\n y'1 3x2 2xy'2y24y2y'0, AG. Must see\n2\nintermediary step between substitution and the\nfinal answer.\n3\n© Cambridge University Press & Assessment 2024 Page 7 of 15\n\n3(b) 6x2xy''2y' B1 Differentiates 3x2 2xy'.\n2y'24y2y''24y'(2y)y'0 B1 B1 Differentiates 2y24y2y'.\nB1 for 2y'24y2y''.\nB1 for\n48yy'2\n. (Allow alternative using\nquotient rule.)\n6(2)2(2)y''412412 y''48(1)12\n020y''260\nM1 Substitutes (2,1) into their expression (must\n2 2\nhave derivatives y' and y'').\ny''13 A1 OE\n10\n5\n© Cambridge University Press & Assessment 2024 Page 8 of 15",
      "source_pages": [
        7,
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p07.png",
        "../answer-assets/9231_s24_ms_23-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q04",
      "question_id": "9231-2024-mj-23-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) N1 1 B1 Forms sum of the areas of the rectangles. Need\n22 (N 1)2 \nto see this sum written explicitly, may be in\nr2\nr2 summation form, cannot be embedded in\nworking.\nN N 1 1 M1 A1 Evaluates integral, correct limits for M1.\n x2dxx1  \n \n2 2 2 N\nN N 1 M1 Compares with integral correctly (using >).\nr2  x2dx1\n2 N2\nr1\nN 3 1 1 A1 AG\nr2   \n2 N N2\nr1\n5\n\n4(b) 32 42 N2 M1* Forms the sum of the areas of appropriate\nrectangles. Allow addition of one extra rectangle\nor shift one left or right if limits on the integral\nare adjusted accordingly.\nN N 1 1 DM1 Compares with integral, correct limits and\n x2dxx1  \n  correct inequality symbol (<).\n2 2 2 N\nN 7 1 A1\nr2  \n4 N\nr1\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 15\n\n4(c)  3 B1 Needs to clear (using > or words) is that value is\nr2  lower bound.\n2\nr1\n 7 B1 FT Needs to clear (using < or words) is that value is\nr2  upper bound. Must have scored final A mark in\n4\nr1 part (b). FT on fully correct alternative to part\n4(b).\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p09.png",
        "../answer-assets/9231_s24_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q05",
      "question_id": "9231-2024-mj-23-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) m2 10m250m5 M1 Auxiliary equation.\nxe5tAtB A1 Complementary function. Allow ‘x’ missing.\nx psintqcostx' pcostqsintx''psintqcost M1 A1 Particular integral and its derivatives. For M1,\nboth derivatives must be present and of the\ncorrect form (only allow sign errors).\np10q25p338 q10p25q0 M1 Substitutes and equates coefficients.\np12 q5 A1\nxe5tAtB12sint5cost A1 Must have ‘x’.\n7\n© Cambridge University Press & Assessment 2024 Page 10 of 15\n\n5(b) x12sint5cost 13 12sint 5 cost  M1 Considering PI only.\n13 13\n12sint5cost 13sin(t0.395...) A1 A1 A1 for R and A1 for .Condone 22.6.\nWithhold final A1 mark if working with\nincorrect PI. CWO.\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p10.png",
        "../answer-assets/9231_s24_ms_23-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q06",
      "question_id": "9231-2024-mj-23-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "6(a) z4 z4n 1  M1 Uses sum of geometric series.\nz4 z4(z4)z4(z4)n1 \nz4 1\nz4 z4n 1  z2 z4n2 z2 A1 Divides numerator and denominator by z2, AG.\n \nz4 1 z2 z2 z2\n2\n\n6(b) z2 z2 2isin2 B1 Simplifies denominator.\nz4n2 z2 cos(4n2)isin(4n2)cos2isin2 M1 A1 Applies de Moivre’s theorem to numerator.\n (Used at least once for M1.) Accept\nz2 z2 2isin2\nz2 cos22isincossin2.\nn cos(4n2)cos2 M1 Equates imaginary parts.\nsin4r\n2sin2\nr1\nn cos2cos(4n2) A1 AG. Must see evidence of eliminating i from the\nsin4r\ndenominator.\n2sin2\nr1\n5\n© Cambridge University Press & Assessment 2024 Page 11 of 15",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q07",
      "question_id": "9231-2024-mj-23-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) d x 9 x M1 Applies product rule and\ndx  2\nx2 9\n2\ncosh1\n3   d  cosh1 x  1\ndx 3 x2 9\nx2 1 9 A1 OE\n x2 9\n2 x2 9 2 2 x2 9\nx2 9 1 A1 Puts over common denominator, AG.\n x2 9  x2 9\n2 x2 9 2\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n7(b) dy y B1 Divides through by x.\n  x x2 9\ndx x\ne\nx1d\nelnx  x1\nM1 A1 Finds integrating factor.\nd  x1y   x2 9 M1 Correct form on LHS, d Iyfor their\ndx dx\nintegrating factor I.\nx 9 x M1 A1 Integrates RHS. For M1, RHS must be of the\nx1y x2 9 cosh1 C\n2 2 3 form c x2 9, where c is a non-zero constant.\n1 M1 Substitutes initial conditions.\nC\n3\nx 9 x 1 M1 A1 Divides through by their integrating factor.\nyx\n\nx2 9 cosh1 \n Accept\n2 2 3 3\nx 9 x 1  1\nyx x2 9 ln   x2 9  .\n2 2 3 3  3\n9\nQuestion Answer Marks Guidance",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p12.png",
        "../answer-assets/9231_s24_ms_23-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-23-q08",
      "question_id": "9231-2024-mj-23-q08",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) 12630d or 32630d M1 Substitutes point into x2y3z.\n1 2\nx2y3z13 A1\nx2y3z9 A1\n3\n© Cambridge University Press & Assessment 2024 Page 13 of 15\n\n8(b) x 1  2  x 1  1  M1 Forms vector equation of line.\n           \ny  6  12 or y  6  6\n           \n\nz\n \n0\n \n0\n \nz\n \n0\n \n0\n\n           \ny6 A1 AG. CWO.\n x1 ,z0  1x 1 y1,z0\n6 2 12\n2\n\n8(c) 1 2 3 M1 Uses det(AI) (attempt an expansion).\n1 2 3\n1 1 \n2 12\n(1) 21 2  3 3 1 12 0 A1 Expands determinant, expansion of brackets\n4 2 12 2 shown, AG. CWO. (Must have ‘0’ here or\n3 32 71 23 1 3 30\nabove.)\n4 4 4 2\n3 32  70\n4\n0, 7,1 B1\n2 2\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n8(d) i j k 1   3  M1 A1 Uses vector product (or equations) to find\n4\n    corresponding eigenvectors.\n0: 1 2 3  3\n2\n\n\n18\n\n   \n1 2 1 1 2 0   1 1 2 1   11 \ni j k 21 6 i j k 3 6 A1 A1\n 2     2   \n 7\n2\n: 5\n2\n2 3 2\n2\n1~\n\n6\n\n1\n2\n: 3\n2\n2 3 3\n2\n~\n\n6\n\n       \n1 3 2 3  7 4  1 1 5 2 3  7 4   7 \n 3 6 6 M1 A1 Or correctly matched permutations of columns.\n  Do not accept\nThus P 18 6 6\n \nn\n  0 0 0 \n 11 1 7 \n \nD 0 7 0 .\n0 0 0   2 \n \n\n0 0\n1\n \nand D  0 7n 0  2\n 2  M1 is for matching their eigenvectors correctly\n\n0 0\n1n with their eigenvalues.\n 2  M0 if a column of zeros appears in P.\n6\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
        13,
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_23-p13.png",
        "../answer-assets/9231_s24_ms_23-p14.png",
        "../answer-assets/9231_s24_ms_23-p15.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q01",
      "question_id": "9231-2024-mj-31-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 Along line of centres, PCLM: 5mv mv mucos M1 Must include correct masses.\nB A\nNEL: v\nB\nv\nA\n 1\n2\nucos M1 Signs consistent with PCLM equation.\nu u A1\nv  cos, v  cos\nB 4 A 4\nPerpendicular to line of centres: speed of A is usin B1\n1\nm\n\n\n\n\nu\ncos\n 2 usin2\n\n\n\n1\n5m \nu\ncos\n 2 M1 E\nm\nq\na\nu\nss\na\ne\nte\ns .\nf inal kinetic energies, 3 terms, correct\n2   4   2 4 \n \ncos2  4 , cos 2 , tan 1 A1\n5 5 2\n6\n© Cambridge University Press & Assessment 2024 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q02",
      "question_id": "9231-2024-mj-31-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) In equilibrium: 2Tcosmg M1\n 5a   B1\nHooke’s law: T    2a  \n2a  2  4\nEquate and use cos 3: 10mg A1\n5 3\n3\n\n2(b) 1 20  2 B1 80\nEPE loss = 2 4a   3 a4a  27 mga\n1 8a 80 M1 A1 All 3 terms required, dimensionally correct, their\nEnergy equation: mv2 mg ' mga'\n.\n2 3 27\n4 ga A1 Any equivalent form.\nv\n3 3\n4\n© Cambridge University Press & Assessment 2024 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q03",
      "question_id": "9231-2024-mj-31-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "3 usin605g, and ucos60 , or usin6015g , and ucos60 B1\nusin605g M1* Accept equivalent for t15.\nIf  is direction of velocity at t 5, tan\nucos60\nusin605g usin6015g M1dep Multiply two expressions involving relevant\nFor perpendicular directions,   1\nvelocities and equate to – 1.\nucos60 ucos60\nSimplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000 M1 Simplify to quadratic in u (may see g ).\n4 4\nu5 3g A1 OE. Accept 50 3 or 86.6.\n5\n© Cambridge University Press & Assessment 2024 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q04",
      "question_id": "9231-2024-mj-31-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) Frictional force F and normal reaction R at point of contact of ring with plane. M1 Only allow cos/sin errors for T and W\nResolve parallel to plane: F TcosWsin components, sign errors.\nAccept equations for vertical and horizontal (both\nneeded).\nMoments about O: FaTasin B1\nCombine and substitute for : M1 Expression for T in terms of W .\nT  1W A1 CAO\n3\nAlternative solution for question 4(a)\nMoments about point where ring touches plane: TasinTacosWasin M1 A1 Only allow cos/sin errors, sign errors.\nMust be dimensionally correct.\nRearrange and substitute for : M1 Expression for T in terms of W .\nT  1W A1 CAO\n3\n4\n\n4(b) Resolve perpendicular to plane: RTsinWcos M1 Only allow cos/sin errors for T and W\ncomponents, sign errors.\nUse F R and combine to reach an equation in  only. M1 From part (a), F TcosWsin or\nF Tsin.\n 1 A1\n7\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q05",
      "question_id": "9231-2024-mj-31-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) For A: F T ma2 M1 Only allow sign errors.\nA\nF mg  1mg B1 Accept with g replaced by 10.\nA 5\nCombine: T  1mg 4 mg M1 To reach an equation in T and mg only.\n5 25\nAccept with g replaced by 10.\nT  1 mg A1 CAO\n25\n4\n\n5(b) For B: F T km2a2 M1 Only allow sign errors.\nB\nF kmg 1kmg and combine to find k M1 To reach an equation in k only.\nB 5\nk  1 A1\n3\n3\n© Cambridge University Press & Assessment 2024 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p10.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q06",
      "question_id": "9231-2024-mj-31-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) dv 1 dv 1 *M1 Separate variables and attempt to integrate both\n2  2v12 et so  etdt\ndt 10 2v12 20 sides.\nWhere p and q are constants.\np\nqet  A\n2v1\n1 1 A1 AEF\n  et  A\n22v1 20\n 3  DM1 Substituting the boundary condition and obtain a\nt0, v3, A\n\n\n value.\n 20\n1 5et *M1 A1 Find v in terms of t . AEF.\nv \n2 3et 1\n5\n\n6(b) Integrate: x ptqln(ret s) B *M1\n1 5 A1 AEF\nx t ln(3et 1) B\n2 3\n5 DM1 Substituting the boundary condition and obtain a\nt 0, x1, B1 ln2\nvalue.\n3\n1 5 (3et 1) A1 AEF\nx1 t ln\n2 3 2\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-31-q07",
      "question_id": "9231-2024-mj-31-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) Energy: 1mu2  1mv2 mga1cos *M1 m must be present, dimensionally correct, no\n2 2 missing terms.\nAllow sin instead of cos.\nAllow sign errors.\nmv2 B1 No reaction when P loses contact.\nN2L: mgcos\na\nEliminate v2 DM1\nu2 2ag A1 AG\ncos\n3ag\n4\n\n7(b) Vertical component of velocity of P when it leaves the sphere: vsin *B1 Must not come from u.\n 55ag \n \n \n216\n \nV2 vsin2 2ga1cos DM1 Use of 'v2 u2 2as.\nAllow sin for cos.\nAllow sign errors.\n847ag A1 AEF\nV \n216\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 13\n\n7(c) 1 847ag 55ag  M1\nt    \n \ng 216 216\n \n1  a a A1\n847  55 1.48\n6 6g g\n2\n© Cambridge University Press & Assessment 2024 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_31-p12.png",
        "../answer-assets/9231_s24_ms_31-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q01",
      "question_id": "9231-2024-mj-32-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 Along line of centres, PCLM: 5mv mv mucos M1 Must include correct masses.\nB A\nNEL: v\nB\nv\nA\n 1\n2\nucos M1 Signs consistent with PCLM equation.\nu u A1\nv  cos, v  cos\nB 4 A 4\nPerpendicular to line of centres: speed of A is usin B1\n1\nm\n\n\n\n\nu\ncos\n 2 usin2\n\n\n\n1\n5m \nu\ncos\n 2 M1 E\nm\nq\na\nu\nss\na\ne\nte\ns .\nf inal kinetic energies, 3 terms, correct\n2   4   2 4 \n \ncos2  4 , cos 2 , tan 1 A1\n5 5 2\n6\n© Cambridge University Press & Assessment 2024 Page 6 of 13",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q02",
      "question_id": "9231-2024-mj-32-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) In equilibrium: 2Tcosmg M1\n 5a   B1\nHooke’s law: T    2a  \n2a  2  4\nEquate and use cos 3: 10mg A1\n5 3\n3\n\n2(b) 1 20  2 B1 80\nEPE loss = 2 4a   3 a4a  27 mga\n1 8a 80 M1 A1 All 3 terms required, dimensionally correct, their\nEnergy equation: mv2 mg ' mga'\n.\n2 3 27\n4 ga A1 Any equivalent form.\nv\n3 3\n4\n© Cambridge University Press & Assessment 2024 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q03",
      "question_id": "9231-2024-mj-32-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "3 usin605g, and ucos60 , or usin6015g , and ucos60 B1\nusin605g M1* Accept equivalent for t15.\nIf  is direction of velocity at t 5, tan\nucos60\nusin605g usin6015g M1dep Multiply two expressions involving relevant\nFor perpendicular directions,   1\nvelocities and equate to – 1.\nucos60 ucos60\nSimplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000 M1 Simplify to quadratic in u (may see g ).\n4 4\nu5 3g A1 OE. Accept 50 3 or 86.6.\n5\n© Cambridge University Press & Assessment 2024 Page 8 of 13",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q04",
      "question_id": "9231-2024-mj-32-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) Frictional force F and normal reaction R at point of contact of ring with plane. M1 Only allow cos/sin errors for T and W\nResolve parallel to plane: F TcosWsin components, sign errors.\nAccept equations for vertical and horizontal (both\nneeded).\nMoments about O: FaTasin B1\nCombine and substitute for : M1 Expression for T in terms of W .\nT  1W A1 CAO\n3\nAlternative solution for question 4(a)\nMoments about point where ring touches plane: TasinTacosWasin M1 A1 Only allow cos/sin errors, sign errors.\nMust be dimensionally correct.\nRearrange and substitute for : M1 Expression for T in terms of W .\nT  1W A1 CAO\n3\n4\n\n4(b) Resolve perpendicular to plane: RTsinWcos M1 Only allow cos/sin errors for T and W\ncomponents, sign errors.\nUse F R and combine to reach an equation in  only. M1 From part (a), F TcosWsin or\nF Tsin.\n 1 A1\n7\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q05",
      "question_id": "9231-2024-mj-32-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) For A: F T ma2 M1 Only allow sign errors.\nA\nF mg  1mg B1 Accept with g replaced by 10.\nA 5\nCombine: T  1mg 4 mg M1 To reach an equation in T and mg only.\n5 25\nAccept with g replaced by 10.\nT  1 mg A1 CAO\n25\n4\n\n5(b) For B: F T km2a2 M1 Only allow sign errors.\nB\nF kmg 1kmg and combine to find k M1 To reach an equation in k only.\nB 5\nk  1 A1\n3\n3\n© Cambridge University Press & Assessment 2024 Page 10 of 13",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p10.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q06",
      "question_id": "9231-2024-mj-32-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) dv 1 dv 1 *M1 Separate variables and attempt to integrate both\n2  2v12 et so  etdt\ndt 10 2v12 20 sides.\nWhere p and q are constants.\np\nqet  A\n2v1\n1 1 A1 AEF\n  et  A\n22v1 20\n 3  DM1 Substituting the boundary condition and obtain a\nt0, v3, A\n\n\n value.\n 20\n1 5et *M1 A1 Find v in terms of t . AEF.\nv \n2 3et 1\n5\n\n6(b) Integrate: x ptqln(ret s) B *M1\n1 5 A1 AEF\nx t ln(3et 1) B\n2 3\n5 DM1 Substituting the boundary condition and obtain a\nt 0, x1, B1 ln2\nvalue.\n3\n1 5 (3et 1) A1 AEF\nx1 t ln\n2 3 2\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p11.png"
      ]
    },
    {
      "id": "9231-2024-mj-32-q07",
      "question_id": "9231-2024-mj-32-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7(a) Energy: 1mu2  1mv2 mga1cos *M1 m must be present, dimensionally correct, no\n2 2 missing terms.\nAllow sin instead of cos.\nAllow sign errors.\nmv2 B1 No reaction when P loses contact.\nN2L: mgcos\na\nEliminate v2 DM1\nu2 2ag A1 AG\ncos\n3ag\n4\n\n7(b) Vertical component of velocity of P when it leaves the sphere: vsin *B1 Must not come from u.\n 55ag \n \n \n216\n \nV2 vsin2 2ga1cos DM1 Use of 'v2 u2 2as.\nAllow sin for cos.\nAllow sign errors.\n847ag A1 AEF\nV \n216\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 13\n\n7(c) 1 847ag 55ag  M1\nt    \n \ng 216 216\n \n1  a a A1\n847  55 1.48\n6 6g g\n2\n© Cambridge University Press & Assessment 2024 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_32-p12.png",
        "../answer-assets/9231_s24_ms_32-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q01",
      "question_id": "9231-2024-mj-33-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 1 1 1 M1 Masses must be included. Allow sign errors.\nAlong line of centres, PCLM: 2m ucosmu 2m ucosm u\n2 2 4\n5 A1\ncos\n8\n1 1 1  M1 Allow sign errors, e must be on correct side.\nNEL: ucos ue  ucosu \n2 4 2 \n3 A1 AEF\ne\n7\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_33-p06.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q02",
      "question_id": "9231-2024-mj-33-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) 2mgx B1 2mgx\nHooke’s law: T  seen anywhere.\na a\nmga B1 RHS seen anywhere.\n2 May be in terms of radius or extended length, for\n T \na mga mga\na x\n4 example 2 , 2 .\n1 r\nl a\n4\nmga M1 Equate two expressions for T and obtain a\nEquate: 2mgx  2 , 4x2 3axa2 0 simplified homogeneous quadratic equation\na 3a 4l2 5al0,\nx\n4 4r2 3ara2 0,\n2k2 3k20\na A1 Single correct answer only.\nx\n4\n4\n\n2(b)  T kmg B1 kmg seen anywhere in an equation. This may be\nseen in part (a). Note that no response in part (b)\ncan earn B1 if kmg seen in part (a).\n2mgx 1 B1 CWO. Part (a) needs to be correct.\nT  , k \na 2\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_33-p07.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q03",
      "question_id": "9231-2024-mj-33-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) mu2 B1 Condone r used consistently instead of a\nAt lowest point, T mg  throughout this question.\na\nmv2 B1\nWhen string makes angle  with upward vertical, Smgcos\na\n1 1 M1 Must include m. Allow sin instead of cos for\nEnergy: mu2  mv2 mga1cos\n2 2 this mark, allow sign errors.\nEliminate u2 and v2 M1 Need to see at least one line of working.\nS T 3mg1cos A1 AG\n5\n\n3(b) When string goes slack, S = 0 so T 3mg1cos M1 May use v2 agcos substituted into energy\nequation.\nmu2 2 A1\nBut T mg mg4mg 5mg, so cos\na 3\n2\n© Cambridge University Press & Assessment 2024 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_33-p08.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q04",
      "question_id": "9231-2024-mj-33-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4 Consider one situation: B1\nkmg  1  2 3 Accept unsimplified.\nA to B: Loss in EPE = a2   a   = kmga\n2a  2   8\n \n1 mga 3  3  M1A1 KE, GPE, EPE terms present.\nEnergy: mV2  sin kmga ( V2  gak1  Must be dimensionally correct.\n2 2 8  4 \nMust include sin or cos in GPE.\nConsider a second situation: B1\nkmg   a 2 15 Accept unsimplified.\nA to C: Loss in EPE = a2      = kmga\n2a   4  32\n \n1 1  2 mg5a 15  15  M1 KE, GPE, EPE terms present.\nEnergy: m  V   sin kmga  V2  agk2  Must be dimensionally correct.\n2 2  4 32  4 \nMust include sin or cos in GPE.\nThird possible situation: (B1) This may be used in combination with either of\nkmg  a 2  a 2 3 the first two situations.\nB to C : Loss in EPE =        = kmga Mark to the candidate’s benefit,\n2a  2  4  32\n \n1 1  2 1 1  2 mg3a 3  1  (M1) KE, GPE, EPE terms present.\nEnergy: m V  m V  sin kmga  V2  ag6k  Must be dimensionally correct .\n2 2  2 4  4 32  4 \nMust include sin or cos in GPE.\nEliminate V2 from two energy equations to obtain expression involving only k, a M1 At least one of the energy equations must have\nscored M1.\nand possibly sin\n9 A1\nk \n4\n7\n© Cambridge University Press & Assessment 2024 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_33-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q05",
      "question_id": "9231-2024-mj-33-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) M1 A1 Note that moments about OB is M0 (y 6a).\nOBC OAC ABC\nMoments equation about OC with all terms\npresent, allow sign error, dimensionally correct.\n1\nArea 24a18a 9ax 216a2 9ax All correct for A1.\n2\nCentre of mass 8a 1x x\nfrom OC 3\nMoments about OC\n 216a2 9ax  x 216a28a9ax 1 x\n3\n576a2 x2 x24a A1 Accept any equivalent form.\nx  or\n72a3x 3\nAlternative solution to question 5(a)\nConsider system as equivalent to particles at (0, 18a), (x, 0) and (24a, 0) B1\nThen the x-coordinate of the centre of mass is at 1x24a M1 A1\n3\n3\n© Cambridge University Press & Assessment 2024 Page 10 of 15\n\n5(b) 18a6a 18a y y 18a M1 A1 Either way up (their value for x may be\ntan or or or\nx x 1 5ax 1 2a 1x substituted in).\n2\nx2 30ax144a2 0 M1 Obtain homogeneous (quadratic) equation\nNote that if simplified form of x is used, equation\nOr, with simplified form, 30ax24a will be linear.\nx6a A1 Single correct answer only.\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2024-mj-33-q06",
      "question_id": "9231-2024-mj-33-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Components of velocity are ucos, usingt B1\n2 M1 Square and add components of velocity and\nucos2 usingt2\n\n\n \n3\n4 u\n\n  equate to   3 u   2 .\n4 \n7 A1 AG\nu2 100usin25000\nAt least one correct line of working seen.\n16\n3\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n6(b) Let  be angle of direction of motion with horizontal at t5, then B1 Either way up.\nusin5g\ntan\nucos\n usin5g M1 Must be – 1 not +1.\ntan tan1, so tan   1 FT their expression for tan.\n ucos \nu50sin A1\nUse in result from part (a) to form equation in u or sin M1\n4 A1 Both.\nu2 1600, u40 and sin\n5\nAlternative method for question 6(b)\n3 M1\nucos usin\n4\n4 4 A1\ntan or sin\n3 5\n3 4 M1 A1 Allow sign error.\n ucos u50\n4 5\n4 A1 Both seen.\nu40 and sin\n5\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n6(b) Alternative method for question 6(b)\n3 M1\nucos usin\n4\n4 4 A1\ntan or sin\n3 5\nUse in result from part (a) to form equation in u M1\n 1000 A1\nu40\n\nand\n\n 7 \n4 A1 Both seen.\nu40 only and sin\n5\n5\n© Cambridge University Press & Assessment 2024 Page 13 of 15",
      "source_pages": [
        11,
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_33-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-33-q07",
      "question_id": "9231-2024-mj-33-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) dv B1 No marks in this part if suvat used.\nm m10v\ndt\nMust have sight of m (for example in F = ma).\nln10v t A or lnv10t A *M1 A1 Separate variables and integrate to obtain a ln\nterm. Constant may be omitted.\nConstant needed for A1\nUse t 0, v50: Aln 40 DM1 Find constant, dependent on previous M1.\nMay use limits instead.\n0.1vet 4et M1 Remove all logs .\nv1040et A1 Correct work only .\n6\n\n7(b) x10t40et B M1 No marks in this part if suvat used in part (a) or\npart (b).\nIntegrate their answer to part (a).\nConstant may be omitted.\nUse t 0, x0: B40 M1 Use initial condition in their expression for x in\nterms of t .\nx10t40et 40 A1\n3\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n7(c) 1 M1 No marks in this part if suvat used in part (a) part\nWhen v15, et  , t 2.08 or ln 8\n(b) or part (c).\n8\nFind value of t from their answer to part (a).\nx 55.8 (metres) A1 Note 35 + 10 ln 8 scores A0.\n2\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_33-p14.png",
        "../answer-assets/9231_s24_ms_33-p15.png"
      ]
    },
    {
      "id": "9231-2024-mj-41-q01",
      "question_id": "9231-2024-mj-41-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 s2  941.5 B1 72.423.\n13\ns2 M1 A1 Correct formula with a t value.\nCI: 42.82.160 2.160 seen in a CI formula for A1.\n14\n[37.9, 47.7] A1 (37.9, 47.7) A1\n37.9'x'47.7 A1 (ignore symbol used inside\nthe inequality)\nCondone [47.7, 37.9] A1\nFinal answer 42.84.91 A0\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 15",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-41-q02",
      "question_id": "9231-2024-mj-41-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 H 0 : practical results and written results are equal B1 Or H 0 : m P m W 0, H 1 : m P m W 0, oe\nH : practical results are greater than written results\n1 (m is population median).\nThere are 10 + signs; B(15, 0.5) soi M1\n(P(X 10)1PX 9 =) 1 – 0.849 = 0.151 A1 P(X ⩽ 5) = 0.151 (implies M1)\nCompare with 0.1: ‘0.151’ > 0.1 accept H M1 ‘0.151’ must come from a valid binomial\n0\ncalculation Correct ft conclusion for their 0.151\nand 0.1 Condone Reject H\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 0.151 and 0.1\nInsufficient evidence to suggest that practical results are greater than written results A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nlanguage. Not ‘prove’.\nCondone ‘no sufficient’ ‘not enough’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’.\n5\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_41-p07.png"
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    },
    {
      "id": "9231-2024-mj-41-q03",
      "question_id": "9231-2024-mj-41-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Signed differences M1 Allow one error.\n0.4 1.1 0.8 0.5 1 .2 0.9 1 .9 0.3 2.2 1 .6 0.6 1 .0\nRanks: 2 8 5 3 9 6 1 1 1 12 1 0 4 7 M1 Attempt at ranking.\n(P = 54) P = 24 A1 All ranks must be correct.\n+ -\nH : population median = 22.0 B1 Accept m used as ‘population median’\n0\nH : population median  22.0\n1\nCritical value from table = 17 B1\n‘24’ > ‘17’ Accept H M1 ‘24’ must come from a ranking of signed\n0\ndifferences, ‘17’ must be critical value from tables\n(7, 9, 10, 13, 17 or 21).\nCorrect ft conclusion for their 24 and their 17.\nCondone Reject H .\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 24 and their 17.\nInsufficient evidence to support (population) median not being 22.0 A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nInsufficient evidence against the manager’s claim language. Not ‘prove’.\nCondone ‘no sufficient’ ‘not enough’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’.\n7\n© Cambridge University Press & Assessment 2024 Page 8 of 15\n\n3(b) Distribution is symmetric about the population median B1 Need to see population or underlying distribution\nUnderlying/population distribution is symmetric about the median mentioned.\nUnderlying distribution is symmetric B0.\nPopulation distribution is symmetric B0.\n‘data’ implies B0 but condone ‘population data’\n‘mean’ implies B0.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_41-p09.png"
      ]
    },
    {
      "id": "9231-2024-mj-41-q04",
      "question_id": "9231-2024-mj-41-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) t2 43t42t12t243t3 M1 ‘Two’ terms obtained, correct denominator(s).\nG (t) = , G’ (t) = or\nY Y\n43t4 43t8\n2t 12t2\n\n43t4 43t5\nE(Y) = G’ (1) = 14 M1 A1 t = 1 in their expression.\nY\nCWO\n3\n\n4(b) t B1\nG (t) =\nX 43t2\n1\n© Cambridge University Press & Assessment 2024 Page 9 of 15\n\n4(c) G X (t) = t43t2  t  1 3 t   2  t  1 3 t3 9 t2 4 27 t3    M1 Expansion as far as t3 OR term in t4 calculated.\n16 4  16 2 16 64 \nUse P(X = 4) = their coefficient of t4 M1 Find the numerical value of their coefficient of t4.\n27 A1 Accept 0.105.\n256\n3\n© Cambridge University Press & Assessment 2024 Page 10 of 15",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_41-p10.png"
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    },
    {
      "id": "9231-2024-mj-41-q05",
      "question_id": "9231-2024-mj-41-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) M1 A1 Calculate E values to at least 2dp, allow one error.\nCompany Poor Satisfactory Good\nThese must be seen.\nFor A1, working to at least 2dp.\nP (18) (43) (64)\n25 40.625 59.375\nQ (22) (22) (31)\n15 24.375 35.625\nChi-squared contributions: M1\nCompany Poor Satisfactory Good\nAt least 2 correct values or expressions seen.\nOr may be implied by AWRT 6.56\nP 1.96 0.1388 0.3603\nQ 3.2667 0.2314 0.6004\nTest statistic = 6.56 A1 AWRT 6.56.\nH : Quality (of items) is independent of company B1 Condone ‘no association/relationship between….’\n0\nH : Quality (of items) is not independent of company ‘association/relationship between….’.\n1\n6.56 > 5.991 Reject H M1 ‘6.56’ from their attempt at chi-squared, 5.991\n0\nmust be correct.\nConsistent signs in comparison.\nFT conclusion from their 6.56.\nCondone ‘Accept H ’.\n1\n‘Reject H ’ can be implied by a conclusion that is\n0\nconsistent with their 6.56.\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n5(a) There is sufficient evidence to suggest that the quality of brushes is not A1 Correct work only, conclusion in context with\nindependent of company. level of uncertainty in language. Not ‘prove’.\nAccept ‘enough evidence’.\nCondone ‘some evidence’.\nDo not accept ‘there is insufficient…’.\n7\n\n5(b) P produces better quality brushes (than Q) B1 Condone P is better (than Q).\nor Q produces worse quality brushes (than P)\nP has fewer poor quality brushes than expected AND Q has more poor quality\nbrushes than expected\n1\n© Cambridge University Press & Assessment 2024 Page 12 of 15",
      "source_pages": [
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        12
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2024-mj-41-q06",
      "question_id": "9231-2024-mj-41-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Signed differences: M1 Allow one error, must be signed differences.\n4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0\n1 14.52  M1\nd 14.5, d2 52.81 s2  52.81 3.532\n9 10\n \nH :   1 B1\n0 B A\nAccept H :  1, H :  1.\nH 1 :  B  A 1 0 d 1 d\n1.451 M1 The – 1 must be present in the numerator.\nt \ns2\n10\n0.757 A1\nCritical value is 1.383: ‘0.757’ < 1.383 Accept H M1 ‘0.757’ must come from a paired sample t-\n0\ncalculation, 1.383 must be correct\nCorrect ft conclusion for their 0.757 and 1.383\nCondone Reject H\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 0.757 and 1.383\nInsufficient evidence to support Jade’s claim A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nlanguage. Not ‘prove’\nCondone ‘no sufficient’ ‘not enough’\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’\n7\n© Cambridge University Press & Assessment 2024 Page 13 of 15\n\n6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1\n‘data’ implies B0.\n1",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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        "../answer-assets/9231_s24_ms_41-p13.png",
        "../answer-assets/9231_s24_ms_41-p14.png"
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    },
    {
      "id": "9231-2024-mj-41-q07",
      "question_id": "9231-2024-mj-41-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) \n\nx 4x2\n0x2,\nB1\nfx4\n  0 otherwise.\n(E(X) =)  2 1 4x2 x4 dx 14x3 1x516 ( = 1.067)\n04 43 5  15\n(E( X))  2 1 4x1.5 x7.2 dx 18x2.5 2x4.5 2  32 2 1.0057 M1 Integration, correct powers required, ignore limits.\n04 45 9  0  45 \nVar( X)EX  E  X 2 used M1 Correct formula used following attempt at\nintegration of both terms.\n16 32\n2\n2\n0.0553\nA1\n2\n1\n0\n1\n2\n2\n5\n 15 45  \n4\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n7(b) F(x) = 1  8x2 x4 0x2 B1\n16\nG(y) =\n1 \n8y\ny2\n[0 y4]\nM1 Change variable in their F(x).\n16\n1 M1 Differentiate their G(y) .\ngy 82y 0 y4\n16\n 1 A1 Complete and correct.\n 82y 0y4,\ngy16\n  0 otherwise.\n4\n\n7(c) 1  8mm2\n\n1 M1 Equate their G(y) to 1\n2\n.\n16 2\n[m2 8m80,] m42 2 A1 Single answer in an exact form.\n2\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2024-mj-42-q01",
      "question_id": "9231-2024-mj-42-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 s2  941.5 B1 72.423.\n13\ns2 M1 A1 Correct formula with a t value.\nCI: 42.82.160 2.160 seen in a CI formula for A1.\n14\n[37.9, 47.7] A1 (37.9, 47.7) A1\n37.9'x'47.7 A1 (ignore symbol used inside\nthe inequality)\nCondone [47.7, 37.9] A1\nFinal answer 42.84.91 A0\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-42-q02",
      "question_id": "9231-2024-mj-42-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 H 0 : practical results and written results are equal B1 Or H 0 : m P m W 0, H 1 : m P m W 0, oe\nH : practical results are greater than written results\n1 (m is population median).\nThere are 10 + signs; B(15, 0.5) soi M1\n(P(X 10)1PX 9 =) 1 – 0.849 = 0.151 A1 P(X ⩽ 5) = 0.151 (implies M1)\nCompare with 0.1: ‘0.151’ > 0.1 accept H M1 ‘0.151’ must come from a valid binomial\n0\ncalculation Correct ft conclusion for their 0.151\nand 0.1 Condone Reject H\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 0.151 and 0.1\nInsufficient evidence to suggest that practical results are greater than written results A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nlanguage. Not ‘prove’.\nCondone ‘no sufficient’ ‘not enough’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’.\n5\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-42-q03",
      "question_id": "9231-2024-mj-42-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) Signed differences M1 Allow one error.\n0.4 1.1 0.8 0.5 1 .2 0.9 1 .9 0.3 2.2 1 .6 0.6 1 .0\nRanks: 2 8 5 3 9 6 1 1 1 12 1 0 4 7 M1 Attempt at ranking.\n(P = 54) P = 24 A1 All ranks must be correct.\n+ -\nH : population median = 22.0 B1 Accept m used as ‘population median’\n0\nH : population median  22.0\n1\nCritical value from table = 17 B1\n‘24’ > ‘17’ Accept H M1 ‘24’ must come from a ranking of signed\n0\ndifferences, ‘17’ must be critical value from tables\n(7, 9, 10, 13, 17 or 21).\nCorrect ft conclusion for their 24 and their 17.\nCondone Reject H .\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 24 and their 17.\nInsufficient evidence to support (population) median not being 22.0 A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nInsufficient evidence against the manager’s claim language. Not ‘prove’.\nCondone ‘no sufficient’ ‘not enough’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’.\n7\n© Cambridge University Press & Assessment 2024 Page 8 of 15\n\n3(b) Distribution is symmetric about the population median B1 Need to see population or underlying distribution\nUnderlying/population distribution is symmetric about the median mentioned.\nUnderlying distribution is symmetric B0.\nPopulation distribution is symmetric B0.\n‘data’ implies B0 but condone ‘population data’\n‘mean’ implies B0.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
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        "../answer-assets/9231_s24_ms_42-p09.png"
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    },
    {
      "id": "9231-2024-mj-42-q04",
      "question_id": "9231-2024-mj-42-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) t2 43t42t12t243t3 M1 ‘Two’ terms obtained, correct denominator(s).\nG (t) = , G’ (t) = or\nY Y\n43t4 43t8\n2t 12t2\n\n43t4 43t5\nE(Y) = G’ (1) = 14 M1 A1 t = 1 in their expression.\nY\nCWO\n3\n\n4(b) t B1\nG (t) =\nX 43t2\n1\n© Cambridge University Press & Assessment 2024 Page 9 of 15\n\n4(c) G X (t) = t43t2  t  1 3 t   2  t  1 3 t3 9 t2 4 27 t3    M1 Expansion as far as t3 OR term in t4 calculated.\n16 4  16 2 16 64 \nUse P(X = 4) = their coefficient of t4 M1 Find the numerical value of their coefficient of t4.\n27 A1 Accept 0.105.\n256\n3\n© Cambridge University Press & Assessment 2024 Page 10 of 15",
      "source_pages": [
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        10
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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        "../answer-assets/9231_s24_ms_42-p10.png"
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    },
    {
      "id": "9231-2024-mj-42-q05",
      "question_id": "9231-2024-mj-42-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) M1 A1 Calculate E values to at least 2dp, allow one error.\nCompany Poor Satisfactory Good\nThese must be seen.\nFor A1, working to at least 2dp.\nP (18) (43) (64)\n25 40.625 59.375\nQ (22) (22) (31)\n15 24.375 35.625\nChi-squared contributions: M1\nCompany Poor Satisfactory Good\nAt least 2 correct values or expressions seen.\nOr may be implied by AWRT 6.56\nP 1.96 0.1388 0.3603\nQ 3.2667 0.2314 0.6004\nTest statistic = 6.56 A1 AWRT 6.56.\nH : Quality (of items) is independent of company B1 Condone ‘no association/relationship between….’\n0\nH : Quality (of items) is not independent of company ‘association/relationship between….’.\n1\n6.56 > 5.991 Reject H M1 ‘6.56’ from their attempt at chi-squared, 5.991\n0\nmust be correct.\nConsistent signs in comparison.\nFT conclusion from their 6.56.\nCondone ‘Accept H ’.\n1\n‘Reject H ’ can be implied by a conclusion that is\n0\nconsistent with their 6.56.\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n5(a) There is sufficient evidence to suggest that the quality of brushes is not A1 Correct work only, conclusion in context with\nindependent of company. level of uncertainty in language. Not ‘prove’.\nAccept ‘enough evidence’.\nCondone ‘some evidence’.\nDo not accept ‘there is insufficient…’.\n7\n\n5(b) P produces better quality brushes (than Q) B1 Condone P is better (than Q).\nor Q produces worse quality brushes (than P)\nP has fewer poor quality brushes than expected AND Q has more poor quality\nbrushes than expected\n1\n© Cambridge University Press & Assessment 2024 Page 12 of 15",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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        "../answer-assets/9231_s24_ms_42-p12.png"
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    },
    {
      "id": "9231-2024-mj-42-q06",
      "question_id": "9231-2024-mj-42-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Signed differences: M1 Allow one error, must be signed differences.\n4.1 1 .1 0.4 0.2 3.9 2.6 0.6 3.5 0.1 0\n1 14.52  M1\nd 14.5, d2 52.81 s2  52.81 3.532\n9 10\n \nH :   1 B1\n0 B A\nAccept H :  1, H :  1.\nH 1 :  B  A 1 0 d 1 d\n1.451 M1 The – 1 must be present in the numerator.\nt \ns2\n10\n0.757 A1\nCritical value is 1.383: ‘0.757’ < 1.383 Accept H M1 ‘0.757’ must come from a paired sample t-\n0\ncalculation, 1.383 must be correct\nCorrect ft conclusion for their 0.757 and 1.383\nCondone Reject H\n1\n‘Accept H ’ can be implied by a conclusion that is\n0\nconsistent with their 0.757 and 1.383\nInsufficient evidence to support Jade’s claim A1 Correct work only, ignoring their hypotheses,\nconclusion in context with level of uncertainty in\nlanguage. Not ‘prove’\nCondone ‘no sufficient’ ‘not enough’\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…..’\n7\n© Cambridge University Press & Assessment 2024 Page 13 of 15\n\n6(b) Distribution of population differences is normal B1 Underlying distribution of differences normal B1\n‘data’ implies B0.\n1",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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        "../answer-assets/9231_s24_ms_42-p14.png"
      ]
    },
    {
      "id": "9231-2024-mj-42-q07",
      "question_id": "9231-2024-mj-42-q07",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 7,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) \n\nx 4x2\n0x2,\nB1\nfx4\n  0 otherwise.\n(E(X) =)  2 1 4x2 x4 dx 14x3 1x516 ( = 1.067)\n04 43 5  15\n(E( X))  2 1 4x1.5 x7.2 dx 18x2.5 2x4.5 2  32 2 1.0057 M1 Integration, correct powers required, ignore limits.\n04 45 9  0  45 \nVar( X)EX  E  X 2 used M1 Correct formula used following attempt at\nintegration of both terms.\n16 32\n2\n2\n0.0553\nA1\n2\n1\n0\n1\n2\n2\n5\n 15 45  \n4\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n7(b) F(x) = 1  8x2 x4 0x2 B1\n16\nG(y) =\n1 \n8y\ny2\n[0 y4]\nM1 Change variable in their F(x).\n16\n1 M1 Differentiate their G(y) .\ngy 82y 0 y4\n16\n 1 A1 Complete and correct.\n 82y 0y4,\ngy16\n  0 otherwise.\n4\n\n7(c) 1  8mm2\n\n1 M1 Equate their G(y) to 1\n2\n.\n16 2\n[m2 8m80,] m42 2 A1 Single answer in an exact form.\n2\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2024-mj-43-q01",
      "question_id": "9231-2024-mj-43-q01",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1 M1 A1 Attempt at ranking from 1 to 19.\n21.4 4 20.9 2\nA1 for all correct.\n24.6 15 23.5 11\n25.3 18 24.8 16\n22.7 8 21.9 6\n20.8 1 23.4 10\n21.5 5 24.0 13\n22.9 9 23.8 12\n21.3 3 24.1 14\n22.3 7 25.1 17\n25.8 19\nTest statistic: 70 A1 If ranks reversed, then sum of ranks = 110, and then 9(9 + 10 + 1) – 110 = 70.\nH 0 : population median of X = population median of B1 Accept H 0 : m X m Y H 1 : m X m Y .\nY\nDo not accept any other notation.\nH population median of X < population median of\n1:\nY\nCritical value for m = 9, n = 10 is 61 B1 Using normal approximation implies last 3 marks are B0M0A0, max 4 out of 7.\n© Cambridge University Press & Assessment 2024 Page 6 of 15\n\n1 ‘70’ > ‘61’ accept H M1 ‘70’ must come from ranks 1 – 19, ‘61’ must be a critical value from the correct\n0\ntable.\nft conclusion for their 70 and their 61.\nCondone Reject H .\n1\n‘Accept H ’ can be implied by a conclusion in context that is consistent with their\n0\n70 and their 61.\nInsufficient evidence to suggest that population A1 Correct work only, ignoring their hypotheses, conclusion in context with level of\nmedians of X and Y are not equal uncertainty in language. Not ‘prove’ or equivalent.\nor Condone ‘no sufficient’ ‘not enough’.\nInsufficient evidence to suggest that population Do not accept statements such as ‘there is sufficient evidence to suggest…..’.\nmedian of X is less than median of Y\nor\nInsufficient evidence to support the interviewer’s\nbelief\n7\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s24_ms_43-p06.png",
        "../answer-assets/9231_s24_ms_43-p07.png"
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    },
    {
      "id": "9231-2024-mj-43-q02",
      "question_id": "9231-2024-mj-43-q02",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) 1 M1\nx  68.6664.2266.44\n2\nx1266.44797.28 A1 May come from use of incorrect t value.\nAccept AWRT 797.\nts 1 M1 Must be with a t value (not z value).\n 68.6664.22\n12 2\nWith t  2.718 gives s2 8.0055 A1 Accept AWRT 8.01; may be recovered by final answer exactly 53059(.34).\ns 2.8294.\n1\n x2  M1 Must see an expression for x2.\nUse s2  x2   to find an\n11 12 \n \nexpression for x2\nx2 53059 A1 AWRT 53100 but not from wrong working.\n6\n\n2(b) Population is normally distributed B1 Underlying distribution is normal B1.\nPulse rates are normal B1.\nPopulation mean is normally distributed B0.\n(Underlying) data is normal B0.\n1\n© Cambridge University Press & Assessment 2024 Page 8 of 15",
      "source_pages": [
        8
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      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-mj-43-q03",
      "question_id": "9231-2024-mj-43-q03",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 M1 A1 Calculate expected values, at least 4 correct to 3sf.\nBus Bus Bus\nThese must be seen.\ncompany company company\nA1 for all correct to at least 3sf.\nA B C\nArrived 22 (17.28) 22 (21.60) 10 (15.12)\nearly\nArrived 30 (39.68) 52 (49.60) 42 (34.72)\non time\nArrived 28 (23.04) 26 (28.80) 18 (20.16)\nlate\nM1 At least 4 correct values (to 2sf) or expressions.\n1.2893 0.00741 1.7338\nMay be implied by 8.600 – 8.615.\n2.3615 0.1161 1.5265\n1.0678 0.2722 0.2314\nTest statistic = 8.61 A1 8.600 – 8.615\nUnsupported 8.600 – 8.615 scores M0A0 M1A1.\nH : Reliability of buses is independent of bus B1 Condone ‘No association/relationship between…’\n0\ncompany ‘Association/relationship between…’\nH : Reliability of buses is not independent\n1\nof/dependent on bus company\n‘8.61’ < 9.488 Accept H M1 ‘8.61’ from an attempt at chi-squared, 9.488 must be correct.\n0\nft conclusion from their 8.61.\nCondone ‘Reject H ’.\n1\n‘Accept H ’ can be implied by a conclusion in context that is consistent with their\n0\n8.61.\n© Cambridge University Press & Assessment 2024 Page 9 of 15\n\n3 There is insufficient evidence to support/ suggest A1 Correct work only, ignoring their hypotheses, conclusion in context with level of\nthat reliability of buses is dependent/not uncertainty in language. Not ‘prove’ or equivalent.\nindependent of bus company\nCondone ‘no sufficient’ ‘not enough’.\nDo not accept statements such as ‘there is sufficient evidence to suggest…..’\n7\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_43-p10.png"
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    },
    {
      "id": "9231-2024-mj-43-q04",
      "question_id": "9231-2024-mj-43-q04",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 B1 Accept 0.03125 or 0.0313.\nc \n32\n1\n\n4(b)\nG (t) =\nct1t5\n, G’ (t) =\nM1 Must use product rule OR differentiate their expansion of G\nX X\nAllow one error in their differentiation.\nc  5t1t4 1t5 Condone missing c.\nE(X) = G’ (1) = c80323.5 A1 Correct work only.\nX\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 15\n\n4(c)\nG (t) =\nc2t21t10 *M1 Or\nY\nc2 t2 10t3 45t4 120t5 210t6 252t7 210t8 120t9 45t10 10t11t12\nG' (t) = c2  10t21t9 2t1t10 DM1\nY Differentiate their G (t) twice, allow one error.\nY\nG'' (t) =\nY\nc2  90t21t8 40t1t9 21t10 )\nG' (1) = 7, G’’ (1) = 44.5 M1 A1 Use correct formula and substitute values for t = 1 (may be in terms of c).\nY Y\nVar(Y) = G’’ (1) + G’ (1) \n G’12\n= 2.5\nA1 for correct work only.\nY Y\nAlternative solution for question 4(c)\nVar (Y) = 2Var(X) *M1\nDifferentiate G (t) twice DM1 Note that G' (t) has already been found in part (b).\nX X\nVar(X)G'' (1)G' (1)\n\nG'\n12 M1 Use correct formula and substitute values for t = 1.\nX X X\n1.25  2 = 2.5 A1\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n4(d) P(Y = 5) = coefficient of t5 in expansion of their M1 Identify required term.\nc2t21t10\nRequired term is c2120\nExpansion = c2t2 110t45t2 120t3  \n15 A1 0.117, 120 or any equivalent fraction.\n128 1024\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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        "../answer-assets/9231_s24_ms_43-p12.png"
      ]
    },
    {
      "id": "9231-2024-mj-43-q05",
      "question_id": "9231-2024-mj-43-q05",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) x2 M1 Attempt at differentiating both parts of CDF to obtain linear expressions.\n6\nMay be implied by first A1.\n8x\n12\nSketch A1 Correct shape (Two line segments touch the x-axis and meet at a point above the x-\naxis).\nA1 Correct shape with 2 and 8 and at least one of 4 or 1/3 correctly labelled.\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n5(b) 4 8 B1 FT FT their PDF, with correct limits.\n 1 xx2dx 1 x8xdx\n26 412\n1 1 x3 x2  1 4x2 1 x3 M1 Attempt at integration with their PDF (not CDF) no limits needed.\n   \n6 3 12 3 May be implied by correct answer.\n14 A1 Accept 4.67 or any equivalent fraction.\n3\n3\n\n5(c) LQ: 1 x22  1, x 2 3 B1 Allow 3.73.\n12 4\n8x2\n3\nB1 Allow 5.55.\nUQ: 1  , x8 6\n24 4\nIQR = UQ – LQ, with values found for both M1\nquartiles\nA1 Exact answer required.\n6 6 3\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 15",
      "source_pages": [
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        13
      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s24_ms_43-p13.png"
      ]
    },
    {
      "id": "9231-2024-mj-43-q06",
      "question_id": "9231-2024-mj-43-q06",
      "subject": "9231",
      "year": 2024,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) 1  10802  M1 152\ns 2  23480  (= 3.102) 9\nx 49  50  10\n3\n1  9402  Both.\ns 2  22220  (= 3.333)\ny 39 40\n \n3.102 3.333 M1\ns2   0.1454\n50 40\ns2  2137 0.1454 A1\n14700\n1080 940 M1 B1 Use of correct formula with a z value.\nCI:  1.645 s2 or\nB1 for 1.645 ( accept 1.64 or 1.65) used in CI formula.\n50 40\n940 1080\n 1.645 s2\n40 50\n1.27, 2.53 or [–2.53, –1.27] A1 (1.27, 2.53) A1.\n1.27  2.53 A1 (ignore symbol used inside the inequality).\nY X\nCondone [2.53, 1.27] or [–1.27, –2.53] A1.\nFinal answer 1.90.627 A0.\nPooled estimate: max score M1 M0A0 M1B1A0.\nFirst M1 for correct form of pooled estimate.\n6\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n6(b) 1.91.1 M1 1.91.1\nz  z  or 7.87 implies M1.\ns s\n2.098 A1 AWRT 2.10\n‘2.098’ > 1.28(2) Reject H M1 ‘2.098’ from their attempt at z, 1.28(2) must be correct.\n0\nConsistent signs in comparison.\nor\nΦ2.098\n0.9821 > 0.9 Reject H\nFT conclusion from their 2.098.\n0\nCondone ‘Accept H ’.\n1\nor 1 \nΦ2.098\n= 0.0179 < 0.1 Reject H 0 ‘Reject H 0 ’ can be implied by a conclusion in context that is consistent with their\n2.098.\nSufficient evidence to suggest that the lengths of A1 Correct work only, conclusion in context with level of uncertainty in language. Not\ntails of wallabies in region Y are, on average, ‘prove’ or equivalent.\nmore than 1.1 cm longer than the tails of\nwallabies in region X Accept ‘enough evidence’.\nCondone ‘some evidence’.\n4 Note that it is possible to use part (a) but it must be made clear explicitly that\nrejection at 5% level implies rejection at 10% level.\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
        14,
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      ],
      "source_pdf": "_source-pdfs/2024-May-June/ms/9231_s24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-may-june/9231_s24_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2024-on-11-q01",
      "question_id": "9231-2024-on-11-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "1(a) k 0 B1 [Stretch parallel to the x-axis, scale factor k\n  (k0)]. (Allow without identification.)\n0 1\n1 k B1 [Shear, x-axis fixed, with (0,1) mapped to\n \n0 1 (k,1).] (Allow without identification.)\n1 kk 0 k k M1 A1 Correct order for M1, must have identified\nM=  =  which matrix gives which transformation, AG.\n0 10 1 0 1\n4\n\n1(b) k kx kx+ky B1 x X\n  =  Transforms   to  \n0 1y  y  y Y \nkx+ky=x M1 X x\nSets  = \nY  y\n1−k A1\ny= x oe\nk\n3\n\n1(c) 11 −k B1 (An alternative is possible.)\nM−1=\n \nk0 k \n1\n© Cambridge University Press & Assessment 2024 Page 6 of 15\n\n1(d) |k|=3k2 M1 Uses that detM=k. Without modulus is SC\nB1.\nk 0k =1 A1\n3\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_11-p07.png"
      ]
    },
    {
      "id": "9231-2024-on-11-q02",
      "question_id": "9231-2024-on-11-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 d ( tan−1x ) 1 B1 Differentiates once.\n= so true when n=1.\ndx 1+x2\nAssume that d d x k k ( tan−1x ) =P k (x)( 1+x2)−k , where degP k (x)=k−1. B1 S d t e a g te P s k i ( n x d ) u = ct k iv − e 1 h . y pothesis. Must have\ndk+1( tan−1x ) M1 A1 Differentiates kth derivative using the product\n=P\n'(x)( 1+x2)−k\n−2kxP\n(x)( 1+x2)−k−1\nrule.\ndxk+1 k k\n= ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k A1 Writes in the form P (x)( 1+x2)−k−1 .\nk k k+1 k+1\nSo true when n=k+1. By induction, true for all positive integers n. A1 Attempts to show degree of P (x) is at most k\nk+1\n(condone not showing coefficient of xk is non-\nzero) and states conclusion.\n6\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_11-p07.png"
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    },
    {
      "id": "9231-2024-on-11-q03",
      "question_id": "9231-2024-on-11-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) y= x4 B1 Uses correct substitution.\ny+2y 3 4 −1=016y3 =(1−y)4 M1 Substitutes and obtains an equation not\ninvolving radicals.\n16y3 =1−4y+6y2 −4y3+ y4 M1 Uses binomial expansion.\ny4 −20y3 +6y2 −4y+1=0 A1 Must be an equation.\n4 +4 +4 +4 =20 B1\n5\n\n3(b) +++=−2 B1\nx5+2x4−x=05+5+5+5 =−2(20)+(−2) M1 Multiplies original equation by x and substitutes.\n−42 A1\n3\n\n3(c) 8+8+8+8 =202−2(6) M1 Uses formula for sum of squares. Or alternative\ncomplete method eg another substitution.\n388 A1 CAO.\n2\n© Cambridge University Press & Assessment 2024 Page 8 of 15",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_11-p08.png"
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    },
    {
      "id": "9231-2024-on-11-q04",
      "question_id": "9231-2024-on-11-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 5k  1 1  M1 A1 Finds partial fractions. (Don’t allow a\n=k\n\n−\n substitution of a value for k.)\n(5r+k)(5r+5+k) 5r+k 5r+5+k\nn 5k  1 1 1 1 1 1  M1 Writes at least three terms, including last.\n =k  − + − + + −  (Allow any value of k.)\n(5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k\nr=1\n 1 1  A1\n=k\n\n−\n\n5+k 5n+5+k\n4\n\n4(b) k 1 5 M1 A1\n= 3k =5+kk =\n5+k 3 2\n2\n\n4(c) n2 5k n2 5k n−1 5k M1 Or applies the method of differences again.\n = −\n(5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k)\nr=n r=1 r=1\n 1 1   1 1   1 1  A1FT FT on their value of k (must be substituted in).\n=k − −k − =k − \n5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k\n1 1\n= −\n2n+1 2n2 +3\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_11-p09.png"
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    },
    {
      "id": "9231-2024-on-11-q05",
      "question_id": "9231-2024-on-11-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) r4 =6r2sincos M1 Substitutes x=rcos, y=rsin and applies\nsin2=2sincos.\nr2 =3sin2 A1 AG.\n2\n\n5(b) B1 Correct position and symmetrical about = 1π.\n4\nB1 Single correct loop.\n=0\n3 B1 States maximum distance or labels sketch.\n( ) ( )\nAllow 3,1π but not 1π, 3 . Allow 3sf.\n4 4\n3\n\n5(c) 3 π 2sin2d= 3  −1cos2  π 2 M1 Forms 1 2 r2 d. (Allow with wrong limits.)\n2 0 2 2 0\n3 A1\n2\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 15\n\n5(d) 1 1 B1\ny=32sin22sin\nsin 1 2 2cos+sin−1 2 2cos2sin=0 M1 A1 Sets dy =0.\nd\n2tan M1 Applies suitable trigonometric identity.\nsin2cos+cos2sin=0tan2=−tan =−tan\n1−tan2 Accept sin3=0.\n= 1π A1\n3\n5 A1 AEF.\n34\n=1.40\n3\n22\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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        11
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2024-on-11-q06",
      "question_id": "9231-2024-on-11-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 B1 Vertical asymptotes.\nx= , x=3\n2\ny=2 B1 Horizontal asymptote.\n2\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n6(b) dy (2x2 −7x+3)(8x+1)−(4x2 +x+1)(4x−7) M1 dy\n= Finds . Allow top line only for M1.\ndx ( 2x2 −7x+3 )2 dx\n−3x2 +2x+1=0 M1 Sets equal to 0 and forms equation.\n( −1,1) , (1,−3) A1 A1\n3 5\n4\nxy\n\n6(c) B1 Axes and asymptotes. Clear identification (label\nor clear intersection with axes at correct place).\nB1 x3 correctly approaching asymptotes, not too\ntruncated.\nB1 1 x3 correct.\n2\nB1 x 1 correct.\n2\n( 0,1) B1 States coordinates of intersection with axis. May\n3\nbe seen on their graph.\n5\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n6(d) B1FT FT from sketch in (c). At least two branches.\nk3 B1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-11-q07",
      "question_id": "9231-2024-on-11-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k  6  −2 M1 A1 Finds common perpendicular. Allow one error.\n   \n2 1 1 = −3 ~ 1\n   \n   \n1 −4 2 −9  3 \n−2(1)+(3)+3(−2)=−5 M1 Substitutes point on l .\n1\n2x−y−3z=5 A1 CAO.\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 15\n\n7(b) i j k 6 M1 A1 Finds the normal to  .\n2\n \n1 −4 2 = 4\n \n \n1 1 −2 5\n 2  6 M1 Uses dot product of normal vectors.\n    −7\n−1 4 = 14 77coscos=\n   \n    14 77\n−3 5\n77.7 A1 No ISW. Accept 1.36 rad.\n4\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n7(c) 1+2  1+   −2  M1 A1 Finds PQ.\n     \nOP= 3+ ,OQ= −2−4 PQ= −5−4−\n     \n     \n−2+  9+2 11+2−\n −2  2  −2   6  M1 Uses that dot product of PQ with line direction\n       \n−5−4− 1 =0 or −5−4− =k −3\n        is zero, or, alternatively, PQ is a multiple of the\n       \n11+2− 1 11+2− −9 common perpendicular.\n−6+6=0 A1 Deduces one equation.\n −2   1  A1 Deduces second equation.\n   \n−5−4− −4 =021+42=0\n   \n   \n11+2−  2 \n 3  −1 M1 Solves for  or  and substitutes into OP or\n   \n=1OP= 4 or =−2OQ= 6\n    OQ\n   \n−1  5 \n−1  6  A1 FT OE. FT using their common perpendicular. Must\n    have\"r=\".\nr= 6 +t −3\n   \n   \n 5  −9\n7\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2024-on-12-q01",
      "question_id": "9231-2024-on-12-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 u =4=31+1 B1 Shows base case\n1\nAssume that it is true for n=k, so u =3k +1. B1 States inductive hypothesis.\nk\nThen u =3(3k +1)−2=3k+1+1 M1 A1 Substitutes into recursion formula.\nk+1\n[So, it is also true for n=k+1]. Hence, by induction, u =3n +1for all positive integers. A1 States conclusion.\nn\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-12-q02",
      "question_id": "9231-2024-on-12-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) i j k 24 M1 A1 Finds vector perpendicular to the plane.\n \n1 −1 −4 = −4\n \n \n2 5 −4 7\n \n24(1)−4(3)+7(−1)=d 24x−4y+7z=5 M1 A1 Uses point in the plane.\n4\n\n2(b) 24  5  M1A1FT Uses dot product of 5i−5j−2k and\n    126 their normal.\n−4 −5 = 641 54coscos=\n   \n641 54\n   \n7 −2\n   \nAcute angle between l and Π is 90−=42.6 A1 0.744 radians\n2\n3\n© Cambridge University Press & Assessment 2024 Page 6 of 16",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-12-q03",
      "question_id": "9231-2024-on-12-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) 2 +2 +2 +2 =(+++)2 −2(+++++) M1 Substitutes into formula for sum of\nsquares.\n3=22−2(+++++)\n+++++= 1 A1\n2\n2\n\n3(b) 2(++)+2(++)+2(++)+2(++) M1 A1 Factorises and substitutes.\n=2(2−)+2(2−)+2(2−)+2(2−)\n2(3)−4=2 A1\n3\n\n3(c)(i) 6S −12S +3S +2S +24=06S −12(4)+3(3)+2(2)+24=0 M1 Sums and substitutes.\n4 3 2 1 4\nA1\nS = 11 A1\n4 6\n3\n\n3(c)(ii) 6S −12S +3S +2S +6S =0 6S −12(11)+3(4)+2(3)+12=0 M1 Multiplies equation through by x, sums\n5 4 3 2 1 5 6\nand substitutes.\nS −2S + 1S +1S +S =0\n5 4 2 3 3 2 1\nS −2(11)+1(4)+1(3)+2=0\n5 6 2 3\nS =−4 A1\n5 3\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_12-p07.png"
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    },
    {
      "id": "9231-2024-on-12-q04",
      "question_id": "9231-2024-on-12-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 2 3 0 −2 −2 4  M1 A1 Multiplying two matrices correctly,\n−2 −1 1   −2 −1 1  correct dimensions.\n \n2 1 3\n\n−1 3\n\n=\n \n−1 −1\n\n 1 1 3    1 1 3 \n3 2 5 0 0 −2 0\n    \n1 2 3 0 −2  0 −2\n−2 −1 1   −1 −3 −4 \nOr   2 1 3  −1 3  =   −1 3 \n 1 1 3   12 9 21 \n3 2 5 0 0 0 0\n    \n 3 −7 B1 Convincingly completing matrix\n=   multiplication, AG.\n−9 3 \n3\n\n4(b)  3 −7x 3x−7y  B1 x X\n   =   Transforms   to  .\n−9 3 y −9x+3y y Y \n−9x+3mx=m(3x−7mx) M1 A1 Uses y=mx and Y =mX .\n−9+3m=3m−7m2 7m2 =9 A1\ny= 3 x and y=− 3 x A1\n7 7\n5\n\n4(c) Stretch B1\nparallel to the x-axis, scale factor 3. B1\n2\n© Cambridge University Press & Assessment 2024 Page 8 of 16\n\n4(d) 11 0 B1\nM−1 =\n \n30 3\n1 3 −71 0  3 −71 0 M1 Correct order\nN=    or N=   3 \n3−9 3 0 3 −9 3 0 1\n 1 −7 A1\n=\n \n−3 3 \n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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        "../answer-assets/9231_w24_ms_12-p09.png"
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    },
    {
      "id": "9231-2024-on-12-q05",
      "question_id": "9231-2024-on-12-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) xf(1) −xf(2) +xf(2) −xf(3) + +xf(n) −xf(n+1) M1 Writes at least three terms, including last.\n= xf(1) −xf(n+1) A1\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 16\n\n5(b) 0x1 B1 Accept 0x1\nn\nu =1−xln(n+1)\nr\nr=1\n B1 Without wrong working.\nu =1 [for x1]\nr\nr=1\n B1\nu =0 for x=1\nr\nr=1\n3\n\n5(c) Uses x2log x r =r2 M1\nS =1−(n+1)2 =−n2−2n A1\nn\nN M1 Substitutes formulae from MF19.\n−n2 −2n=−1 N(N+1)(2N+1)−N(N+1)\n6\nn=1\n−1 N(N +1)(2N +7) A1\n6\n4\n© Cambridge University Press & Assessment 2024 Page 10 of 16",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2024-on-12-q06",
      "question_id": "9231-2024-on-12-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 +1=0 has no real roots. B1\ny=1 B1 Horizontal asymptote.\n2\n\n6(b) yx2 + y = x2 +3(1− y)x2 − y+3=0 M1 A1 Forms quadratic in x or uses\n2\ny =1+ .\nx2 +1\n−4(1−y)(3−y) 0 M1 Uses that discriminant is 0 or\n2\n0 2.\nx2 +1\n1 y3 A1 Explanation of why y 1\nAG.\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 16\n\n6(c) dy ( x2 +1 )(2x)− ( x2 +3 )(2x) M1 Differentiates\n= =0\ndx ( x2 +1 )2\n(0,3) A1\nAlternative method for question 6(c)\nWhen y =3 2x2 =0 M1 Using inequality from (b)\n(0,3) A1\n2\ny\n\n6(d) B1 Axes and correct asymptote labelled.\nB1 Correct shape and position.\nB1 States (0,3)coordinates of intersection\nwith axes, may been seen on diagram.\nx\n3\n© Cambridge University Press & Assessment 2024 Page 12 of 16\n\n6(e) B1FT FT from sketch in (d)\n1\n3\nx2 +1 1 M1 Finds critical points.\n=\nx2 +3 2\nx2 =1x=1 A1\n−1x1 A1\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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        12,
        13
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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        "../answer-assets/9231_w24_ms_12-p13.png"
      ]
    },
    {
      "id": "9231-2024-on-12-q07",
      "question_id": "9231-2024-on-12-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) r =a ( x + y) r2 =ax+ay M1 Uses x=rcos and y =rsin to\nr r\neliminate \nx2 −ax+ y2 −ay =0 M1 A1 OE Obtains Cartesian equation.\n(using r2 = x2 + y2.)\n(x−a)2 +(y−a)2 = a2  Centre (a,a) and radius a B1\n2 2 2 2 2 2\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 16\n\n7(b) B1 Circle\nB1 Correct position, passing throughO.\nO =0\na 2 B1 States maximum distance or labels sketch.\n3\n\n7(c) cos+sin−=0 M1\ncos1.25+sin1.25−1.25=0.01 cos1.26+sin1.26−1.26=−0.002 A1 Shows sign change.\n2\n© Cambridge University Press & Assessment 2024 Page 14 of 16\n\n7(d) a2   2d+a2 4 3π (cos+sin)2 d M1 Uses area formula on both curves.\n2 2\n0 \nA1 Forms area of smaller region enclosed by\nC and C with correct limits.\n1 2\n 3π M1 Applies relevant identities for the circle\na2  2d+ a2 4 1+2sincosd\n2 2 integral to produce integrable form.\n0 \n 3π A1 For correct integration of the circle part.\na2 13 +a2 +sin24\n2 3  0 2   \na23 + a2 ( 3π+ 1 −−sin2 ) = a2 ( 3 + 3π−+ 1cos2 ) A1 Integrates spiral and substitutes correct\n6 2 4 2 2 3 4 2 limits.\nAG.\na2 π− a2 ( 3 + 3π−+ 1cos2 ) M1 Forms area of larger region enclosed by\n2 2 3 4 2 C and C .\n1 2\na2 ( −3 + 1 π+− 1cos2 ) A1\n2 3 4 2\n© Cambridge University Press & Assessment 2024 Page 15 of 16\n\n7(d) Alternative method for question 7(d)\na2   (cos+sin)2 d−a2   2d M1 Uses area formula on both curves.\n2 −1π 2 0\n4\nA1 Forms area of larger region enclosed by\nC and C with correct limits.\n1 2\n  M1 Applies relevant identities for the circle\na2  1+sin2 d−a2  2d\n2 −1π 2 0 integral to produce integrable form.\n4\na2− 1cos2 − a2 13  A1 For correct integration of the circle part.\n2 2 −π 4 2 3  0\na2 (− 1cos2+ 1 π)− a23 = a2 ( −3 + 1 π+− 1cos2 ) A1 Integrates spiral and substitutes correct\n2 2 4 6 2 3 4 2 limits.\nAnswer not given.\na2 π− a2 ( −3 + 1π+− 1cos2 ) M1 Forms area of smaller region enclosed by\n2 2 3 4 2 C and C .\n1 2\na2 ( 3 + 3π−+ 1cos2 ) A1 AG\n2 3 4 2\n7\n© Cambridge University Press & Assessment 2024 Page 16 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-13-q01",
      "question_id": "9231-2024-on-13-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "1(a) k 0 B1 [Stretch parallel to the x-axis, scale factor k\n  (k0)]. (Allow without identification.)\n0 1\n1 k B1 [Shear, x-axis fixed, with (0,1) mapped to\n \n0 1 (k,1).] (Allow without identification.)\n1 kk 0 k k M1 A1 Correct order for M1, must have identified\nM=  =  which matrix gives which transformation, AG.\n0 10 1 0 1\n4\n\n1(b) k kx kx+ky B1 x X\n  =  Transforms   to  \n0 1y  y  y Y \nkx+ky=x M1 X x\nSets  = \nY  y\n1−k A1\ny= x oe\nk\n3\n\n1(c) 11 −k B1 (An alternative is possible.)\nM−1=\n \nk0 k \n1\n© Cambridge University Press & Assessment 2024 Page 6 of 15\n\n1(d) |k|=3k2 M1 Uses that detM=k. Without modulus is SC\nB1.\nk 0k =1 A1\n3\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2024-on-13-q02",
      "question_id": "9231-2024-on-13-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 d ( tan−1x ) B1 Differentiates once.\n1\n= so true when n=1.\ndx 1+x2\nAssume that d d x k k ( tan−1x ) =P k (x)( 1+x2)−k , where degP k (x)=k−1. B1 S d t e a g te P s k i ( n x d ) u = ct k iv − e 1 h . y pothesis. Must have\ndk+1( tan−1x ) M1 A1 Differentiates kth derivative using the product\n=P '(x)( 1+x2)−k −2kxP (x)( 1+x2)−k−1 rule.\ndxk+1 k k\n= ( P '(x)( 1+x2) −2kxP (x) )( 1+x2)−k−1 so degP (x)=k A1 Writes in the form P (x)( 1+x2)−k−1 .\nk k k+1 k+1\nSo true when n=k+1. By induction, true for all positive integers n. A1 Attempts to show degree of P (x) is at most k\nk+1\n(condone not showing coefficient of xk is non-\nzero) and states conclusion.\n6\n© Cambridge University Press & Assessment 2024 Page 7 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-13-q03",
      "question_id": "9231-2024-on-13-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) y= x4 B1 Uses correct substitution.\ny+2y 3 4 −1=016y3 =(1−y)4 M1 Substitutes and obtains an equation not\ninvolving radicals.\n16y3 =1−4y+6y2 −4y3+ y4 M1 Uses binomial expansion.\ny4 −20y3 +6y2 −4y+1=0 A1 Must be an equation.\n4 +4 +4 +4 =20 B1\n5\n\n3(b) +++=−2 B1\nx5+2x4−x=05+5+5+5 =−2(20)+(−2) M1 Multiplies original equation by x and substitutes.\n−42 A1\n3\n\n3(c) 8+8+8+8 =202−2(6) M1 Uses formula for sum of squares. Or alternative\ncomplete method eg another substitution.\n388 A1 CAO.\n2\n© Cambridge University Press & Assessment 2024 Page 8 of 15",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_13-p08.png"
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    },
    {
      "id": "9231-2024-on-13-q04",
      "question_id": "9231-2024-on-13-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) 5k  1 1  M1 A1 Finds partial fractions. (Don’t allow a\n=k\n\n−\n substitution of a value for k.)\n(5r+k)(5r+5+k) 5r+k 5r+5+k\nn 5k  1 1 1 1 1 1  M1 Writes at least three terms, including last.\n =k  − + − + + −  (Allow any value of k.)\n(5r+k)(5r+5+k) 5+k 10+k 10+k 15+k 5n+k 5n+5+k\nr=1\n 1 1  A1\n=k − \n5+k 5n+5+k\n4\n\n4(b) k 1 5 M1 A1\n= 3k =5+kk =\n5+k 3 2\n2\n\n4(c) n2 5k n2 5k n−1 5k M1 Or applies the method of differences again.\n = −\n(5r+k)(5r+5+k) (5r+k)(5r+5+k) (5r+k)(5r+5+k)\nr=n r=1 r=1\n 1 1   1 1   1 1  A1FT FT on their value of k (must be substituted in).\n=k − −k − =k − \n5+k 5n2 +5+k 5+k 5n+k 5n+k 5n2 +5+k\n1 1\n= −\n2n+1 2n2 +3\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_13-p09.png"
      ]
    },
    {
      "id": "9231-2024-on-13-q05",
      "question_id": "9231-2024-on-13-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) r4 =6r2sincos M1 Substitutes x=rcos, y=rsin and applies\nsin2=2sincos.\nr2 =3sin2 A1 AG.\n2\n\n5(b) B1 Correct position and symmetrical about = 1π.\n4\nB1 Single correct loop.\n=0\n3 B1 States maximum distance or labels sketch.\n( ) ( )\nAllow 3,1π but not 1π, 3 . Allow 3sf.\n4 4\n3\n\n5(c) 3 π 2sin2d= 3  −1cos2  π 2 M1 Forms 1 2 r2 d. (Allow with wrong limits.)\n2 0 2 2 0\n3 A1\n2\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 15\n\n5(d) 1 1 B1\ny=32sin22sin\nsin 1 2 2cos+sin−1 2 2cos2sin=0 M1 A1 Sets dy =0.\nd\n2tan M1 Applies suitable trigonometric identity.\nsin2cos+cos2sin=0tan2=−tan =−tan\n1−tan2 Accept sin3=0.\n= 1π A1\n3\n5 A1 AEF.\n34\n=1.40\n3\n22\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2024-on-13-q06",
      "question_id": "9231-2024-on-13-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 B1 Vertical asymptotes.\nx= , x=3\n2\ny=2 B1 Horizontal asymptote.\n2\n© Cambridge University Press & Assessment 2024 Page 11 of 15\n\n6(b) dy (2x2 −7x+3)(8x+1)−(4x2 +x+1)(4x−7) M1 dy\n= Finds . Allow top line only for M1.\ndx ( 2x2 −7x+3 )2 dx\n−3x2 +2x+1=0 M1 Sets equal to 0 and forms equation.\n( −1,1) , (1,−3) A1 A1\n3 5\n4\nxy\n\n6(c) B1 Axes and asymptotes. Clear identification (label\nor clear intersection with axes at correct place).\nB1 x3 correctly approaching asymptotes, not too\ntruncated.\nB1 1 x3 correct.\n2\nB1 x 1 correct.\n2\n( 0,1) B1 States coordinates of intersection with axis. May\n3 be seen on their graph.\n5\n© Cambridge University Press & Assessment 2024 Page 12 of 15\n\n6(d) B1FT FT from sketch in (c). At least two branches.\nk3 B1\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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      ]
    },
    {
      "id": "9231-2024-on-13-q07",
      "question_id": "9231-2024-on-13-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k  6  −2 M1 A1 Finds common perpendicular. Allow one error.\n   \n2 1 1 = −3 ~ 1\n   \n   \n1 −4 2 −9  3 \n−2(1)+(3)+3(−2)=−5 M1 Substitutes point on l .\n1\n2x−y−3z=5 A1 CAO.\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 15\n\n7(b) i j k 6 M1 A1 Finds the normal to  .\n2\n \n1 −4 2 = 4\n \n \n1 1 −2 5\n 2  6 M1 Uses dot product of normal vectors.\n    −7\n−1 4 = 14 77coscos=\n   \n    14 77\n−3 5\n77.7 A1 No ISW. Accept 1.36 rad.\n4\n© Cambridge University Press & Assessment 2024 Page 14 of 15\n\n7(c) 1+2  1+   −2  M1 A1 Finds PQ.\n     \nOP= 3+ ,OQ= −2−4 PQ= −5−4−\n     \n     \n−2+  9+2 11+2−\n −2  2  −2   6  M1 Uses that dot product of PQ with line direction\n       \n−5−4− 1 =0 or −5−4− =k −3\n        is zero, or, alternatively, PQ is a multiple of the\n       \n11+2− 1 11+2− −9\ncommon perpendicular.\n−6+6=0 A1 Deduces one equation.\n −2   1  A1 Deduces second equation.\n   \n−5−4− −4 =021+42=0\n   \n   \n11+2−  2 \n 3  −1 M1 Solves for  or  and substitutes into OP or\n   \n=1OP= 4 or =−2OQ= 6\n    OQ\n   \n−1  5 \n−1  6  A1 FT OE. FT using their common perpendicular. Must\n    have\"r=\".\nr= 6 +t −3\n   \n   \n 5  −9\n7\n© Cambridge University Press & Assessment 2024 Page 15 of 15",
      "source_pages": [
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        15
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2024-on-21-q01",
      "question_id": "9231-2024-on-21-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 1 5 6 M1A1 Evaluates determinant.\n2 2 k 2 k 2\nk 2 2 = −5 +6 =8−5(8k+6)+6(4k+6)\n4 8 −3 8 −3 4\n−3 4 8\nk  7 A1\n8\nThree planes intersect at a unique point. B1\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-21-q02",
      "question_id": "9231-2024-on-21-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) dy\n=\ndy\n\ndt\n=−\n1\n\n( −t2)\n=\nt2 M1A1 Uses chain rule, AG.\ndx dt dx 1−t2 1−t2\n2\n\n2(b) d dy 2t 1−t2 +t3( 1−t2)−1 2 M1A1 Applies Quotient rule.\n =\ndtdx 1−t2\nd2y\n=\n2t 1−t2 +t3( 1−t2)−1 2 ( −t2) =−2t3( 1−t2)−1\n2\n−t5( 1−t2)−3\n2\nM1A1 Uses chain rule.\ndx2 1−t2\n=−t3( 1−t2)−3 2 ( 2 ( 1−t2) +t2 ) =−t3( 1−t2)−3 2 ( 2−t2)\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-21-q03",
      "question_id": "9231-2024-on-21-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)\nA=2π\nln1\n5\n2\nex 1+e2x dx=2π\n1\n5\n2\nu 1+u2(1) du\nM1A1 Correct formula and applies substitution, AG.\nln4 4 u\n3 3\n2\n\n3(b) u=sinhvdu=coshvdv B1\nA=2π ln5 1+sinh2vcoshvdv A=2π ln5 cosh2vdv M1A1 Applies substitution andcosh2x=1+sinh2x.\nln3 ln3 Need limits for A1.\nπ ln5 (cosh2v+1)dv M1 Applies 2cosh2x=cosh2x+1.\nln3\nln5 A1 Correct integration.\n=π1sinh2v+v\n \n2 ln3\nπ (312+ln5−40−ln3 ) =π (904+ln5) A1 AG.\n50 18 225 3\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-21-q04",
      "question_id": "9231-2024-on-21-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) −11 1 8 1 −2 M1 Multiplies matrix with eigenvector.\n    \n0 −2 0 1 = −2\n    \n    \n−16 1 131 −2\n=−2 A1\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 14\n\n4(b) −11− 1 8 M1 Sets determinant equal to zero.\n0 −2− 0 =0\n−16 1 13−\n(−11−)(−2−)(13−)+128(−2−)=0 A1 Expands determinant, AG.\n(−2−)( 2 −2−15 ) =0\n3 −19−30=0\n=5,−3 B1\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n4(c) A3−19A−30I=0 B1 States that A satisfies its characteristic equation.\n30A−1 =A2 −19I M1 Multiplies through by A−1.\n −7 −5 16 −26 −5 16 M1A1 M1 for substituting A2.\n  1  \nA2 = 0 4 0 A−1 = 0 −15 0\n   \n30\n   \n−32 −5 41 −32 −5 22\n4\n© Cambridge University Press & Assessment 2024 Page 9 of 14",
      "source_pages": [
        7,
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_21-p09.png"
      ]
    },
    {
      "id": "9231-2024-on-21-q05",
      "question_id": "9231-2024-on-21-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 6m2−5m+1=0m=1,1 M1 Auxiliary equation, two distinct real roots.\n2 3\nx= Ae 1 2 t +Be 1 3 t A1 Complementary function.\nx= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives.\np=1 −10p+q=1 12p−5q+r=1 M1 Substitutes and equates coefficients.\np=1 q=11 r=44 A1\nx= Ae 1 2 t +Be 1 3 t +t2 +11t+44 A1 General solution. Need to see ‘x=’. FT on CF.\nx'=1Ae 1 2 t +1Be 1 3 t +2t+11 M1 Differentiates.\n2 3\nA+B+44=12 1 A+1B+11=−6 A=−38, B=6 M1A1 Uses initial conditions.\n2 3\nx=−38e 1 2 t +6e 1 3 t +t2 +11t+44 A1 Need to see ‘x=’.\n10\n© Cambridge University Press & Assessment 2024 Page 10 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-21-q06",
      "question_id": "9231-2024-on-21-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a)   1(1)x dx (1)(1) N 1 + (1)(1) N 2 + + (1)(1)N N −1 + (1)(1) N N M1A1 Forms the sum of the areas of the rectangles. M1\n  0 2   N 2 N 2 N 2 N 2 for correct number of rectangles.\n= 1  N (1) N n = (1 2 )N 1 (1 2 −1) = (1 2 )(1 2 )N 1 = 1 . M1A1 Applies  N rn = r ( rN −1 ) , AG.\nN n=1 2 N(1)N 1 −N N −N(1)N 1 2N ( 2N 1 −1 ) n=1 r−1\n2 2\n4\n\n6(b)  1(1)x dx (1) + (1)(1) N 1 + + (1)(1)N N −2 + (1)(1)N N −1 M1A1 Forms the sum of the areas of appropriate\n2 N N 2 N 2 N 2 rectangles.\n0\n1  N−1 (1) N n = 1 2 −1 = 1 = 2N 1 M1A1 Applies  N−1 rn = rN −1 .\nN n=0 2 N(1)N 1 −N 2N ( 1−(1) N 1 ) 2N ( 2N 1 −1 ) n=0 r−1\n2 2\n4\n\n6(c) 1 M1 c\n2N − 1 = 1 10−32N 103 Simplifies U N −L N to .\n( 1 ) ( 1 ) 2N N\n2N 2N −1 2N 2N −1\nLeast value of N is 500 A1 CWO.\n2\n\n6(d) L  1 U  1 ln2 1 M1A1FT Forms inequality. FT on their U .\nN 2ln2 N 2U 2L N\nN N\n0.693= 1 ln2 1 =0.694 A1A1 CWO. Must have used N =500.\n2U 2L\nN N\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 14",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_21-p11.png"
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    },
    {
      "id": "9231-2024-on-21-q07",
      "question_id": "9231-2024-on-21-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) dy + y =x B1 Divides through by x2 +16.\ndx x2 +16\n1 M1A1 Finds integrating factor\ne\n\nx2+16\ndx\n=e\nsinh−1(\n4\nx)\nA1 AG.\n= 1x+1 x2 +16\n4 4\n4\n\n7(b) d ( ( )) M1A1 Correct form on LHS and RHS.\ny x+ x2 +16 =x2+x x2+16\ndx\ny ( x+ x2 +16 ) =1x3+1 ( x2+16 )3 2 +C M1A1 Integrates RHS. RHS of the correct form.\n3 3\n6 ( 3+ 25 ) = 27+ 25 25+C M1 Substitutes initial conditions into their expression.\n3 3\ny ( x+ x2 +16 ) =1x3+1 ( x2 +16 )3 2 −8 A1 OE.\n3 3 3\n6\n© Cambridge University Press & Assessment 2024 Page 12 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-21-q08",
      "question_id": "9231-2024-on-21-q08",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) z+z−1 =2cos B1 For LHS.\n( z+z−1)7 = ( z7 +z−7) +7 ( z5 +z−5) +21 ( z3+z−3) +35 ( z+z−1) M1A1 Expands and groups. A0 if not grouped clearly.\n27cos7=2cos7+7(2cos5)+21(2cos3)+35(2cos) M1 Substitutes zn +z−n =2cosn.\ncos7= 1 cos7+ 7 cos5+21cos3+35cos A1\n64 64 64 64\n5\n\n8(b) 1π 1π 1π M1A1 Applies integration by parts.\nI =4 cosn−1cosd=cosn−1sin4 +(n−1)4 cosn−2sin2d\nn  \n0 0 0\nI =cosn−1sin 1 4 π +(n−1) 1 4 π cosn−2 ( 1−cos2 ) d M1 Applies sin2=1−cos2\nn  \n0 0\nI\n=2−n\n2 +(n−1)I −(n−1)I nI\n=2−n\n2 +(n−1)I\nA1 AG.\nn n−2 n n n−2\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(c) 9I =2−9 2 +8I = 1 ( 2−1 2 ) +8I M1A1 Applies reduction formula from (b).\n9 7 16 7\n1π M1 Applies identity from (a). Allow\nI =4 1 cos7+ 7 cos5+ 21cos3+35cosd\n7 64 64 64 64 missing/incorrect limits.\n0\n1π A1\nI = 1 1sin7+7sin5+ 21sin3+35sin4\n7 647 5 3  0\n9I = 1 ( 2−1 2 ) +1 ( 1 ( −2−1 2 ) +7 ( −2−1 2 ) +21 ( 2−1 2 ) +35 ( 2−1 2 )) A1 Check their exact answer when I 9 is the subject,\n9 16 8 7 5 3\nlike terms collected. ISW. 0.402237\nI = 2867 ( 2−1 2 ) = 2867 2\n9 5040 10080\n5\n© Cambridge University Press & Assessment 2024 Page 14 of 14",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2024-on-22-q01",
      "question_id": "9231-2024-on-22-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 7 M1A1 Applies formula.\n 1 7\n\ndx=\n\ncosh−1(x−5)\n\n (x−5)2 −12 6\n6\n=cosh−1(2)−cosh−1(1) M1 Uses limits.\n=ln\n(\n2+ 3\n) A1\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2024-on-22-q02",
      "question_id": "9231-2024-on-22-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) d\n( 4y2)\n=8yy'\nB1 Differentiates y2 correctly.\ndx\nd xy'+ y B1 Differentiates ln(xy) correctly.\n(4ln(xy))=4 =4y'y−1+4x−1\ndx xy\n8 (1) y'+4(2)y'+4 (1) =0y'=−1 B1 Substitutes (2,1) into\n2 2 6 2\n8y'y+4y'y−1+4x−1=0, AG.\n3\n© Cambridge University Press & Assessment 2024 Page 6 of 13\n\n2(b) EITHER Solution 1 (B1B1 Differentiates 8y'y+4y'y−1+4x−1=0.\n8(y')2 +8y''y−4(y')2 y−2+4y−1y''−4x−2 =0 B1 for 8(y')2 +8y''y and B1 for other terms.\n8(−1)2 +8y''(1)−16(−1)2 +8y''−1=0 M1) Substitutes ( 2,1) .\n6 2 6 2\nOR Solution 2 (B1B1 B1 for x−1( 2y+ y−1)−2( 2y'−y−2y' ) and B1\ny'=−x−1(\n2y+\ny−1)−1\nfor\nx−2(\n2y+\ny−1)−1\n.\ny''=x−1( 2y+ y−1)−2( 2y'−y−2y' ) +x−2( 2y+ y−1)−1\ny''=1(1+2)−2(−1)(2−4)+1(1+2)−1 M1) Substitutes ( 2,1) and y'=−1.\n2 6 4 2 6\n12y''−11=0 y''= 11 A1\n9 108\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
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    },
    {
      "id": "9231-2024-on-22-q03",
      "question_id": "9231-2024-on-22-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 x=e2t −t2, y=2et +2(t−1)et =2tet B1B1 Differentiates x and y with respect to t.\nx2 + y2 = ( e2t −t2)2 +4t2e2t =e4t +2t2e2t +t4 = ( e2t +t2)2 M1A1 Factorises x2 + y2.\n1 1 M1A1 Applies correct formula for arc length.\n e2t +t2dt =1e2t +1t3\n0 2 3  0\n= 1e2 −1 A1\n2 6\n7\n© Cambridge University Press & Assessment 2024 Page 7 of 13",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2024-on-22-q04",
      "question_id": "9231-2024-on-22-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) cos6=Re(cos+isin)6 =cos6−15cos4sin2+15cos2sin4−sin6 M1A1 Expands and takes real part.\nsin6=Im(cos+isin)6 =6cos5sin−20cos3sin3+6cossin5 M1A1 Takes imaginary part\ncos6−15cos4sin2+15cos2sin4−sin6 sin−6 M1 Divides numerator and denominator by\ncot6= \n6cos5sin−20cos3sin3+6cossin5 sin−6 sin6.\ncot6−15cot4+15cot2−1 A1 AG\ncot6=\n6cot5−20cot3+6cot\n6\n\n4(b) x=cot, cot6=1 B1 Applies identity given in (b).\n6= 1+k M1 Solves cot6=1\n4\ncot ( 1 π ) A1 Gives one correct solution.\n24 7.60\ncot ( 5 π ) ,cot (3π ) ,cot (13π ) ,cot (17π ) ,cot (7π ) A1 Gives other five solutions.\n24 8 24 24 8 1.30, 0.41, –0.13, –0.77, –2.41\nWithhold A1 mark for repeated roots.\n4\n© Cambridge University Press & Assessment 2024 Page 8 of 13",
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      "html": "9231-topic-25-complex-numbers/answers.html",
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        "../answer-assets/9231_w24_ms_22-p08.png"
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    },
    {
      "id": "9231-2024-on-22-q05",
      "question_id": "9231-2024-on-22-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 3m2 +2m+1=0m=−1 2i M1 Auxiliary equation. M1 for correct type of\n3 3 roots (complex).\ny=e−1 3 x( Acos 2 x+Bsin 2 x ) A1 Complementary function.\n3 3\ny= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives.\np=1 4p+q=0 6p+2q+r=0 M1 Substitutes and equates coefficients.\nq=−4 r=2 A1\ny=e−1 3 x( Acos 2 x+Bsin 2 x ) +x2 −4x+2 A1FT General solution. Must have ‘y=’. FT on\n3 3 CF.\ny'=e−1 3 x( − 2 Asin 2 x+ 2Bcos 2 x ) −1e−1 3 x( Acos 2 x+Bsin 2 x ) +2x−4 *M1 Differentiates. GS must be in correct form.\n3 3 3 3 3 3 3\nA+2=0 2B−1A−4=0\n A=−2,B=5 2\nDM1A1 Uses initial conditions.\n3 3\ny=e−1 3 x( −2cos 2 x+5 2sin 2 x ) +x2 −4x+2 A1 Must have ‘ y=’.\n3 3\n10\n© Cambridge University Press & Assessment 2024 Page 9 of 13",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_22-p09.png"
      ]
    },
    {
      "id": "9231-2024-on-22-q06",
      "question_id": "9231-2024-on-22-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a)   1 e1−xdx (1) e1−0 n + (1) e1−1 n + + (1) e1−n n −1 M1A1 Forms the sum of the areas of the rectangles.\n  0   n n n M1 for correct number of rectangles.\ne n−1 e1−e−1  e−1 M1A1 Applies sum of geometric progression, AG.\n= e− n r =  =\nn r=0 n 1−e−1 n  n ( 1−e−1 n )\n4\n\n6(b)  1 e1−x dx(1)e1−1 n +(1)e1− n 2 +(1)e1−n n −1 +(1)e1− n n M1A1 Forms the sum of the areas of appropriate\nn n n n rectangles. M1 for correct number rectangles.\n0\ne−1 e 1 e−1 M1A1 Applies sum of geometric progression.\n= − + =\nn ( 1−e−1 n ) n n n ( e 1 n −1 ) e−1 n (e−1) e1−1 n ( 1−e−1)\nOE. E.g. or .\nn ( 1−e−1 n ) n ( 1−e−1 n )\n4\n\n6(c) e−1 M1 c\nU −L = Simplifies U −L to , where c is a\nn n n n n n\nnonzero constant.\ne−1 A1 AG. CWO.\n→0 as n→\nn\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 13\n\n6(d) z ( 1−e−1 z ) =z ( 1− ( 1−1+ 1 − 1 + )) =1− 1 + 1 + M1A1 Substitutes power series for e−1 z. A1 for\nz 2z2 6z3 2z 6z2\nenough terms.\nlim(U )=e−1 B1\nn\nn→\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_22-p10.png",
        "../answer-assets/9231_w24_ms_22-p11.png"
      ]
    },
    {
      "id": "9231-2024-on-22-q07",
      "question_id": "9231-2024-on-22-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) d sech2x M1A1 Applies chain rule.\n(ln(tanhx))=\ndx tanhx\n1 2 A1 AG.\n= = .\nsinhxcoshx sinh2x\n3\n\n7(b) dy 2y B1 Divides through by sinh(2x).\n+ =1\ndx sinh(2x)\nd M1A1 Correct form on LHS and RHS.\n(ytanhx)=tanhx\ndx\nytanhx=ln(coshx)+C M1A1 Integrates RHS.\n5 (3) =ln5+C M1 Substitutes initial conditions.\n5 4\nytanhx=ln(coshx)+3−ln5 A1 Accept equivalent exact form.\n4\n7\n© Cambridge University Press & Assessment 2024 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_22-p11.png"
      ]
    },
    {
      "id": "9231-2024-on-22-q08",
      "question_id": "9231-2024-on-22-q08",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) −2− 0 0 B1 Sets determinant equal to zero.\n0 7− 9 =0\n4 1 7−\n(−2−) ( (7−)2 −9 ) =0(−2−)( 49−14+2 −9 ) M1A1 Expands, AG\n3−122 +12+80=0\n=−2,4,10 B1\n4\n\n8(b) A3−12A2 +12A+80I=0A4 −12A3+12A2 +80A=0 M1A1 Substitutes A and multiplies through by A.\nA4 =12 ( 12A2 −12A−80I ) −12A2 −80A=132A2 −224A−960I M1A1 Substitutes A3.\n4\n© Cambridge University Press & Assessment 2024 Page 12 of 13\n\n8(c) i j k  72   2  *M1A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\n=−2: 0 9 9 = 36 1\n   \n   \n4 1 9 −36 −1\ni j k  0   0  i j k  0  0 A1A1\n       \n=4: 0 3 9 = 36 ~ 3 =10: −12 0 0 = 108 ~ 3\n       \n       \n4 1 3 −12 −1 0 −3 9  36  1\n 2 0 0 625 0 0  DM1A1 Or correctly matched permutations of\n    columns.\nThus P= 1 3 3 and D= 0 1 0\n   \n   \n−1 −1 1  0 0 2401 Correctly matched but eigenvalues not\nchanged scores M1 A0.\nM0 if a column of zeros appears in P.\n6\n© Cambridge University Press & Assessment 2024 Page 13 of 13",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_22-p13.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q01",
      "question_id": "9231-2024-on-23-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 1 5 6 M1A1 Evaluates determinant.\n2 2 k 2 k 2\nk 2 2 = −5 +6 =8−5(8k+6)+6(4k+6)\n4 8 −3 8 −3 4\n−3 4 8\nk  7 A1\n8\nThree planes intersect at a unique point. B1\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q02",
      "question_id": "9231-2024-on-23-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) dy\n=\ndy\n\ndt\n=−\n1\n\n( −t2)\n=\nt2 M1A1 Uses chain rule, AG.\ndx dt dx 1−t2 1−t2\n2\n\n2(b) d dy 2t 1−t2 +t3( 1−t2)−1 2 M1A1 Applies Quotient rule.\n =\ndtdx 1−t2\nd2y\n=\n2t 1−t2 +t3( 1−t2)−1 2 ( −t2) =−2t3( 1−t2)−1\n2\n−t5( 1−t2)−3\n2\nM1A1 Uses chain rule.\ndx2 1−t2\n=−t3( 1−t2)−3 2 ( 2 ( 1−t2) +t2 ) =−t3( 1−t2)−3 2 ( 2−t2)\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 14",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p06.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q03",
      "question_id": "9231-2024-on-23-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)\nA=2π\nln1\n5\n2\nex 1+e2x dx=2π\n1\n5\n2\nu 1+u2(1) du\nM1A1 Correct formula and applies substitution, AG.\nln4 4 u\n3 3\n2\n\n3(b) u=sinhvdu=coshvdv B1\nA=2π ln5 1+sinh2vcoshvdv A=2π ln5 cosh2vdv M1A1 Applies substitution andcosh2x=1+sinh2x.\nln3 ln3 Need limits for A1.\nπ ln5 (cosh2v+1)dv M1 Applies 2cosh2x=cosh2x+1.\nln3\nln5 A1 Correct integration.\n=π1sinh2v+v\n \n2 ln3\nπ (312+ln5−40−ln3 ) =π (904+ln5) A1 AG.\n50 18 225 3\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p07.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q04",
      "question_id": "9231-2024-on-23-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) −11 1 8 1 −2 M1 Multiplies matrix with eigenvector.\n    \n0 −2 0 1 = −2\n    \n    \n−16 1 131 −2\n=−2 A1\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 14\n\n4(b) −11− 1 8 M1 Sets determinant equal to zero.\n0 −2− 0 =0\n−16 1 13−\n(−11−)(−2−)(13−)+128(−2−)=0 A1 Expands determinant, AG.\n(−2−)( 2 −2−15 ) =0\n3 −19−30=0\n=5,−3 B1\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n4(c) A3−19A−30I=0 B1 States that A satisfies its characteristic equation.\n30A−1 =A2 −19I M1 Multiplies through by A−1.\n −7 −5 16 −26 −5 16 M1A1 M1 for substituting A2.\n  1  \nA2 = 0 4 0 A−1 = 0 −15 0\n   \n30\n   \n−32 −5 41 −32 −5 22\n4\n© Cambridge University Press & Assessment 2024 Page 9 of 14",
      "source_pages": [
        7,
        8,
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p07.png",
        "../answer-assets/9231_w24_ms_23-p08.png",
        "../answer-assets/9231_w24_ms_23-p09.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q05",
      "question_id": "9231-2024-on-23-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 6m2 −5m+1=0m=1,1 M1 Auxiliary equation, two distinct real roots.\n2 3\nx= Ae 1 2 t +Be 1 3 t A1 Complementary function.\nx= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives.\np=1 −10p+q=1 12p−5q+r=1 M1 Substitutes and equates coefficients.\np=1 q=11 r=44 A1\nx= Ae 1 2 t +Be 1 3 t +t2 +11t+44 A1 General solution. Need to see ‘x=’. FT on CF.\nx'=1Ae 1 2 t +1Be 1 3 t +2t+11 M1 Differentiates.\n2 3\nA+B+44=12 1 A+1B+11=−6 A=−38, B=6 M1A1 Uses initial conditions.\n2 3\nx=−38e 1 2 t +6e 1 3 t +t2 +11t+44 A1 Need to see ‘x=’.\n10\n© Cambridge University Press & Assessment 2024 Page 10 of 14",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2024-on-23-q06",
      "question_id": "9231-2024-on-23-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a)   1(1)x dx (1)(1) N 1 + (1)(1) N 2 + + (1)(1)N N −1 + (1)(1) N N M1A1 Forms the sum of the areas of the rectangles. M1\n  0 2   N 2 N 2 N 2 N 2 for correct number of rectangles.\n= 1  N (1) N n = (1 2 ) N 1 (1 2 −1) = (1 2 )(1 2 ) N 1 = 1 . M1A1 Applies  N rn = r ( rN −1 ) , AG.\nN n=1 2 N(1) N 1 −N N −N(1) N 1 2N ( 2N 1 −1 ) n=1 r−1\n2 2\n4\n\n6(b)  1(1)x dx(1)+(1)(1) N 1 + +(1)(1)N N −2 +(1)(1)N N −1 M1A1 Forms the sum of the areas of appropriate\n2 N N 2 N 2 N 2 rectangles.\n0\n1  N−1 (1) N n = 1 2 −1 = 1 = 2N 1 M1A1 Applies  N−1 rn = rN −1 .\nN n=0 2 N(1) N 1 −N 2N ( 1−(1) N 1 ) 2N ( 2N 1 −1 ) n=0 r−1\n2 2\n4\n\n6(c)\n(\n2N 1\n1 ) − (\n1\n1 ) = 2\n1\nN 10−32N 103\nM1\nSimplifies U N −L N to N\nc\n.\n2N 2N −1 2N 2N −1\nLeast value of N is 500 A1 CWO.\n2\n\n6(d) L  1 U  1 ln2 1 M1A1FT Forms inequality. FT on their U .\nN 2ln2 N 2U 2L N\nN N\n0.693= 1 ln2 1 =0.694 A1A1 CWO. Must have used N =500.\n2U 2L\nN N\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 14",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_23-p11.png"
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    },
    {
      "id": "9231-2024-on-23-q07",
      "question_id": "9231-2024-on-23-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) dy y B1 Divides through by x2 +16.\n+ =x\ndx x2 +16\n1 M1A1 Finds integrating factor\ne\n\nx2+16\ndx\n=e\nsinh−1(\n4\nx)\nA1 AG.\n= 1x+1 x2 +16\n4 4\n4\n\n7(b) d ( ( )) M1A1 Correct form on LHS and RHS.\ny x+ x2 +16 =x2+x x2+16\ndx\ny ( x+ x2 +16 ) =1x3+1 ( x2+16 )3 2 +C M1A1 Integrates RHS. RHS of the correct form.\n3 3\n( ) M1 Substitutes initial conditions into their expression.\n6 3+ 25 = 27+ 25 25+C\n3 3\ny ( x+ x2 +16 ) =1x3+1 ( x2 +16 )3 2 −8 A1 OE.\n3 3 3\n6\n© Cambridge University Press & Assessment 2024 Page 12 of 14",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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      ]
    },
    {
      "id": "9231-2024-on-23-q08",
      "question_id": "9231-2024-on-23-q08",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) z+z−1 =2cos B1 For LHS.\n( z+z−1)7 = ( z7 +z−7) +7 ( z5 +z−5) +21 ( z3+z−3) +35 ( z+z−1) M1A1 Expands and groups. A0 if not grouped clearly.\n27cos7=2cos7+7(2cos5)+21(2cos3)+35(2cos) M1 Substitutes zn +z−n =2cosn.\ncos7= 1 cos7+ 7 cos5+21cos3+35cos A1\n64 64 64 64\n5\n\n8(b) 1π 1π 1π M1A1 Applies integration by parts.\nI =4 cosn−1cosd=cosn−1sin4 +(n−1)4 cosn−2sin2d\nn  \n0 0 0\nI =cosn−1sin 1 4 π +(n−1) 1 4 π cosn−2 ( 1−cos2 ) d M1 Applies sin2=1−cos2\nn  \n0 0\nI\n=2−n\n2 +(n−1)I −(n−1)I nI\n=2−n\n2 +(n−1)I\nA1 AG.\nn n−2 n n n−2\n4\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n8(c) 9I =2−9 2 +8I = 1 ( 2−1 2 ) +8I M1A1 Applies reduction formula from (b).\n9 7 16 7\n1π M1 Applies identity from (a). Allow\nI =4 1 cos7+ 7 cos5+ 21cos3+35cosd\n7 64 64 64 64 missing/incorrect limits.\n0\n1π A1\nI = 1 1sin7+7sin5+ 21sin3+35sin4\n7 647 5 3  0\n9I = 1 ( 2−1 2 ) +1 ( 1 ( −2−1 2 ) +7 ( −2−1 2 ) +21 ( 2−1 2 ) +35 ( 2−1 2 )) A1 Check their exact answer when I 9 is the subject,\n9 16 8 7 5 3\nlike terms collected. ISW. 0.402237\nI = 2867 ( 2−1 2 ) = 2867 2\n9 5040 10080\n5\n© Cambridge University Press & Assessment 2024 Page 14 of 14",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_23-p14.png"
      ]
    },
    {
      "id": "9231-2024-on-31-q01",
      "question_id": "9231-2024-on-31-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1  1  M1 For one equation with one error.\nUse equation of trajectory with point (56, H) or 84, H\n 2 \nA1 Both correct.\n5g 1 5g\nH =112− 562 or H =168− 842\n2u2 2 2u2\nEliminate to find u or H M1\nu=35 A1\nH = 48 A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p06.png"
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    },
    {
      "id": "9231-2024-on-31-q02",
      "question_id": "9231-2024-on-31-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 mv2 B1\nAt B, T+mgsin=\na\n1 1 M1A1\nEnergy A to B: mu2 − mv2 =mga(cos+sin)\n2 2\nSubstitute for u and  to find T: M1\n 4 3\nT =mg5−2 −3 \n 5 5\n8 A1\nT = mg\n5\n5\n© Cambridge University Press & Assessment 2024 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p06.png"
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    },
    {
      "id": "9231-2024-on-31-q03",
      "question_id": "9231-2024-on-31-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 2mg M1 Equilibrium position.\nHooke’s law: T = extension and T =mg\n2\nExtension = 1 m A1\n1 2mg B1\nEPE loss = \n(1+d)2\n2 2\nGain in GPE = mg(2+1+d) B1\n1 M1\nEquate: mg(1+d)2 =mg(3+d)\n2\nd = 5 A1 SC: 3 marks for final answer of 5+1 .\nSC: 2 marks for final answer of 5+k , k 1 .\n6\n\n3(b) 1 1 2mg M1 GPE, KE, EPE terms.\nEnergy equation: mV2 +mg(1+d)=  (1+d)2\n2 2 2\nV2 =g ( d2 −1 ) V = 40 =2 10 A1\nAlternatively: M1\nUsing KE and GPE from 2 m below O to point O\n1\nmV2 =2mg\n2\nV2 =4g V = 40 =2 10 A1\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p07.png"
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    },
    {
      "id": "9231-2024-on-31-q04",
      "question_id": "9231-2024-on-31-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) B1 Correct volumes and distances for large and small.\nLarge Small Object\nM1 Moments equation with 3 terms, dimensionally\ncorrect.\n2  2  4 \nVolume a2h   a  kh a2h1− k\n3   9  A1 Correct, unsimplified.\nCentre of 1 1\nmass from h kh x\nAB\n2 2\nMoments about AB:\n2\n 4  1 2  1\na2h1− k y = a2h h−\n\na\n\nkh kh\n 9  2 3  2\n( 9−4k2)\nh\nA1\ny =\n2(9−4k)\n4\n\n4(b)\ny\n( 9−4k2)\nh 3\nB1 FT FT their part (a)\ntan= : =\na 2(9−4k)a 2\n8 M1\nUse h= a and simplify to quadratic in k: 32k2 −36k+9=0\n3\n3 3 A1\nk = ,\n8 4\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p08.png"
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    },
    {
      "id": "9231-2024-on-31-q05",
      "question_id": "9231-2024-on-31-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) dx  4  M1 Separate variables, obtain RHS in integrable form.\n= −1dt\nx t+1 \nln x =4ln t+1−t+ A A1\nt=0, x=5: A=ln5 M1\nx=5(t+1)4 e−t A1\n4\n\n5(b) v=(3−t)5(t+1)3 e−t M1\nAcceleration = dv =5e−t ( −(t+1)3 +(3−t)3(t+1)2 −(3−t)(t+1)3)\ndt\nAcceleration =\n5e−t(t+1)2(5−t)(1−t) AEF\nF=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) M1\nAt t=3, magnitude of force is 640e−3 N A1 31.9 N\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p09.png"
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    },
    {
      "id": "9231-2024-on-31-q06",
      "question_id": "9231-2024-on-31-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) At P:  T cos=T cos+0.05g B1\n1 2\nAt Q:  T cos=0.04g\n2 B1 OR: whole system: T cos=0.09g\n1\nT =0.15g =1.5 N B1\n1\n3\n\n6(b) T sin−T sin=0.050.82 M1 Allow sin/cos mix\n1 2\nT\n2\nsin=0.041.42\nM1\nT sin=0.050.82 +0.041.42 A1\n1\n5\n2 =12.5,  = 2\n2\n3\n\n6(c) T cos=0.04g and T sin=0.041.42 M1 From part (a) and part (b)\n2 2\n7\nDivide: tan=\n4\n=60.3 A1\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_31-p10.png"
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    },
    {
      "id": "9231-2024-on-31-q07",
      "question_id": "9231-2024-on-31-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) When P strikes plane, velocity is →ucos, M1 3\nu\n5\n3 3\nBefore impact: parallel to inclined plane ucos, perpendicular to plane usin\n5 5\n3 3 A1\nAfter impact: components ucos (parallel) and eusin (perpendicular)\n5 5\n3 3 M1\nSince velocity is vertical after impact, tan= ucos/ eusin\n5 5\ntan=1/etan, e tan2 =1 A1 AG\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 12\n\n7(b) (usin)2 8u2 M1A1 Note: alternative methods.\nGreatest height of P before impact: H = =\n2g 25g\n3 M1\nAfter impact, vertical speed of P is u\n(cos)2 +e2(sin)2\n5\n3 M1\nUse V2 =U2 +2as to greatest height, equal to H\n16\n9 u2 ( (cos)2 +e2(sin)2) =2g 3 H\n25 16\n1 e 1 M1\nUse part (a): tan= , cos= , sin=\ne 1+e 1+e\nSubstitute to find e\n1 A1\n3e2 +2e−1=0, e=\n3\n6\n© Cambridge University Press & Assessment 2024 Page 12 of 12",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2024-on-32-q01",
      "question_id": "9231-2024-on-32-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 100 B1 100\nHooke’s law: T = x T = (r−0.8)\n0.8 0.8\nN2L: T =2(0.8+x)52 B1 T =2r52\nEquate and solve: 50(0.8+x)=125x, x=0.533 B1 8\n15\n3\n© Cambridge University Press & Assessment 2024 Page 6 of 19",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_32-p06.png"
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    },
    {
      "id": "9231-2024-on-32-q02",
      "question_id": "9231-2024-on-32-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 1 5mg  a 2  B1 Both terms seen.\nExtension when P comes to rest is x, EPE loss =    −x2\n2 a  2 \n \n1 5mg  a 2  a  M1 At least one EPE term and a GPE term,\nEnergy:    −x2=mg −x dimensionally correct.\n2 a  2  2 \n \nSolve: 20x2 −8ax−a2 =0 M1 Obtain homogeneous quadratic equation in x and a\nMust come from an energy equation involving two\nEPE terms.\nNote that the correct case simplifies to a linear\n5mga \nequation  +x=mg and this scores M1.\n2a 2 \na 9 A1\nx=− , so distance of P below O is a\n10 10\n© Cambridge University Press & Assessment 2024 Page 7 of 19\n\n2 Alternative method for question 2\nDistance of P below O when it comes to rest comes to rest is h B1\nEPE loss = =\n1\n\n5mg\n\n\n\n\n\na\n\n2 −(h−a)2 \n\n2 a  2 \n \nEnergy:\n1\n\n5mg\n\n\n\n\n\na\n\n2 −(h−a)2 \n=mg\n\n\n3\na−h\n\n\nM1 A\ndi\nt\nm\nle\ne\na\nn\ns\ns\nt\ni o\no\nn\nn\na\ne\nl l\nE\ny\nP\nc\nE\no r\nt\nr\ne\ne\nr\nc\nm\nt.\nand a GPE term,\n2 a  2  2 \n \n20h2 −48ah+27a2 =0 M1 Obtain homogeneous quadratic/linear equation in\nh and a. Must come from an energy equation\ninvolving two EPE terms.\n9 A1\n(10h−9a)( 2h−3a)=0, h= a\n10\n4\n© Cambridge University Press & Assessment 2024 Page 8 of 19",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2024-on-32-q03",
      "question_id": "9231-2024-on-32-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) PCLM along line of centres: −mv=m2ucos−m3usin B1 Must include m, must have minus sign on RHS,\naccept positive or negative v.\nIf velocity of B after collision is included, it must\nbe equated to zero before this mark is awarded.\nNEL: v=eu(3sin+2cos) M1 Must have plus sign on RHS,\naccept positive or negative v (sign of v does not\nneed to be consistent with PCLM equation)\nIf velocity of B after collision is included, it must\nbe equated to zero before this mark is awarded.\n4 A1 Correct work only, except possibly missing m.\nEliminate v: 6sin=8cos, tan=\n3\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 19\n\n3(b) Only change in KE is along line of centres M1\nLoss = 1 m ( (2ucos)2 +(3usin)2) − 1 mv2\n2 2\n1 36 144 12 6 2 72 A1 (Note that tan= 2 leads to final answer 36mu2)\nmu2 + − −  = mu2 3 13\n2 25 25  5 5  25\n \n2\nAlternative method for question 3(b)\nAlternative method, using total KE M1\n1 1\nLoss in KE = [\nm(2u)2\n+\nm(3u)2\n]\n2 2\nOr equivalent, with all necessary terms present\n− 1 mv2 + 1 m(2usin)2 + 1 m(3ucos)2\n \n2 2 2 \n13 181 72 A1\nmu2 − mu2 = mu2\n2 50 25\n2\n\n3(c) 6 8 M1\n[Components of velocity of A after collision are  u  u so] angle between\n5 5\nline of centres and A’s direction is .\nAngle of deflection = 180 – 2tan-1 4/3 = 73.7 A1FT FT their answer to part (a)\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 19",
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      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2024-on-32-q04",
      "question_id": "9231-2024-on-32-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) In this question, allow equivalent marks for resolutions in different directions and B1\nmoments about other points. Apply the guidance given in the main scheme.\nTcos+F =2W\n→Tsin=R B1\nMoments about A: M1 All relevant terms included, dimensionally correct,\nTcos3asin+Tsin3acos=W3asin+Wkasin forces must be resolved if appropriate.\nAllow sin/cos mix, allow sign errors.\nLHS: any equivalent expression, for example\n3Tasin2, 3Tasin(180−2).\nOR\nAll relevant terms included, dimensionally correct,\nMoments about C:\nforces must be resolved if appropriate.\nR3acos=F3asin+W(ka−3a)sin\nAllow sin/cos mix, allow sign errors.\nT M1 1\n[6aTcos=(3+k)aW and (3cos+sin)=2W give]  and giveand give Use F = R and eliminate T and W to obtain an\n3 3\n1 expression in k and , dependent on a\n12cos=(3+k)(cos+ sin)\n3 dimensionally correct moments equation.\nk=5 A1\n5\n© Cambridge University Press & Assessment 2024 Page 11 of 19\n\n4(b) A complete method to find F in terms of W M1 Any complete method to find F.\nFor example, substitute into moments equation to\n2\nobtain T [T = 13 W, R=2W] .\n3\n2 A1 Correct.\nF = W\n3\n2\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_w24_ms_32-p12.png"
      ]
    },
    {
      "id": "9231-2024-on-32-q05",
      "question_id": "9231-2024-on-32-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a)\nUse correct equation of trajectory:\n4\na=3a\n1\n−\ng (3a)2\n\n\n1+\n1\n\nM1 No (implied) sight of trajectory equation M0.\n5 3 2u2  9\n4 5ga2 5ga 1 A1 At least one step of intermediate working must be\na=a− , = , u2 =25ga seen.\n5 u2 u2 5\nAG\n2\n© Cambridge University Press & Assessment 2024 Page 12 of 19\n\n5(b) For P, time of flight T and range R B1\nFor Q, time of flight ½ T and range R\n 2usin  10a Time of flight for P or Q.\nT = =\n \n g  g\n2 3 B1 Range for P.\n[From motion of P, R= 25ag =] 1 5a\ng 10\n1 2R M1 Obtain an expression for vcos. May involve u\nFor Q: →R=vcos T, vcos=\nand .\n2 T\n2 M1 Obtain an expression for vsin. May involve u\n1 1 1  1\n 0=vsin T − g  T  , vsin= gT and .\n2 2 2  4\n2 2 M1\n2R 1   5 \nSquare and add: v2 =   +  gT   =90ag+ ag \n T  4   8 \n725 A1 1\nv2 = ag gT\n8 tan= 4 = 1\n2R 12\nT\n© Cambridge University Press & Assessment 2024 Page 13 of 19\n\n5(b) Alternative method for question 5(b)\nFor P, time of flight T and range R M1\nFor Q, time of flight ½ T and range R\nHorizontal motion for P and Q\nT\nR=ucosT and R=(vcos)\n2 Both.\nVertical motion for P and Q M1\ngT gT 1\nusin= and vsin= Both, may come from using s=ut+ at2.\n2 4 2\nEquate two expressions for R: A1 6\nvcos= u\nvcos=2ucos\n10\nEquate two expressions for vertical motion: A1\n1 1\nvsin= usin vsin= u\n2 2 10\n 1   29  M1\nSquare and add: v2 =u2 4cos2+ sin2\n\n= u2\n\n 4   8 \n29 725 A1\n25ag = ag\n8 8\n6\n© Cambridge University Press & Assessment 2024 Page 14 of 19",
      "source_pages": [
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        13,
        14
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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        "../answer-assets/9231_w24_ms_32-p14.png"
      ]
    },
    {
      "id": "9231-2024-on-32-q06",
      "question_id": "9231-2024-on-32-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) For P to lowest point L: M1*\n1 1 Dimensionally correct, all terms present, allow\nEnergy: mv2 = mu2 +mga(1−cos)\nsign errors, allow cos/sin error.\n2 2\n28\n[v2 = ag−2agcos]\n9\nFrom L to string goes slack: M1*\n4a\n1 1 4a\nEnergy: mv2 = mw2 + mg(1+cos) Dimensionally correct, all terms present, with ,\n9\n2 2 9\nallow sign errors, allow cos/sin error.\n 20 26 \nw2 = ag− agcos\n \n 9 9 \nBoth equations correct, allow unsimplified. A1\nmw2 B1\n4a\nWhen string goes slack: mgcos=\n9\nEquate expressions for w2 to find a value for cos DM1\n2 A1\ncos=\n3\n6\n© Cambridge University Press & Assessment 2024 Page 15 of 19\n\n9231/32 Cambridge International AS & A Level – Mark Scheme October/November 2024\nPUBLISHED\nQuestion Answer Marks Guidance\nAlternative method for question 6(a)\nFor P from start to string goes slack: M2 4\nDimensionally correct, all terms present, with a\n9\n1 1  4 \nEnergy: mw2 = mu2 +mga (1−cos)− (1+cos) Allow sign errors, allow cos/sin error\n \n2 2  9  5  4\nRHS may appear with a−  1+  acos\n9  9\n4\nAllow M1 if is missing or if an attempt at both\n9\nheights, but all other conditions are met.\n 10 5 13  20 26  A1 Correct, allow unsimplified.\n\nw2 = ag+2ag\n\n− cos\n\n= ag− agcos\n\n 9 9 9  9 9 \nmw2 B1\n4a\nWhen string goes slack: mgcos=\n9\nEquate expressions for w2 to find a value for cos DM1\n2 A1\ncos=\n3\n6\n© Cambridge University Press & Assessment 2024 Page 16 of 19\n\n6(b) mv2 M1 EITHER equation, dimensionally correct, allow\nTension before: T −mg = sign error only\n1 a\nmv2\nTension after: T −mg =\n2 4\na\n9\n 16  25 A1 EITHER tension correct\nv2 = ag, T = mg T =5mg\n  9   1 9 2\nFind the other tension (from a valid equation) and find ratio of tensions M1 mv2\nEquation must be of the form T −mg =\nr\nRatio is 5 : 9 A1 5\nAny equivalent ratio, allow\n9\n4\n© Cambridge University Press & Assessment 2024 Page 17 of 19",
      "source_pages": [
        15,
        16,
        17
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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        "../answer-assets/9231_w24_ms_32-p17.png"
      ]
    },
    {
      "id": "9231-2024-on-32-q07",
      "question_id": "9231-2024-on-32-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) dv B1 Use of suvat means 0 marks in this part\nm =mg−0.1mv2\nNote that no mg term means 0 marks in this part.\ndt\ndv\nMust see m, may be cancelled before a = used\ndt\ndv =10−0.1v2 = 1 ( 100−v2) M1*\ndt 10\ndv 1\n= dt\n100−v2 10\nSeparate variables and integrate. May see partial\nfractions, but integral is on Formula sheet,\nv+10 allow missing +A for M1 only\nln\n \n=2t+A\n10−v\nA1 Must see modulus sign\nUse t=0, v=0, A=0 DM1\nRemove logs to obtain v in terms of t M1\n10 ( e2t −1 ) A1 10 ( 1−e−2t)\nv= aef v=\ne2t +1 1+e−2t\n6\n© Cambridge University Press & Assessment 2024 Page 18 of 19\n\n7(b)\nv\ndv\n=\n1 ( 100−v2) M1* Use of suvat means 0 marks in this part\ndx 10\nSeparate variables and integrate, allow missing\n+A for M1 only.\n1 ( ) 1\n− ln 100−v2 = x+B\n2 10\nA1 For A1, allow missing modulus sign\n1 DM1\nUse x=0, v=0, B=− ln100\n2\nRemove logs to obtain v2 in terms of x M1\nx A1 AEF\n−\nv2 =100(1−e 5) Allow 10g instead of 100.\n5\n© Cambridge University Press & Assessment 2024 Page 19 of 19",
      "source_pages": [
        18,
        19
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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        "../answer-assets/9231_w24_ms_32-p19.png"
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    },
    {
      "id": "9231-2024-on-33-q01",
      "question_id": "9231-2024-on-33-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1  1  M1 For one equation with one error.\nUse equation of trajectory with point (56, H) or 84, H\n 2 \nA1 Both correct.\n5g 1 5g\nH =112− 562 or H =168− 842\n2u2 2 2u2\nEliminate to find u or H M1\nu=35 A1\nH = 48 A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_33-p06.png"
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    },
    {
      "id": "9231-2024-on-33-q02",
      "question_id": "9231-2024-on-33-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 mv2 B1\nAt B, T+mgsin=\na\n1 1 M1A1\nEnergy A to B: mu2 − mv2 =mga(cos+sin)\n2 2\nSubstitute for u and  to find T: M1\n 4 3\nT =mg5−2 −3 \n 5 5\n8 A1\nT = mg\n5\n5\n© Cambridge University Press & Assessment 2024 Page 6 of 12",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_33-p06.png"
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    },
    {
      "id": "9231-2024-on-33-q03",
      "question_id": "9231-2024-on-33-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 2mg M1 Equilibrium position.\nHooke’s law: T = extension and T =mg\n2\nExtension = 1 m A1\n1 2mg B1\nEPE loss = \n(1+d)2\n2 2\nGain in GPE = mg(2+1+d) B1\n1 M1\nEquate: mg(1+d)2 =mg(3+d)\n2\nd = 5 A1 SC: 3 marks for final answer of 5+1 .\nSC: 2 marks for final answer of 5+k , k 1 .\n6\n\n3(b) 1 1 2mg M1 GPE, KE, EPE terms.\nEnergy equation: mV2 +mg(1+d)=  (1+d)2\n2 2 2\nV2 =g ( d2 −1 ) V = 40 =2 10 A1\nAlternatively: M1\nUsing KE and GPE from 2 m below O to point O\n1\nmV2 =2mg\n2\nV2 =4g V = 40 =2 10 A1\n2\n© Cambridge University Press & Assessment 2024 Page 7 of 12",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_33-p07.png"
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    },
    {
      "id": "9231-2024-on-33-q04",
      "question_id": "9231-2024-on-33-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) B1 Correct volumes and distances for large and small.\nLarge Small Object\nM1 Moments equation with 3 terms, dimensionally\ncorrect.\n2  2  4 \nVolume a2h   a  kh a2h1− k\n3   9  A1 Correct, unsimplified.\nCentre of 1 1\nmass from h kh x\nAB\n2 2\nMoments about AB:\n2\n 4  1 2  1\na2h1− k y = a2h h−\n\na\n\nkh kh\n 9  2 3  2\n( 9−4k2)\nh\nA1\ny =\n2(9−4k)\n4\n\n4(b)\ny\n( 9−4k2)\nh 3\nB1 FT FT their part (a)\ntan= : =\na 2(9−4k)a 2\n8 M1\nUse h= a and simplify to quadratic in k: 32k2 −36k+9=0\n3\n3 3 A1\nk = ,\n8 4\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 12",
      "source_pages": [
        8
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_33-p08.png"
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    },
    {
      "id": "9231-2024-on-33-q05",
      "question_id": "9231-2024-on-33-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5(a) dx  4  M1 Separate variables, obtain RHS in integrable form.\n= −1dt\nx t+1 \nln x =4ln t+1−t+ A A1\nt=0, x=5: A=ln5 M1\nx=5(t+1)4 e−t A1\n4\n\n5(b) v=(3−t)5(t+1)3 e−t M1\nAcceleration = dv =5e−t ( −(t+1)3 +(3−t)3(t+1)2 −(3−t)(t+1)3)\ndt\nAcceleration =\n5e−t(t+1)2(5−t)(1−t) AEF\nF=2acceleration, so at F\n=10e−t(t+1)2(5−t)(1−t) M1\nAt t=3, magnitude of force is 640e−3 N A1 31.9 N\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 12",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_33-p09.png"
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    },
    {
      "id": "9231-2024-on-33-q06",
      "question_id": "9231-2024-on-33-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) At P:  T cos=T cos+0.05g B1\n1 2\nAt Q:  T cos=0.04g\n2 B1 OR: whole system: T cos=0.09g\n1\nT =0.15g =1.5 N B1\n1\n3\n\n6(b) T sin−T sin=0.050.82 M1 Allow sin/cos mix\n1 2\nT sin=0.041.42 M1\n2\nT sin=0.050.82 +0.041.42 A1\n1\n5\n2 =12.5,  = 2\n2\n3\n\n6(c) T cos=0.04g and T sin=0.041.42 M1 From part (a) and part (b)\n2 2\n7\nDivide: tan=\n4\n=60.3 A1\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 12",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-33-q07",
      "question_id": "9231-2024-on-33-q07",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) When P strikes plane, velocity is →ucos, M1 3\nu\n5\n3 3\nBefore impact: parallel to inclined plane ucos, perpendicular to plane usin\n5 5\n3 3 A1\nAfter impact: components ucos (parallel) and eusin (perpendicular)\n5 5\n3 3 M1\nSince velocity is vertical after impact, tan= ucos/ eusin\n5 5\ntan=1/etan, e tan2 =1 A1 AG\n4\n© Cambridge University Press & Assessment 2024 Page 11 of 12\n\n7(b) (usin)2 8u2 M1A1 Note: alternative methods.\nGreatest height of P before impact: H = =\n2g 25g\n3 M1\nAfter impact, vertical speed of P is u\n(cos)2 +e2(sin)2\n5\n3 M1\nUse V2 =U2 +2as to greatest height, equal to H\n16\n9 u2 ( (cos)2 +e2(sin)2) =2g 3 H\n25 16\n1 e 1 M1\nUse part (a): tan= , cos= , sin=\ne 1+e 1+e\nSubstitute to find e\n1 A1\n3e2 +2e−1=0, e=\n3\n6\n© Cambridge University Press & Assessment 2024 Page 12 of 12",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2024-on-41-q01",
      "question_id": "9231-2024-on-41-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 1  16542  B1 3392\ns 2 = 54850−  =2.769\nx 49 50 1225\n \n4.95 2.769 M1 A1\ns2 = + =0.1379\n60 50\n1654 M1 Correct formula with a z value.\nCI: 35.6− 1.96s\n50\nB1 1.96 seen.\n1.79, 3.25 A1 Accept inequality.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-41-q02",
      "question_id": "9231-2024-on-41-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) 2 4 −8 5 1 0 7 −3 6 −12 9 1 4 1 1 M1 Signed differences, allow one error.\n1 3 −7 4 9 6 −2 5 −11 8 1 2 1 0 M1 Attempt at ranking.\n(Sum of + ranks = 58) Test statistic = 20 A1\nH : population median for teacher 2 = population median for teacher 1 B1\n0\nH : population median for teacher 2 > population median for teacher 1 Allow use of m for population median.\n1\nCritical value, from tables, is 17 B1\n‘20’ > 17, accept H M1\n0\nInsufficient evidence to support the claim A1 Correct work only except possibly hypotheses, in\nInsufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1 context, level of uncertainty in language.\n7\n© Cambridge University Press & Assessment 2024 Page 6 of 11\n\n2(b) New test statistic is ‘13’ B1 FT Their 20 minus 7.\n13 < 17, so conclusion is now ‘reject H ’, (sufficient evidence to support claim) B1 Must be 13 and 17.\n0\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
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    },
    {
      "id": "9231-2024-on-41-q03",
      "question_id": "9231-2024-on-41-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 17.098 8.122 3.086 B1 One correct.\nB1 All correct.\n2\n\n3(b) M1\n0 1 2 3 4 or more\nLast two or three columns combined.\n10 18 35 21 16\n14.957 28.418 26.997 17.098 12.53\nContributions to test statistic are: M1\n1.6428 3.8192 2.3724 0.8905 0.9609(7) May be implied by awrt 9.69\nTest statistic is 9.69 A1 9.686\nH : Po(1.9) is a good fit for the data B1\n0\nH : Po(1.9) is not a good fit for the data\n1\nCritical value is 7.779, compare ‘9.69’ > 7.779 reject H M1 4 degrees of freedom\n0\nSufficient evidence to suggest that Po(1.9) is a not a good fit for the data/ A1 Correct work only, including hypotheses, in\nSufficient evidence to reject/not support the statistician’s claim context, level of uncertainty in language.\n6\n© Cambridge University Press & Assessment 2024 Page 7 of 11",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-41-q04",
      "question_id": "9231-2024-on-41-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) Correct sketch. B1 Label 1 and 5 on x-axis, ignore labels, if any, on\ny-axis.\n1\n\n4(b) kx4 k B1 For either part correct.\nwith limits 0 and 1 gives\n4 4\n x2 \nk5x−  with limits 1 and 5 gives 8k\n2\n \nk 4 B1 AG\n+8k =1, k =\n4 33\n2\n\n4(c)  M1 Integrate both parts.\n0 x0\n\n A1 Middle two parts correct AEF.\n 1\nx4 0≤x1\n A1 0 and 1 correct.\n33\nF(x)=\n 4  5x− 1 x2 − 17 1≤x≤5\n \n33 2 4 \n\n\n1 x5.\n\n\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 11\n\n4(d) 1 m 1 M1* Or integrate m to 5.\n kx3dx+ k(5−x)dx=\nOr use F(x) = 0.5.\n0 1 2\n4 1 1 1 1 4 25 1  1 M1 Must be a quadratic in m.\n+5m− m2 −5+ = or −5m+ m2 =\n   \n334 2 2 2 33 2 2  2\n4  1 17 1\nor 5m− m2 − =\n \n33 2 4  2\n4m2 −40m+67=0 *DM1\n1( ) A1\nm= 10− 33 =2.13\n2\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2024-on-41-q05",
      "question_id": "9231-2024-on-41-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 8 14 7 1 B1 All correct.\nP(0 heads) = , P(1 head) = , P(2 heads) = , P(3 heads) =\n30 30 30 30\n8 14 7 1 M1 Cubic polynomial.\nG (t) = + t+ t2 + t3\nX\n30 30 30 30\nA1 FT FT their probabilities that sum to one.\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 11\n\n5(b) 25 10 1 B1\nG (t) = + t+ t2\nY\n36 36 36\nG Z (t) = 1   8 + 14 t+ 7 t2 + 1 t3   ( 25 +1 0t +t2) M1 With attempt to multiply.\n303630 30 30 30 \n1 ( 200+430t+323t2 +109t3+17t4 +t5) M1 A1 Obtain quintic polynomial.\n1080\n4\n\n5(c) 1 M1 Differentiate and substitute t = 1.\nG ’ (1) = (430+646+327+68+5)\nZ\n1080\n1476 41 A1\n= =1.37\n1080 30\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 11",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-41-q06",
      "question_id": "9231-2024-on-41-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6 1 64.22  M1 851\ns 2 = 519.46−  =0.607857\nx 7 8 1400\n \n17\n1 71.12  , both.\ns 2 = 567.13−  =0.680 25\ny 8 9\n \n80.680 + 70.607857 M1 A1 1939\ns2 = =0.64633 Correct formula for pooled variance used, .\n9+8−2 3000\nH :  − =0 B1\n0 X Y\nH :  − 0\n1 X Y\n8.025−7.9 0.125 M1 A1 Dependent on pooled variance being used\nt = = =0.320\n1 1 0.3906\ns +\n8 9\n‘0.320’ < 1.341 accept H M1 Compare with 1.341 and conclusion\n0\nInsufficient evidence to support Ansal’s suspicion A1 Correct work only, except possibly B1, in\nInsufficient evidence to suggest that wingspan from region X is greater than context, level of uncertainty in language\nwingspan from region Y.\n8\n© Cambridge University Press & Assessment 2024 Page 11 of 11",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2024-on-42-q01",
      "question_id": "9231-2024-on-42-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 x=30.6 x2 =120.52 M1\n139\n1 30.62  With their x, x2, .\ns2 = 120.52−  =0.49643 280\n7 8\n \n30.6 s2 M1 Use correct formula with a t value.\nCI = 1.895\n8 8\nB1 1.895 seen.\n3.35, 4.30 A1 Accept with inequality signs or open brackets.\n.\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_42-p06.png"
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    },
    {
      "id": "9231-2024-on-42-q02",
      "question_id": "9231-2024-on-42-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) 1 B1\n[G (t) = + pt+qt2, G’ (t) = p+2qt, ] p+2q=1.1\nX X\n5\n1 4 M1\n+ p+q=1, p+q= and solve\n5 5\n 1  3 A1\np= , q=\n \n 2  10\n3 A1\nG’’ (t) = 2q = , Var(X) = 0.6+1.1−1.12 =0.49\nX\n5\n4\n© Cambridge University Press & Assessment 2024 Page 6 of 13\n\n2(b) 1 1 3  2  1  M1 Wrong p and q (or swapped) or missing t, M1A0.\nG Z (t) =  + t+ t2  t1+ t2  p and q must have their numerical values from\n5 2 10  3  2 \npart (a).\n1 ( 4t+10t2 +8t3+5t4 +3t5) A1 Accept answer in any equivalent form.\n30\n2\n\n2(c) 16 8 B1 FT FT their fully expanded form of PGF.\n= oe\n30 15\n1\n\n2(d) (Z =) 2 B1 Correct work only, must see correct fully\nexpanded polynomial form of PGF.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2024-on-42-q03",
      "question_id": "9231-2024-on-42-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) 40+86+105+64+20 315 2.1 B1 Must see either 315 or 2.1.\nx = = =2.1, p= =0.42\nAG\n150 150 5\n1\n© Cambridge University Press & Assessment 2024 Page 7 of 13\n\n3(b) B1 Calculate expected frequencies (must be seen)\nNumber of 0 1 2 3 4 5\nat least 2 correct to at least 2 decimal places.\nseeds that\ngerminate\nB1 At least 4 correct to at least 2 decimal places.\nNumber of 12 40 43 35 16 4\npots\nExpected 9.846 35.6475 51.627 37.3845 13.536 1.9605\nfrequency\nCombine last two columns M1 20, 15.50, may be implied by answer 3.90 – 3.91.\nChi-squared contributions: M1 At least 2 correct, may be implied by answer 3.90\n0.4712 0.5314 1.4416 0.1521 1.3088 – 3.91.\nTest statistic = 3.905 A1 accept 3.90 – 3.910\nH : Binomial B(5, 0.42) fits the data B1 Allow ‘Binomial’ for ‘B(5, 0.42)’\n0\nH : Binomial B(5, 0.42) does not fit the data Allow ‘Number of seeds that germinate can be\n1\nmodelled by B(5, 0.42)’\nCritical value is 6.251 B1 Must come from combined columns.\nAllow 7.779.\n‘3.905’ < ‘6.251’ Accept H M1 Reject H , not significant.\n0 1\nInsufficient evidence to suggest that B(5, 0.42) is a not a good fit (to the data) A1 Correct work only, including hypotheses, level of\nuncertainty in language.\n9\n© Cambridge University Press & Assessment 2024 Page 8 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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        "../answer-assets/9231_w24_ms_42-p08.png"
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    },
    {
      "id": "9231-2024-on-42-q04",
      "question_id": "9231-2024-on-42-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)  0 x2, M1 Integrate 1 ( x3−3x2 +3x )+c oe.\n\n63\n\n 1\nF(X)= ( (x−13−1 ) 2≤x≤5, A1 2\n63   Correct F(x) (with c =− evaluated).\n\n63\n\n1 x5.\n\n A1 0 and 1 correct, inequalities correct.\n3\n\n4(b) G(y) = 1 ( y0.75 −1 ) M1 Use transformation correctly on function .\nMay see\n63\n1 ( y0.25 +1 )3 − 1 ( y0.25 +1 )2 + 1 ( y0.25 +1 ) − 2\n63 21 21 63\n 1 M1 Differentiate and change variable in domain.\ny−0.25 1≤y≤256,\n\n84\ng(y)= A1 All correct and simplified.\n\n0 otherwise.\n\n3\n\n4(c) 1 ( y0.75 −1 ) =0.5 M1 Equate their G(y) to 0.5 and attempt to solve.\n63\ny = 104 A1 AWRT 104.\n2\n© Cambridge University Press & Assessment 2024 Page 9 of 13\n\n4(d) 2  56 1 y ( y−0.25) dy= 2  56 1 ( y0.75) dy=  1  4 y1.75  M1 Use their y g(y), integrate, ignore limits.\n \n84 84 84 7 \n1 1\n111 A1 5461\n49\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 13",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    {
      "id": "9231-2024-on-42-q05",
      "question_id": "9231-2024-on-42-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5 1  30.82  B1 237\ns 2 = 29.0−  =0.055765\nx 34 35 4250\n \n1488\n1  62.42 \ns 2 = 76.8−  =0.201763 7375\ny 59 60\n \nAllow B1 for correct expression for pooled\nvariance.\n0.055765 0.201763 M1 Pooled variance M0A0.\ns2 = + =0.004956\nA1\n35 60\nH :  = B1\n0 x y\nH  \n1: x y\n30.8 62.4 M1\n−\n35 60\nz=\nA1 Pooled variance M0A0.\ns\n0.25\n− = −2.27\n0.004956\nCorresponding probability is 0.9885 M1 FT their z .\nOne tail: 100(1−'0.9885')=1.15 M1 Allow for 1 – ‘0.9885’.\n1.15 A1 Allow ⩾.\n9\n© Cambridge University Press & Assessment 2024 Page 11 of 13",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_42-p11.png"
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    },
    {
      "id": "9231-2024-on-42-q06",
      "question_id": "9231-2024-on-42-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 Signed differences from 51.0 M1 Allow up to 2 errors.\n0.3 1.0 2.4 - 1.8 - 1.7 0.1 1.2 - 3.8\n2.0 - 2.5 - 1.6 - 0.7 - 0.2 0.6 - 1.9 1.3\n0.8 1.4 - 3.1 - 2.1 - 0.4 0.9\nM1 Attempt at ranking of signed differences from\n3 9 19 -15 -14 1 10 -22\n51.0.\n17 -20 -13 -6 -2 5 -16 11\n7 12 -21 -18 -4 8\nSmaller sum is 102 A1 151\nH : population median is 51.0 (seconds) B1 Allow m.\n0\nH population median is less than 51.0 (seconds)\n1:\nNormal approximation: mean =126.5 B1 Both.\nvariance = 948.75\n'102'+0.5−126.5 M1 Allow no continuity correction.\n948.75\n−0.779 A1 Accept 0.779.\nAccept p-value = 0.218.\n(A0 for z = – 0.795 or p = 0.213 from no\ncontinuity correction).\n'−0.779' −1.645 accept H M1 Or ‘0.218’ > 0.05 or ‘0.782’ < 0.95\n0\n© Cambridge University Press & Assessment 2024 Page 12 of 13\n\n6 Insufficient evidence to suggest that (population) median is less than 51.0 seconds/ A1 Correct work only, except possibly hypotheses, in\nInsufficient evidence to support the claim context, level of uncertainty in language.\nAlternative approach using critical values.\n9\n© Cambridge University Press & Assessment 2024 Page 13 of 13",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_42-p13.png"
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    },
    {
      "id": "9231-2024-on-43-q01",
      "question_id": "9231-2024-on-43-q01",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 1  16542  B1 3392\ns 2 = 54850−  =2.769\nx 49 50 1225\n \n4.95 2.769 M1 A1\ns2 = + =0.1379\n60 50\n1654 M1 Correct formula with a z value.\nCI: 35.6− 1.96s\n50\nB1 1.96 seen.\n1.79, 3.25 A1 Accept inequality.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        6
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_43-p06.png"
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    },
    {
      "id": "9231-2024-on-43-q02",
      "question_id": "9231-2024-on-43-q02",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) 2 4 −8 5 1 0 7 −3 6 −12 9 1 4 1 1 M1 Signed differences, allow one error.\n1 3 −7 4 9 6 −2 5 −11 8 1 2 1 0 M1 Attempt at ranking.\n(Sum of + ranks = 58) Test statistic = 20 A1\nH : population median for teacher 2 = population median for teacher 1 B1\n0\nH : population median for teacher 2 > population median for teacher 1 Allow use of m for population median.\n1\nCritical value, from tables, is 17 B1\n‘20’ > 17, accept H M1\n0\nInsufficient evidence to support the claim A1 Correct work only except possibly hypotheses, in\nInsufficient evidence that the scores of Teacher 2 are higher than those of Teacher 1 context, level of uncertainty in language.\n7\n© Cambridge University Press & Assessment 2024 Page 6 of 11\n\n2(b) New test statistic is ‘13’ B1 FT Their 20 minus 7.\n13 < 17, so conclusion is now ‘reject H ’, (sufficient evidence to support claim) B1 Must be 13 and 17.\n0\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        6,
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_43-p07.png"
      ]
    },
    {
      "id": "9231-2024-on-43-q03",
      "question_id": "9231-2024-on-43-q03",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 17.098 8.122 3.086 B1 One correct.\nB1 All correct.\n2\n\n3(b) M1\n0 1 2 3 4 or more\nLast two or three columns combined.\n10 18 35 21 16\n14.957 28.418 26.997 17.098 12.53\nContributions to test statistic are: M1\n1.6428 3.8192 2.3724 0.8905 0.9609(7) May be implied by awrt 9.69\nTest statistic is 9.69 A1 9.686\nH : Po(1.9) is a good fit for the data B1\n0\nH : Po(1.9) is not a good fit for the data\n1\nCritical value is 7.779, compare ‘9.69’ > 7.779 reject H M1 4 degrees of freedom\n0\nSufficient evidence to suggest that Po(1.9) is a not a good fit for the data/ A1 Correct work only, including hypotheses, in\nSufficient evidence to reject/not support the statistician’s claim context, level of uncertainty in language.\n6\n© Cambridge University Press & Assessment 2024 Page 7 of 11",
      "source_pages": [
        7
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_43-p07.png"
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    },
    {
      "id": "9231-2024-on-43-q04",
      "question_id": "9231-2024-on-43-q04",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) Correct sketch. B1 Label 1 and 5 on x-axis, ignore labels, if any, on\ny-axis.\n1\n\n4(b) kx4 k B1 For either part correct.\nwith limits 0 and 1 gives\n4 4\n x2 \nk5x−  with limits 1 and 5 gives 8k\n2\n \nk 4 B1 AG\n+8k =1, k =\n4 33\n2\n\n4(c)  M1 Integrate both parts.\n0 x0\n\n A1 Middle two parts correct AEF.\n 1\nx4 0≤x1\n A1 0 and 1 correct.\n33\nF(x)=\n 4  5x− 1 x2 − 17 1≤x≤5\n \n33 2 4 \n\n\n1 x5.\n\n\n3\n© Cambridge University Press & Assessment 2024 Page 8 of 11\n\n4(d) 1 m 1 M1* Or integrate m to 5.\n kx3dx+ k(5−x)dx=\nOr use F(x) = 0.5.\n0 1 2\n4 1 1 1 1 4 25 1  1 M1 Must be a quadratic in m.\n+5m− m2 −5+ = or −5m+ m2 =\n   \n334 2 2 2 33 2 2  2\n4  1 17 1\nor 5m− m2 − =\n \n33 2 4  2\n4m2 −40m+67=0 *DM1\n1( ) A1\nm= 10− 33 =2.13\n2\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_43-p09.png"
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    },
    {
      "id": "9231-2024-on-43-q05",
      "question_id": "9231-2024-on-43-q05",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 8 14 7 1 B1 All correct.\nP(0 heads) = , P(1 head) = , P(2 heads) = , P(3 heads) =\n30 30 30 30\n8 14 7 1 M1 Cubic polynomial.\nG (t) = + t+ t2 + t3\nX\n30 30 30 30\nA1 FT FT their probabilities that sum to one.\n3\n© Cambridge University Press & Assessment 2024 Page 9 of 11\n\n5(b) 25 10 1 B1\nG (t) = + t+ t2\nY\n36 36 36\nG Z (t) = 1   8 + 14 t+ 7 t2 + 1 t3   ( 25 +1 0t +t2) M1 With attempt to multiply.\n303630 30 30 30 \n1 ( 200+430t+323t2 +109t3+17t4 +t5) M1 A1 Obtain quintic polynomial.\n1080\n4\n\n5(c) 1 M1 Differentiate and substitute t = 1.\nG ’ (1) = (430+646+327+68+5)\nZ\n1080\n1476 41 A1\n= =1.37\n1080 30\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 11",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w24_ms_43-p10.png"
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    },
    {
      "id": "9231-2024-on-43-q06",
      "question_id": "9231-2024-on-43-q06",
      "subject": "9231",
      "year": 2024,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6 1 64.22  M1 851\ns 2 = 519.46−  =0.607857\nx 7 8 1400\n \n17\n1 71.12  , both.\ns 2 = 567.13−  =0.680 25\ny 8 9\n \n80.680 + 70.607857 M1 A1 1939\ns2 = =0.64633 Correct formula for pooled variance used, .\n9+8−2 3000\nH :  − =0 B1\n0 X Y\nH :  − 0\n1 X Y\n8.025−7.9 0.125 M1 A1 Dependent on pooled variance being used\nt = = =0.320\n1 1 0.3906\ns +\n8 9\n‘0.320’ < 1.341 accept H M1 Compare with 1.341 and conclusion\n0\nInsufficient evidence to support Ansal’s suspicion A1 Correct work only, except possibly B1, in\nInsufficient evidence to suggest that wingspan from region X is greater than context, level of uncertainty in language\nwingspan from region Y.\n8\n© Cambridge University Press & Assessment 2024 Page 11 of 11",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9231_w24_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2024-oct-nov/9231_w24_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_w24_ms_43-p11.png"
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    },
    {
      "id": "9231-2025-mj-11-q01",
      "question_id": "9231-2025-mj-11-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) (2−3r)(5−3r)=9r2 −21r+10 B1\n9 (1n(n+1)(2n+1) ) −21 (1n(n+1) ) +10n M1 Substitutes formulae for r2 and r.\n6 2\n=3n3−6n2 +n A1\n3\n\n1(b) 1 1 1 1  B1 Finds partial fractions.\n=\n\n−\n\n(2−3r)(5−3r) 32−3r 5−3r\nn 1 1 1 1 1 1 1  1 1 1  M1 A1 Writes at least three terms, including first and\n =  −  +  −  +  −  last, to demonstrate cancelation.\n(2−3r)(5−3r) 3−1 2 3−4 −1 3−7 −4\nr=1\n1 1 1 \n+ +  − \n32−3n 5−3n\n1 1 A1\n=− +\n6 3(2−3n)\n4\n\n1(c)  1 1 B1 Allow FT from (b) if of similar form.\n =−\n(2−3r)(5−3r) 6\nr=1\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_11-p09.png"
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    },
    {
      "id": "9231-2025-mj-11-q02",
      "question_id": "9231-2025-mj-11-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) y=x3−1 B1 Uses correct substitution.\n(y+1)\n3\n3\n+2(y+1)1\n3\n+1=02(y+1)1\n3 =−y−2\nM1\nSubstitutes and obtains an equation not\n8(y+1)=−(y+2)3 8y+8=−y3−6y2 −12y−8 involving radicals.\ny3 +6y2 +20y+16=0 A1\n3\n\n2(b) ( 3−1 )2 + ( 3−1 )2 + ( 3−1 )2 =62 −2(20) M1 Uses\n2 +2 +2 =(++)2 −2(++).\n−4 A1 FT\n2\n\n2(c) ( 3−1 )3 + ( 3−1 )3 + ( 3−1 )3 =−6(−4)−20(−6)−16(3) M1 Uses their equation from (a) or a valid formula\nfor the sum of cubes.\n96 A1\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 18",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_11-p10.png"
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    },
    {
      "id": "9231-2025-mj-11-q03",
      "question_id": "9231-2025-mj-11-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) u =5=61−1 B1 Shows base case.\n1\nAssume that it is true for n=k, so u =6k −1. B1 States inductive hypothesis.\nk\nThen u =6(6k −1)+5=6k+1−1. M1 A1 Substitutes into recursion formula.\nk+1\nSo, it is also true for n=k+1. Hence, by induction, u =6n −1is true for all positive A1 States conclusion.\nn\nintegers.\n5\n\n3(b) u 62n −1 ( 6n −1 )( 6n +1 ) M1 A1 Uses (a) and difference of two squares.\n2n = = =6n +1\nu 6n −1 6n −1\nn\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-11-q04",
      "question_id": "9231-2025-mj-11-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) A rotation, followed by a shear. B2 Award B1 if given in the wrong order.\n[anticlockwise, centred at the origin, through an\nangle], [x-axis fixed, with\n(0,1)\nmapped to\n(2,1).]\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 18\n\n4(b) cos+2sin 2cos−sin B1\nM= \n sin cos \ncos+2sin 2cos−sinx xcos+2xsin−ysin+2ycos B1 x X\n  =  Transforms   to  .\n sin cos y  xsin+ ycos  y Y \nxsin+ ycos= y AND xcos+2xsin− ysin+2ycos=x M1* Uses X =x and Y = y.\nxsin\nxcos+2xsin+ (2cos−sin)=x\n1−cos A1 Combines equations correctly to eliminate y or\nx.\n(cos+2sin)(1−cos)+2sincos−sin2=1−cos DM1 Simplifies.\nA1\ncos−cos2+2sin−sin2=1−cossin+cos=1\n= 1π A1\n2\n7\n© Cambridge University Press & Assessment 2025 Page 12 of 18",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2025-mj-11-q05",
      "question_id": "9231-2025-mj-11-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Spiral shape on [0, 2π], with correct orientation,\nstarting at pole and with one other point of\nintersection with the initial line.\n=0\nB1 r strictly increasing, correct gradient at points of\nintersection with the initial line.\n2\n\n5(b) 1 2π 2e 1 4 d B1 1 2π r2d\n2 2\n0 0\n= 142e 1 4  2π −1 2π 8e 1 4 d M1 A1 Integrates by parts.\n2  2\n0 0\n= 142e 1 4  2π −44e 1 4  2π +16 2π e 1 4 d=22e 1 4 −16e 1 4 +64e 1 4  2π M1 A1 Integrates by parts again.\n2     \n0 0 0 0\n= ( 8π2 −32π+64 ) e 1 2 π −64 A1\n6\n© Cambridge University Press & Assessment 2025 Page 13 of 18\n\n5(c) 1 B1 Uses y=rsin\ny=e8 sin\ndy =e 1 8 cos+sin ( 1e 1 8 +e 1 8  ) M1 A1 Differentiates.\nd 8\ne 1 8 0cos+ (1+1 ) sin=0 A1 AG.\n8\n5cos5+\n(5+1 )\nsin5=−0.1399\nB1 Shows sign change.\n8\nand 5.05cos5.05+\n(5.05+1 )\nsin5.05=0.1336\n8\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_11-p13.png",
        "../answer-assets/9231_s25_ms_11-p14.png"
      ]
    },
    {
      "id": "9231-2025-mj-11-q06",
      "question_id": "9231-2025-mj-11-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "6(a) B1 Finds direction vectors of two lines in the plane.\nAB=2j+4k AC=i−j+k\ni j k  6   3  M1 A1 Finds normal to the plane ABC.\n   \n0 2 4 = 4 ~ 2\n   \n   \n1 −1 1 −2 −1\n3(1)+2(0)−(−2)=53x+2y−z=5 M1 A1 Substitutes point.\n5\n© Cambridge University Press & Assessment 2025 Page 14 of 18\n\n6(b) i j k 2t+4 1 M1 A1 Finds normal to the plane ABD.\n   \n0 2 4 = 0 ~ 0\n   \n   \n0 0 t+2  0  0\n 3  1 M1 Uses dot product of normal vectors in correct\n    formula.\n2 0 = 14cos\n   \n   \n−1 0\n(t −2)\n=36.7\nA1 0.641 rad.\n4\n\n6(c) i j k 2t−2 t−1 M1 A1 Finds common perpendicular.\n   \n0 2 4 = −4 ~ −2\n   \n   \n−1 1 t+1  2   1 \n 1  t−1 M1 A1 Uses formula for perpendicular distance.\n1     t+2\n−1 −2 =\n   \nt2 −2t+6    t2 −2t+6\n 1   1 \nt+2 = 2(t+2)2 =2 ( t2 −2t+6 ) M1 Sets equal to 2 and solves for t.\nt2 −2t+6\nt2 −8t+8=0t=2 ( 2+ 2 ) ,t=2 ( 2− 2 ) A1 A1 t =6.83,t =1.17.\n7\n© Cambridge University Press & Assessment 2025 Page 15 of 18",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_11-p14.png",
        "../answer-assets/9231_s25_ms_11-p15.png"
      ]
    },
    {
      "id": "9231-2025-mj-11-q07",
      "question_id": "9231-2025-mj-11-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1, x=4 B1 Vertical asymptotes.\n2\ny=1 B1 Horizontal asymptote.\n2\n\n7(b) dy (2x2 −7x−4)(4x−5)−(2x2 −5x)(4x−7) M1 dy\n= Finds .\ndx ( 2x2 −7x−4 )2 dx\n4x2 +16x−20=0 M1 Sets equal to 0 and forms equation.\n( −5,25) ( 1,1) A1 A1\n27 3\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 18\n\n7(c) B1 Axes and asymptotes labelled.\nB1 Correct shape and position.\nx\n(0,0), (5,0 ) B1 States coordinates of intersections with axes.\n2 May be seen on their diagram.\n3\ny\n\n7(d) B1 Correct shape and position.\nx\n1\n© Cambridge University Press & Assessment 2025 Page 17 of 18\n\n7(e) 2x2 −5x 1 2x2 −5x 1 M2 Finds critical points, award M1 for each case.\n= or =−\n2x2 −7x−4 9 2x2 −7x−4 9\n16x2 −38x+4=0 or 20x2 −52x−4=0\nx=19 + 3 33, x=19− 3 33 or x=13+ 3 21, x=13− 3 21 A1 A1\n16 16 16 16 10 10 10 10\n13 − 3 21 x19 − 3 33, 19+ 3 33x13+ 3 21 A1\n10 10 16 16 16 16 10 10\n5\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
        16,
        17,
        18
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_11-p16.png",
        "../answer-assets/9231_s25_ms_11-p17.png",
        "../answer-assets/9231_s25_ms_11-p18.png"
      ]
    },
    {
      "id": "9231-2025-mj-12-q01",
      "question_id": "9231-2025-mj-12-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a) (2−3r)(5−3r)=9r2 −21r+10 B1\n9 (1n(n+1)(2n+1) ) −21 (1n(n+1) ) +10n M1 Substitutes formulae for r2 and r.\n6 2\n=3n3−6n2 +n A1\n3\n\n1(b) 1 1 1 1  B1 Finds partial fractions.\n=\n\n−\n\n(2−3r)(5−3r) 32−3r 5−3r\nn 1 1 1 1 1 1 1  1 1 1  M1 A1 Writes at least three terms, including first and\n =  −  +  −  +  −  last, to demonstrate cancelation.\n(2−3r)(5−3r) 3−1 2 3−4 −1 3−7 −4\nr=1\n1 1 1 \n+ +  − \n32−3n 5−3n\n1 1 A1\n=− +\n6 3(2−3n)\n4\n\n1(c)  1 1 B1 Allow FT from (b) if of similar form.\n =−\n(2−3r)(5−3r) 6\nr=1\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_12-p09.png"
      ]
    },
    {
      "id": "9231-2025-mj-12-q02",
      "question_id": "9231-2025-mj-12-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) y=x3−1 B1 Uses correct substitution.\n(y+1)\n3\n3\n+2(y+1)1\n3\n+1=02(y+1)1\n3 =−y−2\nM1\nSubstitutes and obtains an equation not\n8(y+1)=−(y+2)3 8y+8=−y3−6y2 −12y−8 involving radicals.\ny3 +6y2 +20y+16=0 A1\n3\n\n2(b) ( 3−1 )2 + ( 3−1 )2 + ( 3−1 )2 =62 −2(20) M1 Uses\n2 +2 +2 =(++)2 −2(++).\n−4 A1 FT\n2\n\n2(c) ( 3−1 )3 + ( 3−1 )3 + ( 3−1 )3 =−6(−4)−20(−6)−16(3) M1 Uses their equation from (a) or a valid formula\nfor the sum of cubes.\n96 A1\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 18",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_12-p10.png"
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    },
    {
      "id": "9231-2025-mj-12-q03",
      "question_id": "9231-2025-mj-12-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) u =5=61−1 B1 Shows base case.\n1\nAssume that it is true for n=k, so u =6k −1. B1 States inductive hypothesis.\nk\nThen u =6(6k −1)+5=6k+1−1. M1 A1 Substitutes into recursion formula.\nk+1\nSo, it is also true for n=k+1. Hence, by induction, u =6n −1is true for all positive A1 States conclusion.\nn\nintegers.\n5\n\n3(b) u 62n −1 ( 6n −1 )( 6n +1 ) M1 A1 Uses (a) and difference of two squares.\n2n = = =6n +1\nu 6n −1 6n −1\nn\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_12-p11.png"
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    },
    {
      "id": "9231-2025-mj-12-q04",
      "question_id": "9231-2025-mj-12-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) A rotation, followed by a shear. B2 Award B1 if given in the wrong order.\n[anticlockwise, centred at the origin, through an\nangle], [x-axis fixed, with\n(0,1)\nmapped to\n(2,1).]\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 18\n\n4(b) cos+2sin 2cos−sin B1\nM= \n sin cos \ncos+2sin 2cos−sinx xcos+2xsin−ysin+2ycos B1 x X\n  =  Transforms   to  .\n sin cos y  xsin+ ycos  y Y \nxsin+ ycos= y AND xcos+2xsin− ysin+2ycos=x M1* Uses X =x and Y = y.\nxsin\nxcos+2xsin+ (2cos−sin)=x\n1−cos A1 Combines equations correctly to eliminate y or\nx.\n(cos+2sin)(1−cos)+2sincos−sin2=1−cos DM1 Simplifies.\nA1\ncos−cos2+2sin−sin2=1−cossin+cos=1\n= 1π A1\n2\n7\n© Cambridge University Press & Assessment 2025 Page 12 of 18",
      "source_pages": [
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        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2025-mj-12-q05",
      "question_id": "9231-2025-mj-12-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Spiral shape on [0, 2π], with correct orientation,\nstarting at pole and with one other point of\nintersection with the initial line.\n=0\nB1 r strictly increasing, correct gradient at points of\nintersection with the initial line.\n2\n\n5(b) 1 2π 2e 1 4 d B1 1 2π r2d\n2 2\n0 0\n= 142e 1 4  2π −1 2π 8e 1 4 d M1 A1 Integrates by parts.\n2  2\n0 0\n= 142e 1 4  2π −44e 1 4  2π +16 2π e 1 4 d=22e 1 4 −16e 1 4 +64e 1 4  2π M1 A1 Integrates by parts again.\n2     \n0 0 0 0\n= ( 8π2 −32π+64 ) e 1 2 π −64 A1\n6\n© Cambridge University Press & Assessment 2025 Page 13 of 18\n\n5(c) 1 B1 Uses y=rsin\ny=e8 sin\ndy =e 1 8 cos+sin ( 1e 1 8 +e 1 8  ) M1 A1 Differentiates.\nd 8\ne 1 8 0cos+ (1+1 ) sin=0 A1 AG.\n8\n5cos5+\n(5+1 )\nsin5=−0.1399\nB1 Shows sign change.\n8\nand 5.05cos5.05+\n(5.05+1 )\nsin5.05=0.1336\n8\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-12-q06",
      "question_id": "9231-2025-mj-12-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "6(a) B1 Finds direction vectors of two lines in the plane.\nAB=2j+4k AC=i−j+k\ni j k  6   3  M1 A1 Finds normal to the plane ABC.\n   \n0 2 4 = 4 ~ 2\n   \n   \n1 −1 1 −2 −1\n3(1)+2(0)−(−2)=53x+2y−z=5 M1 A1 Substitutes point.\n5\n© Cambridge University Press & Assessment 2025 Page 14 of 18\n\n6(b) i j k 2t+4 1 M1 A1 Finds normal to the plane ABD.\n   \n0 2 4 = 0 ~ 0\n   \n   \n0 0 t+2  0  0\n 3  1 M1 Uses dot product of normal vectors in correct\n    formula.\n2 0 = 14cos\n   \n   \n−1 0\n(t −2)\n=36.7\nA1 0.641 rad.\n4\n\n6(c) i j k 2t−2 t−1 M1 A1 Finds common perpendicular.\n   \n0 2 4 = −4 ~ −2\n   \n   \n−1 1 t+1  2   1 \n 1  t−1 M1 A1 Uses formula for perpendicular distance.\n1     t+2\n−1 −2 =\n   \nt2 −2t+6    t2 −2t+6\n 1   1 \nt+2 = 2(t+2)2 =2 ( t2 −2t+6 ) M1 Sets equal to 2 and solves for t.\nt2 −2t+6\nt2 −8t+8=0t=2 ( 2+ 2 ) ,t=2 ( 2− 2 ) A1 A1 t =6.83,t =1.17.\n7\n© Cambridge University Press & Assessment 2025 Page 15 of 18",
      "source_pages": [
        14,
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-12-q07",
      "question_id": "9231-2025-mj-12-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1, x=4 B1 Vertical asymptotes.\n2\ny=1 B1 Horizontal asymptote.\n2\n\n7(b) dy (2x2 −7x−4)(4x−5)−(2x2 −5x)(4x−7) M1 dy\n= Finds .\ndx ( 2x2 −7x−4 )2 dx\n4x2 +16x−20=0 M1 Sets equal to 0 and forms equation.\n( −5,25) ( 1,1) A1 A1\n27 3\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 18\n\n7(c) B1 Axes and asymptotes labelled.\nB1 Correct shape and position.\nx\n(0,0), (5,0 ) B1 States coordinates of intersections with axes.\n2 May be seen on their diagram.\n3\ny\n\n7(d) B1 Correct shape and position.\nx\n1\n© Cambridge University Press & Assessment 2025 Page 17 of 18\n\n7(e) 2x2 −5x 1 2x2 −5x 1 M2 Finds critical points, award M1 for each case.\n= or =−\n2x2 −7x−4 9 2x2 −7x−4 9\n16x2 −38x+4=0 or 20x2 −52x−4=0\nx=19 + 3 33, x=19− 3 33 or x=13+ 3 21, x=13− 3 21 A1 A1\n16 16 16 16 10 10 10 10\n13 − 3 21 x19 − 3 33, 19+ 3 33x13+ 3 21 A1\n10 10 16 16 16 16 10 10\n5\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
        16,
        17,
        18
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_12.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_12-p16.png",
        "../answer-assets/9231_s25_ms_12-p17.png",
        "../answer-assets/9231_s25_ms_12-p18.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q01",
      "question_id": "9231-2025-mj-13-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "1(a) 14 0 B1 Stretch parallel to the x-axis, scale factor k\n  (k 0).\n 0 1\n 1 − 3 B1 Rotation anticlockwise about the origin through\n 2 2  1π.\nangle\n 3 1  3\n \n2 2\n 1 − 3 14 0  7 − 3 M1 Correct order.\nM= 2 2 \n\n= 2 \n 3 1   0 1  7 3 1 \n 2 2   2 \n 14 − 3 A1 AG\n2M= .\n \n14 3 1 \n4\n© Cambridge University Press & Assessment 2025 Page 9 of 19\n\n1(b)  7 − 3 x  7x− 3 y  B1 x X\n\n 7 3 1\n2 \n  y  \n=\n 7 3x+\n2\n1 y\n\n\nTransforms \ny\n\n\nto \nY\n\n\n 2   2 \n7 3x+1mx=m ( 7x− 3mx ) M1 Uses y =mx and Y =mX .\n2 2\n3\n7 3+1m=7m− m2\n2 2\n3m2 −13m+14 3=0 A1 Correct quadratic AEF\n7 A1 Correct solutions to the quadratic\nm=2 3 m=\n3\nA1 Correct invariant lines. If M0 then SCB1 for two\ny=2 3x y= 7 x\n3 correct lines.\n5\n\n1(c)  1 3  1 3 M1 A1 SCB1 for finding the inverse of 2M\nM−1 = 1  2 2 = 1 14 14 \n14  2 \n−7 3 7  − 3 1 \n2\n© Cambridge University Press & Assessment 2025 Page 10 of 19",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s25_ms_13-p10.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q02",
      "question_id": "9231-2025-mj-13-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 2025+47−2=2070=4546is divisible by 46. B1 Checks base case.\nAssume that 2025k +47k −2 is divisible by 46 for some positive integer k. B1 States inductive hypothesis.\nThen 2025k+1+47k+1−2=(2024+1)2025k +(46+1)47k −2 M1 A1 Separates 2025k +47k −2 or considers\ndifference.\nis divisible by 46 because A1 Shows that it is divisible by 46.\n20242025k +4647k =\nis divisible by 46\nHence, [by induction], true for every positive integer n. A1 Must reference the statement to be proven.\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p11.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q03",
      "question_id": "9231-2025-mj-13-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) 0=2 +2 +2 +2 +2(7) M1\nUses\n(+++)2 =2 +2 +2 +2 +\n2(+++++)\n2 +2 +2 +2 =−14 A1\n2\n\n3(b) S +7S +3S +88=0 M1 Uses given equation.\n4 2 1\nS =10 A1FT FT from (a)\n4\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 19\n\n3(c) ( r+2)2 =r2 +22r+4 B1 Expands.\n10 ( ) M1 Collects like terms and uses results from parts\n ( r+2)2 + ( r+2)2 + ( r+2)2 + ( r+2)2 (a) and (b).\nr=1\n10 A1FT FT from (a) and (b) evaluated coefficients.\n=( 4r2 −28r+10 )\nr=1\n=4 ( 10(10+1)( 2(10)+1 )) −14(10)(10+1) M1 Applies formulae from MF19.\n6\n+10(10)\n=100 A1 Evaluated. CAO\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p11.png",
        "../answer-assets/9231_s25_ms_13-p12.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q04",
      "question_id": "9231-2025-mj-13-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) w −w =(r+1)(r+2) (r+10)−r(r+1)(r+2) (r+9) M1 Correct expression seen for the difference.\nr+1 r\n(r+1)(r+2) (r+9)(r+10−r)=10(r+1)(r+2) (r+9) A1 Must see the (r + 10 – r) AG.\n2\n\n4(b) 1 (w −w )+ 1 (w −w )+ + 1 (w −w ) M1 A1 Writes at least three terms, including last. Can\n10 2 1 10 3 2 10 n+1 n\nbe implied by correct final answer.\n= 1 (w −w )= 1 (n+1)(n+2) (n+10)−10!= 1 (n+1)(n+2) (n+10)−9! A1 ISW\n10 n+1 1 10 10 10\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 19\n\n4(c) v +v + +v = xw 2 −xw 1 +xw 3 −xw 2 + +xw n+1 −xw n = xw n+1 −xw 1 M1\n1 2 n\nFor −1x1,sum to infinity is −xw 1 =−x10! A1\nFor x=1, sum to infinity is 0. A1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p12.png",
        "../answer-assets/9231_s25_ms_13-p13.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q05",
      "question_id": "9231-2025-mj-13-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k  6  M1 A1 Finds vector perpendicular to Π.\n \n1 −2 −1 = −1\n \n \n3 2 −2  8 \n6(2)−(3)+8(−2)=−7 M1 Substitutes point on Π.\n6x−y+8z=−7 A1\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 19\n\n5(b) 4  6  4+6t M1 A1FT Uses OP and multiple of their normal toΠ.\n     \nOF =OP+PF = 2 +t −1 = 2−t\n     \n     \n9  8  9+8t\n6(4+6t)−(2−t)+8(9+8t)=−7101t+94=−7 M1 Substitutes into the equation forΠ.\n−2 A1\n \nt=−1OF = 3\n \n \n 1 \n4\n\n5(c)  3   6  M1 A1FT Uses dot product of 3i+5j−k and their normal\n    5\n5 −1 = 35 101coscos= in a correct formula.\n   \n    35 101\n−1  8 \nAcute angle between l and Π is 90−=4.8 A1 Mark final answer. No ISW\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p13.png",
        "../answer-assets/9231_s25_ms_13-p14.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q06",
      "question_id": "9231-2025-mj-13-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x=−a B1 States vertical asymptote.\n(x+a)(x−a)+a2 +a a2 +a M1 Finds oblique asymptote.\ny= =x−a+\nx+a x+a\ny=x−a\nA1\n3\n© Cambridge University Press & Assessment 2025 Page 14 of 19\n\n6(b) dy M1 Differentiates\n=1−(a2 +a)(x+a)−2 =0 or x2+2ax−a=0\ndx\nA1 Sets correct derivative to zero or ACF of\nquadratic equation\nA1 CAO. No equivalent forms.\nx=−a a2 +a\n3\n\n6(c) y B1FT Axes shown and asymptotes labelled.\nFT on asymptotes from (a)\nB1 Lower branch correct.\nB1 Upper branch with (0,1) stated or 1 marked on\n1\nvertical axis and no others.\nx\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n6(d)\nxy\nB1FT FT from sketch in (c)\n1\n1\n\n6(e) x2 +a M1 Finds critical points.\n=a\nx+a\nx2−ax+a−a2 =0\nx2 +a B1 Quadratic equation sufficient for B1. ISW\nOr =−ax2 +ax+a+a2 =0\nx+a x2 +a\nMay state that =−a has no real solutions\nx+a\nsince a0.\na2 −4 ( a−a2) =5a2 −4a0 M1 Uses that discriminant is positive to get a range\nof values for a.\nCan not be 0 so M0.\na 4. A1\n5\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 19",
      "source_pages": [
        14,
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p14.png",
        "../answer-assets/9231_s25_ms_13-p15.png",
        "../answer-assets/9231_s25_ms_13-p16.png"
      ]
    },
    {
      "id": "9231-2025-mj-13-q07",
      "question_id": "9231-2025-mj-13-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) −esinsin+esincos2=0 M1 dr\nFinds and sets equal to 0.\nd\ncos2−sin=0 A1\n1−sin2−sin=0 M1 Makes it a quadratic in sin and attempts to\nsolve\n1( ) A1 Correct solution. Condone 0.666\nsin= −1+ 5\n2\n(1.208,0.666) A1 Must be to 3 decimal places\n5\n\n7(b) x=e1 2 sincos2 3  M1 Uses x=rcos\ndx dx\n=−3e 1 2 sincos 1 2sin+ 1cos 5 2e 1 2 sin=0 Finds and sets equal to 0.\nd 2 2 d\n−3sin+cos2=0 A1\n−3sin+1−sin2=0 M1 Makes it a quadratic in sin and attempts to\nsolve.\n1( ) A1 Correct solution. Condone 0.308\nsin= −3+ 13\n2\n(1.136,0.308) A1 If already penalised for rounding to 3 decimal\nplaces in part (a) then allow to 3 s.f.\n5\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n7(c) B1 Closed loop passing through O and one other\npoint on the initial line. Does not go into 2nd and\n3rd quadrants. More in the 1st than 4th quadrant.\nB1 −1π0.666correct. Tangent needs to\n2\nhave positive gradient at the intersection (1,0).\nB1 0.666 1πcorrect.\n2\n\nGraph should approach the line =\n2\nO =0\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n7(d) π B1 Condone absence of dθ but must include the\n12 esincosd\n2 −π limits.\n2\nπ M1 Recognises integral.\n= 1esin2\n2 \n−π\n2\n= 1\n( e−e−1) A1 Must be exact. ISW after correct answer seen.\n2\n3\n© Cambridge University Press & Assessment 2025 Page 19 of 19",
      "source_pages": [
        17,
        18,
        19
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_13-p17.png",
        "../answer-assets/9231_s25_ms_13-p18.png",
        "../answer-assets/9231_s25_ms_13-p19.png"
      ]
    },
    {
      "id": "9231-2025-mj-14-q01",
      "question_id": "9231-2025-mj-14-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) n n(n+1) M1 Uses formula for r.\n2r+n=2 +n\n2\nr=1\n=n(n+2) A1 Or n2 +2n\n2\n\n1(b) 1 1 r2 +2r+2−r2 −1 2r+1 B1 Puts over common denominator.\n− = =\nr2 +1 r2 +2r+2 ( r2 +1 )( r2 +2r+2 ) ( r2 +1 )( r2 +2r+2 )\n1\n\n1(c) 1 1 1 1 1 1 1 1 M1 Shows enough complete terms to demonstrate\n− + − + − + + −\n2 5 5 10 10 17 n2 +1 n2 +2n+2 cancellation.\n1 1 A1\n−\n2 n2 +2n+2\n2\n\n1(d) 1 B1 FT\n2\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_14-p09.png"
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    },
    {
      "id": "9231-2025-mj-14-q02",
      "question_id": "9231-2025-mj-14-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 d M1 d\n(xlnx)=1+lnx Finds (xlnx).\ndx dx\nd2 A1 Checks base case.\n(xlnx)=x−1 =(−1)2(2−2)!x1−2\ndx2\nAllow\nd2\n(xlnx)=x−1 =(−1)2(0)!x1−2\ndx2\ndk B1 States inductive hypothesis.\nAssume that (xlnx)=(−1)k(k−2)!x1−k\ndxk\ndk+1 M1 Differentiates kth derivative.\n(xlnx)=(−1)k(k−2)!(1−k)x1−k−1\ndxk+1\n=(−1)k+1((k+1)−2)!x1−(k+1) A1\ndn A1 States conclusion.\nSo, (xlnx)=(−1)n(n−2)!x1−n is also true for n=k+1.\ndxn\n[Hence, by induction], true for every integer n 2.\n6\n© Cambridge University Press & Assessment 2025 Page 10 of 18",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_14-p10.png"
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    },
    {
      "id": "9231-2025-mj-14-q03",
      "question_id": "9231-2025-mj-14-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 3,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) AB=−5i+j−2k AC=i+2k B1 Finds direction vectors of two lines in the plane.\ni j k  2  M1 A1 Finds normal to the plane ABC.\n \n−5 1 −2 = 8\n \n \n1 0 2 −1\n \n2(0)+8(2)−1(3)=132x+8y−z=13 M1 A1 Substitutes point.\n5\n\n3(b) 13 M1 Divides their d by magnitude of their normal vector.\n=1.57\n22 +82 +12\nA1 FT 13\nOr oe\n69\n2\n\n3(c) 2 0 M1 Dot product of normal and line direction.\n    13\n8 2 = 69 13coscos=\n   \n69 13\n   \n-1 3\n   \nA1 FT\n90−=25.7ο A1 0.449 radians CAO\n3\n© Cambridge University Press & Assessment 2025 Page 11 of 18",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_14-p11.png"
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    },
    {
      "id": "9231-2025-mj-14-q04",
      "question_id": "9231-2025-mj-14-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) 1   M1 A1 Finds determinant.\n1     1\n 1  = − + =1−2 −2 −2 +2\n 1  1  \n  1\n1−2 −2 −2 +2=02 +2 +2 =1+2(1)=3 M1 A1 Sets determinant=0 and uses =1, AG.\n4\n\n4(b) 2 +2 +2 =b2 −2c3=b2 −2c *M1 A1 Uses 2 +2 +2 =(++)2 −2(++).\n3+3+3 =−b(3)−c(−b)+33=−3b+cb+3 *M1 A1 Uses cubic equation or formula for 3+3+3.\nc=3, b=3 DM1 A1 Solves simultaneous equations. WWW\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_14-p12.png"
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    },
    {
      "id": "9231-2025-mj-14-q05",
      "question_id": "9231-2025-mj-14-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "5(a) 3 0 B1 One-way stretch in the x-direction, scale factor 3.\n \n0 1\n0 1 B1 Reflection in the line y= x.\n \n1 0\n0 13 0 0 1 B1 Correct order and final answer.\nM=  = \n1 00 1 3 0\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 18\n\n5(b) Reflection in the line y= x B1\nfollowed by\nB1\na [one-way] stretch in the x-direction, scale factor1.\n3\nB1 Allow scale or factor.\n3\n\n5(c) 0 1 B1FT Inverse of their M.\nM−1 = 3  This may be seen in part (b)\n1 0\n1 2 0 11 2 1 2 M1 A1 Multiplies on the left by the inverse of their M.\nN=M−1  = 3  = 3  May also use simultaneous equations.\n3 2 1 03 2 1 2\n3\n\n5(d) 1 2x  x+2y  B1 x X \n  =  Transforms   to  .\n3 2y 3x+2y y Y \n3x+2mx=m(x+2mx) M1 Uses y=mx and Y =mX .\n3+2m=m(1+2m)2m2 −m−3=0 A1 Forms quadratic in m.\n3 A1\nm= m=−1\n2\ny= 3x and y=−x A1\n2\n5\n© Cambridge University Press & Assessment 2025 Page 13 of 18",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s25_ms_14-p13.png"
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    },
    {
      "id": "9231-2025-mj-14-q06",
      "question_id": "9231-2025-mj-14-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) B1 Spiral shape on [0, 2π], with correct orientation,\nstarting at pole and with one other point of intersection\nwith the initial line.\n=0\nB1 r increasing, correct gradient at points of intersection\nwith the initial line.\nMaximum distance of C from the pole is a. B1\n3\n\n6(b) 1a2 2π tan2(1 ) d M1 Uses 1r2d with correct limits.\n2 8 2\n0\n= 1a2 2π sec2(1 ) −1d= 1a28tan (1 ) − 2π M1 A1 Uses tan2 A=sec2 A−1.\n2 0 8 2  8  0\n= 1a2(8−2π)=a2(4−π) A1 Accept 0.858a2 if M1 A1 earned.\n2\n4\n© Cambridge University Press & Assessment 2025 Page 14 of 18\n\n6(c) y=atan (1 ) sin B1 Uses y=rsin seen or implied.\n8\ndy\n=a\n( tan(1)cos+1sinsec2(1))\n=0\nM1 A1 Sets derivative equal to zero.\nd 8 8 8\ncos (1 ) sin (1 ) cos+1sin=0sin (1 ) cos+1sin=0 M1 Uses sin2A=2sinAcosAin their derivative.\n8 8 8 4 4\n4sin (1 ) cos+sin=0 A1 AG.\n4\n4sin (1(4.95) ) cos4.95+sin4.95=−0.0822 and B1 Shows sign change.\n4\n4sin\n(1(5) )\ncos5+sin5=0.118\n4\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_14-p14.png",
        "../answer-assets/9231_s25_ms_14-p15.png"
      ]
    },
    {
      "id": "9231-2025-mj-14-q07",
      "question_id": "9231-2025-mj-14-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) y=1 B1\n1\n© Cambridge University Press & Assessment 2025 Page 15 of 18\n\n7(b) x2 +x−4= yx2 + yx+2y M1\n(y−1)x2 +(y−1)x+2y+4[=0]\nCorrectly forms quadratic equation in x.\n(y−1)2 −4(y−1)(2y+4) 0(y−1)(y−1−8y−16) 0 M1 Uses discriminant to obtain condition for real roots.\n(y−1)(7y+17) 0−17 y 1. A1\n7\nExplains why y1 −17 y1. A1 AG.\n7\n4\n\n7(c) dy 12x+6 M1 A1 Sets derivative equal to 0.\n= =0\ndx ( x2 +x+2 )2\nAlternative\nx2 +x−4 Sets y=−17 and forms equation for x.\n=−17 7\nx2 +x+2 7\n( −1,−17) B1\n2 7\n3\n© Cambridge University Press & Assessment 2025 Page 16 of 18\n\n7(d) y B1 Axes and asymptotes drawn and labelled.\nB1 Correct shape and position.\nx\n(0,−2), ( 1 ( −1− 17 ) ,0 ) , ( 1 ( −1+ 17 ) ,0 ) B1 Coordinates of the intersections with the axes.\n2 2 Allow (0, -2), (1.56, 0), (-2.56, 0)\n3\n© Cambridge University Press & Assessment 2025 Page 17 of 18\n\n7(e) y B1FT Symmetrical about x=0, FT their graph from (d).\nB1 Correct shape, not smooth at (0,−2).\nx\n|x|2 +|x|−4 M1 Finds critical point(s).\n=−1 3|x|2 +3|x|−6=0\n|x|2 +|x|+2 2\n|x|=1 A1\n−1x1 A1\n5\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
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        17,
        18
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_14.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s25_ms_14-p16.png",
        "../answer-assets/9231_s25_ms_14-p17.png",
        "../answer-assets/9231_s25_ms_14-p18.png"
      ]
    },
    {
      "id": "9231-2025-mj-21-q01",
      "question_id": "9231-2025-mj-21-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 z3 =27−27i=27 2e−i1 4 π B1 Modulus and argument of z3 (does not need to be in\nexponential form).\nz =32 1 6e−i 1 1 2 π M1 A1 Finds one root. Cube root must be taken for M1.\n1\nz =32 1 6ei 1 7 2 π, z =32 1 6e−i 1 9 2 π A1FT A1FT Finds other two roots. FT on modulus. Withhold one\n2 3 answer mark for omission of i.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_21-p09.png"
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    },
    {
      "id": "9231-2025-mj-21-q02",
      "question_id": "9231-2025-mj-21-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) I = 1 (1−x)n sinhxdx=(1−x)ncoshx 1 + 1 n(1−x)n−1coshxdx M1 A1 Applies integration by parts. Appropriate choice of\nn   parts for M1.\n0 0 0\nI =−1+n(1−x)n−1sinhx 1 +n(n−1) 1 (1−x)n−2 sinhxdx M1 Applies integration by parts again. Appropriate choice\nn   of parts for M1.\n0 0\nI =−1+n(n−1)I A1 AG, withhold for no evidence of substitution or\nn n−2\nincorrect use of limits.\n4\n\n2(b) I = 1 sinhxdx=coshx1 =cosh1−1. B1 Finds I 0 .\n0 0 0\nI =−1+2I M1 Uses reduction formula from (a).\n2 0\nI =2cosh1−3 A1 Accept e+e−1−3.\n2\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-21-q03",
      "question_id": "9231-2025-mj-21-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 z−z−1=2isin B1 Use of z−z−1=2isin\n( z−z−1)5 = ( z5 −z−5) −5 ( z3−z−3) +10 ( z−z−1) M1 A1 Expands and groups. No clear grouping is M1 A0.\n25isin5=2isin5−5(2isin3)+10(2isin) M1 A1FT Substituteszn −z−n =2isinn on RHS. FT on signs of\ncoefficients only. Substitutingzn −z−n =isinnon\nRHS is M1 A0. Omission of i is M0.\n16 A1 Cancelation of i justified.\n cosec5= .\nsin5−5sin3+10sin\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_21-p10.png"
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    },
    {
      "id": "9231-2025-mj-21-q04",
      "question_id": "9231-2025-mj-21-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 e 1 + 1 e 2 + + 1 e n−1 M1 Forms sum of the areas of the rectangles.\n1 2 n−1\nn A1 Compares with integral. Wrong comparison\n 1 e x dx\n1 x (missing/wrong inequality or wrong limits) is A0.\nn n M1 A1 Correct integration. Attempt at integration by parts is\n 1 e xdx=2e x\n1 x   1 M0 A0.\nn n ( ) A1 AG, must have obtained all previous marks.\n 1 e r 2e x + 1 e n = 2+ 1 e n −2e\nr   1 n n\nr=1\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 18\n\n4(b) 1 e 2 + 1 e 3 + + 1 e n M1 Forms the sum of the areas of appropriate rectangles.\n2 3 n\nn n A1 Compares with integral.\n 1 e xdx=2e x\n1 x   1\nn n M1 A1 Adds term from r=1.\n 1 e r 2e x +e=2e n −e\nr   1\nr=1\n4\n© Cambridge University Press & Assessment 2025 Page 11 of 18",
      "source_pages": [
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        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-mj-21-q05",
      "question_id": "9231-2025-mj-21-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 6m2 +3m+6=0m=−1  1i 15 M1 Auxiliary equation. M1 for correct type (complex) of\n4 4 roots.\nx=e−1 4 t( Acos 15t+Bsin 15t ) A1 Complimentary function.\n4 4 Allow ‘x=’ missing. Using x instead of t is A0.\nx=ke−t x'=−ke−t x''=ke−t B1 Particular integral and its derivatives.\n6ke−t −3ke−t +6ke−t =e−t M1 Substitutes and equates coefficients.\nk =1 A1\n9\nx=e−1 4 t( Acos 15t+Bsin 15t ) +1e−t A1 FT General solution. FT on CF and their value of k. Must\n4 4 9 have ‘x=’.\nx'= 15e−1 4 t( −Asin 15t+Bcos 15t ) −1e−1 4 t( Acos 15t+Bsin 15t ) −1e−t *M1 Using product rule.\n4 4 4 4 4 4 9\n0= A+1 0= 15 B−1 A−1 A=−1, B= 1 DM1 A1 Substitutes initial conditions and forms simultaneous\n9 4 4 9 9 3 15 equations.\nx=1e−1 4 t( −1cos 15t+ 1 sin 15t ) +1e−t A1 Must have ‘x=’.\n3 3 4 15 4 9\n10\n© Cambridge University Press & Assessment 2025 Page 12 of 18",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_21-p12.png"
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    },
    {
      "id": "9231-2025-mj-21-q06",
      "question_id": "9231-2025-mj-21-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) eu −e−u 2 B1 Writes in exponential from.\ntanhu= sechu=\neu +e−u eu +e−u\neu −e−u  2 ( eu +e−u)2 − ( eu −e−u)2 4 M1 Writes over common denominator.\n1−  = =\neu +e−u  ( eu +e−u)2 ( eu +e−u)2\n=sech2u A1 Withold this mark if working with LHS and RHS\nsimultaneously.\n3\n\n6(b) du M1 A1 Differentiates implicitly.\nu=sech−1tsechu=t−tanhusechu =1\ndt\ndu 1 1 M1 A1 Uses identity from (a), AG.\n=− =−\ndt tanhusechu t 1−t2\nFirst alternative method for question 6(b)\n1 du 1 M1 A1 Differentiates implicitly.\nu=sech−1tcoshu= sinhu =−\nt dt t2\ndu 1 1 1 M1 A1 Uses cosh2u−sinh2u=1, AG.\n=− =− =−\ndt t2 cosh2u−1 t2 t−2 −1 t 1−t2\n© Cambridge University Press & Assessment 2025 Page 13 of 18\n\n6(b) Second alternative method for question 6(b)\n1 1 M1 A1 1\nu=sech−1tcoshu= u=cosh−1\n \nDifferentiatescosh−1\n \n.\nt t t\ndu\n=\n1\n\n( −t−2)\n=−\n1 M1 A1 Applies formula with chain rule and simplifies, AG.\ndt (1)2 −1 t 1−t2\nt\nThird alternative method for question 6(b)\n1 1 1   1+ 1−t2  M1 A1 Uses logarithmic form of sech−1t\nsechu=tcoshu= u=ln + −1=ln \nt  t t2   t \n   \ndu  −1− 1−t2  t  1 M1 A1 Differentiates logarithmic form using chain and\n=  =− quotient rules, AG.\ndt   t2 1−t2    1+ 1−t2   t 1−t2\n4\n\n6(c) dy 1 dx 1 B1 B1\n=sech−1t− , =\ndt 1−t2 dt 1−t2\ndy = dy  dt =− 1−t2 + ( 1−t2) sech−1t M1 A1 Uses chain rule, AG.\ndx dt dx\n4\n© Cambridge University Press & Assessment 2025 Page 14 of 18\n\n6(d) d dy t 1−t2 M1 A1 Applies product rule. Might see\ndt  dx   = 1−t2 − t 1−t2 −2tsech−1t 2t2 −1 −2tsech−1t.\nt 1−t2\nd2y\n=t 1−t2 −\n( 1−t2)3\n2\n−2t ( 1−t2) sech−1t\nM1 A1 Uses chain rule.\ndx2 t\n4\n© Cambridge University Press & Assessment 2025 Page 15 of 18",
      "source_pages": [
        13,
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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      ]
    },
    {
      "id": "9231-2025-mj-21-q07",
      "question_id": "9231-2025-mj-21-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7   x+5 dx= 1ln ( x2 +10x+61 ) M1 A1 Finds integrating factor using  f'(x) dx=ln f(x).\n x2 +10x+61 2 f(x)\n−1ln ( x2+10x+61 ) 1 A1\ne 2 =\nx2 +10x+61\nd  y  1 M1 Correct form on LHS using their integrating factor.\n =\n \ndx  x2 +10x+61 x2 +10x+61\nx2 +10x+61=(x+5)2 +36 B1 Completes the square.\n 1 x+5 *M1 A1 Integrates RHS. Attempting to integrate by parts is\n dx=sinh−1   +C M0.\n (x+5)2 +62  6 \nC=−sinh−1(8) =−ln (8+5) =−ln3 DM1 A1 Substitutes initial conditions.\n6 6 3\nx2 +1\ny\n0x+61\n=sinh−1 \n \nx\n6\n+5\n \n−sinh−1(4\n3\n) A1\nAccept x2 +1\ny\n0x+61 =ln\n\n  \nx\n1\n+\n8\n5\n+\nx2 +\n1\n1\n8\n0x+61 \n   .\n10\n© Cambridge University Press & Assessment 2025 Page 16 of 18",
      "source_pages": [
        16
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_21-p16.png"
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    },
    {
      "id": "9231-2025-mj-21-q08",
      "question_id": "9231-2025-mj-21-q08",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) Ae=eA3e=A(Ae) M1 Multiples by A2on LHS. Trying to invert an\neigenvector or missing eigenvector is M0.\n A3e=A(e)=2(Ae)=3e A1\n2\n\n8(b) −1− 3 4 M1 Forms det(A−I)=0. Do not accept checking each\n0 1− 0 =0 given value is an eigenvalue. Do not accept row\noperations on A.\n0 −2 5−\n(−1−)((1−)(5−)−0 )\n−3(0)+4(0)=0\n(−1−)(1−)(5−)=0=−1,1 and 5. A1 AG.\n2\n\n8(c) i j k −8 1 M1 A1 Uses vector product (or equations) of rows of A−I\n    to find corresponding eigenvectors.\n=−1: 0 3 4 = 0 0\n   \n   \n0 2 0  0  0\ni j k 20 5 i j k 16 2 A1 A1\n       \n=1: −2 3 4 = 8 ~ 2 =5: −6 3 4 = 0 ~ 0\n       \n       \n0 −2 4  4  1 0 −4 0 24 3\n1 5 2 −3 0 0 M1 A1 Or correctly matched permutations of columns.\n    A column of zeros is M0. Repeated eigenvalues in D is\nThus P= 0 2 0 and D= 0 −1 0\n    M0.\n   \n0 1 3  0 0 3\n6\n© Cambridge University Press & Assessment 2025 Page 17 of 18\n\n8(d) (A−2I)3 =A3−6A2 +12A−8I B1 Expands.\n=5A2+A−5I−6A2+12A−8I M1 Substitutes A3 =5A2+A−5I .\n=−A2+13A−13I A1\nAlternative method for 8(d)\n(+3)(+1)(−3)=3+2 −9−9=0 B1 Finds characteristic equation of A−2I.\n(A−2I)3 =−(A−2I)2 +9(A−2I)+9I M1 SubstitutesA−2I and makes (A−2I)3 the subject.\n=−A2+13A−13I A1\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
        17,
        18
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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        "../answer-assets/9231_s25_ms_21-p18.png"
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    },
    {
      "id": "9231-2025-mj-22-q01",
      "question_id": "9231-2025-mj-22-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 z3 =27−27i=27 2e−i1 4 π B1 Modulus and argument of z3 (does not need to be in\nexponential form).\nz =32 1 6e−i 1 1 2 π M1 A1 Finds one root. Cube root must be taken for M1.\n1\nz =32 1 6ei 1 7 2 π, z =32 1 6e−i 1 9 2 π A1FT A1FT Finds other two roots. FT on modulus. Withhold one\n2 3 answer mark for omission of i.\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_22-p09.png"
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    },
    {
      "id": "9231-2025-mj-22-q02",
      "question_id": "9231-2025-mj-22-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) I = 1 (1−x)n sinhxdx=(1−x)ncoshx 1 + 1 n(1−x)n−1coshxdx M1 A1 Applies integration by parts. Appropriate choice of\nn   parts for M1.\n0 0 0\nI =−1+n(1−x)n−1sinhx 1 +n(n−1) 1 (1−x)n−2 sinhxdx M1 Applies integration by parts again. Appropriate choice\nn   of parts for M1.\n0 0\nI =−1+n(n−1)I A1 AG, withhold for no evidence of substitution or\nn n−2\nincorrect use of limits.\n4\n\n2(b) I = 1 sinhxdx=coshx1 =cosh1−1. B1 Finds I 0 .\n0 0 0\nI =−1+2I M1 Uses reduction formula from (a).\n2 0\nI =2cosh1−3 A1 Accept e+e−1−3.\n2\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-22-q03",
      "question_id": "9231-2025-mj-22-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 z−z−1=2isin B1 Use of z−z−1=2isin\n( z−z−1)5 = ( z5 −z−5) −5 ( z3−z−3) +10 ( z−z−1) M1 A1 Expands and groups. No clear grouping is M1 A0.\n25isin5=2isin5−5(2isin3)+10(2isin) M1 A1FT Substituteszn −z−n =2isinn on RHS. FT on signs of\ncoefficients only. Substitutingzn −z−n =isinnon\nRHS is M1 A0. Omission of i is M0.\n16 A1 Cancelation of i justified.\n cosec5= .\nsin5−5sin3+10sin\n6\nQuestion Answer Marks Guidance",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_22-p10.png"
      ]
    },
    {
      "id": "9231-2025-mj-22-q04",
      "question_id": "9231-2025-mj-22-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 e 1 + 1 e 2 + + 1 e n−1 M1 Forms sum of the areas of the rectangles.\n1 2 n−1\nn A1 Compares with integral. Wrong comparison\n 1 e x dx\n1 x (missing/wrong inequality or wrong limits) is A0.\nn n M1 A1 Correct integration. Attempt at integration by parts is\n 1 e xdx=2e x\n1 x   1 M0 A0.\nn n ( ) A1 AG, must have obtained all previous marks.\n 1 e r 2e x + 1 e n = 2+ 1 e n −2e\nr   1 n n\nr=1\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 18\n\n4(b) 1 e 2 + 1 e 3 + + 1 e n M1 Forms the sum of the areas of appropriate rectangles.\n2 3 n\nn n A1 Compares with integral.\n 1 e xdx=2e x\n1 x   1\nn n M1 A1 Adds term from r=1.\n 1 e r 2e x +e=2e n −e\nr   1\nr=1\n4\n© Cambridge University Press & Assessment 2025 Page 11 of 18",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-mj-22-q05",
      "question_id": "9231-2025-mj-22-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5 6m2 +3m+6=0m=−1  1i 15 M1 Auxiliary equation. M1 for correct type (complex) of\n4 4 roots.\nx=e−1 4 t( Acos 15t+Bsin 15t ) A1 Complimentary function.\n4 4 Allow ‘x=’ missing. Using x instead of t is A0.\nx=ke−t x'=−ke−t x''=ke−t B1 Particular integral and its derivatives.\n6ke−t −3ke−t +6ke−t =e−t M1 Substitutes and equates coefficients.\nk =1 A1\n9\nx=e−1 4 t( Acos 15t+Bsin 15t ) +1e−t A1 FT General solution. FT on CF and their value of k. Must\n4 4 9 have ‘x=’.\nx'= 15e−1 4 t( −Asin 15t+Bcos 15t ) −1e−1 4 t( Acos 15t+Bsin 15t ) −1e−t *M1 Using product rule.\n4 4 4 4 4 4 9\n0= A+1 0= 15 B−1 A−1 A=−1, B= 1 DM1 A1 Substitutes initial conditions and forms simultaneous\n9 4 4 9 9 3 15 equations.\nx=1e−1 4 t( −1cos 15t+ 1 sin 15t ) +1e−t A1 Must have ‘x=’.\n3 3 4 15 4 9\n10\n© Cambridge University Press & Assessment 2025 Page 12 of 18",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-22-q06",
      "question_id": "9231-2025-mj-22-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) eu −e−u 2 B1 Writes in exponential from.\ntanhu= sechu=\neu +e−u eu +e−u\neu −e−u  2 ( eu +e−u)2 − ( eu −e−u)2 4 M1 Writes over common denominator.\n1−  = =\neu +e−u  ( eu +e−u)2 ( eu +e−u)2\n=sech2u A1 Withold this mark if working with LHS and RHS\nsimultaneously.\n3\n\n6(b) du M1 A1 Differentiates implicitly.\nu=sech−1tsechu=t−tanhusechu =1\ndt\ndu 1 1 M1 A1 Uses identity from (a), AG.\n=− =−\ndt tanhusechu t 1−t2\nFirst alternative method for question 6(b)\n1 du 1 M1 A1 Differentiates implicitly.\nu=sech−1tcoshu= sinhu =−\nt dt t2\ndu 1 1 1 M1 A1 Uses cosh2u−sinh2u=1, AG.\n=− =− =−\ndt t2 cosh2u−1 t2 t−2 −1 t 1−t2\n© Cambridge University Press & Assessment 2025 Page 13 of 18\n\n6(b) Second alternative method for question 6(b)\n1 1 M1 A1 1\nu=sech−1tcoshu= u=cosh−1\n \nDifferentiatescosh−1\n \n.\nt t t\ndu\n=\n1\n\n( −t−2)\n=−\n1 M1 A1 Applies formula with chain rule and simplifies, AG.\ndt (1)2 −1 t 1−t2\nt\nThird alternative method for question 6(b)\n1 1 1   1+ 1−t2  M1 A1 Uses logarithmic form of sech−1t\nsechu=tcoshu= u=ln + −1=ln \nt  t t2   t \n   \ndu  −1− 1−t2  t  1 M1 A1 Differentiates logarithmic form using chain and\n=  =− quotient rules, AG.\ndt   t2 1−t2    1+ 1−t2   t 1−t2\n4\n\n6(c) dy 1 dx 1 B1 B1\n=sech−1t− , =\ndt 1−t2 dt 1−t2\ndy = dy  dt =− 1−t2 + ( 1−t2) sech−1t M1 A1 Uses chain rule, AG.\ndx dt dx\n4\n© Cambridge University Press & Assessment 2025 Page 14 of 18\n\n6(d) d dy t 1−t2 M1 A1 Applies product rule. Might see\ndt  dx   = 1−t2 − t 1−t2 −2tsech−1t 2t2 −1 −2tsech−1t.\nt 1−t2\nd2y\n=t 1−t2 −\n( 1−t2)3\n2\n−2t ( 1−t2) sech−1t\nM1 A1 Uses chain rule.\ndx2 t\n4\n© Cambridge University Press & Assessment 2025 Page 15 of 18",
      "source_pages": [
        13,
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_22-p13.png",
        "../answer-assets/9231_s25_ms_22-p14.png",
        "../answer-assets/9231_s25_ms_22-p15.png"
      ]
    },
    {
      "id": "9231-2025-mj-22-q07",
      "question_id": "9231-2025-mj-22-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7   x+5 dx= 1ln ( x2 +10x+61 ) M1 A1 Finds integrating factor using  f'(x) dx=ln f(x).\n x2 +10x+61 2 f(x)\n−1ln ( x2+10x+61 ) 1 A1\ne 2 =\nx2 +10x+61\nd  y  1 M1 Correct form on LHS using their integrating factor.\n =\n \ndx  x2 +10x+61 x2 +10x+61\nx2 +10x+61=(x+5)2 +36 B1 Completes the square.\n 1 x+5 *M1 A1 Integrates RHS. Attempting to integrate by parts is\n dx=sinh−1   +C M0.\n (x+5)2 +62  6 \nC=−sinh−1(8) =−ln (8+5) =−ln3 DM1 A1 Substitutes initial conditions.\n6 6 3\nx2 +1\ny\n0x+61\n=sinh−1 \n \nx\n6\n+5\n \n−sinh−1(4\n3\n) A1\nAccept x2 +1\ny\n0x+61 =ln\n\n  \nx\n1\n+\n8\n5\n+\nx2 +\n1\n1\n8\n0x+61 \n   .\n10\n© Cambridge University Press & Assessment 2025 Page 16 of 18",
      "source_pages": [
        16
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_22-p16.png"
      ]
    },
    {
      "id": "9231-2025-mj-22-q08",
      "question_id": "9231-2025-mj-22-q08",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) Ae=eA3e=A(Ae) M1 Multiples by A2on LHS. Trying to invert an\neigenvector or missing eigenvector is M0.\n A3e=A(e)=2(Ae)=3e A1\n2\n\n8(b) −1− 3 4 M1 Forms det(A−I)=0. Do not accept checking each\n0 1− 0 =0 given value is an eigenvalue. Do not accept row\noperations on A.\n0 −2 5−\n(−1−)((1−)(5−)−0 )\n−3(0)+4(0)=0\n(−1−)(1−)(5−)=0=−1,1 and 5. A1 AG.\n2\n\n8(c) i j k −8 1 M1 A1 Uses vector product (or equations) of rows of A−I\n    to find corresponding eigenvectors.\n=−1: 0 3 4 = 0 0\n   \n   \n0 2 0  0  0\ni j k 20 5 i j k 16 2 A1 A1\n       \n=1: −2 3 4 = 8 ~ 2 =5: −6 3 4 = 0 ~ 0\n       \n       \n0 −2 4  4  1 0 −4 0 24 3\n1 5 2 −3 0 0 M1 A1 Or correctly matched permutations of columns.\n    A column of zeros is M0. Repeated eigenvalues in D is\nThus P= 0 2 0 and D= 0 −1 0\n    M0.\n   \n0 1 3  0 0 3\n6\n© Cambridge University Press & Assessment 2025 Page 17 of 18\n\n8(d) (A−2I)3 =A3−6A2 +12A−8I B1 Expands.\n=5A2+A−5I−6A2+12A−8I M1 Substitutes A3 =5A2+A−5I .\n=−A2+13A−13I A1\nAlternative method for 8(d)\n(+3)(+1)(−3)=3+2 −9−9=0 B1 Finds characteristic equation of A−2I.\n(A−2I)3 =−(A−2I)2 +9(A−2I)+9I M1 SubstitutesA−2I and makes (A−2I)3 the subject.\n=−A2+13A−13I A1\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
        17,
        18
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_22-p17.png",
        "../answer-assets/9231_s25_ms_22-p18.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q01",
      "question_id": "9231-2025-mj-23-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 f '(x)=−(x+2)−2e (x+2)−1 B1 Finds first derivative.\nf\n''(x)=(x+2)−4e(x+2)-1 +2(x+2)−3 e(x+2)-1\n=\n2x+5\ne\n(x+2)−1 B1 Finds second derivative.\n(x+2)4\nf(0)=e 1 2 f '(0)=−1e 1 2 f ''(0)= 5 e 1 2 *M1 Evaluates derivatives at zero. Must have\n4 16 attempted second derivative using the\nproduct rule.\ne (x+2)−1 =e 1 2 −1e 1 2x+ 5 e 1 2x2 DM1 A1\n4 32\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p09.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q02",
      "question_id": "9231-2025-mj-23-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) et −e−t 2 B1\ntanht= secht=\net +e−t et +e−t\n2 M1 Writes in a single fraction over common\net −e−t  4 e2t +e−2t +2\n  + = =1 denominator, AG.\net +e−t  ( et +e−t)2 ( et +e−t)2\nA1 Withold A1 mark if they used mixed\nvariables within a single line of working.\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 17\n\n2(b) dx B1\n=tanht\ndt\ndy cosht M1 A1\n= =secht\ndt 1+sinh2t Applies 1+sinh2t=cosh2t.\n1 1 M1 Substitutes their derivatives into a correct\n tanh2t+sech2t dt = 1dt\n0 0 2 2\ndx dy\nformula for arc length eg. + dt\n   \ndt  dt \n=1 A1\n5\nQuestion Answer Marks Guidance",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p09.png",
        "../answer-assets/9231_s25_ms_23-p10.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q03",
      "question_id": "9231-2025-mj-23-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) d ( 9y2)\n=18yy'\nB1 Differentiates y2 correctly.\ndx\nd ( −3sinh−1(xy) ) =−3 xy'+ y M1 A1 Differentiates sinh−1(xy) correctly.\ndx 1+x2y2\n18 y'−3 4y'+1 3 =0 y'=−1 A1 Substitutes( 4,1) and shows at least one\n3 5 2 3\n3\nintermediate step, AG.\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 17\n\n3(b) 18yy''+18(y')2 B1 Differentiates 18yy'\n 1+x2y2 (xy''+2y')−(xy'+ y)2( 1+x2y2)−1 2(xy)  B1 B1 Differentiates −3 xy'+ y .\n \n−3 1+x2y2\n 1+x2y2 \n \n \n5(4y''−1)−(5)2(3)(4) M1\nSubstitutes ( 4,1) . and y=−\n1\n18 y''+18−33 3 5 3 =0 3 2\n3 4 25\n9\n45(4y''−1)−60\n18 y''+18− =0\n3 4 25\n−6 y''+87 =0 y''= 29 A1\n5 10 4\n5\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        10,
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p10.png",
        "../answer-assets/9231_s25_ms_23-p11.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q04",
      "question_id": "9231-2025-mj-23-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4 m2 +m−2=0m=1,−2 M1 Auxiliary equation. M1 for correct type of\nroots.\nx=Aet +Be−2t A1 Complementary function. Allow “x=”\nmissing. Using x instead of t is A0.\nx= pt2 +qt+rx'=2pt+qx''=2p B1 Particular integral and its derivatives.\np=−1 2p−2q=1 2p+q−2r=−1 M1 Substitutes and equates coefficients.\np=−1 q=−3 r=−5 A1\n2 4\nx= Aet +Be−2t −t2 − 3t− 5 A1 FT General solution. FT on CF and their values\n2 4\nof p, q and r. Must have “x=”.\nx'= Aet −2Be−2t −2t− 3 *M1 Differentiates.\n2\nA+B−5 =0 A−2B−3 =0  A= 4, B=− 1 DM1 A1 Uses initial conditions.\n4 2 3 12\nx= 4et − 1 e−2t −t2 − 3t− 5 A1 Must have “x=”.\n3 12 2 4\n10\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p12.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q05",
      "question_id": "9231-2025-mj-23-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) (cos5+isin5)=(c+is)5 M1 Uses binomial theorem.\ncos5=c5−10s2c3+5s4c A1\n=c5−10 ( 1−c2) c3+5 ( 1−c2)2 c M1 Uses s2 =1−c2.\n=16c5−20c3+5c A1\n1 sec5 M1 Divides simplified numerator and\nsec5= \n16c5 −20c3+5c sec5 denominator by c5.\nsec5 A1 AG.\n=\n5sec4−20sec2+16\n6\n\n5(b)\nx5 = 2 ( 5x4 −20x2 +16 ) \nx5\n= 2\nB1 Relates with equation in part (a). Can be\n3 5x4 −20x2 +16 3 implied by M1 mark.\nsec5= 2 cos5= 3 M1 Solves cos5= 3.\n3 2 2\nx=sec ( 1 π ) A1 Gives one correct solution.\n30\nsec (11π ) ,sec (13π ) ,sec (23π ) ,sec (25π ) A1 Gives four other solutions. Allow different\n30 30 30 30\nvalues of q as long as all five solutions are\nfound.\n-1.35, -1.15, 1.01, 2.46, 4.81\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p13.png"
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    },
    {
      "id": "9231-2025-mj-23-q06",
      "question_id": "9231-2025-mj-23-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) 1  1 1 1  1 n2 n2 n2  M1 A1 Forms sum of the areas of the rectangles.\n + + + =  + + + \nn  1 n 2 2 +1 n 22 2 +1 n n 2 2 +1  n 12 +n2 22 +n2 n2 +n2 \n  1 1 dx=tan−1x 1 = 1π M1 Evaluates integral.\n x2 +1   0 4\n0 A1 CAO\n1 n n2 n n A1 AG. Clear and convincing comparison with\n =  1π\nn n2 +r2 n2 +r2 4 the integral.\nr=1 r=1\n5\n\n6(b)   *M1 A1 Forms the sum of the areas of appropriate\n1 1 1 1 1 n2 n2 n2 \n + + + =  + + +  rectangles. M0 for more than n rectangles.\nn02 +1 12 +1 (n−1)2 +1 n 02 +n2 12 +n2 (n−1)2 +n2 \n n2 n2 \n1 n n 1 n n 1 DM1 A1 Compares with integral. Correctly adjust for\nn + n2 +r2 − 2n  1 4 π n2 +r2 1 4 π− 2n 0th and last rectangles. (Allow sign errors\nr=1 r=1\nonly for M1). Limits must match area\ncovered by chosen rectangles.\n4\n\n6(c) 1π B1\n4\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p14.png"
      ]
    },
    {
      "id": "9231-2025-mj-23-q07",
      "question_id": "9231-2025-mj-23-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "7   2x+6 dx=ln ( x2 +6x+5 ) M1 A1 Finds integrating factor using\n x2 +6x+5 f'(x)\n dx=ln f(x).\nf(x)\n−ln ( x2+6x+5 ) 1 A1\ne =\nx2 +6x+5\nd  y  4 M1 Correct form on LHS.\n =\ndxx2 +6x+5 x2 +6x+5\n4 1 1 B1 Partial fractions. Or completing the square.\n= −\nx2 +6x+5 x+1 x+5\n 4  x+1 *M1 A1 Integrates RHS. (RHS must be of the correct\n  x2 +6x+5 dx=ln x+5   +C form.)\n0=ln1+C DM1 Substitutes initial conditions.\n5\ny =ln   x+1 −ln1 =ln   5x+5   A1 Accept y= ( x2 +6x+5 ) ln   5x+5 .\nx2 +6x+5 x+5 5  x+5   x+5 \n9\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
      "source_pages": [
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p15.png"
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    },
    {
      "id": "9231-2025-mj-23-q08",
      "question_id": "9231-2025-mj-23-q08",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) 3 3 8 M1 A1 Sets determinant equal to zero and forms 3\n2\nterm quadratic equation.\na 3 4 =08a2 −3a−9 =0\n2\n0 a −1\na= 3 ( 1 17 ) A1\n16\n3\n\n8(b) ( −3) (−3)(+1)=3−72 +9 =0 B1 Finds characteristic equation.\n2 2 2\n9I= 7A2 −A3  9B= 7A−A2 M1 Using C-H Theorem and replaces  with A.\n2 2 2 2\nB2 = 7I− 2A M1 A1 Multiplies by B2 = A−2\n9 9\nCAO\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n8(c) Eigenvalues of A are 3,3 and −1. B1 Upper diagonal matrix or characteristic\n2\nequation.\ni j k −9 1 M1 A1 Uses vector product (or equations) to find\n2\n    corresponding eigenvectors.\n= 3 2 : 0 3 8 = 0   0 \n0 3 2 5 2   0    0  \ni j k 12 2 i j k −20  2  A1 A1\n       \n=3: −3 3 8 = 6 ~ 1 =−1: 5 3 8 = −10 ~ 1\n2     2    \n       \n0 0 4  0  0 0 4 4  10  −1\n1 2 2  2 0 0  M1 A1 Or correctly matched permutations of\n3\n    columns. Column of zeros or repeated\nThus P= 0 1 1 and D= 0 1 0\n   3  column in P, or repeated eigenvalues in D is\n   \n0 0 −1\n\n0 0 −1\n M0.\n7\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
        16,
        17
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_23-p16.png",
        "../answer-assets/9231_s25_ms_23-p17.png"
      ]
    },
    {
      "id": "9231-2025-mj-24-q01",
      "question_id": "9231-2025-mj-24-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 2 3 M1 Expands determinant and forms quadratic\nk 5 6 =45−12k−2(9k−42)+3 ( 2k2 −35 ) expression.\n7 2k 9\n6k2 −30k+24=0k=1,4 M1 A1 Solves quadratic. Need both values for A1.\n3\n\n1(b) x+ 2y+3z =1, M1 A1 For M1 must use all three equations and reduce to\n3y+3z=1 e.g. y = x. Accept explicit solution using a\nx+ 5y+6z =2, \n3x+3z=1 parameter. Accept row echelon form (with clear\n7x+2y+9z =3,\n1 2 3 1\n \noperations on all three rows) 0 3 3 1 .\n \n \n0 0 0 0\nThree planes meeting in a common line. B1 Sheaf.\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p09.png"
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    },
    {
      "id": "9231-2025-mj-24-q02",
      "question_id": "9231-2025-mj-24-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 x2 −2x+5=4+(x−1)2 B1 Completes the square. OE.\n5 5 M1 A1 Applies formula.\n2 1  x−12\n dx=  sinh−1  \n 4+(x−1)2   2 \n1 1\n=sinh−1(3) −sinh−1(0) M1 Uses limits.\n4\n( ) M1 A1 Uses logarithmic form.\n=ln 3+ 9 +1 =ln2\n4 16\n6\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p10.png"
      ]
    },
    {
      "id": "9231-2025-mj-24-q03",
      "question_id": "9231-2025-mj-24-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3 m2+4m+5=0m=−2i M1 Auxiliary equation. M1 for correct type of roots.\ny=e−2x(Acosx+Bsinx) A1 Complementary function. Allow ‘y=’ missing.\nAccept y=e−2x(Acosx+Bisinx).\ny=ke3x  y'=3ke3x  y''=9ke3x B1 Particular integral and its derivatives.\n9k+12k+5k =13k = 1 M1 A1 Substitutes and cancels e3x.\n2\ny=e−2x(Acosx+Bsinx)+ 1e3x A1 FT General solution. Must have ‘y=’. FT on their CF\n2\n(The PI must be correct).\nA+k =1 A=1−k = 1 B1 FT Uses y(0)=1, FT on their 1−k. Their CF must be\n2\ncorrect.\ny'=e−2x(−Asinx+Bcosx)−2e−2x(Acosx+Bsinx)+ 3e3x M1 Finds y'. For M1, their CF must have the correct\n2\nform.\nB−2A+ 3 =0B=−1 A1 Uses y'(0)=0.\n2 2\ny= 1e−2x(cosx−sinx)+ 1e3x A1 Must have ‘y=’.\n2 2\n10\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p11.png"
      ]
    },
    {
      "id": "9231-2025-mj-24-q04",
      "question_id": "9231-2025-mj-24-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 4,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) x=0(t−1)(t2 +1)=0 B1 Factorises or considers other roots of x=0.\n1\n© Cambridge University Press & Assessment 2025 Page 11 of 17\n\n4(b) dx dy B1\n=3t2 −2t+1 =(t+1)et\ndt dt\ndy dy dt dy (t+1)et M1 A1 Applies chain rule, AG.\n=   =\ndx dt dx dx 3t2 −2t+1\n3\n\n4(c) d  (t+1)et  ( 3t2 −2t+1 )( (t+1)et +et) −(t+1)et(6t−2) *M1 Differentiates dy with respect to t.\n = dx\ndt 3t2 −2t+1 ( 3t2 −2t+1 )2\nd2y ( 3t2 −2t+1 ) (t+2)et −(t+1)(6t−2)et M1 A1 Applies chain rule.\n=\ndx2 ( 3t2 −2t+1 )3\ny(0)=e, y'(0)=e, y''(0)=−1e A1 Evaluates derivatives at t=1.\n4\ny''(0) DM1 A1 Finds Maclaurin’s series. For M1, must have\ny= y(0)+ y'(0)x+ x2 =e+ex−1ex2\n2! 8 attempted second derivative.\n6\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p11.png",
        "../answer-assets/9231_s25_ms_24-p12.png"
      ]
    },
    {
      "id": "9231-2025-mj-24-q05",
      "question_id": "9231-2025-mj-24-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 5,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a)\nsin7=Im\n( (c+is)7) M1 Uses binomial expansion and takes imaginary part.\n=−s7 +21c2s5−35c4s3+7c6s A1\n=−s7 +21 ( 1−s2) s5−35 ( 1−s2)2 s3+7 ( 1−s2)3 s M1 Applies c2 =1−s2.\n( 1−s2)3 =−s6 +3s4 −3s2 +1 B1 No sight of ( 1−s2)3 fully expanded is B0.\nsin7=−64s7 +112s5−56s3+7s A1 AG.\n5\n\n5(b) x=sin sin0 sin7=0 M1 Solves sin7=0 and excludes sin=0. For M1\nthe roots from sin=0 must be excluded at some\nstage of their working. (This could be implied by\ntheir final answers.)\nx=sin (1π ) A1 One correct root.\n7\nsin ( −1π ) , sin ( 2π ) , sin ( 3π ) A1 Exactly five other correct roots. Allow different\n7 7 7 values of q which give these roots. Accept\n−sin\n(1π )\n, sin\n(2π )\n, sin\n(3π )\n.\n7 7 7\nJust listing correct values of q without any\nreference of x is M1 SCB1 (2/3).\n3\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p13.png"
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    },
    {
      "id": "9231-2025-mj-24-q06",
      "question_id": "9231-2025-mj-24-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6 dy 1 B1 Divides through by x.\n− y=2xtan−1x\ndx x\ne\n−x−1dx\n=x−1\nM1 A1 Finds integrating factor.\nd *M1 Correct form on LHS and attempt to integrate RHS\n(yx−1)=2tan−1x\ndx by parts. Treating tan−1x as cotx is M0.\nyx−1=2xtan−1x−2   x dx=2xtan−1x−ln ( 1+x2) +C A1 A1 −ln ( 1+x2) =lncos ( tan−1x ) .\n1+x2\nSC B1 for xtan−1x−1ln ( 1+x2) +C\n2\n1π=2 (1π ) −ln2+CC=ln2 DM1 Finds C, substitutes given conditions into their\n2 4 solution.\ny=2x2tan−1x−xln ( 1+x2) +xln2 DM1 A1 Divides through by their coefficient of y.\n9\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p14.png"
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    },
    {
      "id": "9231-2025-mj-24-q07",
      "question_id": "9231-2025-mj-24-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 7,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) Eigenvalues of A are 1, 2 and −3. B1 Upper diagonal matrix or characteristic equation.\ni j k −4 1 M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\n=1: 0 1 5 = 0 0\n   \n   \n0 0 −4 0  0\ni j k 35 7 A1\n   \n=2: −1 7 11 = 5 1\n   \n   \n0 0 5  0  0\ni j k −20  1  A1\n   \n=−3: 4 7 11 = −20 1\n   \n   \n0 5 5  20  −1\n1 7 1  1 0 0  M1 A1 Or correctly matched permutations of columns.\n    OE.\nThus P= 0 1 1 and D= 0 64 0\n    A column of zeros or a repeated column is M0.\n   \n0 0 −1 0 0 729\n7\n\n7(b) (−1)(−2)(+3)=3 −7+6=0 B1 Characteristic equation. ‘=0’ can be implied later.\nA3 =7A−6I M1 Substitutes for A and makes A3 the subject.\n(Could substitute for A at the end).\nA6 =(7A−6I)(7A−6I)=49A2 −84A+36I M1 A1 Expands (7A−6I)(7A−6I).\n4\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
      "source_pages": [
        15
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_24-p15.png"
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    },
    {
      "id": "9231-2025-mj-24-q08",
      "question_id": "9231-2025-mj-24-q08",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 8,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "8(a) B1 Correct shape and position. Ignore x0 on their\nsketch.\nx\ny=1 B1 States the equation of the asymptote. Ignore\ny=−1.\n2\n\n8(b) N M1 A1 Compares sum with integral. Limits on integral\nN\ntanhr tanhxdx must be present and correct for M1.\n0\nr=1\n N tanhx=ln(coshx) N =ln(coshN) A1 Evaluates integral, AG. Must see ln(coshx).\n \n0 0\n3\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n8(c)(i)  1 2 ln3 dy 2 1ln3 M1 A1 Correct formula for surface area. Missing limits is\nS =2π  y 1+  dx=2π2 tanh(x) 1+sech4(x)dx M1 A0. y outside of integral is M0.\n dx 0\n0\nu= 1+sech4x  du =−2 ( 1+sech4x )−1 2sech4xtanhx M1 A1 Finds du . Sign error gets M1 A0.\ndx dx\nx=1ln3coshx= 2 3 sechx= 1 3u= 1+ 9 M1 A1 Finds u when x= 1ln3. Need to see working to\n 2 3  2 16 2\njustify given limit. Working without evidence of\nevaluating sechx is M1 A0.\n 5 4 u  2 u2 A1 AG, must have gained all previous marks.\nS =−π u du=π du\n u2 −1  u2 −1\n2 5\n4\n7\n\n8(c)(ii) = 1π   2 2+ 1 − 1 du=πu+ 1ln(u−1)−1ln(u+1) 2 M1 A1 Uses partial fractions or u2 =1+ 1 .\n2  u−1 u+1  2 2  5 u2 −1 u2 −1\n5 4\n4\n  2−1 1  ( ( )) A1 OE.\n=π   2+1 2 ln   2+1    −5 4 −1 2 ln  9      =π 2−5 4 +ln3+ 1 2 ln 3−2 2\n3\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
        16,
        17
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_24.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-31-q01",
      "question_id": "9231-2025-mj-31-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) dv B1 Correct use of N2L.\nx3 +4x=8v\ndx\n(x3+4x) dx=8v dv M1 Attempt to integrate by separating variables.\ndv\nMust have v .\n4v2 = 1x4 +2x2 +c dx\n4\nMust include +c.\nv2 = 1 x4 + 1x2 +1  = ( 1x2 +1 )2 A1 Use initial conditions to find c and rearrange to\n16 2  4  obtain an expression for v in terms of x.\nv= 1x2 +1 x2 +4\n4 Condone v= .\n4\nCondone omitted explanation for choosing\npositive v.\n3\n\n1(b) dx  1 M1 Separate variables and integrate totan−1 form.\n= 1x2 +1 leading to  dx=dt .\ndt 4  1x2 +1 Only FT their v of the form ax2 +b.\n4\nt=4   1 dx=2tan−1(1x )+C\n x2 +4 2\nC=0 so x=2tan (1t ) A1 Must have +C and statement that C=0.\n2\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-31-q02",
      "question_id": "9231-2025-mj-31-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a)\n10=35t−\ng 352( 1+t2) B1 Use trajectory equation to set up quadratic in\n2242 t=tan using (35,10).\n6125t2−20160t+11885=0 M1 Form three-term quadratic in t and equate to 0.\nOE, for example 1225t2−4032t+2377=0.\ntan=0.769leading to =37.6. A1 If B0M0 earned, award B1 for correct answer.\ntan=2.52 leading to =68.4. A1 If B0M0 earned, award B1 for correct answer.\n4\n\n2(b) g M1 Complete method to find the range, x, using the\nUse trajectory equation with y = 0: 0=xtan− x2(1+tan2)\n2242 larger of their two values found in part 2(a).\n39.5 A1 Accept 39.4≤x≤39.5.\nAlternative method for question 2(b)\nTime of flight, T =4.8sin [= 4.4629…] M1 Complete method to find the range, x, using the\nx=24Tcos larger of their two values found in part 2(a).\n39.5 A1 Accept 39.4≤x≤39.5.\nAlternative method for question 2(b)\nu2sin2 M1 Complete method to find the range, x, using the\nUse range formula: x=\nlarger of their two values found in part 2(a).\ng\n39.5 A1 Accept 39.4≤x≤39.5.\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-31-q03",
      "question_id": "9231-2025-mj-31-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) 32x M1 λx\n8= leading to x=0.5. Use T = to find x. May be implied by\n2 l\ncorrectly derived value for sin=0.6.\nHence, PA =2.5 and sin=1.5 =0.6. A1 AG, shown convincingly.\n2.5\n2\n\n3(b) ( ) *M1 For either equation (resolving horizontally or\n R=1.6g−8cos\nvertically). Allow sin/cos wrong way round.\n() mr2 =8sin+R Correct number of terms and correct signs.\nDimensionally correct. Numbers or algebraic.\nA1 Both equations correct.\n8(3)+0.5 ( 1.6g−8(4)) =1.61.52 DM1 Eliminate R to obtain equation involving 2\n5 5\nwith no other unknown quantities.\n2 =4 leading to =2 A1\n60 A1 19.0985…\nNumber of revolutions per minute = =19.1 .\nAccept final answer of 19.\nπ\n60\nAllow final answer of .\nπ\n5\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-31-q04",
      "question_id": "9231-2025-mj-31-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Allow one error for cone or hemisphere.\nVolume CoM above base\nSOI, may be seen in moments equation.\nCone 18πr3 3r\n2\nHemisphere 16πr3 3r\n3 4\nTotal object 18πr3 −16πr3 y\n3\n( 18πr3−16πr3) y =18πr3(3r)−16πr3(3r) M1 A moments equation with 3 terms, allow sign\n3 2 3 4 errors, must be dimensionally correct.\nA1 Fully correct equation, does not need to be\nsimplified.\ny = 69r A1\n38\n4\n\n4(b) When on point of toppling: M1 3r their y\nFor attempt at tan using or .\n3r\ntan=\ntheir y\n\n\n= 3\n2\n8\n3\n\n\ntheir y 3r\nA1 FT FT their y .\n=58.8o A1 CAO\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_31-p12.png"
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    },
    {
      "id": "9231-2025-mj-31-q05",
      "question_id": "9231-2025-mj-31-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "5 At v , T =2gsin   =14g  M1 Allow T =mgsin.\nmax 25\n14x 14g A1 Find extension at maximum velocity.\n= leading to x=1.\n0.5 25 5\n140.32 B1 FT One correct EPE term seen anywhere.\nEPE at top: =1.26 FT only on their x found using a correct method\n20.5\n(dependent on first M mark).\n14(\"their x\") 2\nEPE at bottom: =0.56\n20.5\n140.32 14(\"their x\") 2 M1 Dimensionally correct, including KE, GPE, and\n2gsin(0.8+0.5+\"their x\")+ = + 12v2 at least one EPE.\n20.5 20.5 2\nA1 FT Correct equation.\nFT only on their x found using a correct method\n(dependent on first M mark).\nv2 =0.91g =9.1 leading to v=3.02ms–1. A1 v=3.016620625\n6\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_31-p13.png"
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    },
    {
      "id": "9231-2025-mj-31-q06",
      "question_id": "9231-2025-mj-31-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) 912mv2 = 1mv2 3v =v  B1 KE equation linking v and v .\n2 2 A 2 B A B A B\nNeeds to be KE equation with 1 and m.\n2\n2mu=2mv +mv B1 Conservation of momentum.\nA B\nMust see m present.\nv = 2u v =6u B1 AG, shown convincingly.\n A 5  B 5 If a quadratic equation is formed, must show\nmethod for solving and obtain a single answer.\n3\n\n6(b) Let w denote the speed after the second collision. B1 OE, for example 5cos=2cos.\nPerpendicular to wall: v cos=wcos\nB\nParallel to wall: e v sin=wsin B1 OE, for example 2sin= 5sin.\nB\ne2v2sin2=w2sin2 576sin2=144sin2 *M1 Obtain an equation involving only one of , .\nB 625 125 \nEliminate  or  by squaring both sides and\nv2cos2=w2cos2 36cos2=144cos2\nB 25 125  adding.\n576sin2+36 ( 1−sin2 ) =144 DM1 Use sin2+cos2=1 to reduce to equation\n625 25 125\nwith just sine or cosine of 1 variable.\nsin2= 5 sin=1 5 A1 Award for any of the following:\n 9 3\ncos= 2 or cos= 1 5or sin= 2.\n3 3 3\nsin(+)=1 A1 Exact answer.\n6\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_31-p14.png"
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    },
    {
      "id": "9231-2025-mj-31-q07",
      "question_id": "9231-2025-mj-31-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) mu2 B1 Both equations correct.\nR = Use of N2L towards centre at A and B.\nA a\nMay see u2 = 6ag already substituted.\nmv2 5\nR\nB\n=\na\n−mgcos\nMay see k or cos.\nmu2 mv2 *M1 Use of R = 1R .\n= −mgcos leading to v2 = 1u2 +agk. B 2 A\n2a a 2 Condone use of R = 1R for this mark.\nA 2 B\nTwo terms required in their expression for R .\nB\nMay see u2 = 6ag already substituted.\n5\nMay see k or cos.\n1mu2 = 1mv2 +magk leading to v2 =u2 −2agk . *M1 Energy equation, dimensionally correct.\n2 2\n6ag−2agk = 3ag+agk DM1 Dependent on both previous M marks.\n5 5\nCombine N2L and energy equations to obtain an\nequation that simplifies to involve k only.\nk = 1 A1\n5\n5\n\n7(b) v2 =0 leading to u2 = 2ag u= 2ag B1 FT FT their 2agk .\n 5  5\n1\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c) 2vsin M1 Complete method to find v2.\nVertically B to C: t =\ng Must use 2asin and not 2a.\nMay quote range formula.\n2vsin\nHorizontally B to C: 2asin=vcost =vcos FT their k from part 7(a).\ng\ncos=their k, sin= 1−their k2\nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\nAlternative method for question 7(c)\nEquation of trajectory: (M1) Complete method to find v2.\n ( )2 Must use 2asin and not 2a.\n0=2asin (tan)−\ng(2asin)2\n 0= 4 6a ( 2 6 ) −\ng\n5\n4 6\n FT their k from part 7(a).\n2(vcos)2 \n\n5\n2\n2\n5\nv2 \n cos=their k, sin= 1−their k2\n \nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(c) Alternative method for question 7(c)\ngx2 dy gx M1 Complete method to find v2.\ny=xtan− leading to =tan−\n2v2cos2 dx v2cos2 Must use asin and not a.\nFT their k from part 7(a).\ngasin cos=their k, sin= 1−their k2\nReaches maximum when x=asin: 0=tan−\nv2cos2\nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
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        16,
        17
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-mj-32-q01",
      "question_id": "9231-2025-mj-32-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) dv B1 Correct use of N2L.\nx3 +4x=8v\ndx\n(x3+4x) dx=8v dv M1 Attempt to integrate by separating variables.\ndv\nMust have v .\n4v2 = 1x4 +2x2 +c dx\n4\nMust include +c.\nv2 = 1 x4 + 1x2 +1  = ( 1x2 +1 )2 A1 Use initial conditions to find c and rearrange to\n16 2  4  obtain an expression for v in terms of x.\nv= 1x2 +1 x2 +4\n4 Condone v= .\n4\nCondone omitted explanation for choosing\npositive v.\n3\n\n1(b) dx  1 M1 Separate variables and integrate totan−1 form.\n= 1x2 +1 leading to  dx=dt .\ndt 4  1x2 +1 Only FT their v of the form ax2 +b.\n4\nt=4   1 dx=2tan−1(1x )+C\n x2 +4 2\nC=0 so x=2tan (1t ) A1 Must have +C and statement that C=0.\n2\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_32-p09.png"
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    },
    {
      "id": "9231-2025-mj-32-q02",
      "question_id": "9231-2025-mj-32-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a)\n10=35t−\ng 352( 1+t2) B1 Use trajectory equation to set up quadratic in\n2242 t=tan using (35,10).\n6125t2−20160t+11885=0 M1 Form three-term quadratic in t and equate to 0.\nOE, for example 1225t2−4032t+2377=0.\ntan=0.769leading to =37.6. A1 If B0M0 earned, award B1 for correct answer.\ntan=2.52 leading to =68.4. A1 If B0M0 earned, award B1 for correct answer.\n4\n\n2(b) g M1 Complete method to find the range, x, using the\nUse trajectory equation with y = 0: 0=xtan− x2(1+tan2)\n2242 larger of their two values found in part 2(a).\n39.5 A1 Accept 39.4≤x≤39.5.\nAlternative method for question 2(b)\nTime of flight, T =4.8sin [= 4.4629…] M1 Complete method to find the range, x, using the\nx=24Tcos larger of their two values found in part 2(a).\n39.5 A1 Accept 39.4≤x≤39.5.\nAlternative method for question 2(b)\nu2sin2 M1 Complete method to find the range, x, using the\nUse range formula: x=\nlarger of their two values found in part 2(a).\ng\n39.5 A1 Accept 39.4≤x≤39.5.\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-32-q03",
      "question_id": "9231-2025-mj-32-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) 32x M1 λx\n8= leading to x=0.5. Use T = to find x. May be implied by\n2 l\ncorrectly derived value for sin=0.6.\nHence, PA =2.5 and sin=1.5 =0.6. A1 AG, shown convincingly.\n2.5\n2\n\n3(b) ( ) *M1 For either equation (resolving horizontally or\n R=1.6g−8cos\nvertically). Allow sin/cos wrong way round.\n() mr2 =8sin+R Correct number of terms and correct signs.\nDimensionally correct. Numbers or algebraic.\nA1 Both equations correct.\n8(3)+0.5 ( 1.6g−8(4)) =1.61.52 DM1 Eliminate R to obtain equation involving 2\n5 5\nwith no other unknown quantities.\n2 =4 leading to =2 A1\n60 A1 19.0985…\nNumber of revolutions per minute = =19.1 .\nAccept final answer of 19.\nπ\n60\nAllow final answer of .\nπ\n5\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-32-q04",
      "question_id": "9231-2025-mj-32-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) M1 Allow one error for cone or hemisphere.\nVolume CoM above base\nSOI, may be seen in moments equation.\nCone 18πr3 3r\n2\nHemisphere 16πr3 3r\n3 4\nTotal object 18πr3 −16πr3 y\n3\n( 18πr3−16πr3) y =18πr3(3r)−16πr3(3r) M1 A moments equation with 3 terms, allow sign\n3 2 3 4 errors, must be dimensionally correct.\nA1 Fully correct equation, does not need to be\nsimplified.\ny = 69r A1\n38\n4\n\n4(b) When on point of toppling: M1 3r their y\nFor attempt at tan using or .\n3r\ntan=\ntheir y\n\n\n= 3\n2\n8\n3\n\n\ntheir y 3r\nA1 FT FT their y .\n=58.8o A1 CAO\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2025-mj-32-q05",
      "question_id": "9231-2025-mj-32-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "5 At v , T =2gsin   =14g  M1 Allow T =mgsin.\nmax 25\n14x 14g A1 Find extension at maximum velocity.\n= leading to x=1.\n0.5 25 5\n140.32 B1 FT One correct EPE term seen anywhere.\nEPE at top: =1.26 FT only on their x found using a correct method\n20.5\n(dependent on first M mark).\n14(\"their x\") 2\nEPE at bottom: =0.56\n20.5\n140.32 14(\"their x\") 2 M1 Dimensionally correct, including KE, GPE, and\n2gsin(0.8+0.5+\"their x\")+ = + 12v2 at least one EPE.\n20.5 20.5 2\nA1 FT Correct equation.\nFT only on their x found using a correct method\n(dependent on first M mark).\nv2 =0.91g =9.1 leading to v=3.02ms–1. A1 v=3.016620625\n6\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2025-mj-32-q06",
      "question_id": "9231-2025-mj-32-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) 912mv2 = 1mv2 3v =v  B1 KE equation linking v and v .\n2 2 A 2 B A B A B\nNeeds to be KE equation with 1 and m.\n2\n2mu=2mv +mv B1 Conservation of momentum.\nA B\nMust see m present.\nv = 2u v =6u B1 AG, shown convincingly.\n A 5  B 5 If a quadratic equation is formed, must show\nmethod for solving and obtain a single answer.\n3\n\n6(b) Let w denote the speed after the second collision. B1 OE, for example 5cos=2cos.\nPerpendicular to wall: v cos=wcos\nB\nParallel to wall: e v sin=wsin B1 OE, for example 2sin= 5sin.\nB\ne2v2sin2=w2sin2 576sin2=144sin2 *M1 Obtain an equation involving only one of , .\nB 625 125 \nEliminate  or  by squaring both sides and\nv2cos2=w2cos2 36cos2=144cos2\nB 25 125  adding.\n576sin2+36 ( 1−sin2 ) =144 DM1 Use sin2+cos2=1 to reduce to equation\n625 25 125\nwith just sine or cosine of 1 variable.\nsin2= 5 sin=1 5 A1 Award for any of the following:\n 9 3\ncos= 2 or cos= 1 5or sin= 2.\n3 3 3\nsin(+)=1 A1 Exact answer.\n6\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-32-q07",
      "question_id": "9231-2025-mj-32-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) mu2 B1 Both equations correct.\nR = Use of N2L towards centre at A and B.\nA a\nMay see u2 = 6ag already substituted.\nmv2 5\nR\nB\n=\na\n−mgcos\nMay see k or cos.\nmu2 mv2 *M1 Use of R = 1R .\n= −mgcos leading to v2 = 1u2 +agk. B 2 A\n2a a 2 Condone use of R = 1R for this mark.\nA 2 B\nTwo terms required in their expression for R .\nB\nMay see u2 = 6ag already substituted.\n5\nMay see k or cos.\n1mu2 = 1mv2 +magk leading to v2 =u2 −2agk . *M1 Energy equation, dimensionally correct.\n2 2\n6ag−2agk = 3ag+agk DM1 Dependent on both previous M marks.\n5 5\nCombine N2L and energy equations to obtain an\nequation that simplifies to involve k only.\nk = 1 A1\n5\n5\n\n7(b) v2 =0 leading to u2 = 2ag u= 2ag B1 FT FT their 2agk .\n 5  5\n1\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c) 2vsin M1 Complete method to find v2.\nVertically B to C: t =\ng Must use 2asin and not 2a.\nMay quote range formula.\n2vsin\nHorizontally B to C: 2asin=vcost =vcos FT their k from part 7(a).\ng\ncos=their k, sin= 1−their k2\nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\nAlternative method for question 7(c)\nEquation of trajectory: (M1) Complete method to find v2.\n ( )2 Must use 2asin and not 2a.\n0=2asin (tan)−\ng(2asin)2\n 0= 4 6a ( 2 6 ) −\ng\n5\n4 6\n FT their k from part 7(a).\n2(vcos)2 \n\n5\n2\n2\n5\nv2 \n cos=their k, sin= 1−their k2\n \nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(c) Alternative method for question 7(c)\ngx2 dy gx M1 Complete method to find v2.\ny=xtan− leading to =tan−\n2v2cos2 dx v2cos2 Must use asin and not a.\nFT their k from part 7(a).\ngasin cos=their k, sin= 1−their k2\nReaches maximum when x=asin: 0=tan−\nv2cos2\nag A1 ag\nv2 = May seev2 = or v2 =5ag .\ncos k\nag M1 Use their energy equation from part 7(a)\nu2 = +2agk\ncos ( v2 =u2 −2agk ) . Must substitute for v2.\nu2 = 27ag, u= 27ag A1 CAO\n5 5\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
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        17
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-mj-33-q01",
      "question_id": "9231-2025-mj-33-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 Tcos=mg B1 B1 for both equations, SOI.\nTsin=m2r\n'5mg' sin=m2asin M1 Combining and using r=asin.\n3\nAllow sin/cos mix in resolving.\nMust have a trigonometric ratio on both sides.\n5g A1 OE, CAO\n=\n3a\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-33-q02",
      "question_id": "9231-2025-mj-33-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) mg M1 Energy equation with all three terms.\n(x−a)2 =mgxsin30−11mgx\n2a 30 Dimensionally correct.\nB1 Two terms correct (need not be in equation).\nA1 Correct energy equation.\n15x2 −34ax+15a2 =0 M1 Simplify to three-term homogeneous quadratic\nequation and attempt to solve.\n(3x−5a)(5x−3a)=0\nx= 5a A1 CAO, must reject x= 3a if seen.\n3 5\n5\n\n2(b) T = 2mg B1\n3\nF =T −mgsin30 = 2mg−1mg M1 Equation for F. Allow sin/cos mix.\n \n3 2 Allow sign errors.\n1mg A1\n6\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_33-p10.png"
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    },
    {
      "id": "9231-2025-mj-33-q03",
      "question_id": "9231-2025-mj-33-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) dv B1 Correct use of N2L with m seen.\n−mg−mkv2 =mv\ndx\ndx=−   v dv leading to x=− 1 ln ( g+kv2)+c. M1 A Co tt r e r m ec p t t f t o o r m in . t e A g l r l a o t w e b m y i s s s e i p n a g r a + ti c ng . variables.\n g+kv2 2k\nv=U when x=0, so c=\n1\nln\n( g+kU2)\n.\nM1\n2k\n1 g+kU2  A1 AG, shown convincingly.\nx= ln  (metres)\n2k\n\ng+kv2\n\n4\n\n3(b) dv *M1 Attempt to integrate by separating variables.\n10+ 1 v2 =−\n40 dt\n 40\ndt=− dv\n400+v2\nt=−40tan−1( 1 v )(+C) A1 Correct integration.\n20 20\nWhen t =0,v=20, so C = 1π DM1 Find C and substitute v=0 to find value for t.\n2\nt =−2tan−1(0)+ 1π\n2\nt=1.57 (seconds) A1 CWO, accept final answer of 1π.\n2\n4\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        11
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2025-mj-33-q04",
      "question_id": "9231-2025-mj-33-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "4 B1 Allow one error in first two columns.\nAreas 2a2 1\n2\na(2a) 2a2 + 1\n2\na(2a)\nCorrect distances about any point.\nSOI, may be seen in moments equation.\nDistances from D −1a 2a x\n2 3\n−2a2(1a)+1 ( 2a2)(2a)= ( 3a2) x *M1 A moments equation with 3 terms, must be\n2 2 3 dimensionally correct. (x =1a or x =8a)\n9 9\nm (1a ) =km(2a) DM1 Moments equation about toppling point D.\n9 2 terms only, OE. Must include k.\nk = 1 A1\n18\nAlternative method for question 4\nB1 Correct ratios including k.\nRatio of masses 2a2 1 2 a(2a) k ( 2a2 + 1 2 a(2a)) Must simply to 2:1:3k.\nSOI, may be seen in moments equation.\n2 (1a ) =1 (2a ) +3k(2a) *M1 A moments equation about D with 3 terms, must\n2 3 be dimensionally correct. Must include k.\nDM1 Using correct distances.\nk = 1 A1\n18\n4\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-33-q05",
      "question_id": "9231-2025-mj-33-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) 1mv2 +mgr cos=1m (3gr ) v2 = 3gr−2grcos M1 Energy equation. All 3 terms. Allow sin/cos error.\n2 2 2  2  Allow sign errors.\nmv2 B1 N2L at B.\nmgcos=\nr\nmgcos=m\n(3g−2gcos ) M1 Combining to eliminate v.\n2\ncos= 1,  =60 A1\n2\n4\n\n5(b) v2 = 1gr B1 FT FT their 60.\n2\n( )2 M1 Allow sign error.\n0= sin'60' 'v2' −2gh\nh= 3 r A1 FT FT their 60and their v2.\n16\nTotal height = 3 r+rcos60=11r A1\n16 16\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-mj-33-q06",
      "question_id": "9231-2025-mj-33-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v and w denote the horizontal components of the velocities for A and B M1 Allow sign errors, m must be present.\nrespectively after the collision.\n2mucos−mucos=mw+mv\nw−v=e(3ucos) M1 Restitution equation: directions must be consistent\nwith those used in momentum equation (signs\nmust be consistent).\nw=1(3eucos+ucos) =1(3e+1)ucos A1 Combining to find equation with w, e, u, . SOI\n \n2 2\n1(3e+1)ucos M1 OE, using perpendicular motion of B.\n2 =tan\nusin\ntan2= 1(3e+1) M1 Obtaining an expression for tan in terms of e.\n2\nA1 CWO, must reject negative term if seen.\ntan= 1(3e+1)\n2\n6\n\n6(b) w=3eucos=ucos M1 Using v = 0 to find a value for e.\ne= 1\n3\ntan=1,  =45 A1\nAlternative method for question 6(b)\nw M1\nv=0, w=ucos and =tan\nusin\nucos=usintan, tan2=1,  =45 A1\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-33-q07",
      "question_id": "9231-2025-mj-33-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) v =Usin45−gT B1\ny\nv =Ucos45\nx\n1\n\n7(b) v B1 SOI\ny =−tan60\nv\nx\n U  M1 Substitute values for trigonometric ratios to obtain\n−gT\n  a linear equation in U and T only.\n2\n =− 3 U − 2gT =− 3U\nU\n \n 2 \nU ( ) 2U( ) U ( ) A1 AG, shown convincingly.\nT = 1+ 3 = 1+ 3 = 2+ 6\n2g 2g 2g\n3\n\n7(c) U B1 D in terms of U and T .\nD= T\n2\nU B1 H in terms of U and T .\nH = T −1gT2\n2 2 With or without modulus sign.\nU M1 OE\n2 T −1 2 gT2 1 2 2−1 2 ( 1 2 U ( 2+ 6 )) 1− 3 Obtain a numerical value, may be un-simplified.\n= = Division or ratios.\nU 1 2 2\nT 2 With or without modulus sign.\n2\nH:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c) Alternative method for question 7(c)\nU  U ( ) B1 D in terms of U and g .\nD=  2+ 6 \n22g \nU  U ( )  U ( ) 2  U2  B1 H in terms of U and g .\nH =  2+ 6 −1g 2+ 6  = −  With or without modulus sign.\n22g  2 2g   2g \n2 M1 OE\nU  U ( )  U ( )\n 2+ 6 −1g 2+ 6  Obtain a numerical value, may be un-simplified.\n22g  2 2g  1− 3 Division or ratios.\n=\nU  U ( ) 2 With or without modulus sign.\n 2+ 6 \n22g \nH:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO\nAlternative method for question 7(c)\nD= U U ( 2+ 6 ) B1 D= U2 ( 1+ 3 )\n2 2g 2g\n2 2 B1\n 6   2 \n U = U −2gH\n   \n2 2\n   \nU2 M1\n()H =\n2g\nH:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(d) 2 3  2  B1 OE, accept un-simplified.\nw = U = U Allow negative sign.\ny 3 2  3 \n2 M1\n U \nw2 =w2 + = 2U2 + 1U2\ny   2    3 2 \nA1 CWO\nw=U 2+1 =U 7\n3 2 6\n3\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-mj-34-q01",
      "question_id": "9231-2025-mj-34-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 1,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 Energy equation: M1 Energy equation with two EPE terms and at least one GPE term,\n20mg(2a)2\n20mgx2\ndimensionally correct, allow sign errors.\n5 +3mga= +mg(a+x) Award equivalent marks if any other x is used.\n2a 5 2a\nB1 One correct EPE term seen anywhere.\nA1 Correct energy equation.\n50x2+5ax−6a2 =0 M1 Dependent on two EPE terms used.\nSimplify to 3-term quadratic with terms in x2, ax and a2 only.\nx= 3 a A1 Single correct answer. If seen, x=−2a must be rejected.\n10 5\nAlternative method for question 1\nEnergy equation: M1 Energy equation with two EPE terms and at least one GPE term,\n20mg(2a)2\n20mg(h−a)2\ndimensionally correct, allow sign errors.\n5 +3mga= +mgh Award equivalent marks if any other h is used.\n2a 5 2a\nB1 One correct EPE term seen anywhere.\nA1 Correct energy equation.\n50h2 −95ah+39a2 =0 M1 Dependent on two EPE terms used.\nSimplify to 3-term quadratic with terms in h2, ah and a2 only.\nh=13a , so the maximum extension is 3 a. A1 Single correct answer.\n10 10\nIf seen, h= 3a leading to x=−2a must be rejected.\n5 5\n5\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_34.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2025-mj-34-q02",
      "question_id": "9231-2025-mj-34-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) dv M1 N2L (m must be seen, possibly as F = ma), no extra terms.\nm =−2mv3\nAttempt to integrate by separating variables, allow sign errors.\ndt\nv−3dv=−2dt Allow errors in coefficient of v−2.\nAllow missing +A.\n−1v−2 =−2t+ A A1 Any correct equivalent form.\n2\nWhen t =0, v=1 leading to A=−1 . M1 Use correct initial conditions to find value for A.\n2\n1 A1 1\nv= AEF, A0 for  .\n4t+1 4t+1\nSC B2 for complete solution except for m not seen.\n4\n\n2(b) dx 1 M1 FT FT their answer to part 2(a).\n= so x= 1 4t+1 [+B]\ndt 4t+1 2 Allow missing +B for this mark.\nWhen t =0, x=0 leading to B=−1 . M1 Use correct initial conditions to find value for B.\n2\nWhen t =6, x=2(metres). A1 CAO\nAlternative method for question 2(b)\n 6 1 6 M1 FT Integration attempt, FT their answer to part 2(a).\nx= dt=1 4t+1 Ignore limits.\n 4t+1\n2 \n0\n0\nx= 1 25−1 1 M1 Substitute correct limits.\n2 2\nx= 2 (metres) A1 CAO\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_34.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2025-mj-34-q03",
      "question_id": "9231-2025-mj-34-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 3,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a)  x  B1 Explicitly stated.\n(→) x=(ucos) t or t =\n \n ucos\n(  ) y=(usin) t−1gt2 B1 Explicitly stated.\n2\n2 M1 Eliminate t.\n x   x \ny=usin  −1g   Must see this substitution.\nucos 2 ucos\ngx2 A1 AG, shown convincingly.\ny=xtan− sec2\nDependent on all 3 previous marks being awarded.\n2u2\n4\n\n3(b) 8=64tan−128 ( 1+tan2 ) M1 May be un-simplified, must be an equation in a single\n5 trigonometric function.\n16tan2−40tan+21=0 M1 Obtain 3-term quadratic.\nAEF, for example 128tan2−320tan+168=0.\ntan= 3, 7 A1 Both answers required.\n4 4\n3\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_34.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-mj-34-q04",
      "question_id": "9231-2025-mj-34-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 4,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "4 For B: T =mg B1 Must be seen explicitly.\n( ) *M1 Allow sin/cos mix, allow sign error.\nFor A:  Rsin=Tcos+mg\nAllow this mark with mg instead of T.\n(→) Rcos+Tsin=mr2 *M1 Allow sin/cos mix, allow sign error.\nAllow this mark with mg instead of T.\nr=htan B1 Seen anywhere.\nCombine equations to find an expression for 2 in terms of g and DM1 mg(1+cos)\nMay see R=3mg or R= .\nh only. sin\ng A1 If T =mg is not seen explicitly, maximum possible mark is\n=2\nh B0 M1 M1 B1 DM1 A0.\n6\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_34.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-34-q05",
      "question_id": "9231-2025-mj-34-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 5,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) Split into rectangle AGED and triangle GFE (DA parallel to EG). M1 All terms present. All terms must be dimensionally correct.\nArea CoM from AD CoM from AB\nMoments about AD: 18a2x =6ah (1h ) + ( 18a2 −6ah )(1h+2a ) Rectangle 6ah 1h 3a\n2 3 2\nTriangle 18a2 −6ah 1h+2a 2a\n3\nAFED 18a2 x y\nA1 Fully correct equation, does not need to be simplified.\nh2 −6ah+36a2 A1 AG, shown convincingly.\nx =\n18a\nMoments about AB: 18a2y =6ah(3a)+ ( 18a2 −6ah )(2a) M1 All terms present. All terms must be dimensionally correct.\ny =2a+1h A1 Must expand brackets and collect like terms.\n3\nAlternative method for question 5(a)\nSplit into triangle AED and triangle AEF. M1 All terms present. All terms must be dimensionally correct.\nArea CoM from AD CoM from AB\nMoments about AD: 18a2x =3ah (1h ) + ( 18a2 −3ah )(2a) AED 3ah 1h 4a\n3 3\nAEF 18a2 −3ah 2a 2a\nAFED 18a2 x y\nA1 Fully correct equation, does not need to be simplified.\nh2 −6ah+36a2 A1 AG, shown convincingly.\nx =\n18a\nMoments about AB: 18a2y =3ah(4a)+ ( 18a2 −3ah )(2a) M1 All terms present. All terms must be dimensionally correct.\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n5(a) Alternative method for question 5(a)\nSubtract triangle EFH from rectangle AFHD M1 All terms present. All terms must be dimensionally correct.\n(H is on DC, FH parallel to AD). Area CoM from AD CoM from AB\nAFHD 36a2 −6ah 3a− 1h 3a\nMoments about AD: 2\n18a2x = ( 36a2 −6ah )( 3a−1h ) − ( 18a2 −6ah )( 4a−1h ) EFH 18a2 −6ah 4a−1 3 h 4a\n2 3\nAFED 18a2 x y\nA1 Fully correct equation, does not need to be simplified.\nh2 −6ah+36a2 A1 AG, shown convincingly.\nx =\n18a\nMoments about AB: M1 All terms present. All terms must be dimensionally correct.\n18a2y = ( 36a2 −6ah )(3a)− ( 18a2 −6ah )(4a)\ny =2a+1h A1 Must expand brackets and collect like terms.\n3\n5\n\n5(b) x B1 Correct expression for tan.\ntan=\n6a− y\nh2 −6ah+36a2 M1 FT Equate their expression for tan (must involve x and y ) to 7 ,\n15\n18a = 7 leading to 5h2−16ah+12a2 =0 . substitute the correct x and their y and simplify to a 3-term\n6a−(2a+1h) 15\n3 h2 −6ah+36a2 \n  (h−2a)(5h−6a)=0  quadratic. May see 15  18a   =7 ( 6a−2a−1 3 h ) .\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_34.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2025-mj-34-q06",
      "question_id": "9231-2025-mj-34-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 6,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) Speeds for A and B along line of centres after the collision are M1 Momentum equation: m must be present.\nv and v , respectively. Allow minus sign between terms on LHS.\n1 2\n5mv +4mv =5mucos\n1 2\nv −v =eucos M1 Restitution equation: directions must be consistent with those used\n2 1\nin momentum equation (signs must be consistent).\nv = 5u(1+e)cos A1 AG, shown convincingly.\n2 9\n3\n\n6(b) v =1u(5−4e)cos B1 May be found in part 6(a) but credit it in part 6(b).\n1 9\nDoes not need to be simplified.\nSpeed of V perpendicular to line of centres = usin. *B1 Seen anywhere.\n1(5m)\n(\nv2\n+(usin)2)\n= 1(4m)v 2\nM1 Equate KEs after collision.\n2 1 2 2 Allow 1m to be missing on both sides but must have the 5 and 4.\n2\nKE of A must have two terms.\nAllow sin/cos mix.\n5 ( (1u(5−4e)cos )2 +(usin)2 ) = 4(5u(1+e)cos )2 DM1 Substitute expressions for speeds AND numerical values for\n2 9 2 9 trigonometric functions. Accept un-simplified.\n 2 2 2 This mark is dependent on the second B1.\n  3    2     3 \n5  1u(5−4e)   +u    = 4  5u(1+e)   5u2(5−4e)2 +20u2 =100u2(1+e)2 OE may be seen.\n2  9  13   13   2  9  13 9 9 \n \n4e2+80e−41=0 DM1 AEF, obtain 3-term quadratic equation in e.\nThis mark is dependent on the second B1.\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_34.pdf",
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    },
    {
      "id": "9231-2025-mj-34-q07",
      "question_id": "9231-2025-mj-34-q07",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 7,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "7 Energy, highest to lowest: M1 Note that assuming the same speed at highest and lowest\n1mu2 −1mv2 =2amg v2 =u2 −4ag points could give the correct final answer but will not score\n2 2   full marks.\nEnergy, start to lowest: One energy equation, all terms present, dimensionally correct,\n1mu2 −1m(4ag)=mg(a−acos) u2 =ag(6−2cos) allow sign error.\n2 2  \nM1 A second energy equation, all terms present, dimensionally\nEnergy, start to highest,\ncorrect, allow sign error\n1m(4ag)−1mv2 =mg(a+acos) v2 =2ag(1−cos)\n2 2  \nA1 One fully correct energy equation.\nmu2 B1 Use of N2L at the lowest point, may see T =mg(7−2cos).\nAt lowest point, T −mg = L\nL a\nmv2 B1 Use of N2L at the highest point, may see T =mg(1−2cos).\nAt highest point, T +mg = H\nH a Must use a different speed.\nT =11T B1 mu2\nL H +mg\na\nSOI, must be correct way around. For example, =11.\nmv2\n−mg\na\nCombine to find equation involving cos only. M1 Must see an energy equation involving cos only.\ncos= 1 A1\n5\n8\n© Cambridge University Press & Assessment 2025 Page 16 of 17",
      "source_pages": [
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    },
    {
      "id": "9231-2025-mj-41-q01",
      "question_id": "9231-2025-mj-41-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1 H : = H :   B1 If in words, must contain ‘population means’.\n0 x y 1 x y\nAllow H : =0, H :  0.\n0 d 1 d\ns2 =103.218... M1 SOI, one unbiased estimate correct to 3sf or\nx\nbetter.\ns2 =171.139...\ny Do not accept biased estimates (102.4, 169.0).\ns2 s2 A1 A0 for pooled s .\nx + y =2.999 Accept 3, 3.0, 3.00.\n120 80\nMay be implied by z=3.03.\n10470−6560 M1 Value of z obtained using correct form with their\nz= 120 80 =3.03\ns2 s2 s x 2 and their s y 2.\nx + y\nUse of pooled s scores M0 A0.\n120 80\nA1 Use of pooled s scores M0 A0.\n'3.03' > 2.576, reject H (significant). M1 Compare their z with 2.576 and appropriate\n0\nOR conclusion (may be in terms of H ).\n1\nP(Z '3.03')=0.001220.005, reject H (significant). Reject H 0 can be implied by an attempt at an\n0\nappropriate conclusion in context.\nMay see 0.002440.01, reject H (significant).\n0\nAllow M1 for comparison of their 3.03 with\n2.326 and appropriate conclusion.\nThere is sufficient evidence to suggest the mean spend of customers is different on A1 Correct conclusion in context from correct\nweekends compared to weekdays. working ignoring hypotheses.\nLevel of uncertainty in language used (for\nexample, not ‘prove’).\nMust refer to mean or average spend.\n7\n© Cambridge University Press & Assessment 2025 Page 9 of 16",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-41-q02",
      "question_id": "9231-2025-mj-41-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "2(a) 5a+ 3a=1, leading to a= 2 . B1 AG, shown convincingly.\n2 13\nUses area = 1, b and c must be eliminated.\nAttempt to solve 2 independent equations in b and c or 3 independent equations in M1 May see any of the following:\na, b and c. 3b−39c= 3a, a=b−5c, b−8c=0,\n2 2\na\n−c= , 5a+3b−39c=1.\n5−8 2\nOR equation of line through (5,a)and (8,0).\nb=16, c= 2 A1 Correct values imply M1 A1.\n39 39\n3\n\n2(b) E(X)= 5 2 xdx+ 8 x(16 − 2 x)dx M1 Attempt to integrate xf(x)with their b and c,\n \n0 13 5 39 39 limits seen. May be in terms of a, b, c.\n5 8 M1 Correct integrated expression with their b and\n1 x2 +8 x2 − 2 x3\n13  0 39 117  5 their c, including limits.\nMay be in terms of a, b, c.\n43 =3.31 A1 AWRT 3.31.\n13\n3\n\n2(c) m M1 Uses 0m5 and area = 0.5, may be in terms\n2 m=0.5 OR  2 dx =0.5\n13 0 13 of a.\nm=13 =3.25 A1\n4\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 16\n\n2(d) 2 x, 16 x− 1 x2(+k) B1 FT Correct expressions for integrating both pieces\n13 39 39\nof PDF, seen anywhere, using their b and c.\nk =−25 B1 FT FT their b and their c to obtain k .\n39\n0 y0 M1 Both parts with y substituted (condone\n\n2 y 0≤y≤25 missing k ), ignore limits.\nG(y)= 13\n1\n3\n6\n9\ny −\n3\n1\n9\ny−\n3\n2\n9\n5 25 y≤64\nA1 FT FT their b, their c and their k .\n\n1 y64 Fully defined including correct domain in terms\nof y. Domain must cover all reals, may overlap\nat boundaries.\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2025-mj-41-q03",
      "question_id": "9231-2025-mj-41-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 01844+1143+211+30+41+50+61  7 B1 175 , AWRT 0.01458.\n= 12000\n \n 20006  480\n1\n\n3(b) a=2000−(1831.3+6.016+0.119+0.001)=162.6 B1 AG, shown convincingly.\nAlternative method for question 3(b)\n6 B1 AG, shown convincingly.\na=  7  ( 1− 7 )5 2000 = 162.6\n1 480 480\n1\n© Cambridge University Press & Assessment 2025 Page 11 of 16\n\n3(c) H : A binomial distribution is a satisfactory model for the data. B1 Must mention binomial and data.\n0\nH : A binomial distribution is not a satisfactory model for the data.\n1\n(1844−1831.3)2 (143−162.6)2 (13−6.136)2 M1 Expression of correct form with cells correctly\n+ + =0.08807+2.3626+7.6784 combined.\n1831.3 162.6 6.136\nMay use 6.1 (found by subtraction).\nOR\nImplied by 2 correct contributions.\n18442 1432 132\n+ + −2000\n1831.3 162.6 6.136\n10.1 A1 AWRT 10.1.\nAllow AWRT 10.3 from using 6.1.\nSC unsupported 10.1 (or 10.3) scores M0 B1.\n[1 degree of freedom so critical value is 7.879.] M1 Compare their 10.1 (or 10.3) with 7.879 and\n'10.1'7.879, reject H (significant). appropriate conclusion (may be in terms of H ).\n0 1\nReject H can be implied by an attempt at an\n0\nappropriate conclusion in context.\nAllow M1 for comparison of their 10.1 (or 10.3)\nwith 10.6 and appropriate conclusion.\nSufficient evidence to suggest that the manager’s model is not satisfactory. A1 Correct conclusion in context from correct\nworking ignoring hypotheses.\nLevel of uncertainty in language used (for\nexample, not ‘prove’).\n5\n\n3(d) Not independent or probability of “success” may not be constant or plausible reason B1 Credit any valid comment in context.\nin context (for example, higher number of boxes with more than 1 broken egg due to\nboxes being dropped).\n1\n© Cambridge University Press & Assessment 2025 Page 12 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_41.pdf",
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    },
    {
      "id": "9231-2025-mj-41-q04",
      "question_id": "9231-2025-mj-41-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) (x−x)2 =40104− 1 6122 =2649.6 M1 Either expression, may be embedded.\n10 May be implied, for example, by 294.4 or\n(y−y)2 =27460−14442 =2818 402.57 seen.\n8\n2649.6+2818 A1 SOI\ns2 = =341.725\np 10+8−2\n(61.2−55.5)2.120 's2' (1 +1)=5.72.128.7686 B1 2.120, SOI.\np 10 8\nM1 Correct form, must be a t value.\nA1 Correct un-simplified confidence interval, SOI.\n−12.9, 24.3 A1 Accept with inequality signs or open brackets\nor −24.3, 12.9 from Y −X .\nCondone 24.3, −12.9.\nDo not accept 5.718.6.\n6\n\n4(b) Because zero lies inside the confidence interval. B1\n1\n© Cambridge University Press & Assessment 2025 Page 13 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-41-q05",
      "question_id": "9231-2025-mj-41-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) H :m =m B1 If in words, must contain ‘population median’.\n0 s c\nAllow H : m = 0, H : m > 0 (or < 0) if defined\nH :m m 0 d 1 d\n1 s c or consistent with working.\nd 21 19 –2 22 –10 4 16 27 35 29 M1 Attempt at differences.\nRank 6 5 –1 7 –3 2 4 8 10 9\nM1 Attempt at (signed) ranks.\n(P=51, Q=4) T =4 A1 Allow errors in positive ranks.\n'4'≤10, reject H (significant). M1 Compare their 4 (from rank sum only) with 10\n0\n& appropriate conclusion (allow in terms of H ).\n1\nReject H can be implied by an attempt at an\n0\nappropriate conclusion in context.\nAllow M1 for comparison of their 4 (from rank\nsum only) with 8 and appropriate conclusion.\nSufficient evidence to suggest that children, on average, complete the cartoon puzzle A1 CWO ignoring hypotheses.\nmore quickly than the seaside puzzle. In context, level of uncertainty in language.\nOR Sufficient evidence to support the researcher’s belief. A0 for T =4 from incorrect ranks/differences.\n6\n\n5(b) \n\nX ~B(10, 0.5)\n\nP(X ≤2)=0.0547 B1 OR P(X ≥8)=0.0547.\n'0.0547'0.05, accept H (not significant). M1 Compare their 0.0547 with 0.05 and appropriate\n0\nconclusion (may be in terms of H ).\n1\nIn part 5(a) H is rejected, in part 5(b) H is accepted, so opposite conclusion reached. A1 CWO, compares with test result in part 5(a).\n0 0\n3\n\n5(c) The order of the tests means that children may do the second test quicker (or slower) B1 Credit any valid comment in context.\nbecause, for example, they have had practice (or are tired).\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 16",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_41-p14.png"
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    },
    {
      "id": "9231-2025-mj-41-q06",
      "question_id": "9231-2025-mj-41-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "6(a) k=1−a B1\n1\n\n6(b) ( 1−at2)−1 =1+at2(+) *M1 Attempt at series for ( 1−at2)−1 with\n1+at2 seen or implied.\nP(Y 2)=1−k(1+a) DM1 Correct un-simplified expression in terms of\ntheir k and/or a.\nP(Y 2)=1−(1−a)(1+a)=a2 A1 AG, shown convincingly.\nAlternative method for question 6(b)\nP(Y =0)=G (0)=k P(Y =1)=G ' (0)=0 *M1 Attempt at G '' (0) with their k .\nY  Y  Y\nP(Y =2)= 1G '' (0)= 1(2ak).\n2 Y 2\nP(Y 2)=1−k−ak DM1 Correct un-simplified expression in terms of\ntheir k and/or a.\nP(Y 2)=1−(1−a)−a(1−a)=a2 A1 AG, shown convincingly.\n3\n\n6(c) G ' (t)= 0.32t M1 Attempt to differentiate G Y (t) with their k ,\nY ( 1−0.2t2)2 may be in terms of a and/or k .\nE(Y)=G ' (1)= 0.32 =0.5 A1 CWO\nY (1−0.2)2\n2\n© Cambridge University Press & Assessment 2025 Page 15 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-42-q01",
      "question_id": "9231-2025-mj-42-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1 H : = H :   B1 If in words, must contain ‘population means’.\n0 x y 1 x y\nAllow H : =0, H :  0.\n0 d 1 d\ns2 =103.218... M1 SOI, one unbiased estimate correct to 3sf or\nx\nbetter.\ns2 =171.139...\ny Do not accept biased estimates (102.4, 169.0).\ns2 s2 A1 A0 for pooled s.\nx + y =2.999 Accept 3, 3.0, 3.00.\n120 80\nMay be implied by z=3.03.\n10470−6560 M1 Value of z obtained using correct form with their\nz= 120 80 =3.03\ns2 s2\ns\nx\n2 and their s\ny\n2.\nx + y\nUse of pooled s scores M0 A0.\n120 80\nA1 Use of pooled s scores M0 A0.\n'3.03' > 2.576, reject H (significant). M1 Compare their z with 2.576 and appropriate\n0\nOR conclusion (may be in terms of H ).\n1\nP(Z '3.03')=0.001220.005, reject H (significant). Reject H 0 can be implied by an attempt at an\n0\nappropriate conclusion in context.\nMay see 0.002440.01, reject H (significant).\n0\nAllow M1 for comparison of their 3.03 with\n2.326 and appropriate conclusion.\nThere is sufficient evidence to suggest the mean spend of customers is different on A1 Correct conclusion in context from correct\nweekends compared to weekdays. working ignoring hypotheses.\nLevel of uncertainty in language used (for\nexample, not ‘prove’).\nMust refer to mean or average spend.\n7\n© Cambridge University Press & Assessment 2025 Page 9 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s25_ms_42-p09.png"
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    },
    {
      "id": "9231-2025-mj-42-q02",
      "question_id": "9231-2025-mj-42-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "2(a) 5a+ 3a=1, leading to a= 2 . B1 AG, shown convincingly.\n2 13\nUses area = 1, b and c must be eliminated.\nAttempt to solve 2 independent equations in b and c or 3 independent equations in M1 May see any of the following:\na, b and c. 3b−39c= 3a, a=b−5c, b−8c=0,\n2 2\na\n−c= , 5a+3b−39c=1.\n5−8 2\nOR equation of line through (5,a)and (8,0).\nb=16, c= 2 A1 Correct values imply M1 A1.\n39 39\n3\n\n2(b) E(X)= 5 2 xdx+ 8 x (16 − 2 x ) dx M1 Attempt to integrate xf(x)with their b and c,\n \n0 13 5 39 39 limits seen. May be in terms of a, b, c.\n5 8 M1 Correct integrated expression with their b and\n1 x2 +8 x2 − 2 x3\n13  0 39 117  5 their c, including limits.\nMay be in terms of a, b, c.\n43 =3.31 A1 AWRT 3.31.\n13\n3\n\n2(c) m M1 Uses 0m5 and area = 0.5, may be in terms\n2 m=0.5 OR  2 dx =0.5\n13 0 13 of a.\nm=13 =3.25 A1\n4\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 15\n\n2(d) 2 x, 16 x− 1 x2(+k) B1 FT Correct expressions for integrating both pieces\n13 39 39\nof PDF, seen anywhere, using their b and c.\nk =−25 B1 FT FT their b and their c to obtain k .\n39\n0 y0 M1 Both parts with y substituted (condone\n\n2 y 0≤y≤25 missing k ), ignore limits.\nG(y)= 13\n1\n3\n6\n9\ny −\n3\n1\n9\ny−\n3\n2\n9\n5 25 y≤64\nA1 FT FT their b, their c and their k .\n\n1 y64 Fully defined including correct domain in terms\nof y. Domain must cover all reals, may overlap\nat boundaries.\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2025-mj-42-q03",
      "question_id": "9231-2025-mj-42-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) 01844+1143+211+30+41+50+61  7 B1 175 , AWRT 0.01458.\n= 12000\n \n 20006  480\n1\n\n3(b) a=2000−(1831.3+6.016+0.119+0.001)=162.6 B1 AG, shown convincingly.\nAlternative method for question 3(b)\n6 B1 AG, shown convincingly.\na=  7  ( 1− 7 )5 2000 = 162.6\n1 480 480\n1\n© Cambridge University Press & Assessment 2025 Page 11 of 15\n\n3(c) H : A binomial distribution is a satisfactory model for the data. B1 Must mention binomial and data.\n0\nH : A binomial distribution is not a satisfactory model for the data.\n1\n(1844−1831.3)2 (143−162.6)2 (13−6.136)2 M1 Expression of correct form with cells correctly\n+ + =0.08807+2.3626+7.6784 combined.\n1831.3 162.6 6.136\nMay use 6.1 (found by subtraction).\nOR\nImplied by 2 correct contributions.\n18442 1432 132\n+ + −2000\n1831.3 162.6 6.136\n10.1 A1 AWRT 10.1.\nAllow AWRT 10.3 from using 6.1.\nSC unsupported 10.1 (or 10.3) scores M0 B1.\n[1 degree of freedom so critical value is 7.879.] M1 Compare their 10.1 (or 10.3) with 7.879 and\n'10.1'7.879, reject H (significant). appropriate conclusion (may be in terms of H ).\n0 1\nReject H can be implied by an attempt at an\n0\nappropriate conclusion in context.\nAllow M1 for comparison of their 10.1 (or 10.3)\nwith 10.6 and appropriate conclusion.\nSufficient evidence to suggest that the manager’s model is not satisfactory. A1 Correct conclusion in context from correct\nworking ignoring hypotheses.\nLevel of uncertainty in language used (for\nexample, not ‘prove’).\n5\n\n3(d) Not independent or probability of “success” may not be constant or plausible reason B1 Credit any valid comment in context.\nin context (for example, higher number of boxes with more than 1 broken egg due to\nboxes being dropped).\n1\n© Cambridge University Press & Assessment 2025 Page 12 of 15",
      "source_pages": [
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        12
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2025-mj-42-q04",
      "question_id": "9231-2025-mj-42-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) (x−x)2 =40104− 1 6122 =2649.6 M1 Either expression, may be embedded.\n10 May be implied, for example, by 294.4 or\n(y−y)2 =27460−14442 =2818 402.57 seen.\n8\n2649.6+2818 A1 SOI\ns2 = =341.725\np 10+8−2\n(61.2−55.5)2.120 's2' (1 +1)=5.72.128.7686 B1 2.120, SOI.\np 10 8\nM1 Correct form, must be a t value.\nA1 Correct un-simplified confidence interval, SOI.\n−12.9, 24.3 A1 Accept with inequality signs or open brackets\nor −24.3, 12.9 from Y −X .\nCondone 24.3, −12.9.\nDo not accept 5.718.6.\n6\n\n4(b) Because zero lies inside the confidence interval. B1\n1\n© Cambridge University Press & Assessment 2025 Page 13 of 15",
      "source_pages": [
        13
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-42-q05",
      "question_id": "9231-2025-mj-42-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) H :m =m B1 If in words, must contain ‘population median’.\n0 s c\nAllow H : m = 0, H : m > 0 (or < 0) if defined\nH :m m 0 d 1 d\n1 s c or consistent with working.\nd 21 19 –2 22 –10 4 16 27 35 29 M1 Attempt at differences.\nRank 6 5 –1 7 –3 2 4 8 10 9\nM1 Attempt at (signed) ranks.\n(P=51, Q=4) T =4 A1 Allow errors in positive ranks.\n'4'≤10, reject H (significant). M1 Compare their 4 (from rank sum only) with 10\n0\n& appropriate conclusion (allow in terms of H ).\n1\nReject H can be implied by an attempt at an\n0\nappropriate conclusion in context.\nAllow M1 for comparison of their 4 (from rank\nsum only) with 8 and appropriate conclusion.\nSufficient evidence to suggest that children, on average, complete the cartoon puzzle A1 CWO ignoring hypotheses.\nmore quickly than the seaside puzzle. In context, level of uncertainty in language.\nOR Sufficient evidence to support the researcher’s belief. A0 for T =4 from incorrect ranks/differences.\n6\n\n5(b) \n\nX ~B(10, 0.5)\n\nP(X ≤2)=0.0547 B1 OR P(X ≥8)=0.0547.\n'0.0547'0.05, accept H (not significant). M1 Compare their 0.0547 with 0.05 and appropriate\n0\nconclusion (may be in terms of H ).\n1\nIn part 5(a) H is rejected, in part 5(b) H is accepted, so opposite conclusion reached. A1 CWO, compares with test result in part 5(a).\n0 0\n3\n\n5(c) The order of the tests means that children may do the second test quicker (or slower) B1 Credit any valid comment in context.\nbecause, for example, they have had practice (or are tired).\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-42-q06",
      "question_id": "9231-2025-mj-42-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "6(a) k=1−a B1\n1\n\n6(b) ( 1−at2)−1 =1+at2(+) *M1 Attempt at series for ( 1−at2)−1 with\n1+at2 seen or implied.\nP(Y 2)=1−k(1+a) DM1 Correct un-simplified expression in terms of\ntheir k and/or a.\nP(Y 2)=1−(1−a)(1+a)=a2 A1 AG, shown convincingly.\nAlternative method for question 6(b)\nP(Y =0)=G (0)=k P(Y =1)=G ' (0)=0 *M1 Attempt at G '' (0) with their k .\nY  Y  Y\nP(Y =2)= 1G '' (0)= 1(2ak).\n2 Y 2\nP(Y 2)=1−k−ak DM1 Correct un-simplified expression in terms of\ntheir k and/or a.\nP(Y 2)=1−(1−a)−a(1−a)=a2 A1 AG, shown convincingly.\n3\n\n6(c) G ' (t)= 0.32t M1 Attempt to differentiate G Y (t) with their k ,\nY ( 1−0.2t2)2 may be in terms of a and/or k .\nE(Y)=G ' (1)= 0.32 =0.5 A1 CWO\nY (1−0.2)2\n2\n© Cambridge University Press & Assessment 2025 Page 15 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2025-mj-43-q01",
      "question_id": "9231-2025-mj-43-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 H : uniform distribution fits the data. B1 Must mention data and uniform distribution (or\n0\nH : uniform distribution does not fit the data. uniform distribution clearly described).\n1\nExpected frequencies are 40, 40, 40. B1\n(38−40)2 (52−40)2 (30−40)2 M1 Expression of correct form.\n+ + =0.1+3.6+2.5 May see individual terms not summed.\n40 40 40\nOR\n382 522 302\n+ + −120\n40 40 40\n6.2 A1 SC unsupported 6.2 scores M0 B1.\nNote: 6.208 scores M0 A0 XP.\n[2 degrees of freedom, so critical value is 5.991.] M1 Compare their 6.2 with 5.991 and appropriate\n'6.2'5.991, reject H (significant). conclusion (may be in terms of H ).\n0 1\nReject H can be implied by an attempt at an\n0\nappropriate conclusion in context.\nSufficient evidence to suggest that the scientist’s claim is incorrect. A1 Correct conclusion in context from correct\nOR Sufficient evidence eye colour is not uniformly distributed. working ignoring hypotheses.\nOR Sufficient evidence to suggest that uniform distribution is not a good fit. Level of uncertainty in language is used (for\nexample, not ‘prove’).\n6\n© Cambridge University Press & Assessment 2025 Page 9 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    },
    {
      "id": "9231-2025-mj-43-q02",
      "question_id": "9231-2025-mj-43-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2 s2 = 1 ( 720−1682) =0.36923 = 24 M1 Correct method to find one unbiased estimate,\nx 39 40  65 may be un-simplified.\ns2 = 1\n( 900−2282)\n=0.56949 =168\ny 59 60  295\nA1 Both unbiased estimates correct to 3sf or better.\ns2 = '0.36923'+ '0.56949' M1 M0 for pooled estimate.\n40 60\ns2 =0.0187 = 359  A1 SOI\n \n19175\nCI= (168−228) 1.645's' M1 Correct form with any z-value, FT their s.\n40 60\nB1 1.645 (accept 1.64 or 1.65) used in CI formula.\n0.175, 0.625 A1 Accept with inequality signs or open brackets.\nCondone 0.625,0.175.\nDo not accept 0.40.225.\n7\n© Cambridge University Press & Assessment 2025 Page 10 of 16",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-43-q03",
      "question_id": "9231-2025-mj-43-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a)  1 kx dx+  8 k(8−x) dx=1  1kx2 1 +8kx−1kx2 8 =1  M1 Linear equation in terms of k formed following\n0 1  2  0  2  1  attempt to integrate and use correct limits.\n1k+(64k−32k)− ( 8k−1k ) =1\n2 2\n25k=1, leading to k = 1 . A1 AG, shown convincingly. No errors seen.\n25\nAlternative method for question 3(a)\n11k+177k =1 M1 Linear equation in terms of k formed by\n2 2\nconsidering areas.\n25k=1, leading to k = 1 . A1 AG, shown convincingly. No errors seen.\n25\n2\n\n3(b) 1 m *M1 Use of definite integrals and attempt at\n kx dx+ k(8−x)dx= 1\n2 integration to form equation with 0.5 and their k.\n0 1\nCould be in terms of k.\nm2−16m+39=0 or DM1 Quadratic equation in m.\nm=3 A1 Must reject second solution if seen. CAO\nAlternative method for question 3(b)\n1(8−m)k(8−m)= 1 *M1 Use areas to form equation with 0.5 and their k.\n2 2\nCould be in terms of k.\n(8−m)2\n=25\nDM1 Quadratic equation in m.\nm=3 A1 Must reject second solution if seen. CAO\n3\n© Cambridge University Press & Assessment 2025 Page 11 of 16\n\n3(c)   1 x2 [0≤x1] M1 Limits not required. Sight of expressions for\nF(x)= 50 0≤x1 and 1≤x≤8 sufficient.\n2 8\n5\nx−\n5\n1\n0\nx2 −\n2\n7\n5\n[1≤x≤8]\n[F(x)=0 for x0 and F(x)=1 for x8] B1 For − 7 .\n25\n  1 y6 [0≤y1] M1 Both parts with y substituted in their F(x).\nG(y)= 50\nLimits not required. Sight of expressions for\n2 8\n5\ny3 −\n5\n1\n0\ny6 −\n2\n7\n5\n[1≤y≤2]\n0≤ y1 and 1≤ y≤2 sufficient.\n[G(y)=0 for x0 and G(y)=1 for y2]\n3 y5 0≤y1 M1 Differentiation of their G(y).\n 25 Limits not required. Sight of expressions for\ng(y)=24 y2 − 3 y5 1≤y≤2\n25 25 0≤ y1 and 1≤ y≤2 sufficient.\n\n0 otherwise.\n\nA1 Fully correct with correct limits.\nAlternative method for question 3(c)\nLet p(y)= 3 y , then q(y)=p−1(y)= y3. M1 Find inverse function.\nq ' (y)=3y2. A1 Differentiate inverse function. May see 3x 2 3 .\n  1 y3(3y2) [0≤y1]   1 x(3x 2 3) [0≤x1] M2 M2 for using q ' (y)f(y) or q ' (x)f(x) twice.\ng(y)= 25 or g(x)= 25\n2 1\n5\n(8− y3)(3y2) [1≤y≤2] 2 1\n5\n(8−x)(3x 2 3) [1≤x≤8] M\nSig\n1\nh\nf\nt\no\no\nr\nf\no\ne\nn\nx\ne\np\ne\nr\nx\nes\np\ns\nr\ni\ne\no\ns\nn\ns\ns\nio\nf\nn\no\n.\nr\nL\n0\nim\n≤\nit\ny\ns\n\nn\n1\not\na\nr\nn\ne\nd\nq u\n1\nir\n≤\ned\ny\n.\n≤ 2\n(g(y)=0 otherwise) (g(x)=0 otherwise)\nor 0≤x1 and 1≤x≤8 sufficient.\n3 y5 0≤y1 A1 Fully correct with correct limits.\n 25\ng(y)=24 y2 − 3 y5 1≤y≤2\n25 25\n\n0 otherwise.\n\n5\n© Cambridge University Press & Assessment 2025 Page 12 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_43.pdf",
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    {
      "id": "9231-2025-mj-43-q04",
      "question_id": "9231-2025-mj-43-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4 H : population medians are equal for under 25s and over 50s (or m =m ). B1 B0 for defining m as median, not population\n0 25 50\nmedian.\nH : population median for under 25s < population median time for over 50s\n1\n(or m m ).\n25 50\nM1 Attempt at ranks using the integers 1 to 14\n198 5 178 1\n(ranks may be reversed).\n212 8 181 2\n217 9 183 3\n229 11 192 4\n235 13 203 6\n242 14 209 7\n223 10\n231 12\nSum of ranks for sample of size 6: R =60 A1 Note: R =30 from reverse ranks.\n6 6\n6 ( 6+8+1 ) −'R ' [=30] M1 Note:6(6+8+1)−'R '=60 from reverse ranks.\n6 6\n[W =min(60,30)=] 30 A1\nCritical value for n=6, m=8 is 31. B1\n'30'≤31, reject H\n0\n(significant). M1 Compare their W with 31 and appropriate\nconclusion (may be in terms of H ).\n1\nReject H may be implied by attempt at\n0\nappropriate conclusion in context.\nSufficient evidence to suggest that older people take longer to react. A1 Correct conclusion in context from correct\nOR Sufficient evidence to support researcher’s claim. working ignoring hypotheses.\nLevel of uncertainty in language used (for\nexample, not ‘prove’).\n8\n© Cambridge University Press & Assessment 2025 Page 13 of 16",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
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    },
    {
      "id": "9231-2025-mj-43-q05",
      "question_id": "9231-2025-mj-43-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) H :  − = k , H :  − k B1 Allow H :  = k , H :  k if defined or\n0 B A 1 B A 0 d 1 d\nconsistent with working.\nDifferences, d : 19, 17, 8, 16, 9, 23, 6, 5. M1 Allow one error.\nd =103, d2 =1641, M1 Sample mean and variance for differences.\nd =12.875, s2 = 1 ( 1641−1 ( 1032)) =44.982 = 2519\nd 7 8  56 \n12.875−k M1 Correct form for t with their d and their s2.\nt= d\ns2 M0 if k is omitted.\nd\n8\nA1\n12.875−k M1 Accept ≥, > or =.\n≥1.895\ns2\nd\n8\n0≤ k≤8.38 A1 Accept k8.38.\n7\n\n5(b) Population of differences are normally distributed. B1 Accept underlying distribution or population.\nB0 for \"population mean normally distributed\".\nB0 for \"(underlying) data is normal\".\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 16",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_43.pdf?download=true",
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    },
    {
      "id": "9231-2025-mj-43-q06",
      "question_id": "9231-2025-mj-43-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) Probabilities: 1 , 7 , 7 for 0, 1, 2 heads. B1\n15 15 15\nG (t)= 1 + 7 t+ 7 t2 B1 FT FT their probabilities provided p=1.\nX 15 15 15\nProbabilities: 8 , 7 for 1, 2 colours. B1 SC B1 for 8 , 7 for 0,1 colours or 0,2 colours.\n15 15 15 15\nG (t)= 8 t+ 7 t2 B1 FT FT their probabilities provided p=1.\nY 15 15\nSC B1 for G (t)= 8 + 7 t from 0,1 colours or\nY 15 15\nG (t)= 8 + 7 t2 from 0,2 colours.\nY 15 15\n4\n\n6(b) X and Y are not independent. B1\n1\n© Cambridge University Press & Assessment 2025 Page 15 of 16\n\n6(c) B1 Correct 3 cases identified correctly where\nx 0 1 2\nprobabilities are not zero from 1,2 colours.\ny 1 2 1\n(x,y)=(0,1), (1,2) or (2,1).\nAward SC B1 for either of the following:\nz=x+y 1 3 3 (x,y)=(0,0), (1,1) or (2,0) from 0,1 colours.\nP(Z =z) 1 7 7 (x,y)=(0,0), (1,2) or (2,0) from 0,2 colours.\n15 15 15\nM1 At least one non-zero probability correct. Allow\nfor 1 , 7 , or 14.\n15 15 15\nA1 All probabilities correct.\nG (t)= 1 t+14t3 A1 FT FT their probabilities provided p=1.\nZ 15 15\nAllow A1 FT for either of the following:\nG (t)= 1 +14t2 from 0,1 colours.\nZ 15 15\nG (t)= 1 + 7 t2 + 7 t3 from 0, 2 colours.\nZ 15 15 15\n\n6(d) G ' (t)= 1 +42t2 M1 Attempt to differentiate G (t) and find G ' (1).\nZ 15 15 Z Z\nG ' (1)= 1 +42 FT their G (t).\nZ 15 15 Z\n43 =2.87 A1 Allow A1 FT for either of the following:\n15\n28 =1.87 from 0,1 colours.\n15\n35 =2.33 from 0,2 colours.\n15\n2\n© Cambridge University Press & Assessment 2025 Page 16 of 16",
      "source_pages": [
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    },
    {
      "id": "9231-2025-mj-44-q01",
      "question_id": "9231-2025-mj-44-q01",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 10.24−10.6 M1 0.52\nt= Must be in correct form, allow in denominator.\n0.52 12\n12\n()1.729 A1 CWO, AWRT ()1.73.\n'−1.729 ' −1.363 (or 1.729 1.363), reject H (significant). M1 Compare their t with 1.363 (consistent signs) and appropriate\n0\nconclusion (may be in terms of H ).\n1\nReject H can be implied by an attempt at an appropriate\n0\nconclusion in context.\nAllow M1 for comparison of their t with 1.356 , 1.796, or\n1.782 (consistent signs) and appropriate conclusion.\nSufficient evidence to support/suggest that the [population] mean A1 Correct conclusion in context from correct working ignoring\nis less than 10.6. hypotheses.\nLevel of uncertainty in language used (for example, not ‘prove’).\nAllow  or mean for population mean.\n4\n© Cambridge University Press & Assessment 2025 Page 9 of 15",
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
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    {
      "id": "9231-2025-mj-44-q02",
      "question_id": "9231-2025-mj-44-q02",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) H : population median=8.00 B1 Allow m, but not . If m defined, must see population.\n0\nH : population median8.00\n1\n–0.34, 0.48, 0.21, –0.02, 0.01, –0.23, 0.25, 0.11, 0.03, 0.16, –0.08 M1 Attempt at (signed) differences.\n–10, 11, 7, –2, 1, –8, 9, 5, 3, 6, –4 M1 Attempt at (signed) ranks, ranks may be reversed.\n(P=42, Q=24) T =24 A1\n'24' > 13, accept H (not significant). M1 Compare their 24 (from rank sum only) with 13 and appropriate\n0\nconclusion (may be in terms of H ).\n1\nAccept H can be implied by an attempt at an appropriate\n0\nconclusion in context.\nAllow M1 for comparison of their 24 (from rank sum only) with\n10 and appropriate conclusion.\nInsufficient evidence to support/suggest that the average sound A1 Correct conclusion in context from correct working ignoring\nlevel is more than 8.00. hypotheses.\nLevel of uncertainty in language used (for example, not ‘prove’).\nMust refer to average/median sound (level).\n6\n\n2(b) It is not known if the population is normal. B1 Must refer to population and/or distribution.\nIt is not known if the (underlying) distribution is normal. B0 for data is normal.\nB0 for (population) differences are normal.\n1\n© Cambridge University Press & Assessment 2025 Page 10 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_44.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
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    },
    {
      "id": "9231-2025-mj-44-q03",
      "question_id": "9231-2025-mj-44-q03",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 H : type of item sold is independent of salesperson. B1 Allow dependent, association or relationship for not independent.\n0\nFor example: H : No association/relationship between…\nH : type of item sold is not independent of salesperson. 0\n1 H : Association/relationship between…\n1\nLaptop Camera Television M1 Calculate expected values (must be seen).\nM1 for at least 4 expected values correct to 3sf or better.\nAvril 31 (28.5) 40 (41.80) 24 (24.70)\nBen 23 (29.1) 45 (42.68) 29 (25.22)\nA1 All expected values correct to 3sf or better.\nCharlie 21 (17.4) 25 (25.52) 12 (15.08)\n(31−28.5)2 (40−41.8)2 (15.08−12)2 M1 At least 2 terms (not including 250) of correct form seen.\n+ ++ May see individual terms not summed.\n28.5 41.8 15.08\nMay be implied by AWRT 3.67 or 3.68.\nOR\n312 402 122\n+ ++ −250\n28.5 41.8 15.08\nTest statistic = 3.67 (or 3.68) A1 AWRT 3.67 or 3.68.\n'3.67' < 7.779, accept H (not significant). M1 Compare their 3.67 (or 3.68) with 7.779 and appropriate\n0\nconclusion (may be in terms of H ).\n1\nAccept H can be implied by an attempt at an appropriate\n0\nconclusion in context.\nAllow M1 for comparison of their 3.67 (or 3.68) with 9.488 and\nappropriate conclusion.\nInsufficient evidence to suggest that type of item sold and A1 Correct conclusion in context from correct working ignoring\nsalesperson are not independent. hypotheses. Condone ‘no sufficient evidence’.\nOR Level of uncertainty in language used (for example, not ‘prove’).\nInsufficient evidence to suggest that there is dependence between Must refer to both item and salesperson.\ntype of item sold and salesperson. Do not accept statements such as ‘there is sufficient evidence to\nsuggest…’.\n7\n© Cambridge University Press & Assessment 2025 Page 11 of 15",
      "source_pages": [
        11
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_44.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-44-q04",
      "question_id": "9231-2025-mj-44-q04",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)  1 kxdx+  2 kx2 dx=1  1kx2 1 +1kx3 2 =1  M1 Linear equation in terms of k formed following attempt to\n0 1  2  0 3  1  integrate (all powers correct) and use of correct limits.\n1k+1k(8−1) =1 May consider area of triangle (11k ) for first region.\n2 3 2\n17k =1, k = 6 A1 AG, shown convincingly. No errors seen.\n6 17\n2\n\n4(b) 0 x0 M1 Attempt to integrate both parts, all powers correct, condone\n missing c.\n3 x2 0≤x1\nF(x)= 17 Sight of expressions for 0≤x1 and 1≤x≤2 sufficient.\n2 x3 + 1 1≤x≤2\n17 17\n 1 x2 A1 Both parts correct with c=\n1\n1\n7\n.\nSight of expressions for 0≤x1 and 1≤x≤2 sufficient.\nA1 All correct, fully defined including domain.\nDomain must cover all reals, may overlap at boundaries.\n3\n\n4(c) F(m)= 1 so 2 m3 + 1 = 1 M1 Use 1≤m≤2 and their F(x) (or areas) to form equation with 0.5.\n2 17 17 2\nm3 =15, m= 3 15 =1.55 A1 FT FT their non-zero c, to obtain m with 1≤m≤2.\n4 4\n2\n\n4(d) 1 2 M1 Correct method and attempt to integrate both parts. All powers\n   E    X 1    =       1 6 7 x    1 x    dx+    1 6 7 x2    1 x    dx = 1 6 7 x1 0 + 1 6 7   1 2 x2  1 2 correct, limits not required.\n0 1\n6 ( 1+2−1) = 15 =0.882 A1 CWO, AWRT 0.882.\n17 2 17\n2\n© Cambridge University Press & Assessment 2025 Page 12 of 15",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_44.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-44-q05",
      "question_id": "9231-2025-mj-44-q05",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) Probabilities: 8 , 12, 6 , 1 for 0, 1, 2, 3 heads B1\n27 27 27 27\n8 +12t+ 6 t2 + 1 t3 B1 FT FT their probabilities to form a 4-term cubic. Must have p=1.\n27 27 27 27\n(2 + 1t )3 scores B2.\n3 3\n2\n\n5(b) ( 8 +12t+ 6 t2 + 1 t3)( 25+10t+ 1 t2) M1 Attempt to multiply their G (t) and their G (t).\n27 27 27 27 36 36 36 X Y\n1 ( 200+380t+278t2 +97t3+16t4 +t5) M1 Expand to form polynomial of degree 5 (need not be simplified).\n972\nA1 CWO\n3\n\n5(c) G' (t)= 1 ( 380+556t+291t2 +64t3+5t4) M1 Attempt at differentiating their G (t) .\nZ 972 Z\nE(Z)=G' (1)=1296 = 4 A1 FT FT their G (t).\nZ 972 3 Z\nG'' (t)= 1 ( 556+582t+192t2 +20t3) M1 Attempt at differentiating their G' (t).\nZ 972 Z\nVar (Z) =1 9 3 7 5 2 0+ 3 4− ( 3 4)2  = 1 2 8 5+ 3 4− ( 3 4)2  M1 Use correct formula using their G' Z (1) and their G' Z ' (1).\n17 =0.944 A1 CWO\n18\n5\n© Cambridge University Press & Assessment 2025 Page 13 of 15",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_44.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-mj-44-q06",
      "question_id": "9231-2025-mj-44-q06",
      "subject": "9231",
      "year": 2025,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) x=30.4 B1\n2.262 B1 2.26 or 2.262 seen.\n10(3.13−2.95) M1 Subtracting, or substituting, to find an equation involving s .\nx\nSubtract: s =\nx (22.262) Allow M1 for any of the following t values seen instead of 2.262:\n1.812, 1.833, 2.228 or 3.690.\ns2 = 0.0158 or s =0.126 A1 FT FT their 2.262 following M1.\nx x\nImplied by AWRT 92.6 following M1.\n30.42 A1 FT FT their 2.262.\nx2 =90.0158307+ =92.6 CWO\n10\n5\n© Cambridge University Press & Assessment 2025 Page 14 of 15\n\n6(b) H :  = B1 If in words, must contain ‘population means’.\n0 Y X\nAllow H :  =0, H :  0 if defined or consistent with\nH :   0 d 1 d\n1 Y X\nworking.\n24.82 M1 May be embedded, for example s2 = 1 ( 76.98−1 ( 24.82)) .\n76.98− y 7 8\n8\n9s2 +7s2 M1 Correct form using their s2 and their s2, may be embedded.\ns2 = x y =0.015155 x y\np 10+8−2\nA1 SOI, can be implied by any t that rounds to 1.02 or 1.03.\n3.04−3.1 M1 Correct form using their s .\nt= =−1.028 p\ns 1 +1\np 10 8\nA1 AWRT 1.02 or 1.03implies M1 A1.\n[18 degrees of freedom, so critical value is 1.746.] M1 Compare their t with 1.746 (consistent signs) and appropriate\n'1.028 '1.746, accept H (not significant). conclusion (may be in terms of H ).\n0 1\nAccept H can be implied by an attempt at an appropriate\n0\nconclusion in context.\nAllow M1 for comparison of their t with 1.740 or 2.120and\nappropriate conclusion.\nInsufficient evidence to support Mona’s claim. A1 Correct conclusion in context from correct working ignoring\nhypotheses.\nLevel of uncertainty in language used (for example, not ‘prove’).\n8\n© Cambridge University Press & Assessment 2025 Page 15 of 15",
      "source_pages": [
        14,
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      ],
      "source_pdf": "_source-pdfs/2025-May-June/ms/9231_s25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-may-june/9231_s25_ms_44.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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        "../answer-assets/9231_s25_ms_44-p15.png"
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    },
    {
      "id": "9231-2025-on-11-q01",
      "question_id": "9231-2025-on-11-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "1(a) 2n2(n+1)2 +2n(n+1)(2n+1)+2n(n+1)+3n M1 A1 Substitutes correct formulae from MF19.\n=2n4 +8n3+10n2 +7n A1 Simplifies.\n3\n\n1(b) 1 1 (r+1)4 −r4 r4 +4r3+6r2 +4r+1−r4 4r3+6r2 +4r+1 M1 A1 Puts over a common denominator and\n− = = = expands, AG.\nr4 (r+1)4 r4(r+1)4 r4(r+1)4 r4(r+1)4\nn 4r3 +6r2 +4r+1 n  1 1  M1 A1 Shows three complete terms, including\n =  −  first and last.\nr4(r+1)4 r4 (r+1)4\n\nr=1 r=1\n1 1 1 1 1\n=1− + − + + −\n24 24 34 n4 (n+1)4\n1 A1\n=1−\n(n+1)4\n5\n\n1(c) 1 B1FT FT from their answer to part (b).\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_11-p09.png"
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    },
    {
      "id": "9231-2025-on-11-q02",
      "question_id": "9231-2025-on-11-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "2(a) Shear, [x-axis fixed] B1\nWith (0,1) mapped to (3,1 ) . B1 Or other suitable example/description.\n2\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 17\n\n2(b) Shear, [y-axis fixed], B1\nwith (1,0) mapped to ( 1,3) . B1 Or other suitable example/description.\n2\n2\n\n2(c) 13 3 M1 Finds AB or uses product of\nAB= 4 2  or detAB=detAdetB determinants. Or suitable reasoning that\n3 1\n 2  area does not change.\ndetAB=1 A1\n2\n\n2(d) 13 3x 13x+ 3 y B1 x X\n 4 2  = 4 2  Transforms   to   .\n 3 2 1 y   3 2 x+ y   y Y \n3x+mx=m (13x+3mx ) M1A1 Uses Y =mX .\n2 4 2\n3+m=13m+3m2m2+3m−1=0 A1\n2 4 2 2\ny=−2x and y= 1 x. A1 If only values of m are given, A1 A0.\n2\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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        "../answer-assets/9231_w25_ms_11-p10.png"
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    },
    {
      "id": "9231-2025-on-11-q03",
      "question_id": "9231-2025-on-11-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 d B1 Checks base case using product rule.\n(xcosx)=cosx−xsinx=(−1)1(xsinx−(2(1)−1)cosx)\ndx\nd2k−1 B1 States inductive hypothesis.\nAssume true for n=k, so (xcosx)=(−1)k(xsinx−(2k−1)cosx)\ndx2k−1\nd2k M1 A1 Differentiates once.\nThen (xcosx)=(−1)k(xcosx+2ksinx)\ndx2k\nd2k+1 M1 A1 Differentiates again.\n(xcosx)=(−1)k(−xsinx+cosx+2kcosx)\ndx2k+1\n=(−1)k+1( xsinx−(2k+1)cosx )\nSo, it is also true for n=k+1. Hence, [by induction], A1 States conclusion.\nd2n−1\n(xcosx)=(−1)n(xsinx−(2n−1)cosx)\nis true for all positive integers.\ndx2n−1\n7\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-11-q04",
      "question_id": "9231-2025-on-11-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) y=2x+1x= 1(y−1) B1 Correct substitution used in equation.\n2\n1 (y−1)4 +1(y−1)3 +1(y−1)2 +1(y−1)+1=0 M1 A1 Expands.\n16 8 4 2\n1 ( y4 −4y3+6y2 −4y+1 ) +1 ( y3−3y2 +3y−1 ) +1 ( y2 −2y+1 ) +1(y−1)+1=0\n16 8 4 2\n1 y4−1y3+1 y2+1y+11=0y4−2y3+4y2+2y+11=0 A1 Simplifies, AG.\n16 8 4 8 16\n4\n© Cambridge University Press & Assessment 2025 Page 11 of 17\n\n4(b) S =22 −2(4) M1 Uses\n2\n(+++)2 =2 +2 +2 +2\n+2(+++++)\nS =−4 A1\n2\n2\n\n4(c) S −2S +4S +2S +44=0 M1 Sums over roots.\n4 3 2 1\nS =−76 A1\n4\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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        "../answer-assets/9231_w25_ms_11-p12.png"
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    },
    {
      "id": "9231-2025-on-11-q05",
      "question_id": "9231-2025-on-11-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k −5  5  M1 A1 Finds perpendicular to .\n1\n   \n0 1 2 = 2 ~ −2\n   \n   \n1 3 1 −1  1 \n5(−3)−2(−1)+(−1)=−14 M1 Uses point on .\n1\n5x−2y+z=−14 A1\n4\n© Cambridge University Press & Assessment 2025 Page 12 of 17\n\n5(b) −2 −5 M1 A1 Uses correct formula for perpendicular\n1     distance from point to .\n −1    2  1\n52 +22 +12    \n 1  −1\n7 A1 Accept 1.28.\n30\n3\n\n5(c) 0 B1  0 \n   \nStates point common to both planes e.g. 7 . Or 0 .\n   \n   \n0 −14\ni j k  0  0 M1 A1 Finds direction of line.\n   \n5 −2 1 = 8  1\n   \n   \n3 2 −1 16 2\n0 0 A1 OE\n   \nr= 7 + 1\n   \n   \n0 2\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
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    },
    {
      "id": "9231-2025-on-11-q06",
      "question_id": "9231-2025-on-11-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) B1* Initial line drawn, pole clear and one\nloop in the first quadrant.\nDB1 Symmetrical with correct shape at\nextremities.\n= 1π. B1 States the equation of the line of\n6\nsymmetry.\n3\n\n6(b) 1 1 3 π sin23d M1 Uses 1\n2\nr2d with correct limits.\n2\n0\n=1 1 3 π 1−cos6d=1−1sin6 1 3 π M1 A1 Applies double angle formula correctly\n4 0 4  6  0 and integrates.\n= 1 π A1 CWO\n12\n4\n\n6(c) y=3sin2−4sin4 B1 Uses y=rsin.\n6sincos−16sin3cos=0 M1* Differentiates correct function and sets\nequal to 0. Might see completing the\nsquare.\nsincos06−16sin2=0 A1\nsin2=3 DM1 Solves quadratic in sin.\n8\ny= 9 . A1\n16\n5\n© Cambridge University Press & Assessment 2025 Page 14 of 17\n\n6(d) r4 =rsin ( 3r2 −4r2sin2 ) M1\n3y y3 Uses y=rsin\nr = −4\nr r3\n( x2 + y2)2 = y ( 3x2 −y2) M1 Uses r2 =x2 + y2\nA1 Rearranges to a form with no fractions.\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2025-on-11-q07",
      "question_id": "9231-2025-on-11-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−3−1 5, x=−3+1 5 B1 Vertical asymptotes.\n2 2 2 2\ny=0 B1 Horizontal asymptote.\n2\n\n7(b) dy (x2 +3x+1)(1)−(x+2)(2x+3) M1 dy\n= Finds .\ndx ( x2 +3x+1 )2 dx\nx2 +4x+5=0 M1 Sets equal to 0 and forms quadratic\nequation.\nDiscriminant is 42 −20=−40 so no real roots. A1 AG\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c) y B1 Axes and asymptotes labelled.\nB1 Correct shape and position.\nB1 States coordinates of intersections with\naxes, (0,2)and (−2,0), may be seen on\ndiagram.\nx\n3\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(d) y B1FT FT from sketch in part 7(c)\nB1 From a correct graph and with correct\nshape at (−2,0).\nx\n2\n\n7(e) x+2 x+2 M2 Finds critical points, award M1 for each\n=2 or =−2\nx2 +3x+1 x2 +3x+1 case.\n−2x2 −5x=0 or 2x2 +7x+4=0\nx=−5, x=0 or x=−7−1 17, x=−7+1 17 A1\n2 4 4 4 4\n−7−1 17x−5, −7+1 17x0 , x−31 5 A1 −7−1 17x−3−1 5,\n4 4 2 4 4 2 2 4 4 2 2\n−3−1 5x−5,\n2 2 2\n−7+1 17x−3+1 5,\n4 4 2 2\n−3+1 5x0.\n2 2\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_11.pdf?download=true",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2025-on-12-q01",
      "question_id": "9231-2025-on-12-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "1(a) n M1 A1 Substitutes formulae. For the 1st A mark\n( r3−r ) =1n2(n+1)2 −1n(n+1) allow any letter for n.\n4 2\nr=1\n= 1n(n+1)(n−1)(n+2) A1 Factorises.\n4\n3\n\n1(b) r+3 2 3 1 M1 A1 Finds partial fractions. M mark is for an\n= − +\nr3−r r−1 r r+1 equation that would lead to values for\nA, B and C.\nn r+3 M1 A1 Shows five complete terms including\n = 2f(1)−3f(2)+ f(3) first two and last two.\n(r−1)r(r+1)\nr=2\n+2f(2)−3f(3)+ f(4)\n+2f(3)−3f(4)+ f(5)\n\n+2f(n−2)−3f(n−1)+ f(n)\n+2f(n−1)−3f(n)+ f(n+1)\n3 2 1 A1 SCB1 if insufficient terms to show all\n= − +\nof the cancelling, but must start at r = 2\n2 n n+1\n5\n\n1(c) 3 B1FT FT from their part (b). Needs at least\n2\ntwo terms in part (b) with one\ninvolving a number and a term in n.\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 19",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-12-q02",
      "question_id": "9231-2025-on-12-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) b=−(++)=−2 B1\n3=(−2)2 −2c M1 Uses formula for sum of squares.\nc= 1 A1\n2\n3\n\n2(b) S +bS +cS +3d =0 M1 Uses original equation or formula for\n3 2 1\nsum of cubes.\nS =−3b−2c−3d =5−3d A1FT Allow FT from part (a) for their value.\n3\nS +bS +cS +dS =0 M1 Uses original equation.\n4 3 2 1\n5+(−2)(5−3d)+ (1)(3)+2d =08d−7 =0 A1FT Allow FT from part (a) for their value\n2 2 and for their value of S .\n3\nd = 7 A1 CAO\n16\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 19",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-12-q03",
      "question_id": "9231-2025-on-12-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) u 5 (given) B1 States base case.\n1\nAssume that u 5 for some positive integer k. B1 States inductive hypothesis.\nk\n6u +5 5u +10−6u −5 M1 Considers 5−u , puts over common\nThen 5−u =5− k = k k k+1\nk+1 u +2 u +2 denominator.\nk k\n5−u A1\n= k 0.\nu +2\nk\nHence, by induction, u 5 for all positive integers n. A1\nn\n5\n\n3(b) 6u +5 6u +5−u 2 −2u −u 2 +4u +5 M1 A1 Considers u n+1 −u n or uses\nu −u = n −u = n n n = n n\nn+1 n u n +2 n u n +2 u n +2 5−u = 5−u n .\nn+1 u +2\nn\n(5−u )(1+u ) A1 Uses u 5.\n= n n . So u 5u −u 0 n\nn n+1 n\nu +2\nn\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-12-q04",
      "question_id": "9231-2025-on-12-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)(i) Stretch B1\nparallel to the x-axis, scale factor k. B1 Allow in the x direction, x axis\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 19\n\n4(a)(ii) Enlargement, B1\ncentre the origin, scale factor k. B1\n2\n\n4(b)  0 1  0 m M1 A1 Multiplies two matrices correctly.\n k 02 −1 1  2 −1 1 Allow one error for the M mark.\n −1 1   =  −k m  \n\n0 m1 1 2\n\n1 1 2\n 1 1  k m\n m m 2m  A1\n \n−2k+m k+m −k+2m\n \n \n 2k+m −k+m k+2m \nk+m −k+2m −2k+m −k+2m −2k+m k+m M1 A1 Finds determinant. Allow two errors for\nm −m +2m the M mark\n−k+m k+2m 2k+m k+2m 2k+m −k+m\n=m(k+m)(k+2m)−m(−k+m)(−k+2m)−m(−2k+m)(k+2m) A1 AG.\nThere must be evidence to show that\n+m(2k+m)(−k+2m)+2m(−2k+m)(−k+m)−2m(2k+m)(k+m)\nthe determinant is equal to zero\n=6m2k+6m2k−12m2k =0\n6\n© Cambridge University Press & Assessment 2025 Page 12 of 19",
      "source_pages": [
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        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2025-on-12-q05",
      "question_id": "9231-2025-on-12-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Initial line drawn and correct shape.\nMaximum distance of C from the pole is 1. B1\n2\n\n5(b) 1 1 8 π tan2d M1 Uses 1 2 r2d with correct limits.\n2\n0\n= 1 1 8 πsin2 d=−1lncos21 8 π M1 A1 Integrates a correct function to get\n2 0 cos2 4 0 −1lncos2 condoning the lack of\n2\n1\nlimits and factor of for M1A1.\n2\n=−1ln 2 =1ln2 A1 OE, must be exact. ISW\n4 2 8\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 19\n\n5(c) 2tan M1 Uses relevant identity for tan2.\nr2 =\n1−tan2\n2\n(y) M1 For the M mark uses either\n2xy\nx2 + y2 = x = y\n1− y2 x2 −y2 r= x2 +y2 or tan= .\nx2 x\nA1 y\nUse both r= x2 +y2 and tan=\nx\ncorrectly.\n( x2 + y2)( x2 −y2) =2xy A1 At least one intermediate step before\nthe answer given.\nx4 −y4 =2xy Needs to be in this form as answer\ngiven.\nx4 −2xy− y4 =0.\n4\n\n5(d) 1( cos1π )( sin1π ) −1ln2 M1 Area of triangle minus their (b).\n2 8 8 8\n1 2−1ln2 A1 By double angle formula or otherwise.\n8 8\n2\nQuestion Answer Marks Guidance",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2025-on-12-q06",
      "question_id": "9231-2025-on-12-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) 1 1 M1 A1 Allow 0.267 or better\n=\n12 +32 +22 14\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 19\n\n6(b) i j k 1 M1 A1 Finds normal to OAB. Allow 1 error for\n  the M mark to be awarded.\n0 −1 2 = 4\n \n \n2 0 −1 2\n \n1 1 M1 Takes dot product of normal vectors.\n    17\n4  3 = 21 14coscos=\n   \n21 14\n   \n2 2\n   \n7.5° A1 Accept 0.131 radians.\nMark final answer of 7.5°or better.\n4\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n6(c)  2   2   2−2  M1 A1 Finds PQ or direction of common\n\n     \nOP= −1 ,OQ= −1+ PQ= −1++ perpendicular.\n     \n     \n−1 2−3 2−3+\n     \n\n 2−2   2  M1 Uses that dot product of PQ with line\n    \n −1++    −1  =0 direction is zero, or, alternatively, PQ\n   \n2−3+ −1 is a multiple of the common\n   \n1\n \nperpendicular 1 .\n \n \n1\n6−6=1 A1 Deduces one equation.\n 2−2   2  A1 Deduces second equation.\n   \n−1++  1 =014−6=7\n   \n   \n2−3+ −3\n   \n\n 3   6  M1 A1 Solves for  and substitutes into OQ\n3   2    \n=\n4\nOQ=−1\n4\n= 1\n4\n−1\n\n. and PQ or\n  −1 4    −1   7    2  \n= OP= 7 −1 .\n12\n12 \n \n−1\n \n 6  1 B1 FT OE. FT using their common\n    perpendicular.\nr= 1 −1 +k 1\n4    Must have r =\n   \n\n−1\n\n1\n8\n© Cambridge University Press & Assessment 2025 Page 16 of 19",
      "source_pages": [
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        16
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_12.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_12.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2025-on-12-q07",
      "question_id": "9231-2025-on-12-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−1 B1 Vertical asymptote.\nx(x+1)+1 M1 Oblique asymptote.\ny=\nx+1\ny=x A1\n3\n\n7(b) dy x2 +2x M1 dy\n= =0 Sets =0.\ndx (x+1)2 dx\n(0,1), (−2,−3) A1 A1 1 mark for each correct coordinate.\nSCB1 if just the correct x values listed\n3\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n7(c) B1FT Axes and asymptotes labelled.\nB1 Left branch correct with all approaches\nto the asymptotes correct.\nB1 Right branch correct with a smooth\nminimum point.\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n7(d) B1FT Symmetrical about x = 0 and follow\nthrough from part (c) right branch\nB1 Correct shape of a curve and smooth at\n(0,1) from a correct graph in part (c)\n2\n\n7(e) x2 +x+1 M1 Finds critical point.\n=3\nx+1\nx2 −2x−2=0\nx=1+ 3 A1 For selecting just the positive value of\nx.\n−1− 3x1+ 3 A1\n3\n© Cambridge University Press & Assessment 2025 Page 19 of 19",
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    },
    {
      "id": "9231-2025-on-13-q01",
      "question_id": "9231-2025-on-13-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "1(a) 2n2(n+1)2 +2n(n+1)(2n+1)+2n(n+1)+3n M1 A1 Substitutes correct formulae from MF19.\n=2n4 +8n3+10n2 +7n A1 Simplifies.\n3\n\n1(b) 1 1 (r+1)4 −r4 r4 +4r3+6r2 +4r+1−r4 4r3+6r2 +4r+1 M1 A1 Puts over a common denominator and\n− = = = expands, AG.\nr4 (r+1)4 r4(r+1)4 r4(r+1)4 r4(r+1)4\nn 4r3 +6r2 +4r+1 n  1 1  M1 A1 Shows three complete terms, including\n =  −  first and last.\nr4(r+1)4 r4 (r+1)4\n\nr=1 r=1\n1 1 1 1 1\n=1− + − + + −\n24 24 34 n4 (n+1)4\n1 A1\n=1−\n(n+1)4\n5\n\n1(c) 1 B1FT FT from their answer to part (b).\n1\nQuestion Answer Marks Guidance",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_13.pdf?download=true",
      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2025-on-13-q02",
      "question_id": "9231-2025-on-13-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "2(a) Shear, [x-axis fixed] B1\nWith (0,1) mapped to (3,1 ) . B1 Or other suitable example/description.\n2\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 17\n\n2(b) Shear, [y-axis fixed], B1\nwith (1,0) mapped to ( 1,3) . B1 Or other suitable example/description.\n2\n2\n\n2(c) 13 3 M1 Finds AB or uses product of\nAB= 4 2  or detAB=detAdetB determinants. Or suitable reasoning that\n3 1\n 2  area does not change.\ndetAB=1 A1\n2\n\n2(d) 13 3x 13x+ 3 y B1 x X\n 4 2  = 4 2  Transforms   to   .\n 3 2 1 y   3 2 x+ y   y Y \n3x+mx=m (13x+3mx ) M1A1 Uses Y =mX .\n2 4 2\n3+m=13m+3m2m2+3m−1=0 A1\n2 4 2 2\ny=−2x and y= 1 x. A1 If only values of m are given, A1 A0.\n2\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2025-on-13-q03",
      "question_id": "9231-2025-on-13-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 d B1 Checks base case using product rule.\n(xcosx)=cosx−xsinx=(−1)1(xsinx−(2(1)−1)cosx)\ndx\nd2k−1 B1 States inductive hypothesis.\nAssume true for n=k, so (xcosx)=(−1)k(xsinx−(2k−1)cosx)\ndx2k−1\nd2k M1 A1 Differentiates once.\nThen (xcosx)=(−1)k(xcosx+2ksinx)\ndx2k\nd2k+1 M1 A1 Differentiates again.\n(xcosx)=(−1)k(−xsinx+cosx+2kcosx)\ndx2k+1\n=(−1)k+1( xsinx−(2k+1)cosx )\nSo, it is also true for n=k+1. Hence, [by induction], A1 States conclusion.\nd2n−1\n(xcosx)=(−1)n(xsinx−(2n−1)cosx)\nis true for all positive integers.\ndx2n−1\n7\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_13.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-13-q04",
      "question_id": "9231-2025-on-13-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) y=2x+1x= 1(y−1) B1 Correct substitution used in equation.\n2\n1 (y−1)4 +1(y−1)3 +1(y−1)2 +1(y−1)+1=0 M1 A1 Expands.\n16 8 4 2\n1 ( y4 −4y3+6y2 −4y+1 ) +1 ( y3−3y2 +3y−1 ) +1 ( y2 −2y+1 ) +1(y−1)+1=0\n16 8 4 2\n1 y4−1y3+1 y2+1y+11=0y4−2y3+4y2+2y+11=0 A1 Simplifies, AG.\n16 8 4 8 16\n4\n© Cambridge University Press & Assessment 2025 Page 11 of 17\n\n4(b) S =22 −2(4) M1 Uses\n2\n(+++)2 =2 +2 +2 +2\n+2(+++++)\nS =−4 A1\n2\n2\n\n4(c) S −2S +4S +2S +44=0 M1 Sums over roots.\n4 3 2 1\nS =−76 A1\n4\n2\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_13.pdf",
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    },
    {
      "id": "9231-2025-on-13-q05",
      "question_id": "9231-2025-on-13-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) i j k −5  5  M1 A1 Finds perpendicular to .\n1\n   \n0 1 2 = 2 ~ −2\n   \n   \n1 3 1 −1  1 \n5(−3)−2(−1)+(−1)=−14 M1 Uses point on .\n1\n5x−2y+z=−14 A1\n4\n© Cambridge University Press & Assessment 2025 Page 12 of 17\n\n5(b) −2 −5 M1 A1 Uses correct formula for perpendicular\n1     distance from point to .\n −1    2  1\n52 +22 +12    \n 1  −1\n7 A1 Accept 1.28.\n30\n3\n\n5(c) 0 B1  0 \n   \nStates point common to both planes e.g. 7 . Or 0 .\n   \n   \n0 −14\ni j k  0  0 M1 A1 Finds direction of line.\n   \n5 −2 1 = 8  1\n   \n   \n3 2 −1 16 2\n0 0 A1 OE\n   \nr= 7 + 1\n   \n   \n0 2\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_13.pdf?download=true",
      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2025-on-13-q06",
      "question_id": "9231-2025-on-13-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "6(a) B1* Initial line drawn, pole clear and one\nloop in the first quadrant.\nDB1 Symmetrical with correct shape at\nextremities.\n= 1π. B1 States the equation of the line of\n6\nsymmetry.\n3\n\n6(b) 1 1 3 π sin23d M1 Uses 1\n2\nr2d with correct limits.\n2\n0\n=1 1 3 π 1−cos6d=1−1sin6 1 3 π M1 A1 Applies double angle formula correctly\n4 0 4  6  0 and integrates.\n= 1 π A1 CWO\n12\n4\n\n6(c) y=3sin2−4sin4 B1 Uses y=rsin.\n6sincos−16sin3cos=0 M1* Differentiates correct function and sets\nequal to 0. Might see completing the\nsquare.\nsincos06−16sin2=0 A1\nsin2=3 DM1 Solves quadratic in sin.\n8\ny= 9 . A1\n16\n5\n© Cambridge University Press & Assessment 2025 Page 14 of 17\n\n6(d) r4 =rsin ( 3r2 −4r2sin2 ) M1\n3y y3 Uses y=rsin\nr = −4\nr r3\n( x2 + y2)2 = y ( 3x2 −y2) M1 Uses r2 =x2 + y2\nA1 Rearranges to a form with no fractions.\n3\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_13.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_13.pdf?download=true",
      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2025-on-13-q07",
      "question_id": "9231-2025-on-13-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "7(a) x=−3−1 5, x=−3+1 5 B1 Vertical asymptotes.\n2 2 2 2\ny=0 B1 Horizontal asymptote.\n2\n\n7(b) dy (x2 +3x+1)(1)−(x+2)(2x+3) M1 dy\n= Finds .\ndx ( x2 +3x+1 )2 dx\nx2 +4x+5=0 M1 Sets equal to 0 and forms quadratic\nequation.\nDiscriminant is 42 −20=−40 so no real roots. A1 AG\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(c) y B1 Axes and asymptotes labelled.\nB1 Correct shape and position.\nB1 States coordinates of intersections with\naxes, (0,2)and (−2,0), may be seen on\ndiagram.\nx\n3\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(d) y B1FT FT from sketch in part 7(c)\nB1 From a correct graph and with correct\nshape at (−2,0).\nx\n2\n\n7(e) x+2 x+2 M2 Finds critical points, award M1 for each\n=2 or =−2\nx2 +3x+1 x2 +3x+1 case.\n−2x2 −5x=0 or 2x2 +7x+4=0\nx=−5, x=0 or x=−7−1 17, x=−7+1 17 A1\n2 4 4 4 4\n−7−1 17x−5, −7+1 17x0 , x−31 5 A1 −7−1 17x−3−1 5,\n4 4 2 4 4 2 2 4 4 2 2\n−3−1 5x−5,\n2 2 2\n−7+1 17x−3+1 5,\n4 4 2 2\n−3+1 5x0.\n2 2\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
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    },
    {
      "id": "9231-2025-on-14-q01",
      "question_id": "9231-2025-on-14-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 1,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1\nLHS=\n(6)1\n= 6 RHS=1+1 = 6\nB1 Checks base case.\n5 5 5 5\nAssume that\n(6)k\n 1+1k for some positive integer k.\nB1 States inductive hypothesis.\n5 5\nThen (6)k+1  (6)( 1+1k ) M1 Moving to the next term and using the inductive\n5 5 5 hypothesis.\n= 6+ 6 k  6+1k =1+ 1(k+1) A1 Convincing algebra to reach the required form.\n5 25 5 5 5\nHence, by induction,\n(6)n\n 1+1nis true for every positive integer n.\nA1 Must follow M1 A1.\n5 5\n5\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_14.pdf?download=true",
      "html": "9231-topic-17-proof-by-induction/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-14-q02",
      "question_id": "9231-2025-on-14-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) =−d B1 SOI\n1 M1 1\n++ =−b Uses ++=−b and replaces  with . OE\n \n1 A1 AG.\n+ =d−b\n\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 20\n\n2(b) d M1 Uses ++=c.\n−d+1− =c\n\n1 1−c A1 AG.\n+ =\n d\n2\n\n2(c)(i) 1+3 M1 Equates expressions given in (a) and (b), with correct\nd−3= d2 −3d−4=0\nuse of coefficients.\nd\nd =4 A1 d =4only for final answer.\n2\n\n2(c)(ii) 2 +2 +2 =b2 −2c M1 2 +2 +2 =(++)2 −2(++) with\ncorrect use of coefficients.\n15 A1\n2\n© Cambridge University Press & Assessment 2025 Page 10 of 20",
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_14.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_14.pdf?download=true",
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    },
    {
      "id": "9231-2025-on-14-q03",
      "question_id": "9231-2025-on-14-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 3,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) n M1 A1 Expands and substitutes formulae.\n( r3+6r2 +11r+6 ) = 1n2(n+1)2 +n(n+1)(2n+1)+11n(n+1)+6n\n4 2\nr=1\n= 1n ( n3+10n2 +35n+50 ) A1\n4\n3\n\n3(b) 2 1 2 1 M1 A1 Finds partial fractions.\n= − +\n(r+1)(r+2)(r+3) r+1 r+2 r+3\nn 2 M1 A1 Shows enough complete terms for cancellation to be\n = f(1)−2f(2)+ f(3) clear and any terms that form part of the final answer.\n(r+1)(r+2)(r+3)\nr=1 1\n+f(2)−2f(3)+ f(4) f(r)= .\nr+1\n+f(3)−2f(4)+ f(5)\n\n+f(n−1)−2f(n)+ f(n+1)\n+f(n)−2f(n+1)+ f(n+2)\n1 1 1 A1 1 1\n= − + Accept −\n6 n+2 n+3 6 (n+2)(n+3)\n5\n\n3(c) 1 B1 FT their result of similar form\n6\n1\n© Cambridge University Press & Assessment 2025 Page 11 of 20",
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    },
    {
      "id": "9231-2025-on-14-q04",
      "question_id": "9231-2025-on-14-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 4,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) B1 Finds direction vectors of two lines in the plane.\nAB=−i−3j−3k AC=−i−4j−6k\ni j k  6  M1 A1 Finds normal to the plane ABC.\n \n−1 −3 −3 = −3\n \n \n−1 −4 −6  1 \n6(3)−3(5)+(5)=86x−3y+z=8 M1 A1 Substitutes point. CAO\n5\n\n4(b)  1   2  −1 B1\n\n     \nCD= −1 − 1 = −2\n     \n     \n 3  −1  4 \ni j k −18 M1 A1 Find common perpendicular.\n \n−1 −3 −3 = 7\n \n \n−1 −2 4  −1 \n−1 −18 M1 A1 Uses formula for shortest distance.\n1     4\n−4  7 = =0.207\n   \n374 374\n   \n−6  −1 \n5\n© Cambridge University Press & Assessment 2025 Page 12 of 20",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_14.pdf",
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      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2025-on-14-q05",
      "question_id": "9231-2025-on-14-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 5,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "5(a)(i)\ncos\n(1π−1π )\n=cos\n(1π )\ncos\n(1π )\n+sin\n(1π )\nsin\n(1π ) B1 Uses cos(A−B)=cosAcosB+sinAsinB ,\n4 6 4 6 4 6\nEvery step must be seen because AG.\n( )\n= 1 2 3+1 2 = 1 6+ 2\n4 4 4\n1\n\n5(a)(ii) sin (1π−1π ) =sin (1π ) cos (1π ) −cos (1π ) sin (1π ) B1 Uses sin(A−B)=sinAcosB−cosAsinB ,\n4 6 4 6 4 6\nEvery step must be seen because AG.\n( )\n= 1 2 3−1 2= 1 6− 2\n4 4 4\n1\n\n5(b) Reflection and rotation. B1\nCorrect order. B1 Reflection then rotation.\nReflection in the x-axis. B1 Accept reflection in y=0.\nA rotation, 1 π [anticlockwise] about the origin. B1 Accept 15°.\n12\n4\n© Cambridge University Press & Assessment 2025 Page 13 of 20\n\n5(c) 1 ( 6+ 2 ) 1 ( 2− 6 ) −1 1 ( 6+ 2 ) 1 ( 6− 2 ) B1 Both correct inverses required.\n4 4  4 4 \n= ,\n 1 ( 6− 2 ) 1 ( 6+ 2 )   1 ( 2− 6 ) 1 ( 6+ 2 )   cos ( π ) sin ( π )\n4 4  4 4   12 12 \n−sin ( π ) cos ( π )\n1 0  −1 1 0   12 12 \n  = \n0 −1 0 −1\n1 ( 6+ 2 ) 1 ( 6− 2 ) B1 Correct order of their inverses.\nM−1 =   1 0   4 4  1 0  cos ( π ) sin ( π )\n0 −1  1 ( 2− 6 ) 1 ( 6+ 2 )  M−1 =  12 12 \n4 4  0 −1 \n\n−sin (\n1\nπ\n2\n) cos (\n1\nπ\n2\n)\n\n1 0  6+ 2 6− 2\nOr M−1 = 4  \n 0 −1 4    2− 6 6+ 2  \n2\n© Cambridge University Press & Assessment 2025 Page 14 of 20\n\n5(d) cos ( π ) sin ( π )  B1\nM= 12 12 \nsin ( π ) −cos ( π )\n 12 12 \ncos ( π) sin ( π) x cos ( π) x+sin ( π) y B1 FT their M\n 12 12   = 12 12  x X\n  sin ( 1 π 2 ) −cos ( 1 π 2 ) y   sin ( 1 π 2 ) x−cos ( 1 π 2 ) y  Transforms  y   to  Y  \nsin\n(π)\nx−mcos\n(π)\nx=m\n(\ncos\n(π)\nx+sin\n(π)\nmx\n) M1 Uses y=mx and Y =mX .\n12 12 12 12\nsin (π) m2 +2mcos (π) −sin (π) =0 A1 AG.\n12 12 12 ( ) ( ) ( )\n6− 2 m2 +2 6+ 2 m− 6− 2 =0\n−2cos( π ) 4cos2( π )+4sin2( π ) M1 Applies quadratic formula andcos2(π) +sin2(π) =1.\nm= 12 12 12 12 12\n2sin( π ) OE.\n12 ( ) ( )\nMay see m=− 2+ 3  6+ 2 .\nm=−cot (π) cosec (π) A1 Or a=−1, b=1\n12 12\n6\n© Cambridge University Press & Assessment 2025 Page 15 of 20",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_14.pdf",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2025-on-14-q06",
      "question_id": "9231-2025-on-14-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 6,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "6(a) B1* Initial line drawn and (1,0) joined to O by a single arc\nlying in both of the first two quadrants and no others.\nDB1 First B1 awarded and also correct shape at extremities\nand r decreasing.\n2\n\n6(b) 1 π cos2 1d M1 Uses 1r2d with correct limits.\n2 2 2\n0\n= 1 π cos+1d= 1sin+π M1 A1 Applies given identity and integrates.\n4 0 4 0\n= 1π A1\n4\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 20\n\n6(c) y=cos1sin B1 Uses y=rsin.\n2\ncos1cos−1sinsin1=0 M1* Differentiates.\n2 2 2\nA1 Correct and equal to zero.\ncos1 ( 2cos2 1−1 ) −sin2 1cos1=0 DM1\n2 2 2 2\ncos102cos21−1−sin21=0\n2 2 2\n2cos2 1−1− ( 1−cos2 1 ) =03cos2 1−2=0 Forms an equation in one trig ratio using correct\n2 2 2 formulae. (May be in terms of sin or tan)\n2 A1 1\ncos2 1= Solving equation to find one of sin2 1=\n2 3 2 3\n1 2 2 1\ntan2 1= sin= cos=\n2 2 3 3\ny=2cos2 1sin1=2 (2) 1−2 = 4 1 A1 Accept 4 3.\n2 2 3 3 3 3 9\n6\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2025-on-14-q07",
      "question_id": "9231-2025-on-14-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 1,
      "variant": "14",
      "question_number": 7,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) x=1.8 B1\n(10x−18)(x+ 7 )−27 27 M1 Must be an attempt to find the constant term.\ny= 10 5 =x+ 7 − 5\n10x−18 10 10x−18\ny=x+0.7 A1\n3\n© Cambridge University Press & Assessment 2025 Page 17 of 20\n\n7(b) dy 100x2 −360x+378 M1 Differentiates.\n=\ndx (10x−18)2\nA1 Forms quadratic equation if using discriminant\n \ndy 54\n50x2 −180x+189=0 or =1+  dy\n dx (10x−18)2  method. (Or writes in a form to show it is positive)\n  dx\n1802 −4(50)(189)=−54000 (or y'0 )  No stationary points M1 A1 Consideration of discriminant or sign of y' with\ncorrect conclusion.\n4\n© Cambridge University Press & Assessment 2025 Page 18 of 20\n\n7(c) B1 Axes and asymptotes labelled.\nB1 Branches correct.\nx\n(0,1), (−0.9,0), (2,0) B1 May be seen on their diagram.\n3\n© Cambridge University Press & Assessment 2025 Page 19 of 20\n\n7(d) y B1 FT from their attempt in (c).\nB1 Everything correct with sharp changes of direction at\nintersections with xaxis.\nx\n2\n\n7(e) 10x2 −11x−18 −29 4441\n =−4x= \n10x−18 20\n \n10x2 −11x−18 M1 Forms an equation to find required critical values for\n=4 x.\n10x−18\n10x2 −51x+54=0 A1\np= 3, q=18 A1 x= 3,x=18\n2 5 2 5\nA1 p,q clearly identified in correct order.\n4\n© Cambridge University Press & Assessment 2025 Page 20 of 20",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_14.pdf",
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      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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        "../answer-assets/9231_w25_ms_14-p20.png"
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    },
    {
      "id": "9231-2025-on-21-q01",
      "question_id": "9231-2025-on-21-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 1 −1 1 M1 A1 Evaluates determinant.\nk 3 1 3 1 k\n1 k 3 = + + =k2 −7\n2 k 1 k 1 2\n1 2 k\nk = 7 A1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_21-p09.png"
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    },
    {
      "id": "9231-2025-on-21-q02",
      "question_id": "9231-2025-on-21-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) dy dy dt 1+sinht M1 A1 Applies chain rule.\n=  =\ndx dt dx cosht\n2\n\n2(b)\nd\nd\nt\n\n \n1+\nco\ns\ns\ni\nh\nnh\nt\nt\n \n= cosht(cosht\nc\n)\no\n−\nsh\n(1\n2\n+\nt\nsinht)sinht = 1\nc\n−\no\ns\ns\ni\nh\nn\n2\nh\nt\nt M1 A1 Finds\nd\nd\nt\n \n\nd\nd\ny\nx\n \n\n.\nd2y d dy dt 1−sinht M1 A1 Applies chain rule, AG.\n=\n \n =\ndx2 dtdx dx cosh3t\n4\n\n2(c) x2 M1 A1 Finds Maclaurin’s series. M1 for y(0), y'(0) and\ny= y(0)+xy'(0)+ y''(0)=1+x+ 1x2\n2! 2 y''(0) evaluated. A0 if 2! not simplified to 2.\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
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    },
    {
      "id": "9231-2025-on-21-q03",
      "question_id": "9231-2025-on-21-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) d (\nx\n( 1+x2)n ) =2nx2( 1+x2)n−1\n+\n( 1+x2)n M1 A1 Uses the product rule to differentiate.\ndx\n=2n ( 1+x2 −1 )( 1+x2)n−1 + ( 1+x2)n *M1 Uses x2 =x2 +1−1.\n1 DM1 Integrates both sides using the limits given.\n\nx\n( 1+x2)n\n=2nI −2nI +I\n  n n−1 n\n0\n2n =(2n+1)I −2nI (2n+1)I =2n +2nI A1 Substitutes limits and rearranges. AG.\nn n−1 n n−1\n5\n\n3(b) I =tan−1x 1 = 1π B1\n−1   0 4\n−I = 1 −2I M1 A1 Applies reduction formula with n=−1.\n−1 2 −2\n−1π= 1 −2I I = 1 +1π A1\n4 2 −2 −2 4 8\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-on-21-q04",
      "question_id": "9231-2025-on-21-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4 5m2+2m+1=0m=−12i M1 Auxiliary equation. Correct type of roots for M1.\n5 5\ny=e−1 5 x( Acos2x+Bsin2x ) A1 Complementary function. Allow ‘ y=’ missing.\n5 5\ny= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives.\np=1 4p+q=5 10p+2q+r=3 M1 Substitutes and equates coefficients.\nq=1 r=−9 A1\ny=e−1 5 x( Acos2x+Bsin2x ) +x2+x−9 A1 FT General solution. FT on CF. Must have ‘ y=’.\n5 5\ny'=e−1 5 x( −2Asin2x+2Bcos2x ) −1e−1 5 x( Acos2x+Bsin2x ) +2x+1 *M1 Differentiates using the product rule, CF must be of the\n5 5 5 5 5 5 5 correct form.\nDM1 A1 Uses initial conditions.\nA−9=0 2B−1 A+1=0 A=9,B=2\n5 5\ny=e−1 5 x( 9cos2x+2sin2x ) +x2 +x−9 A1 Must have ‘ y=’.\n5 5\n10\n© Cambridge University Press & Assessment 2025 Page 11 of 19",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-21-q05",
      "question_id": "9231-2025-on-21-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a)   1( 1x3+x ) dx (1) ( 1(1)3 +(1) ) +(1) ( 1(2)3 +(2) ) + +(1) ( 1(n)3 +(n) ) M1 A1 Forms the sum of the areas of the rectangles. M1 for\n  0 3   n 3 n n n 3 n n n 3 n n correct width and at least three complete terms\nincluding first and last.\nn n n2(n+1)2 n(n+1) M1 A1 Applies formulae from MF19.\n= 1 r3+ 1 r= +\n3n4 n2 12n4 2n2\nr=1 r=1\n= 1 (1+1)2 +1(1+1)= 1 (1+1)(7+ 1) A1 AG.\n12 n 2 n 12 n n\n5\n\n5(b)  1( 1x3+x ) dx (1) ( 1(1)3 + (1) ) + (1) ( 1(2)3 + (2) ) + + (1) ( 1(n−1)3 + (n−1) ) M1 A1 Forms the sum of the areas of appropriate rectangles.\n3 n 3 n n n 3 n n n 3 n n\n0\nn−1 n−1 (n−1)2n2 (n−1)n M1 4\n= 3 1 n4 r3+ n 1 2 r= 12n4 + 2n2 Applies formulae from MF19. Accept L n =U n − 3n .\nr=1 r=1\n= 1 (1−1)2 +1(1−1)= 1 (1−1)(7−1) A1 At least one line of simplification after subbing in\n12 n 2 n 12 n n formula (powers of n simplified).\n4\n\n5(c) 1 16 4 M1 c\nU n −L n = 12   n   = 3n Simplifies U n −L n to n .\n4 A1 AG.\n→0 as\nn→\n3n\n2\n© Cambridge University Press & Assessment 2025 Page 12 of 19",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-on-21-q06",
      "question_id": "9231-2025-on-21-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) a, 2a, −3a B1\n1\n\n6(b)(i) a2 3a −2a−1 B1\n \nP2 = 0 4a2 a \n \n 0 0 9a2 \n \n1\n\n6(b)(ii) P3−7a2P+6a3I=0 B1 States that P satisfies its characteristic equation.\n6a3P−1 =7a2I−P2 M1 Multiplies through by P−1.\na2 3a −2a−1 6a2 −3a 2a+1 A1\n  1  \nP2 = 0 4a2 a P−1 =  0 3a2 −a \n \n6a3\n \n 0 0 9a2   0 0 −2a2 \n   \n3\n© Cambridge University Press & Assessment 2025 Page 13 of 19\n\n6(c) 1 0 0 B1\n \nD= 0 2 0\n \n \n0 0 3\n1 0 0 M1 Applies A=PDP−1.\n \nA=P 0 2 0 P−1\n \n \n0 0 3\na 2 3  6a2 −3a 2a+1 M1 A1 Multiplies two adjacent matrices. For M1, P must\n1     correct.\n= 0 4a −3  0 3a2 −a \n6a3\n\n\n \n0 0 −9a  0 0 −2a2 \n \n1 1 −4a+1 A1 OE.\n\n2a 6a2\n\n0 2 1 \n3a\n \n0 0 3 \n \n5\n© Cambridge University Press & Assessment 2025 Page 14 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2025-on-21-q07",
      "question_id": "9231-2025-on-21-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) d d x ( tanh−1x ) = 1 2   1 1 + −x x    (1−x)(1 (1 )− − ( x 1 ) + 2 x)(−1) M1 A1 Applies tanh−1x= 1 2 ln    1 1 + −x x   .\n1−x 2 1 A1 AG.\n= 1 =\n \n2 1+x(1−x)2 1−x2\nAlternative method for question 7(a)\ndy M1 A1 Implicit differentiation.\ntanhy=xsech2 y =1\ndx\ndy 1 1 A1 AG.\n= =\ndx 1−tanh2 y 1−x2\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n7(b) dy − y =xtanh−1x B1 Divides through by x.\ndx x\ne\n−x−1d\n=e−lnx =x−1\nM1 A1 Finds integrating factor.\nd ( x−1y ) =tanh−1x M1 Correct form on LHS and RHS.\ndx\nx−1y=xtanh−1x−   x dx *M1 A1 Integrates correct function on RHS.\n1−x2\nx−1y=xtanh−1x+1ln ( 1−x2) +C A1 ‘+C’ not required.\n2\n0=1tanh−11+1ln3+CC=−1ln3−1ln3 DM1 Substitutes initial conditions.\n2 2 2 4 4 2 4\nx−1y=xtanh−1x+1ln ( 1−x2) −1ln27 A1 Accept\n2 4 16 x−1y=xtanh−1x+1ln ( 1−x2) −1tanh−11−1ln3.\n2 2 2 2 4\nAccept y=x ( xtanh−1x+1ln ( 1−x2) −1ln27 ) .\n2 4 16\n9\n© Cambridge University Press & Assessment 2025 Page 16 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2025-on-21-q08",
      "question_id": "9231-2025-on-21-q08",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "8(a) zn+1−z B1\nz−1\n1\n\n8(b) (1)n+1\ncos(n+1)−1cos+i\n( (1)n+1\nsin(n+1)−1sin\n) B1\nzn+1−z\n2 2 2 2\n=\nz−1 1cos−1+i1sin\n2 2\n(1)n+1\ncos(n+1)−1cos+i\n( (1)n+1\nsin(n+1)−1sin\n) *M1 Multiplies numerator and denominator by complex\n2 2 2 2\n1cos−1−i1sin\nconjugate of z–1.\n 2 2\n1cos−1+i1sin 1cos−1−i1sin\n2 2 2 2\n−1sin ( (1)n+1 cos(n+1)−1cos ) +(1cos−1) ( (1)n+1 sin(n+1)−1sin ) DM1 A1 Takes imaginary part.\n2 2 2 2 2 2\n5−cos\n4\n−(1)n+2 sincos(n+1)+(1)n+2 cossin(n+1)−(1)n+1 sin(n+1)+ 1sin M1 A1 Uses sin(A−B)=sinAcosB−cosAsinB in\n2 2 2 2\nnumerator, AG.\n5 −cos\n4\n(1)n+2 sinn−(1)n+1\nsin(n+1)+ 1sin\n= 2 2 2\n5 −cos\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n8(b) Alternative method for question 8(b)\nz=1ei B1\n2\nzn+1−z\n(1)n+1 ei(n+1)−(1)ei M1 A1\n= 2 2\nz−1 1ei−1\n2\n(1)n+1 ei(n+1)−(1)ei 1e−i−1 M1 Multiplies numerator and denominator by complex\n2 2  2 conjugate.\n1ei−1 1e−i−1\n2 2\n(1)n+2 ein−(1)n+1 ei(n+1)+ 1ei−1  M1 A1 Takes imaginary part\nIm 2 2 2 4\n 5 −cos \n 4 \n(1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin\n= 2 2 2\n5 −cos\n4\n6\n\n8(c)  n (1)m mcosm= d   (1 2 )n+2 sinn−(1 2 )n+1 sin(n+1)+1 2 sin   M1 Differentiates the result from (b).\n2 d 5−cos \nm=1  4 \n(5 −cos) ( (1)n+2 ncosn−(1)n+1 (n+1)cos(n+1)+ 1cos ) M1 A1\n4 2 2 2\n=\n(5 −cos)2\n4\n( (1)n+2 sinn−(1)n+1\nsin(n+1)+ 1sin\n)\nsin\n2 2 2\n−\n(5 −cos)2\n4\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n8(d)   (1)m mcosm= (5 4 −cos)(1 2 cos)−(1 2 sin)sin M1 n(1 2 )n →0 as n→.\n2 (5 −cos)2\nm=1 4\n5cos−1 A1\n= 8 2\n(5 −cos)2\n4\n2\n© Cambridge University Press & Assessment 2025 Page 19 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_21.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q01",
      "question_id": "9231-2025-on-22-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 s= π e 8 3 +16e 8 3 d= 5 π e3 4d M1 A1 Forms length of C correctly.\n9 3\n0 0\ns= 5e 4 3  π M1 Integrates an expression of the form aeb. One\n4  error in the a or b.\n0\n= 5 ( e 4 3 π −1 ) A1 Must be exact. Allow equivalent form but must\n4 have the e0evaluated to be one.\n4\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-22-q02",
      "question_id": "9231-2025-on-22-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 z4 =8−8i 3=16e−i1 3 π B1 Finds the modulus and argument of z4. Does\nnot have to be in exponential form.\nz =2e−i 1 1 2 π M1 A1 Finds one root. Allow for the method mark for\n1 \nthem to use their r and their .\n4\nz =2ei 1 5 2 π A1FT Finds another root. Follow through only on the\n2 modulus\nz =2e−i 1 7 2 π, z =2ei 1 1 2 1π A1FT Finds other two roots. Follow through only on\n3 4 the modulus. Withhold 1 answer mark for\nomission of i.\n5\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-22-q03",
      "question_id": "9231-2025-on-22-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) d ( x2y ) = x2y'+2xy B1 Differentiates x2y correctly.\ndx\nd ( (x+ y)3) =3(x+ y)2(1+ y') M1 Differentiates (x+ y)3 correctly.\ndx\ny'−4+3(1+ y')=0 y'= 1 A1 Some evidence of substituting (−1,2), into an\n4\nequation set to zero. AG.\n3\n\n3(b) x2y''+2xy'+2xy'+2y B1 B1 Differentiates x2y'+2xy.\n+3(x+ y)2 y''+6(1+ y')2(x+ y)=0 B1 B1 Differentiates 3(x+y)2(1+y').\n(−1)2y''+4(−1)(1)+2(2)+3y''+6(5)2=0 M1 Substitutes (−1,2) and y'= 1 into their\n4 4 4\nexpression with at least a y''in their expression\ny''=−99 A1\n32\n6\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q04",
      "question_id": "9231-2025-on-22-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) z+z−1 =2cos B1 SOI\n( z+z−1)6 = ( z6 +z−6) +6 ( z4 +z−4) +15 ( z2 +z−2) +20 M1 A1 Expands and groups.\n26cos6=2cos6+12cos4+30cos2+20 M1 Substitutes zn +z−n =2cosn.\ncos6= 1 cos6+ 3 cos4+15cos2+ 5 A1 OE. Fractions do not need to be simplified.\n32 16 32 16\n5\n\n4(b)  1 2 π cos62xdx= 1 π( 1 cos6+ 3 cos4+15cos2+ 5 )d M1 Applies (a). Must have the 1 and the correct\n0 2 0 32 16 32 16 2\nlimits for this mark.\nπ A1\n= 1\n\n1 sin6+ 3 sin4+15sin2+ 5\n\n2 192 64 64 16 0\n= 5 π A1\n32\n3\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q05",
      "question_id": "9231-2025-on-22-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) m2 +4m+3=0m=−1,−3 M1 Auxiliary equation. This mark could be implied\nby a correct complementary function.\n[y =]Ae−x +Be−3x A1 Complementary function. Condone this not\nbeing equal to y. x must be used and not another\nvariable.\ny= psinx+qcosx y'= pcosx−qsinx y''=−psinx−qcosx M1 A1 Particular integral and its derivatives.\n−p−4q+3p=0 −q+4p+3q=5 M1 Substitutes their particular integral and their\nderivatives of the correct form and equates\ncoefficients.\np =1 q = 1 A1\n2\ny= Ae−x +Be−3x +sinx+1cosx A1 This must be equal to y. x must be used and not\n2\nanother variable.\n7\n\n5(b) y sinx+ 1cosx= 5 ( 2 sinx+ 1 cosx ) M1 e−x →0and for setting up for Pythagoras and\n2 2 5 5 applying compound angle formula. Can be\nimplied.\n5sin(x+0.464...) A1 A1 A1 for R and A1 for .Condone 26.6. Allow\n2\n=tan−1 1. OE.\n2\nCondone decimals for R to at least 3 sf.\nNo method shown award 0 marks.\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q06",
      "question_id": "9231-2025-on-22-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) dx  1  coshu M1 A1 Applies substitution and sinh2u+1=cosh2u.\n= 1 2coshu\n\ndx= 1 2 du=1 21du\ndu 2  2x2 +1 2  sinh2u+1 2\n= 1 2sinh−1 ( x 2 )+C A1 sinh−1 ( x 2 ) =ln ( x 2+ 2x2 +1 ) Condone\n2\nC missing.\n3\n\n6(b) *M1 A1 Forms sum of the areas of the rectangles. Just\n \n1  1 + 1 + + 1  required the first line of working for the M1A1.\nn 212 +1 222 +1 2n2 +1 \n n2 n2 n2 \n1 n n n \n=  + + + \nn  212 +n2 222 +n2 2n2 +n2 \n1 DM1 Compares with integral.\n 1\n dx\n 2x2 +1\n0\n1 B1 Evaluates integral.\n  1 dx=1 2sinh−1 ( x 2 ) 1 = 1 2ln ( 2+ 3 )\n2  2\n 2x2 +1 0\n0\n1 n n n 1 ( ) A1 AG. Must see\n =  1 2ln 2+ 3\nn r=1 2r2 +n2 r=1 2r2 +n2 2 1  n n = n 1 or equivalent\nn 2r2 +n2 2r2 +n2\nr=1 r=1\nand limits on integral correct.\n5\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n6(c)   *M1 A1 Forms the sum of the areas of appropriate\n1 1 1 1  rectangles.\n+ + +\n \nn  202 +1 212 +1 2 (n−1)2 +1\n n2 n2 \n \n1 n n\n= 1+ + + \nn 212 +n2 2(n−1)2 +n2 \n \n1 DM1 Compares with integral.\n   1 dx=1 2sinh−1 ( x 2 ) 1 = 1 2ln ( 2+ 3 )\n2  2\n 2x2 +1 0\n0\n1 n 1 1 ( ) A1\n+ −  1 2ln 2+ 3\nn 2r2 +n2 3n2 2\nr=1\nn 1 ( ) 1( )\n  1 2ln 2+ 3 + 1 3−1\n2r2 +n2 2 n 3\nr=1\n4\n\n6(d) ( ) B1\n1 2ln 2+ 3\n2\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
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      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q07",
      "question_id": "9231-2025-on-22-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a) d ( 1 x 4−x2 +2sin−1(1 x) ) =−1 x2( 4−x2)−1 2 + 1 ( 4−x2)1 2 +2 ( 4−x2)−1 2 M1 A1 Applies product rule and formula from MF19\ndx 2 2 2 2 for the derivative of sin−1x.\n= 1\n( 4−x2)( 4−x2)−1\n2 + 1\n( 4−x2)1\n2 =\n( 4−x2)1\n2\nA1 As a minimum, working of\n2 2\n1\n( 4−x2)( 4−x2)−1\n2 + 1\n( 4−x2)1\n2 must be seen\n2 2\nbefore the answer. AG.\n3\n\n7(b) 1(2+x)−1dx M1 A1 Finds integrating factor. Allow a sign error in\ne2 = 2+x\nthe integrating factor for the method mark.\nd ( ) M1 A1 Correct form on LHS and simplifies RHS.\ny 2+x = 2+x 2−x = 4−x2\ndx\ny 2+x = 1 x 4−x2 +2sin−1(1 x)+C M1 A1 Integrates RHS.\n2 2\nM1 Substitutes initial conditions.\n1 3= 1 3+1π+C\n2 2 3\ny 2+x = 1 x 4−x2 +2sin−1(1 x)−1π A1 ISW once a correct answer is seen.\n2 2 3\n8\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2025-on-22-q08",
      "question_id": "9231-2025-on-22-q08",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "8(a) 1 −1 2k M1 Finds determinant equal and forms quadratic\nequation. Condone one error in the determinant\nk 1 2 =02k2 +3k+1=0\ncalculation.\n2 −1 1\n2k2+3k+1=0k =−1,−1 A1 A1 Each value is A1.\n2\n3\n\n8(b) x− y− z = 1, M1 A1 Derives contradiction clearly.\n1 x+z =3,\n−1 x+ y+2z =2, 2\n2 x+2z =2,\n2x− y+ z = 3,\nTriangular prism. B1 Accept diagram.\n3\n\n8(c) x− y−2z =1, M1 A1 Derives contradiction.\n−x+ y+2z =2, 1=−2\n2x− y+ z = 3,\nTwo parallel planes. Other plane not parallel. B1\nParallel planes not identical. B1 Accept diagram.\n4\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n8(d) 1− −1 −2  M1 Finds characteristic equation of A.\n \ndet −1 1− 2 =0\n \n \n2 −1 1−\n \n−3+32 −8=0 A1 Must be an equation. Could be implied by\nsubsequent working. Must be simplified.\nA3 =3A2 −8A M1 Substitutes matrix and makes A3 the subject.\nA4 =3A3−8A2 =3 ( 3A2 −8A ) −8A2 M1 Multiplies by A and substitutes for A3.\nA4 =A2 −24A A1\n5\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
        16,
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_22.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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        "../answer-assets/9231_w25_ms_22-p17.png"
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    },
    {
      "id": "9231-2025-on-23-q01",
      "question_id": "9231-2025-on-23-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 3,
      "status": "available",
      "reason": null,
      "text": "1 1 −1 1 M1 A1 Evaluates determinant.\nk 3 1 3 1 k\n1 k 3 = + + =k2 −7\n2 k 1 k 1 2\n1 2 k\nk = 7 A1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_23-p09.png"
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    },
    {
      "id": "9231-2025-on-23-q02",
      "question_id": "9231-2025-on-23-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) dy dy dt 1+sinht M1 A1 Applies chain rule.\n=  =\ndx dt dx cosht\n2\n\n2(b)\nd\nd\nt\n\n \n1+\nco\ns\ns\ni\nh\nnh\nt\nt\n \n= cosht(cosht\nc\n)\no\n−\nsh\n(1\n2\n+\nt\nsinht)sinht = 1\nc\n−\no\ns\ns\ni\nh\nn\n2\nh\nt\nt M1 A1 Finds\nd\nd\nt\n \n\nd\nd\ny\nx\n \n\n.\nd2y d dy dt 1−sinht M1 A1 Applies chain rule, AG.\n=\n \n =\ndx2 dtdx dx cosh3t\n4\n\n2(c) x2 M1 A1 Finds Maclaurin’s series. M1 for y(0), y'(0) and\ny= y(0)+xy'(0)+ y''(0)=1+x+ 1x2\n2! 2 y''(0) evaluated. A0 if 2! not simplified to 2.\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 19",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-23-differentiation/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_23-p09.png"
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    },
    {
      "id": "9231-2025-on-23-q03",
      "question_id": "9231-2025-on-23-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) d (\nx\n( 1+x2)n ) =2nx2( 1+x2)n−1\n+\n( 1+x2)n M1 A1 Uses the product rule to differentiate.\ndx\n=2n ( 1+x2 −1 )( 1+x2)n−1 + ( 1+x2)n *M1 Uses x2 =x2 +1−1.\n1 DM1 Integrates both sides using the limits given.\n\nx\n( 1+x2)n\n=2nI −2nI +I\n  n n−1 n\n0\n2n =(2n+1)I −2nI (2n+1)I =2n +2nI A1 Substitutes limits and rearranges. AG.\nn n−1 n n−1\n5\n\n3(b) I =tan−1x 1 = 1π B1\n−1   0 4\n−I = 1 −2I M1 A1 Applies reduction formula with n=−1.\n−1 2 −2\n−1π= 1 −2I I = 1 +1π A1\n4 2 −2 −2 4 8\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 19",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-23-q04",
      "question_id": "9231-2025-on-23-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4 5m2+2m+1=0m=−12i M1 Auxiliary equation. Correct type of roots for M1.\n5 5\ny=e−1 5 x( Acos2x+Bsin2x ) A1 Complementary function. Allow ‘ y=’ missing.\n5 5\ny= px2 +qx+r y'=2px+q y''=2p B1 Particular integral and its derivatives.\np=1 4p+q=5 10p+2q+r=3 M1 Substitutes and equates coefficients.\nq=1 r=−9 A1\ny=e−1 5 x( Acos2x+Bsin2x ) +x2+x−9 A1 FT General solution. FT on CF. Must have ‘ y=’.\n5 5\ny'=e−1 5 x( −2Asin2x+2Bcos2x ) −1e−1 5 x( Acos2x+Bsin2x ) +2x+1 *M1 Differentiates using the product rule, CF must be of the\n5 5 5 5 5 5 5 correct form.\nDM1 A1 Uses initial conditions.\nA−9=0 2B−1 A+1=0 A=9,B=2\n5 5\ny=e−1 5 x( 9cos2x+2sin2x ) +x2 +x−9 A1 Must have ‘ y=’.\n5 5\n10\n© Cambridge University Press & Assessment 2025 Page 11 of 19",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_23-p11.png"
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    },
    {
      "id": "9231-2025-on-23-q05",
      "question_id": "9231-2025-on-23-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a)   1( 1x3+x ) dx (1) ( 1(1)3 +(1) ) +(1) ( 1(2)3 +(2) ) + +(1) ( 1(n)3 +(n) ) M1 A1 Forms the sum of the areas of the rectangles. M1 for\n  0 3   n 3 n n n 3 n n n 3 n n correct width and at least three complete terms\nincluding first and last.\nn n n2(n+1)2 n(n+1) M1 A1 Applies formulae from MF19.\n= 1 r3+ 1 r= +\n3n4 n2 12n4 2n2\nr=1 r=1\n= 1 (1+1)2 +1(1+1)= 1 (1+1)(7+ 1) A1 AG.\n12 n 2 n 12 n n\n5\n\n5(b)  1( 1x3+x ) dx (1) ( 1(1)3 + (1) ) + (1) ( 1(2)3 + (2) ) + + (1) ( 1(n−1)3 + (n−1) ) M1 A1 Forms the sum of the areas of appropriate rectangles.\n3 n 3 n n n 3 n n n 3 n n\n0\nn−1 n−1 (n−1)2n2 (n−1)n M1 4\n= 3 1 n4 r3+ n 1 2 r= 12n4 + 2n2 Applies formulae from MF19. Accept L n =U n − 3n .\nr=1 r=1\n= 1 (1−1)2 +1(1−1)= 1 (1−1)(7−1) A1 At least one line of simplification after subbing in\n12 n 2 n 12 n n formula (powers of n simplified).\n4\n\n5(c) 1 16 4 M1 c\nU n −L n = 12   n   = 3n Simplifies U n −L n to n .\n4 A1 AG.\n→0 as\nn→\n3n\n2\n© Cambridge University Press & Assessment 2025 Page 12 of 19",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_23-p12.png"
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    },
    {
      "id": "9231-2025-on-23-q06",
      "question_id": "9231-2025-on-23-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) a, 2a, −3a B1\n1\n\n6(b)(i) a2 3a −2a−1 B1\n \nP2 = 0 4a2 a \n \n 0 0 9a2 \n \n1\n\n6(b)(ii) P3−7a2P+6a3I=0 B1 States that P satisfies its characteristic equation.\n6a3P−1 =7a2I−P2 M1 Multiplies through by P−1.\na2 3a −2a−1 6a2 −3a 2a+1 A1\n  1  \nP2 = 0 4a2 a P−1 =  0 3a2 −a \n \n6a3\n \n 0 0 9a2   0 0 −2a2 \n   \n3\n© Cambridge University Press & Assessment 2025 Page 13 of 19\n\n6(c) 1 0 0 B1\n \nD= 0 2 0\n \n \n0 0 3\n1 0 0 M1 Applies A=PDP−1.\n \nA=P 0 2 0 P−1\n \n \n0 0 3\na 2 3  6a2 −3a 2a+1 M1 A1 Multiplies two adjacent matrices. For M1, P must\n1     correct.\n= 0 4a −3  0 3a2 −a \n6a3\n\n\n \n0 0 −9a  0 0 −2a2 \n \n1 1 −4a+1 A1 OE.\n\n2a 6a2\n\n0 2 1 \n3a\n \n0 0 3 \n \n5\n© Cambridge University Press & Assessment 2025 Page 14 of 19",
      "source_pages": [
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        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2025-on-23-q07",
      "question_id": "9231-2025-on-23-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) d d x ( tanh−1x ) = 1 2   1 1 + −x x    (1−x)(1 (1 )− − ( x 1 ) + 2 x)(−1) M1 A1 Applies tanh−1x= 1 2 ln    1 1 + −x x   .\n1−x 2 1 A1 AG.\n= 1 =\n \n2 1+x(1−x)2 1−x2\nAlternative method for question 7(a)\ndy M1 A1 Implicit differentiation.\ntanhy=xsech2 y =1\ndx\ndy 1 1 A1 AG.\n= =\ndx 1−tanh2 y 1−x2\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 19\n\n7(b) dy − y =xtanh−1x B1 Divides through by x.\ndx x\ne\n−x−1d\n=e−lnx =x−1\nM1 A1 Finds integrating factor.\nd ( x−1y ) =tanh−1x M1 Correct form on LHS and RHS.\ndx\nx−1y=xtanh−1x−   x dx *M1 A1 Integrates correct function on RHS.\n1−x2\nx−1y=xtanh−1x+1ln ( 1−x2) +C A1 ‘+C’ not required.\n2\n0=1tanh−11+1ln3+CC=−1ln3−1ln3 DM1 Substitutes initial conditions.\n2 2 2 4 4 2 4\nx−1y=xtanh−1x+1ln ( 1−x2) −1ln27 A1 Accept\n2 4 16 x−1y=xtanh−1x+1ln ( 1−x2) −1tanh−11−1ln3.\n2 2 2 2 4\nAccept y=x ( xtanh−1x+1ln ( 1−x2) −1ln27 ) .\n2 4 16\n9\n© Cambridge University Press & Assessment 2025 Page 16 of 19",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2025-on-23-q08",
      "question_id": "9231-2025-on-23-q08",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "8(a) zn+1−z B1\nz−1\n1\n\n8(b) (1)n+1\ncos(n+1)−1cos+i\n( (1)n+1\nsin(n+1)−1sin\n) B1\nzn+1−z\n2 2 2 2\n=\nz−1 1cos−1+i1sin\n2 2\n(1)n+1\ncos(n+1)−1cos+i\n( (1)n+1\nsin(n+1)−1sin\n) *M1 Multiplies numerator and denominator by complex\n2 2 2 2\n1cos−1−i1sin\nconjugate of z–1.\n 2 2\n1cos−1+i1sin 1cos−1−i1sin\n2 2 2 2\n−1sin ( (1)n+1 cos(n+1)−1cos ) +(1cos−1) ( (1)n+1 sin(n+1)−1sin ) DM1 A1 Takes imaginary part.\n2 2 2 2 2 2\n5−cos\n4\n−(1)n+2 sincos(n+1)+(1)n+2 cossin(n+1)−(1)n+1 sin(n+1)+ 1sin M1 A1 Uses sin(A−B)=sinAcosB−cosAsinB in\n2 2 2 2\nnumerator, AG.\n5 −cos\n4\n(1)n+2 sinn−(1)n+1\nsin(n+1)+ 1sin\n= 2 2 2\n5 −cos\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 19\n\n8(b) Alternative method for question 8(b)\nz=1ei B1\n2\nzn+1−z\n(1)n+1 ei(n+1)−(1)ei M1 A1\n= 2 2\nz−1 1ei−1\n2\n(1)n+1 ei(n+1)−(1)ei 1e−i−1 M1 Multiplies numerator and denominator by complex\n2 2  2 conjugate.\n1ei−1 1e−i−1\n2 2\n(1)n+2 ein−(1)n+1 ei(n+1)+ 1ei−1  M1 A1 Takes imaginary part\nIm 2 2 2 4\n 5 −cos \n 4 \n(1)n+2 sinn−(1)n+1 sin(n+1)+ 1sin\n= 2 2 2\n5 −cos\n4\n6\n\n8(c)  n (1)m mcosm= d   (1 2 )n+2 sinn−(1 2 )n+1 sin(n+1)+1 2 sin   M1 Differentiates the result from (b).\n2 d 5−cos \nm=1  4 \n(5 −cos) ( (1)n+2 ncosn−(1)n+1 (n+1)cos(n+1)+ 1cos ) M1 A1\n4 2 2 2\n=\n(5 −cos)2\n4\n( (1)n+2 sinn−(1)n+1\nsin(n+1)+ 1sin\n)\nsin\n2 2 2\n−\n(5 −cos)2\n4\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n8(d)   (1)m mcosm= (5 4 −cos)(1 2 cos)−(1 2 sin)sin M1 n(1 2 )n →0 as n→.\n2 (5 −cos)2\nm=1 4\n5cos−1 A1\n= 8 2\n(5 −cos)2\n4\n2\n© Cambridge University Press & Assessment 2025 Page 19 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_23.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q01",
      "question_id": "9231-2025-on-24-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 1,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1(a) 1 1 M1 Integrates by parts. Must see first term\nI =xncoshx −n xn−1coshxdx\nn   0 0 xncoshx.\n=xncoshx 1 −n ( xn−1sinhx 1 −(n−1) 1 xn−2sinhxdx ) M1 Integrates by parts again.\n   \n0 0 0\nI =cosh1−n ( sinh1−(n−1)I ) =cosh1−nsinh1+n(n−1)I A1 Substitutes limits, AG. Omission of\nn n−2 n−2 brackets or limits in their working is A0.\n3\n\n1(b) I =cosh1−2sinh1+2I M1 Applies reduction formula.\n2 0\nI =3cosh1−2sinh1−2 A1 Accept 1 ( e+5e−1) −2.\n2 2\n2\n© Cambridge University Press & Assessment 2025 Page 9 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q02",
      "question_id": "9231-2025-on-24-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 2,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) 1+2sinh2 x=6sinh2 xsinhx=1 M1 A1 Sets equal and uses\n2\ncosh2x=1+2sinh2 x.\nAcceptcoshx= 1 5 or tanhx= 1\n2 5\nor substituting exponential form to reach\ne4x −3e2x +1=0 ore2x −3+e−2x =0\nx=sinh−1(1)=ln ( 1 + 1 +1 ) M1 Uses logarithmic form or\n2 2 4 ( )\ne2x = 1 3 5 .\n2\nx=ln ( 1 + 15 ) A1 Must be exact and in the form lna,\n2 2\nwhere a is simplified.\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 19\n\n2(b) B1 C or C correct.\n1 2\nB1 C andintersections with C correct.\n2 1\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q03",
      "question_id": "9231-2025-on-24-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 3,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) sin5=Im(cos+isin)5 =sin5−10sin3cos2+5sincos4 M1 A1 Expands and clearly taking imaginary\npart, enough terms for M1. Accept RHS\n1\nto LHS using 2isin= z− .\nz\n=sin5−10sin3 ( 1−sin2 ) +5sin ( 1−sin2 )2 M1 Applies cos2=1−sin2.\n=16sin5−20sin3+5sin A1 AG, intermediate line of working needed,\ne.g. expanding\n( 1−sin2 )2 =1−2sin2+sin4.\n4\n\n3(b) M1 Applies identify given in (b).\nx=sin, sin5= 1 3\n2\n5= 1π+2kπ or 5= 2π+2kπ M1 Solves sin5= 1 3.\n3 3 2\nsin( 1 π) A1 Gives one correct solution. Accept\n15\nq= 1 .\n15\nsin(2 π),sin(− 5 π),sin(− 4 π),sin(7 π) A1 Gives other solutions. OE. A0 for\n15 15 15 15\nrepeated roots.\nsin( 2 π)=sin(13π)\n15 15\nsin(7 π)=sin(8 π)\n15 15\nsin(− 5 π)=sin(−1π)=sin(5π)=\n15 3 3\nsin(25π)\n15\nsin(− 4 π)=sin(−11π)=sin(19π)\n15 15 15\n4\n© Cambridge University Press & Assessment 2025 Page 12 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q04",
      "question_id": "9231-2025-on-24-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 4,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)  1 1 ( 3x) dx 1 ( (1)3N 1 +(1)3N 2 + +(1)3 N N −1 +(1)3N N) M1 A1 Forms the sum of the areas of the\n2 2 N N N N rectangles.\n0\nM1 for at least three terms including first\n1\nand last, and common ratio 3N . Missing\n“+” is A0.\n= 1  N ( 3N 1 )n = 3N 1 M1 A1 Applies  N rn = r ( rN −1 ) , AG.\n2N\nn=1\nN ( 3N 1 −1 )\nn=1\nr−1\n4\n\n4(b)  1 1 ( 3x) dx 1 ( (1)+(1)3N 1 + +(1)3 N N −2 +(1)3 N N −1) M1 A1 Forms the sum of the areas of appropriate\n2 2 N N N N rectangles.\n0\nM1 for at least three terms including first\nand last.\n= 1  N−1( 3N 1 )n = 1 M1 A1 Applies  N−1 rn = rN −1 .\n( )\n2N\nn=0\nN 3N 1 −1\nn=0\nr−1\n4\n\n4(c) 1 M1 c\n3N 1 1\nSimplifies their U −L to .\nN ( 3N 1 −1 )\n−\nN ( 3N 1 −1 )\n=\nN N N N\n1 A1\n→0 as N →.\nN\n2\n© Cambridge University Press & Assessment 2025 Page 13 of 19",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q05",
      "question_id": "9231-2025-on-24-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 5,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) m2+m+1=0m=−11 3 M1 Auxiliary equation. Must give the correct\n2 2\ntype of roots ab 3.\ny =e−1 2 x( Acos 3 x+Bsin 3 x ) A1 Complementary function. x missing or\n2 2 labelled incorrectly is A0.\ny= psin2x+qcos2x y'=2pcos2x−2qsin2x y''=−4psin2x−4qcos2x M1 A1 Particular integral and its derivatives.\n−4p−2q+ p=1 −4q+2p+q=1 M1 Substitutes and equates coefficients.\np=− 1 q =− 5 A1\n13 13\ny =e−1 2 x( Acos 3 x+Bsin 3 x ) − 1 sin2x− 5 cos2x A1 Must have “y =”.\n2 2 13 13\n7\n\n5(b) y − 1 sin2x− 5 cos2x= 2 ( − 1 sin2x− 5 cos2x ) M1 e−1 2 x →0 and correct use of compound\n13 13 13 26 26\nangle identity. 2 = 26 .\n13 13\n2 sin(2x−1.77) A1 A1 A1 for R and A1 for .Condone 101.3.\n13\nAllow R=0.392 and =cos−1(− 1 ).\n26\nOE.\n3\n© Cambridge University Press & Assessment 2025 Page 14 of 19",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-24-q06",
      "question_id": "9231-2025-on-24-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6 \n\n1+x\ndx=−1ln\n( 1−2x−x2) M1 A1 Finds integrating factor using\n1−2x−x2 2 f '(x)\n dx=ln f(x).\nf(x)\n−1ln ( 1−2x−x2) 1 A1\ne 2 =\n1−2x−x2\nM1 Correct form on LHS and RHS.\nd  y  1\n =\ndx  1−2x−x2  1−2x−x2\n1−2x−x2 =2−(x+1)2 B1 Completes the square.\n 1  x+1 *M1 A1 Integrates RHS. RHS must be of the\n  dx=sin−1   +C 1\n 2−(x+1)2  2  correct form .\nax2 +bx+c\nπ=sin−1 1 +CC = 3π DM1 A1 Substitutes initial conditions.\n2 4\ny  x+1 A1\n=sin−1 + 3π\n \n1−2x−x2  2  4\n10\n© Cambridge University Press & Assessment 2025 Page 15 of 19",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_24.pdf?download=true",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q07",
      "question_id": "9231-2025-on-24-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 7,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "7(a)(i) sec2 1t B1 Differentiates x with respect to t. AG.\nx=4 2 +3cosec2t−3\ntan1t\n2\nx=8cosect+3cosec2t−3=8cosect+3cot2t\n1\n\n7(a)(ii) y=6cosect−4cot2t B1 Differentiates y with respect to t.\n1\n\n7(a)(iii) x2+ y2 = ( 8cosect+3cot2t )2 + ( 6cosect−4cot2t )2 M1 A1 Expands and simplifies x2+ y2. A1 for\nreaching 100cosec2t+25cot4t .\n=100cosec2t+25cot4t =100cosec2t+25 ( cosec2t−1 )2\n=25cosec4t+50cosec2t+25\n=25 ( cosec2t+1 )2 *M1 A1 Factorises x2+ y2. ALT:\n25 ( cot2t+2 )2 .\n5 1 3 π cosec2t+1dt =5−cott+t1 3 π =5 ( − 1 +1π+ 3−1π ) = 5π+10 3 DM1 A1 Applies correct formula for arc length and\n1 6 π 1 6 π 3 3 6 6 3 integrates.\nAnswer in terms of π and 3 for A1.\n6\n© Cambridge University Press & Assessment 2025 Page 16 of 19\n\n7(b) y 6cosect−4cot2t 6sint−4cos2t 6sect−4cott B1 Finds first derivative.\n= = =\nx 8cosect+3cot2t 8sint+3cos2t 8sect+3cott\nd 6cosect−4cot2t M1 dy\nDifferentiates with respect to t.\n \ndt8cosect+3cot2t\n\ndx\nEnough terms for M1.\n( 8csct+3cot2t )( −6cottcsct+8cottcsc2t ) + ( 6csct−4cot2t )( 6csc2tcott+8cottcsct )\n=\n( 8cosect+3cot2t )2\nor\nd 6sint−4cos2t\n \ndt8sint+3cos2t\n\n( 8sint+3cos2t )(6cost+8costsint)− ( 6sint−4cos2t )(8cost−6costsint)\n=\n( 8sint+3cos2t )2\n100costsin2t+50cos3t 50cost ( sin2t+1 ) 50cottcosect ( cosec2t+1 ) A1\n= =\n( 8sint+3cos2t )2 ( 8sint+3cos2t )2 ( 8cosect+3cot2t )2\nd2y d 6sint−4cos2t dt 50sin2tcost ( sin2t+1 ) M1 A1 Applies chain rule.\n=   =\ndx2 dt8sint+3cos2t  dx ( 8sint+3cos2t )3\n50cottcosect ( cosec2t+1 )\n=\n( 8cosect+3cot2t )3\n5\n© Cambridge University Press & Assessment 2025 Page 17 of 19",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_24.pdf",
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      "html": "9231-topic-23-differentiation/answers.html",
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    },
    {
      "id": "9231-2025-on-24-q08",
      "question_id": "9231-2025-on-24-q08",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 2,
      "variant": "24",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a) 5 a 0 M1 Finds determinant.\n20 −5 0 =−20a−25\n2 −3 1\na  −5 A1\n4\nThree planes intersect at a single point. B1\n3\n\n8(b) 5x− 5 y =3, B1 Shows contradiction clearly.\n4\nIf a =−5 then 20x− 5y =2,2=12\n4\n2x− 3y+ z = 1,\nTwo distinct parallel planes. B1\nOther plane not parallel. B1\n3\n© Cambridge University Press & Assessment 2025 Page 18 of 19\n\n8(c) Eigenvalues of A are 5, −5 and 1. B1 Lower diagonal matrix or characteristic\nequation.\ni j k  40   1  M1 A1 Uses vector product (or equations) to find\n    corresponding eigenvectors.\n=5: 20 −10 0 = 80 ~ 2\n   \n   \n2 −3 −4 −40 −1\n   \ni j k  0  0 A1\n   \n=−5: 20 0 0 = −120  2\n   \n   \n2 −3 6 −60 1\n   \ni j k  0  0 A1\n   \n=1: 4 0 0 = 0  0\n   \n   \n20 −6 0 −24 1\n   \n 1 0 0 64 0 0  M1 A1 Or correctly matched permutations of\n    columns.\nThus P = 2 2 0 and D= 0 4 0\n   \n   \n−1 1 1 0 0 16\n   \n7\n© Cambridge University Press & Assessment 2025 Page 19 of 19",
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    },
    {
      "id": "9231-2025-on-31-q01",
      "question_id": "9231-2025-on-31-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct.\nA B\n4mv +mv =4mu\nA B\n−v +v =eu M1 NEL, signs must be consistent with momentum equation.\nA B\nv = 3u, v =u A1 Use momentum condition after the collision to obtain\nA 4 B\nexpressions for v and v in terms of u.\nA B\ne= 1 A1\n4\nAlternative method for question 1\nLet v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct.\nA B\n4mv +mv =4mu\nA B\n−v +v =eu M1 NEL, signs must be consistent with momentum equation.\nA B\nv = 1u(4−e), v = 4u(1+e) A1 Combine to obtain expressions for v and v in terms of u\nA 5 B 5 A B\nand e.\n1u(4−e) A1 Use momentum condition after the collision to solve for e.\n5 =3, e= 1\n4u(1+e) 4\n5\n4\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2025-on-31-q02",
      "question_id": "9231-2025-on-31-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 N cos =mg B1 N2L horizontally and vertically for the first case.\n1 1\n→N sin=m(rsin) 2 5 4 g=r 1 2\n1 1 1 1\nN cos =mg B1 N2L horizontally and vertically for the second case.\n2 2\n→N sin =m(rsin) 2 5 3 g=r 2 2\n2 2 2 2\n5g r2 M1 Combine to find ratio.\n4 = 1\n5g r2\n3 2\n A1\n1 = 1 3\n 2\n2\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_31-p10.png"
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    },
    {
      "id": "9231-2025-on-31-q03",
      "question_id": "9231-2025-on-31-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) dv =−1 ( v2 +4 ) *M1 Forming differential equation and integrating to a term\ndt 2 involving an inverse tangent. Condone omission of +A.\n1tan−1(1v ) =−1t +A Allow M1 for integrating dv = 1 ( v2 +4 ) to obtain a term\n2 2 2 dt 2\ninvolving an inverse tangent.\nA1 CWO, condone omission of +A.\nt=0, v=2 : A=1tan−1(2) =1π DM1 Use initial condition to find constant.\n2 2 8\ndx DM1 Integrate into term involving ln of cosine or sine.\nv= =2tan\n(1π−t )\ndt 4 Accept modulus in ln term.\nOE, for example\nx=2ln ( cos(t−1π)) +B\n4 v=−2tan(t−1π), x=2ln cos(t−1π) +B.\n4 4\nx=0, t=0: B=ln2 A1 Use initial condition and obtain correct expression.\nAEF, may see an expression in terms of sec.\nx=2ln ( cos(t−1π)) +ln2 =ln ( 2cos2(t−1π))\n4  4  Accept modulus in ln term.\n5\n\n3(b)\n2ln\n( cos(t−1π))\n+ln2=0\nM1 Solve equation to find a value for t.\n4\nt = 1π A1 CWO\n2\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-31-q04",
      "question_id": "9231-2025-on-31-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "4 5mge B1\nIn equilibrium, =5mg, e=a.\na\nB1 At least one correct EPE term seen.\nLet x be the length of the string when speed of P is 7ag .\n5\nEnergy equation in subsequent motion:\nM1 Dimensionally correct energy equation with one KE term,\n5mga2 5mg(x−a)2 two EPE terms and at least one GPE term.\n− =mg(e+a−x)+ 1m(7ag)\n2a 2a 2 5\nA1 Fully correct energy equation.\n25x2 −60ax+27a2 =0 M1 Three-term homogeneous quadratic equation in x and a .\nx= 9a only. A1\n5\nAlternative method for question 4\n5mge B1\nIn equilibrium, =5mg, e=a.\na\nLet x be the extension when P is released. B1 At least one correct EPE term seen.\nEnergy equation in subsequent motion:\n5mga2 5mgx2 M1 Dimensionally correct energy equation with one KE term,\n− =mg(e−x)+1m (7ag ) two EPE terms and at least one GPE term.\n2a 2a 2 5\nA1 Fully correct energy equation.\n25x2 −10ax−8a2 =0 M1 Three-term homogeneous quadratic equation in x and a .\n(5x+2a)(5x−4a)=0, x= 4a only. A1\n5\nRequired length of string = 9a.\n5\n6\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2025-on-31-q05",
      "question_id": "9231-2025-on-31-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) x =ka+1h M1 May be seen in a table.\n3\nx = 1(3ka+h) A1 AG, shown convincingly.\n3\ny =a+1(h−a)= 1(2a+h) B1 May be seen in a table.\n3 3\nAlternative method for question 5(a)\nCoordinates A(ka, 0), B(ka+h, h), C(ka, 2a) M1 Complete method to find x.\nx =1(ka+ka+h+ka)= 1(3ka+h)\n3 3 A1 AG, shown convincingly.\ny =1(0+h+2a)= 1(h+2a) B1\n3 3\n3\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms.\nx ( 2ka2 +ah ) = 2ka21ka+ah ( ka+1h ) Area From OD\n2 3\nOACD 2ka2 1ka\n2\nABC ah ka+1h\n3\nShape OABCD 2ka2 +ah x\nA1 Fully correct moments equation. May be seen in part 5(a).\nFor equilibrium, x≤ka, so M1 Use correct condition with their x for OABCD.\na2k2 +ahk+1h2 Allow strict inequality.\n3 ≤ka M1 for simplifying and attempting to solve an inequality.\n2ak+h\nMay be implied by correct final answer.\n(0≤)h≤ 3ka A1 CAO\n4\n\n5(c) Point of toppling means h= 3ka=a. B1\n( )\nx =1a 3+1\n3\n[h=a, meaning triangle ABC is isosceles.] B1\ny =a (by symmetry)\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
        13,
        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_31-p13.png",
        "../answer-assets/9231_w25_ms_31-p14.png"
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    },
    {
      "id": "9231-2025-on-31-q06",
      "question_id": "9231-2025-on-31-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and\nEnergy conservation before the collision: two KE terms.\n1m(17ag)+3mag=1mv2\n2 2 2\nA1\nv= 20ag\n2\n\n6(b) Let w be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation.\nP\nT = 0 must be seen or implied.\nmV2\nN2L: =(T)+mg V = ag\n \na\nEnergy equation: 1mw2 =1mV2+2mga M1 Dimensionally correct energy equation with a GPE term and\n2 P 2 two KE terms.\nA1\nw = 5ag\nP\n3\n\n6(c) Let w the speed of Q after the collision. B1 Momentum conserved (must see masses).\nQ\nmv=kmw −mw\nQ P\nw +w =ev =v B1 NEL with consistent signs.\nP Q\nOR\nOR\nUse conservation of kinetic energy.\n1mv2 =1mw2 +1kmw2\n2 2 P 2 Q\nk=3 B1 Must have e=1 seen or implied.\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
      "source_pages": [
        15
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-on-31-q07",
      "question_id": "9231-2025-on-31-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) dy 2gx M1 Valid attempt at differentiating.\n=tan−\ndx 2u2cos2\ndy B1\n=−1\ndx\n−1= 4− 2 x M1 Form equation with –1 and attempt to find a value for x .\n3 45 A A\nx =105 A1 525\nA 2 52.5 or .\ng\ny = 35 A1 175 dy\nA 4 8.75 or . SC B1 for 35 OE from setting =+1.\n2g 4 dx\nAlternative method for question 7(a)\nHorizontally: v =25cos=15 M1 Both equations, allow sign error.\nH\nVertically: v =25sin−gt=20−gt\nV\nv B1\nAt A: V =−tan45=−1\nv\nH\n20−gt M1 With RHS = –1, find value for t and x or y\n=−1, t= 7 A A\n15 2\nx =105 A1 525\nA 2 52.5 or .\ng\ny = 35 A1 175 dy\nA 4 8.75 or . SC B1 for 35 OE from setting =+1.\n2g 4 dx\n5\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(b) Let v be the velocity before the collision at A. B1\nu =ucos=15=v\nx x\nv =−v =−15 B1\ny x\nB1 FT\nSpeed: v2 +v2 =15 2 (ms–1)\nx y\n3\n\n7(c) Let w be the velocity after the collision. M1 Correct application of NEL.\nw=ev=5 2\n3 A1\ngx2 M1 Substitute values into trajectory equation and solve to find x.\n−35 =xtan45− , 9x2 −x−35 =0. OR\n4 2 ( 5 2 )2 (cos45)2 5 4 Complete method to find x that firstly determines the time of\n3\nthe flight after the particle strikes the barrier.\nOR\n35=−5 21 2t+1gt2, t=3\n4 3 2 2 2\nx=(wcos45)t =53\n \n3 2\nx= 5 (metres) A1\n2\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
        16,
        17
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_31.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-on-32-q01",
      "question_id": "9231-2025-on-32-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 T cos45+T sin45=m2lsin45 B1 Correct horizontal equation.\nA B\nAllow sin/cos mix.\nB0 B0 M0 A0 for using the same tension in\neach string.\nT sin45−T cos45=mg B1 Correct vertical equation.\nA B\nAllow sin/cos mix.\n2T =m2l+ 2 mg M1 Combine, isolating term in T .\nA A\nT = 1m ( 2l+ 2 g ) A1 AEF\nA 2\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-32-q02",
      "question_id": "9231-2025-on-32-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a)\nEPE=\n2mg (1a )2 B1 Correct EPE seen OR correct work done by\n2a 3 friction seen.\nWork done by friction=mg\n(1a )\n3\n2mg (1a )2\n=mg\n(1a ) M1 Equation or inequality involving EPE and\n2a 3 3 work done terms only.\n= 1 A1 CWO\n3\n3\n© Cambridge University Press & Assessment 2025 Page 9 of 17\n\n2(b) Let x be the extension in the string when P comes to rest. M1 Dimensionally correct equation with two EPE\n2mg (1a )2 = 2mg x2 +1mg (1a−x ) terms and one work done term.\n2a 3 2a 2 3 Use of 1mgx as work done term can be\n2\nawarded M1 only.\nA1 Correct energy equation.\nMay be unsimplified.\n18x2 −9ax+a2 =0, (6x−a)(3x−a)=0 x= 1a only. A1 CWO\n  6\nAlternative method for question 2(b)\nLet x be the distance moved by P before coming to rest. M1 Dimensionally correct equation with two EPE\n2mg (1a )2 = 2mg (1a−x )2 +1mgx terms and one work done term.\n2a 3 2a 3 2 Use of 1mg (1a−x ) as work done term can\n2 3\nbe awarded M1 only.\nA1 Correct energy equation.\nMay be unsimplified.\n6x2−ax=0, x=1a A1 CWO\n6\nExtension = 1a−x= 1a\n3 6\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        9,
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2025-on-32-q03",
      "question_id": "9231-2025-on-32-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3(a) Vertical distance of COM of sector from OA: B1 SOI\n( 2acos(1π)) sin(1π)\n4a\nAllow unsimplified.\n4 4 =\n3(1π) 3π\n4\nka2(1a)+ 1πa2  a− 4a = ( ka2 + 1πa2) y M1 3\neq\n-t\nu\ne\na\nr\nt\nm\nio n\nd\n.\ni mensionally correct moments\n2 4  3π 4\n  6ka+3πa−4a=3(4k+π)y  A1 FT Correct moments equation, allow\nunsimplified.\nFT their vertical distance of COM of sector\nfrom OA.\n6k+3π−4 A1 AEF, but do not allow fractions within\ny = a\n3(4k+π) numerator and/or denominator.\n4\n\n3(b) 4a M1 Dimensionally correct moments equation\nka2(1\n2\nka)= 1\n4\nπa2\n 3π   about OB using their horizontal distance of\nCOM of sector from OB.\nk2 = 2 k = 2 =0.816 A1\n 3 3\nAlternative method for question 3(b)\n( 6k2 +3kπ+4 ) a M1 Use of x =ka with their x, which may be\nOn the point of toppling, x =ka: =ka seen in part 3(a).\n3(4k+π)\n6k2 +4=12k2 k = 2 =0.816 A1\n  3\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-32-q04",
      "question_id": "9231-2025-on-32-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "4(a) Horizontal velocities are the same: B1 B0 M0 M0 A0 for using U =U .\nq\nU cos=Ucos\nq\nd =(Usin)T −1gT2 + ( U sin ) T +1gT2 M1 Apply vertical equation of motion on both\n2 q 2 particles and combine.\nUcos  M1 Substitute U and simplify.\nd =(Usin)T + sinT q\n cos \nd =UT(sin+costan) A1 AG, shown convincingly.\n4\n\n4(b) Usin=gT B1 Correct equation for T.\nU2sin M1 Substitute their expression for T in given\nd = (sin+costan) result from part 4(a) OR complete method for\ng\nfinding d in terms of U and g only.\nMay include unevaluated trigonometric\nfunctions.\nNote: d =2UT and U =2gT .\nU2 A1\nd =\ng\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
      "source_pages": [
        12
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-on-32-q05",
      "question_id": "9231-2025-on-32-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) usin=vsin(2) B1 May be implied by correct answer.\n3u= 24v\n \n5 25\nv= 5u B1\n8\n2\n\n5(b) mucos=mvcos(2)+kmw B1 PCLM equation, must include masses.\n4u= 7 v+kw, kw=5u\n \n5 25 8\neucos=w−vcos(2) B1 NEL equation.\n4eu=w− 7 v\n \n5 25\n5 4 5 7 M1 Combine and substitute for v and w to obtain\n= e+ \nequation in e and k only.\n8k 5 8 25\n25 7 A1 25−7k\ne= − AEF, for example .\n32k 32 32k\nNot dependent on first B mark.\n4\n\n5(c) 25 7 M1 Form inequality using their expression for e.\n0≤ − ≤1\nAllow equality or strict inequality.\n32k 32\n7k≤25≤39k M1 Rearrange at least one side to the form\nak≤b or ak≥b.\nAllow equality or strict inequality.\n25≤k≤25 A1 CAO\n39 7\n3\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2025-on-32-q06",
      "question_id": "9231-2025-on-32-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6 A to ground: 1mu2 +2mga=1m(3u)2 u2 = 1ga *M1 One energy equation, all terms present.\n2 2  2 \n*M1 Second energy equation, all terms present.\nA to B: 1mv2+mgacos=1mu2+mga v2 =u2 +2ag(1−cos)\n2 2   A1 At least one correct energy equation involving\ncos .\nB to ground: 1mv2 +mgacos=1m(3u)2 −mga v2 =9u2 −2ag(1+cos)\n2 2  \nmv2 B1 N2L at B.\nmgcos= v2 =agcos\n \na\ngacos= 1ga+2ag(1−cos) DM1 Eliminate u and v.\n2\ncos= 5 A1\n6\nHorizontally at ground: vcos=3ucos. M1 A complete method to find .\n5ga (5)\ncos= 6 6 = ( 5 15 )\n3 1ga 54\n2\nOR\nv =vcos= 5ga (5) , v2 =(vsin)2 +2ga(1+cos)=847ga\nx 6 6 y 216\n847ga\ntan= 216\n5ga (5)\n6 6\n=69.0 A1\n8\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-32-q07",
      "question_id": "9231-2025-on-32-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) dv B1 Must have m.\n−mgkv−mg =m\ndt\n1 1 *M1 Separate variables and integrate to a natural\nt=−  dv\nlogarithm term.\ngkv+\nIgnore modulus signs in this part.\n1\nt=− ln(kv+) +C\ngk\n1 DM1 Use initial condition to find constant OR use\nWhen v = u, t = 0: C = ln(kU +)\ncorrect limits.\ngk\nDependent on having integrated to obtain a\nnatural logarithm term.\n1 1 A1 AG, shown convincingly.\nt=− ln(kv+)+ ln(kU +)\nMust see an intermediate step before reaching\ngk gk\ngiven result.\n1 kU + Not dependent on B1.\nt = ln \ngk  kv+\n4\n© Cambridge University Press & Assessment 2025 Page 15 of 17\n\n7(b) dv M1 Three terms in equation of motion, separating\n−2v−2=v\n5 dx variables and expressing integrand in a form\nthat can be directly integrated.\n v  5 \nx=−5 dv=−51− dv May be in terms of k,U and , for example\n2v+5 2 v+5\n1   \nx=− gk      1− v+ k  k    dv.\nUse of suvat in this part cannot be awarded\nany credit.\nx=−5v+ 25ln(v+5) +C M1 Correct form.\n2 2\nMay be in terms of k,U and , for example\n1   \nx=− v+ lnv+ +C .\ngk gk2  k \nWhen x=0, v=10: C =25− 25ln(15) M1 Use initial condition to find constant OR use\n2\ncorrect limits in a definite integral.\nWhen v=0, x=25− 25ln(15)+ 25ln(5)=11.3 [m] A1\n2 2\nAlternative method for question 7(b)\n2 M1 Rearrange answer to part 7(a) in the form\n 15  − t\nFrom part (a), t= 5ln   , v=15e 5 −5 v=aebt +c.\n2 v+5\nUse of suvat in this part cannot be awarded\nany credit.\nx=−75e− 5 2t −5t +B M1 Integrate to correct form.\n2\nWhen t=0, x=0, B=75, x=−75e− 5 2t −5t+75 M1 Use initial condition to find constant OR use\n2 2 2 correct limits in a definite integral.\nWhen v=0, t =2.5ln3, x=25− 25ln3=11.3 [m] A1\n2\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(b) Alternative method for question 7(b)\nFrom part (a), t= 5 2 ln   v 1 + 5 5    , v=15e− 5 2t −5 M1 R v e = ar a r e a b n t g + e c a . n swer to part 7(a) in the form\nUse of suvat in this part cannot be awarded\nany credit.\nWhen v=0, t=2.5ln3 so x= 2.5ln3( 15e− 5 2t −5 ) dt M1\n0\n2.5ln3 M1 Integrate to correct form.\nx=−75e−\n5\n2t −5t\n 2 \n0\nx=−751−512.5ln3− ( −75) =25−25ln3=11.3 [m] A1\n2 3 2 2\n4\n\n7(c) \"11.3\" \"11.3\" M1 Divide their x from part 7(b) by the value of t\nAverage speed = =\n1 0.0410+0.2 2.5ln3 when v = 0 using the given answer from part\nln  7(a). Condone if their x from part 7(b) is\n100.04  0.2 \nobtained using suvat.\n4.10 [ms–1] A1 4.102, accept values in the range 4.10 – 4.11.\n2\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_32.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_32.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2025-on-33-q01",
      "question_id": "9231-2025-on-33-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct.\nA B\n4mv +mv =4mu\nA B\n−v +v =eu M1 NEL, signs must be consistent with momentum equation.\nA B\nv = 3u, v =u A1 Use momentum condition after the collision to obtain\nA 4 B\nexpressions for v and v in terms of u.\nA B\ne= 1 A1\n4\nAlternative method for question 1\nLet v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct.\nA B\n4mv +mv =4mu\nA B\n−v +v =eu M1 NEL, signs must be consistent with momentum equation.\nA B\nv = 1u(4−e), v = 4u(1+e) A1 Combine to obtain expressions for v and v in terms of u\nA 5 B 5 A B\nand e.\n1u(4−e) A1 Use momentum condition after the collision to solve for e.\n5 =3, e= 1\n4u(1+e) 4\n5\n4\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2025-on-33-q02",
      "question_id": "9231-2025-on-33-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "2 N cos =mg B1 N2L horizontally and vertically for the first case.\n1 1\n→N sin=m(rsin) 2 5 4 g=r 1 2\n1 1 1 1\nN cos =mg B1 N2L horizontally and vertically for the second case.\n2 2\n→N sin =m(rsin) 2 5 3 g=r 2 2\n2 2 2 2\n5g r2 M1 Combine to find ratio.\n4 = 1\n5g r2\n3 2\n A1\n1 = 1 3\n 2\n2\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 17",
      "source_pages": [
        10
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
      "image_paths": [
        "../answer-assets/9231_w25_ms_33-p10.png"
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    },
    {
      "id": "9231-2025-on-33-q03",
      "question_id": "9231-2025-on-33-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) dv =−1 ( v2 +4 ) *M1 Forming differential equation and integrating to a term\ndt 2 involving an inverse tangent. Condone omission of +A.\n1tan−1(1v ) =−1t +A Allow M1 for integrating dv = 1 ( v2 +4 ) to obtain a term\n2 2 2 dt 2\ninvolving an inverse tangent.\nA1 CWO, condone omission of +A.\nt=0, v=2 : A=1tan−1(2) =1π DM1 Use initial condition to find constant.\n2 2 8\ndx DM1 Integrate into term involving ln of cosine or sine.\nv= =2tan\n(1π−t )\ndt 4 Accept modulus in ln term.\nOE, for example\nx=2ln ( cos(t−1π)) +B\n4 v=−2tan(t−1π), x=2ln cos(t−1π) +B.\n4 4\nx=0, t=0: B=ln2 A1 Use initial condition and obtain correct expression.\nAEF, may see an expression in terms of sec.\nx=2ln ( cos(t−1π)) +ln2 =ln ( 2cos2(t−1π))\n4  4  Accept modulus in ln term.\n5\n\n3(b)\n2ln\n( cos(t−1π))\n+ln2=0\nM1 Solve equation to find a value for t.\n4\nt = 1π A1 CWO\n2\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2025-on-33-q04",
      "question_id": "9231-2025-on-33-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "4 5mge B1\nIn equilibrium, =5mg, e=a.\na\nB1 At least one correct EPE term seen.\nLet x be the length of the string when speed of P is 7ag .\n5\nEnergy equation in subsequent motion:\nM1 Dimensionally correct energy equation with one KE term,\n5mga2 5mg(x−a)2 two EPE terms and at least one GPE term.\n− =mg(e+a−x)+ 1m(7ag)\n2a 2a 2 5\nA1 Fully correct energy equation.\n25x2 −60ax+27a2 =0 M1 Three-term homogeneous quadratic equation in x and a .\nx= 9a only. A1\n5\nAlternative method for question 4\n5mge B1\nIn equilibrium, =5mg, e=a.\na\nLet x be the extension when P is released. B1 At least one correct EPE term seen.\nEnergy equation in subsequent motion:\n5mga2 5mgx2 M1 Dimensionally correct energy equation with one KE term,\n− =mg(e−x)+1m (7ag ) two EPE terms and at least one GPE term.\n2a 2a 2 5\nA1 Fully correct energy equation.\n25x2 −10ax−8a2 =0 M1 Three-term homogeneous quadratic equation in x and a .\n(5x+2a)(5x−4a)=0, x= 4a only. A1\n5\nRequired length of string = 9a.\n5\n6\n© Cambridge University Press & Assessment 2025 Page 12 of 17",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
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      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2025-on-33-q05",
      "question_id": "9231-2025-on-33-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) x =ka+1h M1 May be seen in a table.\n3\nx = 1(3ka+h) A1 AG, shown convincingly.\n3\ny =a+1(h−a)= 1(2a+h) B1 May be seen in a table.\n3 3\nAlternative method for question 5(a)\nCoordinates A(ka, 0), B(ka+h, h), C(ka, 2a) M1 Complete method to find x.\nx =1(ka+ka+h+ka)= 1(3ka+h)\n3 3 A1 AG, shown convincingly.\ny =1(0+h+2a)= 1(h+2a) B1\n3 3\n3\n© Cambridge University Press & Assessment 2025 Page 13 of 17\n\n5(b) Taking moments about the x-axis: M1 Moments equation with 3 terms.\nx ( 2ka2 +ah ) = 2ka21ka+ah ( ka+1h ) Area From OD\n2 3\nOACD 2ka2 1ka\n2\nABC ah ka+1h\n3\nShape OABCD 2ka2 +ah x\nA1 Fully correct moments equation. May be seen in part 5(a).\nFor equilibrium, x≤ka, so M1 Use correct condition with their x for OABCD.\na2k2 +ahk+1h2 Allow strict inequality.\n3 ≤ka M1 for simplifying and attempting to solve an inequality.\n2ak+h\nMay be implied by correct final answer.\n(0≤)h≤ 3ka A1 CAO\n4\n\n5(c) Point of toppling means h= 3ka=a. B1\n( )\nx =1a 3+1\n3\n[h=a, meaning triangle ABC is isosceles.] B1\ny =a (by symmetry)\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 17",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_w25_ms_33-p14.png"
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    },
    {
      "id": "9231-2025-on-33-q06",
      "question_id": "9231-2025-on-33-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and\nEnergy conservation before the collision: two KE terms.\n1m(17ag)+3mag=1mv2\n2 2 2\nA1\nv= 20ag\n2\n\n6(b) Let w be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation.\nP\nT = 0 must be seen or implied.\nmV2\nN2L: =(T)+mg V = ag\n \na\nEnergy equation: 1mw2 =1mV2+2mga M1 Dimensionally correct energy equation with a GPE term and\n2 P 2 two KE terms.\nA1\nw = 5ag\nP\n3\n\n6(c) Let w the speed of Q after the collision. B1 Momentum conserved (must see masses).\nQ\nmv=kmw −mw\nQ P\nw +w =ev =v B1 NEL with consistent signs.\nP Q\nOR\nOR\nUse conservation of kinetic energy.\n1mv2 =1mw2 +1kmw2\n2 2 P 2 Q\nk=3 B1 Must have e=1 seen or implied.\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-on-33-q07",
      "question_id": "9231-2025-on-33-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a) dy 2gx M1 Valid attempt at differentiating.\n=tan−\ndx 2u2cos2\ndy B1\n=−1\ndx\n−1= 4− 2 x M1 Form equation with –1 and attempt to find a value for x .\n3 45 A A\nx =105 A1 525\nA 2 52.5 or .\ng\ny = 35 A1 175 dy\nA 4 8.75 or . SC B1 for 35 OE from setting =+1.\n2g 4 dx\nAlternative method for question 7(a)\nHorizontally: v =25cos=15 M1 Both equations, allow sign error.\nH\nVertically: v =25sin−gt=20−gt\nV\nv B1\nAt A: V =−tan45=−1\nv\nH\n20−gt M1 With RHS = –1, find value for t and x or y\n=−1, t= 7 A A\n15 2\nx =105 A1 525\nA 2 52.5 or .\ng\ny = 35 A1 175 dy\nA 4 8.75 or . SC B1 for 35 OE from setting =+1.\n2g 4 dx\n5\n© Cambridge University Press & Assessment 2025 Page 16 of 17\n\n7(b) Let v be the velocity before the collision at A. B1\nu =ucos=15=v\nx x\nv =−v =−15 B1\ny x\nB1 FT\nSpeed: v2 +v2 =15 2 (ms–1)\nx y\n3\n\n7(c) Let w be the velocity after the collision. M1 Correct application of NEL.\nw=ev=5 2\n3 A1\ngx2 M1 Substitute values into trajectory equation and solve to find x.\n−35 =xtan45− , 9x2 −x−35 =0. OR\n4 2 ( 5 2 )2 (cos45)2 5 4 Complete method to find x that firstly determines the time of\n3\nthe flight after the particle strikes the barrier.\nOR\n35=−5 21 2t+1gt2, t=3\n4 3 2 2 2\nx=(wcos45)t =53\n \n3 2\nx= 5 (metres) A1\n2\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
      "source_pages": [
        16,
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_33.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_33.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-on-34-q01",
      "question_id": "9231-2025-on-34-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1 For A: T =mg M1 Both equations.\nFor B vertically: Tcos=0.5mg Allow sin/cos mix.\n=60 A1 Equate and solve for , SOI.\nFor B horizontally: Tsin=0.5m(BRsin)2 M1 Allow sin/cos mix.\nAllow their stated BR.\nBRcos=AR BR=8a B1 SOI\n1 g A1 AEF in terms of g and a.\n=\n2 a\n5\n© Cambridge University Press & Assessment 2025 Page 9 of 18",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-on-34-q02",
      "question_id": "9231-2025-on-34-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 2,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) gx2 B1 Allow equation derived from first principles.\nUse trajectory equation from MF19: y=xtan− .\n2u2cos2 Substitute tan=2, cos2=1 , g=10 and simplify.\n5\n10x2 25x2 AG, shown convincingly.\ny=2x− =2x−\n2u21 u2\n5\n1\n\n2(b) 2582 M1 Substitute correctly and solve equation from part 2(a).\n12=28−\nu2\nu2 =400 u=20 A1\n25D2 M1 Solve the equation from part 2(a) with y=0.\n0=2D− leading to D=0, D=32.\n400\n25D2 M1 Solve the equation from part 2(a) with y=7.\n7=2D− leading to D=4, D=28.\n400\n28≤D≤32 only. A1 Accept strict inequalities.\n5\n© Cambridge University Press & Assessment 2025 Page 10 of 18",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2025-on-34-q03",
      "question_id": "9231-2025-on-34-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 3,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a) Area From AD From AB B1 Areas and distances from AD correct.\nABCD 4a2 a a Does not need to be presented in a table.\nAED a2 1a 2a\n3 3\nBCF ah 2a−1h 4a\n3 3\nShape BFDE 3a2 −ah x y\nTaking moments about AD: M1 Dimensionally correct moments equation with at least\n( 3a2 −ah ) x =4a2a−a21a−ah(2a−1h) three correct terms.\n3 3\n11a3−2a2h+1ah2 1a(h2 −6ah+11a2) h2 −6ah+11a2 A1 AG, shown convincingly.\nx = 3 3 = 3 =\n3a2 −ah a(3a−h) 3(3a−h)\nTaking moments about AB: M1 Dimensionally correct moments equation with at least\n( 3a2 −ah ) y =4a2a−a22a− ah4a three correct terms.\n3 3\n2a(5a−2h) A1\ny =\n3(3a−h)\n© Cambridge University Press & Assessment 2025 Page 11 of 18\n\n3(a) Alternative method for question 3(a)\nArea From AD From AB B1 Areas and distances from AD correct.\nDEF a(2a−h) 1(3a−h) 4a Does not need to be presented in a table.\n3 3\nEBF a2 1(5a−h) 2a\n3 3\nShape BFDE 3a2 −ah x y\nTaking moments about AD: M1 Dimensionally correct moments equation with at least\n( 3a2 −ah ) x =a (2a−h)1(3a−h)+a21(5a−h) two correct terms.\n3 3\n1a(6a2 −5ah+h2 +5a2 −ah) 1a(h2 −6ah+11a2) h2 −6ah+11a2 A1 AG, shown convincingly.\nx = 3 = 3 =\n3a2 −ah a(3a−h) 3(3a−h)\nTaking moments about AB: M1 Dimensionally correct moments equation with at least\n( 3a2 −ah ) y =a(2a−h)4a+a22a two correct terms.\n3 3\n2a(5a−2h) A1\ny =\n3(3a−h)\n5\n\n3(b) x≥a meaning h2 −3ah+2a2≥0 M1 Use condition x≥a and simplify to three-term\nhomogeneous quadratic inequality.\n[0]h≤a, h=2a. A1 Accept h=0.\n2\n© Cambridge University Press & Assessment 2025 Page 12 of 18",
      "source_pages": [
        11,
        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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        "../answer-assets/9231_w25_ms_34-p12.png"
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    },
    {
      "id": "9231-2025-on-34-q04",
      "question_id": "9231-2025-on-34-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 4,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) mw2 B1 Accept equality.\nAt top R≥0, so ≥mg.\na\n1mv2−1mw2≥mga(1−cos) M1 Energy inequality with correct number of terms and\n2 2 correct dimensions, accept equality.\nA1 Fully correct energy inequality, accept equality.\n1mv2≥13mga M1 Substitute for cos and w2.\n2 9\nv≥1 26ag A1 AG, shown convincingly.\n3 If equality used throughout, introduction of inequality\nmust be convincing.\n5\n\n4(b) Maximum reaction is at bottom: M1 Energy equation with correct number of terms and\n1mv2+mga(1+cos)=1mV2 V2 =5ag correct dimensions and correct signs.\n2 2   Allow sin/cos mix.\nmV2 B1\nR −mg =\nmax a\nR =6mg A1\nmax\n3\n© Cambridge University Press & Assessment 2025 Page 13 of 18",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2025-on-34-q05",
      "question_id": "9231-2025-on-34-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) dv B1 m must be seen.\nmv =−0.025v2m−10m\ndx\n v *M1 Separate variables and integrate to a term involving\n dv=−dx\n10+0.025v2\nln ( av2 +b ) or ln\n\n\nav+b\n .\n pv+q\n−0.025x+c= 1ln ( v2 +400 )\n2\nA1 Ignore modulus signs.\nCondone missing +c.\nOE, for example −x+c=20ln ( v2 +400 ) .\nc= 1ln800 DM1 Use initial condition to find constant of integration.\n2\nA1 AEF\nv=20 2e− 2 1 0 x −1\n5\n\n5(b) dv M1 Equation and integration to a term involving tan−1.\nm =−0.025v2m−10m\ndt Accept missing m.\n 1\n−dt= dv\n10+0.025v2 A1 Condone missing +k .\nv\n−t+k=2tan−1\n20\nk=2tan−120 =1π M1 v\n20 2 Use initial condition with a term involving tan−1  .\na\nv=20tan (1π−1t ) A1\n4 2\n4\n© Cambridge University Press & Assessment 2025 Page 14 of 18",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2025-on-34-q06",
      "question_id": "9231-2025-on-34-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 6,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) N2L: T =T +kmg B1\n1 2\n5mg(22a−BP−4a) B1 Apply Hooke’s law to string AP.\nT =\n1 4a\n6mg(BP−8a) B1 Apply Hooke’s law to string BP.\nT =\n2 8a\n5mg(22a−BP−4a) 6mg(BP−8a) M1 Substitute T and T into N2L and obtain linear\n= +kmg 1 2\n4a 8a expression for BP in terms of a and k .\n8BP=114a−4ak\nA1 AG, shown convincingly.\n57a−2ak\nBP=\n4\nAlternative method for question 6(a)\nN2L: T =T +kmg B1\n1 2\n5mgx B1 Apply Hooke’s law to string AP, where x denotes the\nT =\n1 4a extension in string AP.\n6mg(18a−x−8a) 6mg(10a−x) B1 Apply Hooke’s law to string BP.\nT = = \n2 8a 8a\n \n5mgx 6mg(10a−x) M1 Substitute T and T into N2L and obtain linear\n= +kmg 1 2\n4a 8a expression for x in terms of a and k AND substitute\n15a+2ak their x into BP=18a−x.\nx=\n4\nA1 AG, shown convincingly.\n15a+2ak 57a−2ak\nBP=18a−x=18a− =\n4 4\n© Cambridge University Press & Assessment 2025 Page 15 of 18\n\n6(a) Alternative method for question 6(a)\nN2L: T =T +kmg B1\n1 2\n5mg(x−4a) B1 Apply Hooke’s law to string AP, where x denotes the\nT = length of string AP.\n1 4a\n6mg(22a−x−8a) 6mg(14a−x) B1 Apply Hooke’s law to string BP.\nT = = \n2 8a 8a\n \n5mg(x−4a) 6mg(14a−x) M1 Substitute T and T into N2L and obtain linear\n= +kmg 1 2\n4a 8a expression for x in terms of a and k AND substitute\n31a+2ak 31a+2ak 57a−2ak their x into BP=22a−x.\nx= , BP=22a−x=22a− =\n4 4 4\nA1 AG, shown convincingly.\nAlternative method for question 6(a)\nN2L: T =T +kmg B1\n1 2\n5mg(10a−x) B1 Apply Hooke’s law to string AP, where x denotes the\nT =\n1 4a extension in string BP.\n6mgx B1 Apply Hooke’s law to string BP.\nT =\n2 8a\n5mg(10a−x) 6mgx M1 Substitute T and T into N2L and obtain linear\n= +kmg 1 2\n4a 8a expression for x in terms of a and k AND substitute\n25a−2ak 25a−2ak 57a−2ak their x into BP=8a+x.\nx= , BP=8a+x=8a+ =\n4 4 4\nA1 AG, shown convincingly.\n© Cambridge University Press & Assessment 2025 Page 16 of 18\n\n6(a) Alternative method for question 6(a)\nN2L: T =T +kmg B1\n1 2\n5mgx 6mgx B1 Apply Hooke’s law to string AP or BP, where x\nT = A or T = B A\n1 4a 2 8a denotes the extension in string AP and x denotes the\nB\nextension in string BP.\n5mgx 6mgx B1 Apply Hooke’s law to string AP and BP and recognise\nT = A , T = B , x +x =10a\n1 4a 2 8a A B that the sum of the extensions in both strings is 10a.\n5mgx 6mgx M1 Substitute T and T into N2L and obtain linear\nA = B +kmg, 5x −3x =4k 1 2\n4a 8a A B expression in x and x in terms of a and k AND use\nA B\n15a+2ak 25a−2ak\nx A = 4 OR x B = 4 the expression and x A +x B =10a to obtain an\nexpression for x or x only AND substitute their x\n15a+2ak 57a−2ak A B A\nEither BP=18a−x =18a− = into BP=18a−x or their x into BP=8a+x .\nA 4 4 A B B\n25a−2ak 57a−2ak\nOR BP=8a+x =8a+ = A1 AG, shown convincingly.\nA 4 4\n5\n\n6(b) 6mg(18a−8a)2 5mg(14a−4a)2 B1 At least one correct EPE term.\n+18kmga= +8kmga\n28a 24a\nM1 Dimensionally correct energy equation with two EPE\nterms and at least one GPE term (no KE terms).\nA1 Fully correct energy equation.\nk = 5 A1\n2\n4\n© Cambridge University Press & Assessment 2025 Page 17 of 18",
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    },
    {
      "id": "9231-2025-on-34-q07",
      "question_id": "9231-2025-on-34-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 3,
      "variant": "34",
      "question_number": 7,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "7 Perpendicular to the wall: eusin=vsin B1\nPP =uT B1\n1 2\nPP =vkT\n2 3\nd d M1\nsin= sin=\nPP PP\n1 2 2 3\nd d M1 Eliminate PP and PP .\nsin= sin= 1 2 2 3\nuT vkT\neud vd M1\n=\nuT vkT\n1 A1\ne=\nk\n6\n© Cambridge University Press & Assessment 2025 Page 18 of 18",
      "source_pages": [
        18
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_34.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_34.pdf?download=true",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2025-on-41-q01",
      "question_id": "9231-2025-on-41-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) s2 =1 ( '71314'−'844'2) =134 =8.933 M1 Correct expression, implied by AWRT 8.93.\n \n9 10 15\n2.262 B1 2.262 or 2.26 seen.\nM1 Correct form, must be a t-value.\n'844''2.262' '8.933'\n10 10\n[82.3, 86.5] A1 Accept with inequality signs or open brackets.\nCondone [86.5, 82.3].\nDo not accept 84.42.1.\nThe distribution of estimates of angles is normal. B1\nOR\nThe estimates are a random sample from some population.\nOR\nThe estimates are independent.\nOR\nUnderlying distribution is normal.\n5\n\n1(b) Population unlikely to be normal as  is close to right-angle / data B1 Must refer to the context.\nis skewed.\nOR\nEstimates may not be independent, for example due to collusion.\nOR\nNo indication that sample is random.\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 16",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-41-q02",
      "question_id": "9231-2025-on-41-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) e−2.8(2.8)3 B1 Or by subtraction.\np= 100=22.248 Allow 22.2 or 22.3.\n3!\n(29−22.25)2 B1 AWRT 2.05.\nq= =2.05\n22.25\n2\n\n2(b) H : Po(2.8) fits the data. B1 Must mention distribution and data/number of cars.\n0\nH : Po(2.8) does not fit the data.\n1\nOR\nH : Po(2.8) is a satisfactory model for the number of cars (entering\n0\nthe car park).\nH : Po(2.8) is not a satisfactory model for the number of cars\n1\n(entering the car park).\n2.739+0.241+2.152+'2.049'+0.425+3.753+0.037 =11.396 M1 Sum of chi-square contributions using their q.\nMay be implied by 11.4.\n'11.4'12.59, accept H / do not reject H / not significant. M1 Compare their 11.4 with 12.59 and appropriate result (may be in\n0 0\nterms of H ).\n1\nTest result may be implied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nInsufficient evidence to suggest that Po(2.8) is not a good fit to the A1 Correct conclusion from correct working ignoring hypotheses.\ndata. In context.\nOR Level of uncertainty in language is used (for example, not ‘prove’).\nInsufficient evidence to suggest that the manager’s claim is false. Allow ‘Not enough evidence that / to show / conclude that…’\nDo not accept statements such as ‘there is sufficient evidence to\nsuggest…’ or ‘no evidence…..’\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 16",
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      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-41-q03",
      "question_id": "9231-2025-on-41-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "3 (x−x)2\n=90.36=3.24\nB1\n'16.6'2 M1 Can be implied by 0.172 ( = 43 ) or 0.688 ( = 86).\n'55.8'− 250 125\n5\n3.24+0.688 M1 Correct use of formula for pooled variance using their values.\n(=0.302)\n10+5−2\ns=0.550 A1 CAO\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-41-q04",
      "question_id": "9231-2025-on-41-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "4 128(28+1) =203 B1 Both expressions correct. Implied by 203 and 1928.5 seen.\n4\n1 28(28+1)(56+1) =1928.5\n24\nT +0.5−'203' M1 Correct form or expression for CI with any z-value.\nz=\nCondone T or T−0.5 instead of T+0.5 for M1.\n'1928.5'\nT +0.5−'203' M1 Compares their z with 2.576.\n−2.576\n'1928.5' Condone T or T−0.5 instead of T+0.5 for M1.\nT 89.4 A1 Allow T =89.4.\nT =90 A1 All correct with no errors seen, including correct continuity\ncorrection.\n5\n© Cambridge University Press & Assessment 2025 Page 11 of 16",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-41-q05",
      "question_id": "9231-2025-on-41-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 4 M1 Equation in terms of k formed following attempt to integrate with\n1   2 x 3 2   + 1   2 x  9 =1 correct limits.\n163  k 4\n0\n1 2\n(8−0)+ (3−2)=1\n24 k\nOR\n1 2\n+ =1\n3 k\n2 2 A1 AG, no errors seen.\n= , k =3\nk 3\n2\n\n5(b) 1  4 x dx+1 m x−1 2 dx= 1 *M1 Use of 4m9 to form equation equal to 0.5, OE.\n16 3 2 Equation may be in terms of k.\n0 4\nOR\n1 9 x−1 2 dx=0.5\n3\nm\n1+ 2 ( m−2 ) = 1 DM1 Integrate and form an equation in m .\n3 3 2\nOR\n( )\n2 3− m = 1\n3 2\nm= 81 [=5.0625] A1 Accept AWRT 5.06.\n16\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 16\n\n5(c)   1 x 3 2 0≤x4 M1 Attempt to integrate both parts, limits not required.\nF(x)= 24 May be seen in part 5(a).\n 2x 1 2 −1 4≤x≤9\n3\nB1 For constant –1 obtained in expression for 4≤x≤9.\nMay be seen in part 5(a).\n1 y3 0≤y2 M1 For changing to y.\nG(y)=24\nLimits not required.\n 2 y−1 2≤y≤3\n3\n1 y2 0≤y2 M1 For differentiation, limits not required.\n8\ng(y)=2 2≤y≤3\n3 A1 Fully correct with correct domain covering all reals.\n\n0 otherwise\n\nAlternative method for question 5(c)\nUsing chain rule (or inverse function): M1 M1 for changing to y.\nG(y)=F ( y2) so g(y)= d F ( y2) =2yf ( y2) .\ndy A1\n 2y ( 1 y2 ) 0≤y2 M2 Use of g(y)=2yf ( y2) twice, limits not required.\n 16\n M1 for one expression.\ng(y)=\n \n1\n2y  2≤y≤3\n  3 y2 \n  \n1 y2 0≤y2 A1 Fully correct with correct domain covering all reals.\n8\ng(y)=2 2≤y≤3\n3\n\n0 otherwise\n\n5\n© Cambridge University Press & Assessment 2025 Page 13 of 16",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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        "../answer-assets/9231_w25_ms_41-p13.png"
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    },
    {
      "id": "9231-2025-on-41-q06",
      "question_id": "9231-2025-on-41-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) H : = H:   B1 If in words, must contain ‘population means’.\n0 B A 1 B A\nAllow H : =0,H:  0 if defined or consistent with working.\n0 d 1 d\nDifferences: 5, 0, −1, 6,1 , −2, 8, 7, 15 M1 Attempt at signed differences.\ns2 =1 ( '405'−'39'2) =29.5 M1 Correct form for s2, note that s =5.43.\nd 8 9 d d\n'39' M1 Correct form.\n9 =2.393\n'29.5'\n9 A1 AWRT 2.39.\n'2.393'1.860, reject H / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in\n0\nterms of H ).\n1\nAllow M1 for comparison from 2-sample test using 1.746.\nTest result may be implied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nSufficient evidence to suggest new training programme results in A1 Correct conclusion from correct working ignoring hypotheses.\nreduced times. In context.\nOR Level of uncertainty in language is used (for example, not ‘prove’).\nSufficient evidence to support the coach’s belief. Do not accept statements such as “there is insufficient evidence to\nsuggest…”.\n7\n\n6(b) The population of differences may not be normally distributed. B1 Must mention population/distribution and differences.\nThe population of differences may not be symmetrical (and hence Accept “the sample may not be random” or “times may not be\nnot normal). independent”.\n1\n\n6(c) A Wilcoxon matched-pairs signed-rank test. B1 Must refer to pairs / paired.\nOR A paired-sample sign test. Accept “paired Wilcoxon test” or “paired sign test”.\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 16",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-41-q07",
      "question_id": "9231-2025-on-41-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "41",
      "question_number": 7,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) G (t)=a+2at+bt2 B1\nX\nG ' (t)=2a+2bt B1\nX\nE(X)=G ' (1)=2a+2b\nX\n2\n\n7(b)\nG\n''(t)=2b B1\nX\nVar(X)=2b+(2a+2b)−(2a+2b)2 M1 Use correct formula with their expressions.\n2b+(2a+2b)−(2a+2b)2 =2b+(2a+2b)(1−(2a+2b)) A1 AG, shown convincingly.\n=2b+2(a+b)(1−2a−2b)\nAlternative method for question 7(b)\nE ( X2) =02a+122a+22b=2a+4b B1\nVar(X)=(2a+4b)−(2a+2b)2 M1 Use correct formula with their expressions.\n(2a+4b)−(2a+2b)2 =2b+2a+2b−(2a+2b)2 A1 AG, shown convincingly.\n=2b+(2a+2b)(1−(2a+2b))\n=2b+2(a+b)(1−2a−2b)\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 16\n\n7(c) G (t)= ( a+2at+bt2)10 B1 FT FT their G X (t).\nY\nG\n'(t)=20(a+bt)( a+2at+bt2)9 M1 Attempt to differentiate their G\nY\n(t) and evaluate at t =1.\nY\nE(Y)=20(a+b)(3a+b)9\n3a + b =1 , hence E(Y)=20(a+b)=10E(X). A1 3a+b=1 since total probability is 1, OE.\n3\n\n7(d) Binomial *B1 Uses PGF to identify binomial.\n 2 DB1 Identifies binomial with correct parameters.\nB10, \n 3\n2\n© Cambridge University Press & Assessment 2025 Page 16 of 16",
      "source_pages": [
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_41.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_41.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2025-on-42-q01",
      "question_id": "9231-2025-on-42-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) Data has outliers / is skewed / lacks symmetry, suggesting B1 Must be in context.\npopulation/distribution may not be symmetrical. Condone no explicit mention of population/distribution or a definite\nstatement relating to population.\n1\n\n1(b) H : population median is [$]32,500. B1 Allow m, but not . If m defined, must see population.\n0\nH 1 : population median is not [$]32,500. Median not mean or average.\nTest statistic = 4 (or 11) B1 Can be implied by P(X ≤4) or P(X 4).\n(For use of Normal approximation, the value of the test statistic can be\nimplied by 4.5 or 3.5 for 4 and 10.5 or 11.5 for 11).\n[ X ~B(15,0.5)] B1 AWRT 0.0592.\nMight see 0.1184/0.1185/0.118/0.119 from2P(X ≤4) or 2P(X≥11).\nP(X≤4)=0.0592 OR P(X≥11)=0.0592\nUse of N(7.5,3.75) scores B1 for ±1.549.\n'0.0592'0.05 , accept H / do not reject H / not significant. M1 Compare their 0.0592 (must be a tail probability) with 0.05 (or 0.1 if\n0 0\n1-tail test used) OE and appropriate result (may be in terms of H ).\n1\nResult may be implied by an attempt at an appropriate conclusion in\ncontext ignoring hypotheses.\nUse of Normal: compare their ±1.549 with ±1.645, signs must be\nconsistent (1.282 for 1-tail test).\nInsufficient evidence to reject [company’s] claim. A1 Correct conclusion from correct working ignoring hypotheses.\nOR In context.\nInsufficient evidence to suggest that median [salary] is not Level of uncertainty in language is used (for example, not ‘prove’).\n[$]32,500. Allow “Not enough evidence / no sufficient evidence that / to show /\nconclude that…”.\nDo not accept “No evidence…”.\nDo not accept statements such as “there is sufficient evidence to\nsuggest…”.\n5\n© Cambridge University Press & Assessment 2025 Page 9 of 15",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-42-q02",
      "question_id": "9231-2025-on-42-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) s2 = 1 ( '476117'−'2390'2) = 326 =9.8788 M1 Correct expression, implied by AWRT 9.88.\n \n11 12 33\n3.106 B1 3.106 or 3.11 seen.\n'2390''3.106' '9.8788'\nM1 Correct form, must be a t-value.\n12 12\n196, 202 A1 Accept with inequality signs or open brackets.\nCondone [202, 196].\nDo not accept 1993.\n4\n\n2(b) −1.796 B1 1.796 or 1.80 seen.\n'199.17'−k '199.17'−k M1 Correct form, must be a t-value, consistent signs.\n'−1.796' or '1.796' Condone non-strict inequality.\n'9.8788' '9.8788'\n12 12\n200.8 A1 Accept 200.7, 200 or 201 from correct working.\nCondone final answer of the form k < 200.8 OE.\nCondone non-strict inequalities.\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 15",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2025-on-42-q03",
      "question_id": "9231-2025-on-42-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a) m=69.522 B1\nn=0.156 or 0.157 B1 Subtraction gives 0.156.\n2\n\n3(b) H : Po(0.7) fits the data. B1 Must mention distribution and data/number of breakdowns.\n0\nMay see H : Po(0.7) is a satisfactory model for the data.\nH : Po(0.7) does not fit the data. 0\n1\nH : Po(0.7) is not a satisfactory model for the data.\n1\nCombining last 3 columns give 13 and 6.828 or 6.829. M1 Combine expected frequencies less than 5.\n(88−99.317)2 (73−69.552)2 (13−6.829)2 M1 At least 2 expressions (not including 200) of correct form seen.\n+ ++ May see individual terms not summed.\n99.317 69.552 6.829\nMay be implied by 2 contributions correct to 3sf seen.\n[1.2896+0.1740+0.1142+5.5764]\n6.828 gives 5.5790.\nOR\n882 732 262 132\n+ + + −200\n99.317 69.552 24.333 6.829\n7.15 or 7.16 A1 SC B1 for 7.15 or 7.16 following M0.\n'7.15'7.815, accept H / do not reject H / not significant. M1 Compare their 7.15 with 7.815 (9.488 if 2 columns combined, 11.07 if\n0 0\nno columns combined) and appropriate result (may be in terms of H ).\n1\nTest result may be implied by an attempt at an appropriate conclusion\nin context ignoring hypotheses.\nInsufficient evidence to reject expert’s claim. A1 Correct conclusion from correct working ignoring hypotheses.\nOR In context.\nInsufficient evidence to suggest that Po(0.7) is not a good Level of uncertainty in language is used (for example, not ‘prove’)\nAllow ‘Not enough evidence that / to show / conclude that…’.\nfit/model [for the data].\nDo not accept statements such as ‘there is sufficient evidence to\nsuggest…’ or ‘no evidence…’.\n6\n© Cambridge University Press & Assessment 2025 Page 11 of 15",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-42-q04",
      "question_id": "9231-2025-on-42-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) a=−1 B1\n5\nb= 7 B1\n25\n2\n\n4(b) 1 1≤x4, M1 Attempt at differentiating both parts.\n5\n\nf(x)=1 x 4≤x≤6,\n25 A1 All correct, fully defined including domain.\n\n 0 otherwise. Domain must cover all reals, must not overlap at boundaries.\n2\n\n4(c) E ( X2) = 1 4 x2dx+ 1  6 x3dx=1 x3 4 + 1 x4 6 M1 Attempt to integrate x2f(x) with correct limits.\n5 1 25 4 15  1 100  4\nCan be implied by answer of 73 or 14.6.\n5\nVar(X)=73−(529)2 M1 Variance formula used with their E ( X2) .\n5 150\n2.16 A1\n3\n\n4(d) 10th percentile: 1x−1 = 1 M1 Award M1 for a correct method to find at least one of the 10th or 90th\n5 5 10\npercentile.\n90th percentile: 1 x2+ 7 = 9\n50 25 10 FT their a and b.\n10th percentile = 1.5 A1\n90th percentile = 31 A1 AWRT 5.57.\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 15",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-42-q05",
      "question_id": "9231-2025-on-42-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5 H :  = H:   B1 If in words, must have population means.\n0 A B 1 A B\nAllow H : =0 , H : 0.\n0 d 1 d\n'1331848'−'3264'2 [=136] M1 OE, may be implied by s2 =19.43.\n8 A\nBiased estimate of 17 earns M0 unless recovered.\n'964964'−'2406'2 [=158] M1 OE, may be implied by s2 =31.6.\n6 B\nBiased estimate of 26 earns M0 unless recovered.\n'136'+'158' M1 Correct form using their values, may be embedded.\ns2 = = 24.5\np 8+6−2\nA1 OE, for example 49 .\n2\n'3264'−'2406' M1 Correct form using their values.\nt= 8 6 =2.62\n's ' 1+1\np 8 6 A1 AWRT 2.62.\n'2.62'2.179, reject H / significant. M1 Compare their 2.62 with 2.179 and appropriate result (may be in terms\n0\nof H ).\n1\nTest result may be implied by an attempt at an appropriate conclusion\nin context ignoring hypotheses.\nSufficient evidence to suggest that there is a difference in the A1 Correct conclusion from correct working ignoring hypotheses.\nmean tensile strengths of steel rods produced by the two In context.\nmachines. Level of uncertainty in language is used (for example, not ‘prove’).\nAllow “(Some/enough) evidence that / to show / conclude that…”.\nDo not accept statements such as “there is insufficient evidence to\nsuggest…”.\n9\n© Cambridge University Press & Assessment 2025 Page 13 of 15",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2025-on-42-q06",
      "question_id": "9231-2025-on-42-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) G ' (t)= (3−2t)2 +4(3−2t)t   = 3+2t   M1 Attempt to differentiate G X (t).\nX (3−2t)4  (3−2t)3  May be implied by expressions for G ' (1).\n  X\nE(X)=G ' (1)=5 A1\nX\n2(3−2t)3 +6(3+2t)(3−2t)2  24+8t  M1 Attempt to differentiate G ' (t).\nG\n''(t)=\n= \nX\nX (3−2t)6  (3−2t)4  May be implied by expressions for G ''(1).\n  X\nVar(X)=G ''(1)+G ' (1)− ( G ' (1) )2 =32+5−25 M1 Use correct formula using their values.\nX X X\nVar(X)=12 A1 CWO\n \n5\n© Cambridge University Press & Assessment 2025 Page 14 of 15\n\n6(b)\nG (t)=\nt3\n= 1 t3( 1+(−4)( −2t ) + )\nM1 Multiply and attempt at binomial expansion.\nZ (3−2t)4 81 3\nP(Z =4) = their coefficient of t4. M1 Attempt at finding their coefficient of t3 or t4.\nEither seen earns M1.\nOR P(Z =3) = their coefficient of t3.\nP(Z =4)= 8 and P(Z =3)= 1 A1 Both values seen.\n   \n243 81\nP(Z 4)=1−' 8 '−' 1 ' M1 Attempt to find P(Z 4) using their values.\n243 81\nP(Z 4)= 232 A1 AWRT 0.955.\n243 Note: considering the expansions of the probability generating\nfunctions for X and Y separately, then finding P(Z 4)=1−P(Z≤4)\nwith P(Z≤4)=P(X =1,Y =2)+P(X =1,Y =3)+P(X =2,Y =2)\nearns full credit.\n5\n© Cambridge University Press & Assessment 2025 Page 15 of 15",
      "source_pages": [
        14,
        15
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_42.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_42.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2025-on-43-q01",
      "question_id": "9231-2025-on-43-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) s2 =1 ( '71314'−'844'2) =134 =8.933 M1 Correct expression, implied by AWRT 8.93.\n \n9 10 15\n2.262 B1 2.262 or 2.26 seen.\nM1 Correct form, must be a t-value.\n'844''2.262' '8.933'\n10 10\n[82.3, 86.5] A1 Accept with inequality signs or open brackets.\nCondone [86.5, 82.3].\nDo not accept 84.42.1.\nThe distribution of estimates of angles is normal. B1\nOR\nThe estimates are a random sample from some population.\nOR\nThe estimates are independent.\nOR\nUnderlying distribution is normal.\n5\n\n1(b) Population unlikely to be normal as  is close to right-angle / data B1 Must refer to the context.\nis skewed.\nOR\nEstimates may not be independent, for example due to collusion.\nOR\nNo indication that sample is random.\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 16",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-43-q02",
      "question_id": "9231-2025-on-43-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2(a) e−2.8(2.8)3 B1 Or by subtraction.\np= 100=22.248 Allow 22.2 or 22.3.\n3!\n(29−22.25)2 B1 AWRT 2.05.\nq= =2.05\n22.25\n2\n\n2(b) H : Po(2.8) fits the data. B1 Must mention distribution and data/number of cars.\n0\nH : Po(2.8) does not fit the data.\n1\nOR\nH : Po(2.8) is a satisfactory model for the number of cars (entering\n0\nthe car park).\nH : Po(2.8) is not a satisfactory model for the number of cars\n1\n(entering the car park).\n2.739+0.241+2.152+'2.049'+0.425+3.753+0.037 =11.396 M1 Sum of chi-square contributions using their q.\nMay be implied by 11.4.\n'11.4'12.59, accept H / do not reject H / not significant. M1 Compare their 11.4 with 12.59 and appropriate result (may be in\n0 0\nterms of H ).\n1\nTest result may be implied by an attempt at an appropriate\nconclusion in context.\nInsufficient evidence to suggest that Po(2.8) is not a good fit to the A1 Correct conclusion from correct working ignoring hypotheses.\ndata. In context.\nOR Level of uncertainty in language is used (for example, not ‘prove’).\nInsufficient evidence to suggest that the manager’s claim is false. Allow ‘Not enough evidence that / to show / conclude that…’\nDo not accept statements such as ‘there is sufficient evidence to\nsuggest…’ or ‘no evidence…..’\n4\n© Cambridge University Press & Assessment 2025 Page 10 of 16",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-43-chi-squared-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-43-q03",
      "question_id": "9231-2025-on-43-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "3 (x−x)2\n=90.36=3.24\nB1\n'16.6'2 M1 Can be implied by 0.172 ( = 43 ) or 0.688 ( = 86).\n'55.8'− 250 125\n5\n3.24+0.688 M1 Correct use of formula for pooled variance using their values.\n(=0.302)\n10+5−2\ns=0.550 A1 CAO\n4\nQuestion Answer Marks Guidance",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-43-q04",
      "question_id": "9231-2025-on-43-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "4 128(28+1) =203 B1 Both expressions correct. Implied by 203 and 1928.5 seen.\n4\n1 28(28+1)(56+1) =1928.5\n24\nT +0.5−'203' M1 Correct form or expression for CI with any z-value.\nz=\nCondone T or T−0.5 instead of T+0.5 for M1.\n'1928.5'\nT +0.5−'203' M1 Compares their z with 2.576.\n−2.576\n'1928.5' Condone T or T−0.5 instead of T+0.5 for M1.\nT 89.4 A1 Allow T =89.4.\nT =90 A1 All correct with no errors seen, including correct continuity\ncorrection.\n5\n© Cambridge University Press & Assessment 2025 Page 11 of 16",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-44-non-parametric-tests/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2025-on-43-q05",
      "question_id": "9231-2025-on-43-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a) 4 M1 Equation in terms of k formed following attempt to integrate with\n1   2 x 3 2   + 1   2 x  9 =1 correct limits.\n163  k 4\n0\n1 2\n(8−0)+ (3−2)=1\n24 k\nOR\n1 2\n+ =1\n3 k\n2 2 A1 AG, no errors seen.\n= , k =3\nk 3\n2\n\n5(b) 1  4 x dx+1 m x−1 2 dx= 1 *M1 Use of 4m9 to form equation equal to 0.5, OE.\n16 3 2 Equation may be in terms of k.\n0 4\nOR\n1 9 x−1 2 dx=0.5\n3\nm\n1+ 2 ( m−2 ) = 1 DM1 Integrate and form an equation in m .\n3 3 2\nOR\n( )\n2 3− m = 1\n3 2\nm= 81 [=5.0625] A1 Accept AWRT 5.06.\n16\n3\n© Cambridge University Press & Assessment 2025 Page 12 of 16\n\n5(c)   1 x 3 2 0≤x4 M1 Attempt to integrate both parts, limits not required.\nF(x)= 24 May be seen in part 5(a).\n 2x 1 2 −1 4≤x≤9\n3\nB1 For constant –1 obtained in expression for 4≤x≤9.\nMay be seen in part 5(a).\n1 y3 0≤y2 M1 For changing to y.\nG(y)=24\nLimits not required.\n 2 y−1 2≤y≤3\n3\n1 y2 0≤y2 M1 For differentiation, limits not required.\n8\ng(y)=2 2≤y≤3\n3 A1 Fully correct with correct domain covering all reals.\n\n0 otherwise\n\nAlternative method for question 5(c)\nUsing chain rule (or inverse function): M1 M1 for changing to y.\nG(y)=F ( y2) so g(y)= d F ( y2) =2yf ( y2) .\ndy A1\n 2y ( 1 y2 ) 0≤y2 M2 Use of g(y)=2yf ( y2) twice, limits not required.\n 16\n M1 for one expression.\ng(y)=\n \n1\n2y  2≤y≤3\n  3 y2 \n  \n1 y2 0≤y2 A1 Fully correct with correct domain covering all reals.\n8\ng(y)=2 2≤y≤3\n3\n\n0 otherwise\n\n5\n© Cambridge University Press & Assessment 2025 Page 13 of 16",
      "source_pages": [
        12,
        13
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2025-on-43-q06",
      "question_id": "9231-2025-on-43-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) H : = H:   B1 If in words, must contain ‘population means’.\n0 B A 1 B A\nAllow H : =0,H:  0 if defined or consistent with working.\n0 d 1 d\nDifferences: 5, 0, −1, 6,1 , −2, 8, 7, 15 M1 Attempt at signed differences.\ns2 =1 ( '405'−'39'2) =29.5 M1 Correct form for s2, note that s =5.43.\nd 8 9 d d\n'39' M1 Correct form.\n9 =2.393\n'29.5'\n9 A1 AWRT 2.39.\n'2.393'1.860, reject H / significant. M1 Compare their 2.393 with 1.860 and appropriate result (may be in\n0\nterms of H ).\n1\nAllow M1 for comparison from 2-sample test using 1.746.\nTest result may be implied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nSufficient evidence to suggest new training programme results in A1 Correct conclusion from correct working ignoring hypotheses.\nreduced times. In context.\nOR Level of uncertainty in language is used (for example, not ‘prove’).\nSufficient evidence to support the coach’s belief. Do not accept statements such as “there is insufficient evidence to\nsuggest…”.\n7\n\n6(b) The population of differences may not be normally distributed. B1 Must mention population/distribution and differences.\nThe population of differences may not be symmetrical (and hence Accept “the sample may not be random” or “times may not be\nnot normal). independent”.\n1\n\n6(c) A Wilcoxon matched-pairs signed-rank test. B1 Must refer to pairs / paired.\nOR A paired-sample sign test. Accept “paired Wilcoxon test” or “paired sign test”.\n1\n© Cambridge University Press & Assessment 2025 Page 14 of 16",
      "source_pages": [
        14
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2025-on-43-q07",
      "question_id": "9231-2025-on-43-q07",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "43",
      "question_number": 7,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) G (t)=a+2at+bt2 B1\nX\nG ' (t)=2a+2bt B1\nX\nE(X)=G ' (1)=2a+2b\nX\n2\n\n7(b)\nG\n''(t)=2b B1\nX\nVar(X)=2b+(2a+2b)−(2a+2b)2 M1 Use correct formula with their expressions.\n2b+(2a+2b)−(2a+2b)2 =2b+(2a+2b)(1−(2a+2b)) A1 AG, shown convincingly.\n=2b+2(a+b)(1−2a−2b)\nAlternative method for question 7(b)\nE ( X2) =02a+122a+22b=2a+4b B1\nVar(X)=(2a+4b)−(2a+2b)2 M1 Use correct formula with their expressions.\n(2a+4b)−(2a+2b)2 =2b+2a+2b−(2a+2b)2 A1 AG, shown convincingly.\n=2b+(2a+2b)(1−(2a+2b))\n=2b+2(a+b)(1−2a−2b)\n3\n© Cambridge University Press & Assessment 2025 Page 15 of 16\n\n7(c) G (t)= ( a+2at+bt2)10 B1 FT FT their G X (t).\nY\nG\n'(t)=20(a+bt)( a+2at+bt2)9 M1 Attempt to differentiate their G\nY\n(t) and evaluate at t =1.\nY\nE(Y)=20(a+b)(3a+b)9\n3a + b =1 , hence E(Y)=20(a+b)=10E(X). A1 3a+b=1 since total probability is 1, OE.\n3\n\n7(d) Binomial *B1 Uses PGF to identify binomial.\n 2 DB1 Identifies binomial with correct parameters.\nB10, \n 3\n2\n© Cambridge University Press & Assessment 2025 Page 16 of 16",
      "source_pages": [
        15,
        16
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_43.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_43.pdf?download=true",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2025-on-44-q01",
      "question_id": "9231-2025-on-44-q01",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a) x=723, x2 =42261 B1 x=723 implied by AWRT 55.6 for x.\nx2 =42261 implied by AWRT 171 for s2(or 158 if biased).\ns2 = 1 ( '42261'−'723'2) = 2222 =170.9231 M1 Correct form, must be unbiased, s=13.07.\n \n12 13 13 Implied by AWRT 171.\n1.782 B1 1.782 or 1.78 seen.\n'723' '170.9231' M1 Correct form, must be a t-value.\n'1.782'\n13 13\n49.2, 62.1 A1 Accept with inequality signs or open brackets.\nCondone 62.1,49.2.\nDo not accept 55.66.5.\n5\n\n1(b) The population (of car speeds) is normally distributed. B1 Population mean normally distributed earns B0.\nOR (Underlying) data is normal earns B0.\nUnderlying distribution is normal. It is normal distribution earns B0.\n1\n© Cambridge University Press & Assessment 2025 Page 9 of 17",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9231_w25_ms_44.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2025-oct-nov/9231_w25_ms_44.pdf?download=true",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    {
      "id": "9231-2025-on-44-q02",
      "question_id": "9231-2025-on-44-q02",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "2(a) 0 x0 M1 Attempt to integrate both parts, condone missing c , ignore\n domains.\n 2x2 + 4x 0≤x1\n( F(x)= )\n\n9 9\n1(x−2)3 +1 1≤x≤2 A1 Both parts correct with correct c , ignore domains.\n3\n\n1 x2\nB1 Correct domains. Must cover all reals, may overlap at boundaries.\nMust have 0 for x0 and 1 for x2.\n3\n\n2(b) 2m2 +4m=1 M1 Identifies correct interval for their piecewise F(x) to form an\n9 9 2\nequation of the form equation F(m)= 1 OE.\n2\n( ) A1 OE, must clearly state one answer.\nm= 1 −2+ 13\n2\n2\n\n2(c) 0 y0 M1 Substitute x= y2 into both parts of their F(x), ignore domains.\n\n( G(y)= )\n\n  1\n9\n2\n(\ny\ny\n4\n2\n+\n−\n9\n4\n2\ny\n)\n2\n3 + 1 1\n0\n≤\n≤\ny\ny\n≤\n1\n2\nA1 Correct functions in terms of y, AEF.\n3\n B1 Correct domain in terms of y. Domain may overlap at boundaries.\n1 y 2\nCondone missing 0 for y0 but must have 1 for y 2 .\n3\n© Cambridge University Press & Assessment 2025 Page 10 of 17\n\n2(d) G ( 1 ( −2+ 13 )) =0.379 M1 Evaluates G(their m).\n2\n0.3790.5 so the median of Y is greater than the median of X . A1\nAlternative method for question 2(d)\n−2+ 13 M1 Expresses m in terms of m .\nm2 = y x\ny 2\nm2 =m 1 and hence m m . A1 Uses the fact squaring a number between 0 and 1 makes it smaller.\ny x y x\nAlternative method for question 2(d)\nSolves 2y4+4y2 =1. M1 Identifies correct interval for their piecewise G(y) to form\n9 9 2\nequation of the form G(y)= 1 OE.\n2\nm =0.8960.803=m A1 0.803 may be seen in part 2(b).\ny x\n( ) ( )\nA0 for 1 −2+ 13 0.896 without 1 −2+ 13 evaluated.\n2 2\n2\n© Cambridge University Press & Assessment 2025 Page 11 of 17",
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    {
      "id": "9231-2025-on-44-q03",
      "question_id": "9231-2025-on-44-q03",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "3(a) H : = H:   B1 If in words, must have population means.\n0 Y X 1 Y X\nAllow H : =0, H:  0 if d defined or consistent with\n0 d 1 d\nworking.\nDifferences: 3, –1, 6, 2, 3, 3, –1, –2, 3, 3 M1 Attempt at signed differences, SOI.\ns2 = 1 ( '91'−'19'2) =6.1 M1 Correct form, unbiased, implied by s2 =6.1 or s =2.47 seen.\nd 9 10 d d\n'19' M1 Correct form using their s2.\n10 =2.43 d\n'6.1' '696'−'677'\n10 Allow M1 for 10 10 =0.666 from 2-sample t-test.\n's ' 1 + 1\np 10 10\nA1 AWRT 2.43.\n'2.43'2.821, accept H / do not reject H / not significant. M1 Compare their 2.43 with 2.821 and appropriate result (may be in\n0 0\nterms of H ).\n1\nAllow M1 for comparing their 0.666 with 2.552 and appropriate\nresult from 2-sample t-test.\nTest result may be implied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nTheir 2.43 must come from an attempt at standardising.\nInsufficient evidence to support the manufacturer’s claim. A1 Correct conclusion from correct working ignoring hypotheses.\nOR In context.\nInsufficient evidence to support that (on average) machine Y Level of uncertainty in language is used (for example, not ‘prove’).\nextracts more juice than machine X. Allow ‘Not enough evidence / no sufficient evidence that / to show\n/ conclude that…’.\nDo not accept ‘No evidence…’.\nDo not accept statements such as ‘there is sufficient evidence to\nsuggest…’.\n7\n© Cambridge University Press & Assessment 2025 Page 12 of 17\n\n3(b) Population of differences is normally distributed. B1 Population mean normally distributed earns B0.\nOR (Underlying) data is normal earns B0.\nUnderlying distribution of differences is normal. Must mention population/distribution and differences.\nFollowing 2-sample test in part 3(a), allow either of the following:\nPopulations/underlying distributions are normal.\nPopulations have equal variances.\nMust refer to both populations.\n1\n\n3(c) The difference remains the same for pair 4. The set of differences *B1 Comment that recognises that the differences remain unchanged.\nhas not changed.\nOR OR\nRedo the test: Compare their 2.43 from part 3(a) with 2.821.\n2.432.821 Following 2-sample test in part 3(a), compare their 0.653 with\n2.552.\nThe conclusion remains the same. DB1\n2\n© Cambridge University Press & Assessment 2025 Page 13 of 17",
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    {
      "id": "9231-2025-on-44-q04",
      "question_id": "9231-2025-on-44-q04",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) G (t)= 2t+3t2 B1 CAO\nX 5 5\n1\n\n4(b) G (t)= 1 t4(2+3t)4  =t4(2+3t )4 M1 Attempt at 4th power of their G X (t).\nY 625  5 5 \nA1\n2\n\n4(c) Attempt at coefficient of t6intheirG (t). M1 Identify required term, coefficient must a non-zero probability.\nY\nMay be seen in part 4(b).\nP(Y =6)= 216 A1 AWRT 0.346.\n625\n2\n© Cambridge University Press & Assessment 2025 Page 14 of 17\n\n4(d) G '(t)= 4 t3(2+3t)4 + 12 t4(2+3t)3 M1 Attempt at differentiating their G (t) from part 4(b).\nY 625 625 Y\nOR G '(t)= 1 ( 64t3+480t4 +1296t5 +1512t6 +648t7)\nY 625\nG ''(t)= 12 t2(2+3t)4 + 96 t3(2+3t)3 +108t4(2+3t)2 M1 Attempt at differentiating their G '(t).\nY 625 625 625 Y\nOR G ''(t)= 1 ( 192t2 +1920t3+6480t4 +9072t5+4536t6)\nY 625\nG\n'(1)=32\n=6.4, G\n''(1)=888=35.52 B1 WWW\nY 5 Y 25\nVar(Y)=888+32− (32)2 M1 Use correct formula with their values.\n25 5 5\nVar(Y)= 24 =0.96 A1 CWO\n25\nAlternative method for question 4(d)\nE(X)=2+32=8or useG ' (t)= 2+6t. M1\n5 5 5 X 5 5\nE ( X2) = 2+34=14 or useG ''(t)= 6 M1\n5 5 5 X 5\nVar(X)=14− (8)2 or Var(X)= 6+8− (8)2 M1\n5 5 5 5 5\nVar(X)= 6 A1\n25\nVar(Y)=4Var(X)= 24 A1 CWO\n25\n5\n© Cambridge University Press & Assessment 2025 Page 15 of 17",
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    {
      "id": "9231-2025-on-44-q05",
      "question_id": "9231-2025-on-44-q05",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "5 H : performance is independent of age. B1 Allow dependent, association or relationship for not independent.\n0\nH : performance is not independent of age. For example:\n1\nOR H : No association/relationship between…\n0\nH : passing test is independent of age. H : Association/relationship between…\n0 1\nH : passing test is not independent of age. Must refer to “performance” and “age” in H .\n1 0\n34 (27.93) 41 (44.69) 6 (8.38) M1 At least 3 correct to 3sf (must be seen).\n16 (22.07) 39 (35.31) 9 (6.62)\nA1 All correct to 3sf.\n(34−27.93)2 (41−44.69)2 (9−6.62)2 M1 At least 2 expressions (not including 145) of correct form seen.\n+ ++ May see individual terms not summed.\n27.93 44.69 6.62\nMay be implied by 2 contributions correct to 3sf seen.\n1.319+0.305+0.676+1.669+0.386+0.856\nMay be implied by AWRT 5.21.\nOR\n 342 412 92 \n + ++ −145\n27.93 44.69 6.62\n \n5.21 A1 AWRT 5.21.\n'5.21'4.605, reject H / significant. M1 Compare their 5.21 with 4.605 and appropriate result (may be in\n0\nterms of H ).\n1\nTest result may be implied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nSufficient evidence to suggest that performance and age are not A1 Correct conclusion from correct working ignoring hypotheses.\nindependent. Must refer to both performance and age.\nOR Sufficient evidence to suggest that performance and age are In context.\ndependent. Level of uncertainty in language is used (for example, not ‘prove’).\nOR Sufficient evidence to suggest that performance and age are Do not accept statements such as ‘there is insufficient evidence to\nassociated. suggest…’.\nOR Sufficient evidence to support the (driving) instructor’s belief.\n7\n© Cambridge University Press & Assessment 2025 Page 16 of 17",
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    },
    {
      "id": "9231-2025-on-44-q06",
      "question_id": "9231-2025-on-44-q06",
      "subject": "9231",
      "year": 2025,
      "session": "Oct/Nov",
      "session_code": "on",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "6(a) H : population median mark for school A is equal to population B1 Allow m, but not . If m defined, must see population.\n0\nmedian mark for school B.\nH : population median mark for school A is not equal to\n1\npopulation median mark for school B.\n1123(123+147+1) =16666.5 B1 Both expressions correct.\n2\nImplied by 16666.5 and 408329.25 seen.\n1 123147(123+147+1) =408329.25\n12\n15355+0.5−'16666.5' M1 OR attempts to find critical value or p-value using their mean and\nvariance. Condone continuity correction omitted or incorrect.\n'408329.25'\n−2.05 A1 AWRT −2.05 or CV = 15414 or p=0.0201.\nCondone continuity correction omitted or incorrect.\n−2.05−1.96 OR 0.9800.975 OR 0.0200.025 OR M1 Compare their 2.05 with 1.96 (consistent signs) or their 0.980\n15355+0.515414, reject H / significant. with 0.975, or their 0.020/0.040 with 0.025/0.05 or their 15355.5\n0\nwith 15414 (condone continuity correction omitted or incorrect).\nConsistent signs and appropriate result (may be in terms of H ).\n1\nTest result can be implied by an attempt at an appropriate\nconclusion in context.\nSufficient evidence to suggest that there is a difference in average A1 Correct conclusion from correct working, ignoring hypotheses.\nmarks between the students in school A and B. Must consider continuity correction e.g. 15355.5 seen.\nOR In context.\nSufficient evidence to suggest that, on average, the test scores for Level of uncertainty in language is used (for example, not ‘prove’).\nstudents at the two schools are not equal. Do not accept statements such as ‘there is insufficient evidence to\nsuggest…’.\n6\n\n6(b) The (underlying) distributions are symmetrical/similar shape. B1\nOR The populations are symmetrical/similar shape.\nOR No repeated marks.\n1\n© Cambridge University Press & Assessment 2025 Page 17 of 17",
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    {
      "id": "9231-2026-mj-11-q01",
      "question_id": "9231-2026-mj-11-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a)\n© Cambridge University Press & Assessment 2026 Page 9 of 23\n( 4 r − 3 ) (1 2 r + 3 ) = 4 8 r 2 − 2 4 r − 9 B1\n4 8 ( 16 n ( n + 1 ) ( 2 n + 1 ) ) − 2 4 ( 12 n ( n + 1 ) ) − 9 n M1 Substitutes formulae for  r 2 and  r .\n=16n3+12n2 −13n A1 CAO Must be in this form.\n3\n\n1(b) 1 1  1 1  B1 Finds partial fractions. Accept equivalent\n=  −  form.\n(4r−3)(12r+3) 124r−3 4r+1\nr\nn\n= 1\n( 4 r − 3 )\n1\n(1 2 r + 3 )\n=\n1\n1\n2\n 1\n1\n−\n1\n5\n\n+\n1\n1\n2\n 1\n5\n−\n1\n9\n\n+\n1\n1\n2\n 1\n9\n−\n1\n1 3\n M1 Write enough correct sections of terms so\nthat all that are left in the answer appear\nplus one pair that cancel.\n+  +\n1\n1\n2\n\n4 n\n1\n− 3\n−\n4 n\n1\n+ 1\n A1 All necessary complete terms to\ndemonstrate the cancellation. Either two\nat the beginning and one at the end or one\nat the beginning and two at the end.\n=\n1\n1\n2\n−\n1 2 ( 4\n1\nn + 1 )\nB1 Accept any equivalent form.\n4\n\n1(c) \nr\n= 1\n( 4 r − 3 )\n1\n(1 2 r + 3 )\n=\n1\n1\n2\nB1 FT their answer for 1(b) of similar form.\n1\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-11-q02",
      "question_id": "9231-2026-mj-11-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 EITHER\n© Cambridge University Press & Assessment 2026 Page 10 of 23\n3 6 + 9 2 + 8 9 = 9 1 0\n(B1 Checks base case.\nis divisible by 91.\nAssume that 36k + 92k + 89 is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it\nclear it is an assumption.\nB0 if assume true for ALL values of k.\nThen 3 6 (k + 1 ) + 9 2 k + 1 + 8 9 = ( 7 2 8 + 1 ) 3 6 k + ( 9 1 + 1 ) 9 2 k + 8 9 M1\n= 7 2 8  3 6 k + 9 1  9 2 k + ( 3 6 k + 9 2 k + 8 9 ) A1 Separates 36k + 92k +89\nA1 Connects the divisibility with the is divisible by 91 because both 72836k +9192k and are divisible by 91.\nhypothesis.\nHence, [by induction,] 3 6 n + 9 2 n + 8 9 is divisible by 91for every positive integer n. A1) Must have previous three marks and state\nthe result in full.\nOR\nAlternative method for 2\n3 6 + 9 2 + 8 9 = 9 1 0 is divisible by 91. (B1 Checks base case.\nAssume that ( 3 6 k + 9 2 k + 8 9 ) is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it\nclear it is an assumption.\nB0 if assume true for ALL values of k.\nThen 36(k+1) +92k+1+89−(36k +92k +89) M1 Considers difference between consecutive\nterms.\n=\n=\n(\n9\n7\n1\n2\n(\n8\n8\n)\n\n3\n3\n6 k\n6 k\n+\n+\n( 9\n9\n1\n2\n)\nk\n9\n)\n2 k\n36k +92k +89\nA1\nA1 Connects the divisibility with the\nhypothesis.\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2 Hence, [by induction,]\n© Cambridge University Press & Assessment 2026 Page 11 of 23\n3 6 n + 9 2 n + 8 9 is divisible by 91 for every positive integer n. A1) Must have previous three marks and state\nthe result in full.\n6\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-11-q03",
      "question_id": "9231-2026-mj-11-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 3,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a)  k\n0\n0\n1\n B1 Stretch parallel to the x-axis, scale factor\nk.\n 1\n0\n2\n1\n B1 Shear, x-axis fixed, with (0,1) mapped to\n( 2 ,1 ) .\nM =\n 1\n0\n2\n1\n  k\n0\n0\n1\n M1 Correct order.\nM =\n 1\n0\n\n\nk\nk\n+\n+\n2\n1\n\n\n0\n0\n1\n0\n\n\n0\n0\n+\n+\n2\n1\n\n\n1\n1\n\n=\n k\n0\n2\n1\n\n.\nA1 AG Some intermediate step must be\nshown. At least one element not in final\nform.\n4\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b)\n© Cambridge University Press & Assessment 2026 Page 12 of 23\n k\n0\n2\n1\n  x\ny\n\n=\n k x +\ny\n2 y  B1\nTransforms\n x\ny\n\nMust see matrix applied\nto both coordinates.\nkx+2y=x M1 x x\nUses M =  . Must be considering a\ny y\nfixed point.\ny = 12 ( 1 − k ) x A1\n3\n\n3(c) Area of P = d e t M  Area of S 1= 1|k| M1 Uses\n2\nd e t M = k with P and S the correct\nway round.\nk =  2 A1 CAO\n2\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-11.pdf?download=1",
      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2026-mj-11-q04",
      "question_id": "9231-2026-mj-11-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 13 of 23\ny = x − 2  x = y − 12\n 5 y − 32 − 2 y − 12 + 1 = 0   y 32 = 2 y − 5 \n(B1 Correct substitution seen used in equation.\ny 3 = ( 2 y − 5 ) 2 M1 Squares to eliminate fractional powers.\ny 3 − 4 y 2 + 2 0 y − 2 5 = 0 A1) Must be an equation with like terms\ncollected.\nOR\nAlternative for 4(a)\n( 5 x 2 − 2 ) x = − 1 (M1 Squares to eliminate odd powers of x.\n( 5x2 −2 )2 x2 =1\n( 5 y − 1 − 2 ) 2 y − 1 = 1 B1 Correct substitution used in the equation.\ny3 −4y2 +20y−25=0 A1) Must be an equation with positive powers\nand like terms collected.\n3\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b)\n© Cambridge University Press & Assessment 2026 Page 14 of 23\n2 2 2 4    − + − + − = 2 2 2 2 2 2 2 0       − − + − − + − − = B1 FT Using their answer to part (a). Must be\nusing a cubic equation with positive\npowers and like terms collected.\n−4+−4+−4 =42−2(20) M1\n2 (\n4\n2\n4\n2\n4\n2\n(\n2\n2\n2\n2\n2 )\n2 ) 2      \n     \n− +\n−\n−\n−\n+\n+\n−\n−\n=\n− +\n− +\n−\n−\n−\n+ − −\n= − 2 4 A1 FT FT from a cubic equation answer for 4(a)\n3\n\n4(c) S\n− 6\n= 4 S\n− 4\n− 2 0 S\n− 2\n+ 7 5 M1 Using 6 4 4 2 0 2 2 5 0    − − − + − − = Must\nuse S\n0\n= 3\nFT for their cubic from 4(a) only if using\nthis method.\nS\n− 6\n= − 1 0 1 A1 CAO\nS =4S −20S +25S M1 Using\n−8 −6 −4 −2\n8 4 6 2 0 4 2 5 2 0     − − − + − − − =\nFT for their cubic from 4(a) only if using\nthis method.\nS\n− 8\n= 1 7 6 A1 CAO\n4\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_11.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-11.pdf?download=1",
      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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        "../answer-assets/9231_s26_ms_11-p14.png"
      ]
    },
    {
      "id": "9231-2026-mj-11-q05",
      "question_id": "9231-2026-mj-11-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 15 of 23\n − 3\n− 2\n4\n\n−\n − 2\n4\n4\n\n=\n −\n−\n0\n1\n6\n\n(B1 Finds direction of one line to another.\ni\n3\n2\n−\nj\n4\n4\nk\n2\n3\n=\n − 2 0\n− 5\n2 0\n\n~\n − 4\n− 1\n4\n M1 Find common perpendicular by taking\ncross product of the two correct direction\nvectors.\nA1 Correct.\n−1 −4 M1 Uses formula for shortest distance using\n1     10 their common perpendicular and direction\n−6  −1 =\n    vector. Scalar product and magnitudes\n33 33\n   \n 0   4  must be evaluated.\nA1) Correct answer. Accept 1.74\nOR\nAlternative method for 5(a)\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(a) 1\n1\n0\n6\n1\n5\n =\n1\n1\n3\n6\n9\n5\n =\nA1 Solves equations from using\nperpendicular to both lines to find\na n d  \n−40 M1 Finds PQ\n33\n \n−10\n 33\n 40 \n \n33\nM1 Finds length of PQ\nPQ=\n( −40)2\n+\n( −10)2\n+\n(40)2\n33 33 33\n=\n1 0\n3 3\nA1)\n5\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)\n© Cambridge University Press & Assessment 2026 Page 17 of 23\ni\n− 3\n3\nj\n5\n− 4\nk\n3\n2\n=\n 2\n1\n−\n2\n5\n3\n *M1 Finds the normal to\n1\n by taking cross\nproduct of two correct direction vectors.\nA1 Correct.\ni j k −22 *M1 Finds the normal to  by taking cross\n2\n \n−3 −2 4 = 17 product of two correct direction vectors.\n \n \n2 4 3 −8\n  A1 Correct\n2\n1\n2\n5\n3\n1\n2 2\n7\n8\n7 1 8 8 3 7 c o s c o s\n7 1\n2\n8\n0 5\n8 3 7\n \n\n−\n\n\n −\n−\n\n=  =\n−\nDM1 Uses dot product of their normal vectors.\nScalar product and magnitudes must be\nevaluated.\n1 8 0 7 4 .7  − =  A1 Or 1.30 rad. Mark final answer. AWRT\n74.7 (1dp) or 1.3 (1dp)\n6\nQuestion Answer Marks Guidance",
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_11.pdf",
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      "html": "9231-topic-16-vectors/answers.html",
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    },
    {
      "id": "9231-2026-mj-11-q06",
      "question_id": "9231-2026-mj-11-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x = − 1 B1 Vertical asymptote.\n(x+1)(x−2)+4 M1 Oblique asymptote - must show attempt to\ny=\nfind constant term.\n4x+4\ny = 14 x − 12 A1 Must be a linear equation.\n3\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) dy 4x2 +8x−12\n= =0\ndx (4x+4)2\n© Cambridge University Press & Assessment 2026 Page 18 of 23\n4 x 2 + 8 x − 1 2 = 0\nM1\nSets\nd\nd\ny\nx\n= 0 .\n(1,1), (−3,−7) A1 A1 for x-coordinates.\n4 4\nA1 A1 for y-coordinates correctly linked to x-\ncoordinates.\nAllow A1 for one completely correct\npoint.\n3\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c) B1 FT Axes and asymptotes labelled.\nFollow through their two asymptotes from\n(a) for this mark only.\nB1 Left branch correct and also all\napproaches to the asymptotes.\nB1 Right branch correct including a smooth\nminimum in the first quadrant.\n3\n© Cambridge University Press & Assessment 2026 Page 19 of 23\ny = 14 x − 12\nx = − 1\ny\nx\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(d) B1 Must be fully correct, including vertical\nasymptote.\n1\n\n6(e)\n© Cambridge University Press & Assessment 2026 Page 20 of 23\nx 2\n4\n−\nx\nx\n+\n+\n4\n2\n= − 2 or\nx 2\n4\n−\nx\nx\n+\n+\n4\n2\n= 2\nx 2 + 7 x + 1 0 = 0 x 2 − 9 x − 6 = 0\nM2 Finds critical points, award M1 for each\ncase. Must be using equations.\nx=−5,−2 x = 12\n(\n9 − 1 0 5\n)\n, x = 12\n(\n9 + 1 0 5\n) A1 All four correct, exact answers.\n− 5  x  − 2 12\n(\n9 − 1 0 5\n)\n x  12\n(\n9 + 1 0 5\n)\ny\nA1 CAO\n4\nx\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-11.pdf?download=1",
      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    },
    {
      "id": "9231-2026-mj-11-q07",
      "question_id": "9231-2026-mj-11-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "11",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 17,
      "status": "available",
      "reason": null,
      "text": "7(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 21 of 23\ny\n1 t\na\na n 2\n12\ns i n \n\n=\n\n+\n\n(B1 Uses y r s i n  =\nd\nd\ny\n1 t\na\na n 2\n12\nc o s 12 s i n\n1 t\na\na n 2\n12 ( 2\n( 1\na s\nt\ne\na\nc\nn\n2\n2\n2\n)\n)2 \n \n   \n \n=\n\n+\n\n−\n\n+\n\n−\n+\n*M1 Differentiates using product rule\nd\nd\ny\n( a ( 1 t a n 2 ) ) 12 c o s 12\n1\n1\nt a n\na\nt a\n2\nn 2\n32\n( 2 a s e c 2 2 ) s i n    \n\n \n \n=\n+ − +\n+\n\n− A1 or quotient rule. Must be using correct\nfunction.\nc o s\ns e\n1\nc 2 2\nt a n\ns i\n2\nn\nc o s\nc o s 2 ( s i\ns\nn\ni n\n2 c o s 2 )\n0\n  \n \n   \n −\n+\n= −\n+\n=\nDM1 A1 Puts derivative = 0 and rearranges.\nAG. Intermediate steps must be shown.\nc o s ( 0 .5 ) −\nc o s ( 2  0 .5 ) ( s i n\ns i n\n( 2\n(\n\n0 .5\n0 .5\n))\n+ c o s ( 2  0 .5 ) )\n= 0 .2 3 5 and\nc o s ( 0 .6 ) −\nc o s ( 2  0 .6 ) ( s i n\ns i n\n( 2\n(\n\n0 .6\n0 .6\n))\n+ c o s ( 2  0 .6 ) )\n= − 0 .3 7 9\nB1) Shows sign change. Values calculated to\nat least one decimal place.\nOR\nAlternative method for 7(a)\ny 2\n1 t\na\na n 2\ns i n 2 \n\n=\n\n+\n (B1 Uses y r s i n  =\n−2 *M1 Differentiates using product rule\n dy  1   1 \n\n2y\n\n=\n \n2asincos−asin2\n \nsec22\n d 1+tan2 1+tan2\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(a)\n© Cambridge University Press & Assessment 2026 Page 22 of 23\n2 y d\nd\ny ( ( 1 t a n 2 ) 2 a s i n c o s 2 a s i n 2 s e c 2 2 )\n1 t\n1\na n 2\n2\n    \n \n  = + − \n+\n\nA1 or quotient rule. Must be using correct\nfunction.\nsec22sin sin DM1 Puts derivative = 0 and rearranges.\ncos− =cos− =0\n1+tan2 cos2(sin2+cos2)\nA1 AG. Intermediate steps must be shown.\nsin(0.5)\ncos(0.5)− =0.235 and\ncos(20.5)( sin(20.5)+cos(20.5))\nc o s ( 0 .6 ) −\nc o s ( 2  0 .6 ) ( s i n\ns i n\n( 2\n(\n\n0 .6\n0 .6\n))\n+ c o s ( 2  0 .6 ) )\n= − 0 .3 7 9\nB1)\nShows sign change. Values calculated to\nat least one decimal place.\n6\n\n7(b) B1 Initial line and pole clear. Curve in first\nquadrant through\n(\na , 0\n)\nand O.\nCondone pole not labelled but must be\nclearly identifiable.\n(\na , 0\n)\nnot needed.\nB1 Correct shape (gradient at end points and\napproximately symmetrical) Maximum\npoint should lie between 40% and 60% of\nwidth of the arc.\n2\nO ( a , 0 )\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 23 of 23\n12 14\n0\nπ r 2 d a2 14\n0\nπ\n1 t\n1\na n 2\nd  \n\n = \n+\nB1 Using correct formula and correct limits.\nv t a n 2\nd\nd\nv\n2 s e c 2 2  \n\n=  =\nM1 Applies given substitution. Condone\nd\nd\nv\ns e c 2 2 \n\n=\n 1 4 π 1   1 1 M1 Clearly applies\na d= a dv\n2 1+tan2 4 1+v sec22\n0 0\ns e c 2 2 1 t a n 2 2   = +\nAllow sec22=1+v2written explicitly.\na4\n0 1\n1\nv 1 t a\n1\nn 2 2\nd v a4\n0\n( 1 v )\n1(\n1 v 2 )\nd v\n\n=\n \n+ +\n=\n\n\n+ +\nA1 Some working must be shown, including\ncalculations supporting change of limits.\nAG.\n( 1 + v )\n1(\n1 + v 2 )\n= 12\n\nv\n1\n+ 1\n−\nv\n2 v\n−\n+\n1\n1\n *M1 Partial fractions of correct form.\n= 12\n\nv\n1\n+ 1\n−\nv 2\nv\n+ 1\n+\nv 2\n1\n+ 1\n A1\nand separates\n−\nv\nv\n2\n+\n+\n1\n1\n.\na8  ln ( v + 1 ) − 12 ln ( v 2 + 1 ) + t a n − 1 v \n\n0\nDM1 Applies all three formulae from MF19 to\na three-term expression from above.\nFollow through their constants from line\nabove.\nA1 Correct.\n A1 The four previous marks must have been\n  v+1  \naln +tan−1v = 1 πa awarded and logarithms must be\n8    v2 +1    16 combined for this mark to be awarded.\n0\n9\nTrace ID: PC-1563EA0F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-15-polar-coordinates/answers.html",
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    },
    {
      "id": "9231-2026-mj-12-q01",
      "question_id": "9231-2026-mj-12-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "1(a)\n© Cambridge University Press & Assessment 2026 Page 9 of 23\n( 4 r − 3 ) (1 2 r + 3 ) = 4 8 r 2 − 2 4 r − 9 B1\n4 8 ( 16 n ( n + 1 ) ( 2 n + 1 ) ) − 2 4 ( 12 n ( n + 1 ) ) − 9 n M1 Substitutes formulae for  r 2 and  r .\n=16n3+12n2 −13n A1 CAO Must be in this form.\n3\n\n1(b) 1 1  1 1  B1 Finds partial fractions. Accept equivalent\n=  −  form.\n(4r−3)(12r+3) 124r−3 4r+1\nr\nn\n= 1\n( 4 r − 3 )\n1\n(1 2 r + 3 )\n=\n1\n1\n2\n 1\n1\n−\n1\n5\n\n+\n1\n1\n2\n 1\n5\n−\n1\n9\n\n+\n1\n1\n2\n 1\n9\n−\n1\n1 3\n M1 Write enough correct sections of terms so\nthat all that are left in the answer appear\nplus one pair that cancel.\n+  +\n1\n1\n2\n\n4 n\n1\n− 3\n−\n4 n\n1\n+ 1\n A1 All necessary complete terms to\ndemonstrate the cancellation. Either two\nat the beginning and one at the end or one\nat the beginning and two at the end.\n=\n1\n1\n2\n−\n1 2 ( 4\n1\nn + 1 )\nB1 Accept any equivalent form.\n4\n\n1(c) \nr\n= 1\n( 4 r − 3 )\n1\n(1 2 r + 3 )\n=\n1\n1\n2\nB1 FT their answer for 1(b) of similar form.\n1\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2026-mj-12-q02",
      "question_id": "9231-2026-mj-12-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 2,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 EITHER\n© Cambridge University Press & Assessment 2026 Page 10 of 23\n3 6 + 9 2 + 8 9 = 9 1 0\n(B1 Checks base case.\nis divisible by 91.\nAssume that 36k + 92k + 89 is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it\nclear it is an assumption.\nB0 if assume true for ALL values of k.\nThen 3 6 (k + 1 ) + 9 2 k + 1 + 8 9 = ( 7 2 8 + 1 ) 3 6 k + ( 9 1 + 1 ) 9 2 k + 8 9 M1\n= 7 2 8  3 6 k + 9 1  9 2 k + ( 3 6 k + 9 2 k + 8 9 ) A1 Separates 36k + 92k +89\nA1 Connects the divisibility with the is divisible by 91 because both 72836k +9192k and are divisible by 91.\nhypothesis.\nHence, [by induction,] 3 6 n + 9 2 n + 8 9 is divisible by 91for every positive integer n. A1) Must have previous three marks and state\nthe result in full.\nOR\nAlternative method for 2\n3 6 + 9 2 + 8 9 = 9 1 0 is divisible by 91. (B1 Checks base case.\nAssume that ( 3 6 k + 9 2 k + 8 9 ) is divisible by 91 for some positive integer k. B1 States inductive hypothesis, making it\nclear it is an assumption.\nB0 if assume true for ALL values of k.\nThen 36(k+1) +92k+1+89−(36k +92k +89) M1 Considers difference between consecutive\nterms.\n=\n=\n(\n9\n7\n1\n2\n(\n8\n8\n)\n\n3\n3\n6 k\n6 k\n+\n+\n( 9\n9\n1\n2\n)\nk\n9\n)\n2 k\n36k +92k +89\nA1\nA1 Connects the divisibility with the\nhypothesis.\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2 Hence, [by induction,]\n© Cambridge University Press & Assessment 2026 Page 11 of 23\n3 6 n + 9 2 n + 8 9 is divisible by 91 for every positive integer n. A1) Must have previous three marks and state\nthe result in full.\n6\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-17-proof-by-induction/answers.html",
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    },
    {
      "id": "9231-2026-mj-12-q03",
      "question_id": "9231-2026-mj-12-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 3,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a)  k\n0\n0\n1\n B1 Stretch parallel to the x-axis, scale factor\nk.\n 1\n0\n2\n1\n B1 Shear, x-axis fixed, with (0,1) mapped to\n( 2 ,1 ) .\nM =\n 1\n0\n2\n1\n  k\n0\n0\n1\n M1 Correct order.\nM =\n 1\n0\n\n\nk\nk\n+\n+\n2\n1\n\n\n0\n0\n1\n0\n\n\n0\n0\n+\n+\n2\n1\n\n\n1\n1\n\n=\n k\n0\n2\n1\n\n.\nA1 AG Some intermediate step must be\nshown. At least one element not in final\nform.\n4\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b)\n© Cambridge University Press & Assessment 2026 Page 12 of 23\n k\n0\n2\n1\n  x\ny\n\n=\n k x +\ny\n2 y  B1\nTransforms\n x\ny\n\nMust see matrix applied\nto both coordinates.\nkx+2y=x M1 x x\nUses M =  . Must be considering a\ny y\nfixed point.\ny = 12 ( 1 − k ) x A1\n3\n\n3(c) Area of P = d e t M  Area of S 1= 1|k| M1 Uses\n2\nd e t M = k with P and S the correct\nway round.\nk =  2 A1 CAO\n2\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2026-mj-12-q04",
      "question_id": "9231-2026-mj-12-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 4,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 13 of 23\ny = x − 2  x = y − 12\n 5 y − 32 − 2 y − 12 + 1 = 0   y 32 = 2 y − 5 \n(B1 Correct substitution seen used in equation.\ny 3 = ( 2 y − 5 ) 2 M1 Squares to eliminate fractional powers.\ny 3 − 4 y 2 + 2 0 y − 2 5 = 0 A1) Must be an equation with like terms\ncollected.\nOR\nAlternative for 4(a)\n( 5 x 2 − 2 ) x = − 1 (M1 Squares to eliminate odd powers of x.\n( 5x2 −2 )2 x2 =1\n( 5 y − 1 − 2 ) 2 y − 1 = 1 B1 Correct substitution used in the equation.\ny3 −4y2 +20y−25=0 A1) Must be an equation with positive powers\nand like terms collected.\n3\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b)\n© Cambridge University Press & Assessment 2026 Page 14 of 23\n2 2 2 4    − + − + − = 2 2 2 2 2 2 2 0       − − + − − + − − = B1 FT Using their answer to part (a). Must be\nusing a cubic equation with positive\npowers and like terms collected.\n−4+−4+−4 =42−2(20) M1\n2 (\n4\n2\n4\n2\n4\n2\n(\n2\n2\n2\n2\n2 )\n2 ) 2      \n     \n− +\n−\n−\n−\n+\n+\n−\n−\n=\n− +\n− +\n−\n−\n−\n+ − −\n= − 2 4 A1 FT FT from a cubic equation answer for 4(a)\n3\n\n4(c) S\n− 6\n= 4 S\n− 4\n− 2 0 S\n− 2\n+ 7 5 M1 Using 6 4 4 2 0 2 2 5 0    − − − + − − = Must\nuse S\n0\n= 3\nFT for their cubic from 4(a) only if using\nthis method.\nS\n− 6\n= − 1 0 1 A1 CAO\nS =4S −20S +25S M1 Using\n−8 −6 −4 −2\n8 4 6 2 0 4 2 5 2 0     − − − + − − − =\nFT for their cubic from 4(a) only if using\nthis method.\nS\n− 8\n= 1 7 6 A1 CAO\n4\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    },
    {
      "id": "9231-2026-mj-12-q05",
      "question_id": "9231-2026-mj-12-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 15 of 23\n − 3\n− 2\n4\n\n−\n − 2\n4\n4\n\n=\n −\n−\n0\n1\n6\n\n(B1 Finds direction of one line to another.\ni\n3\n2\n−\nj\n4\n4\nk\n2\n3\n=\n − 2 0\n− 5\n2 0\n\n~\n − 4\n− 1\n4\n M1 Find common perpendicular by taking\ncross product of the two correct direction\nvectors.\nA1 Correct.\n−1 −4 M1 Uses formula for shortest distance using\n1     10 their common perpendicular and direction\n−6  −1 =\n    vector. Scalar product and magnitudes\n33 33\n   \n 0   4  must be evaluated.\nA1) Correct answer. Accept 1.74\nOR\nAlternative method for 5(a)\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(a) 1\n1\n0\n6\n1\n5\n =\n1\n1\n3\n6\n9\n5\n =\nA1 Solves equations from using\nperpendicular to both lines to find\na n d  \n−40 M1 Finds PQ\n33\n \n−10\n 33\n 40 \n \n33\nM1 Finds length of PQ\nPQ=\n( −40)2\n+\n( −10)2\n+\n(40)2\n33 33 33\n=\n1 0\n3 3\nA1)\n5\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)\n© Cambridge University Press & Assessment 2026 Page 17 of 23\ni\n− 3\n3\nj\n5\n− 4\nk\n3\n2\n=\n 2\n1\n−\n2\n5\n3\n *M1 Finds the normal to\n1\n by taking cross\nproduct of two correct direction vectors.\nA1 Correct.\ni j k −22 *M1 Finds the normal to  by taking cross\n2\n \n−3 −2 4 = 17 product of two correct direction vectors.\n \n \n2 4 3 −8\n  A1 Correct\n2\n1\n2\n5\n3\n1\n2 2\n7\n8\n7 1 8 8 3 7 c o s c o s\n7 1\n2\n8\n0 5\n8 3 7\n \n\n−\n\n\n −\n−\n\n=  =\n−\nDM1 Uses dot product of their normal vectors.\nScalar product and magnitudes must be\nevaluated.\n1 8 0 7 4 .7  − =  A1 Or 1.30 rad. Mark final answer. AWRT\n74.7 (1dp) or 1.3 (1dp)\n6\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-16-vectors/answers.html",
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    {
      "id": "9231-2026-mj-12-q06",
      "question_id": "9231-2026-mj-12-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x = − 1 B1 Vertical asymptote.\n(x+1)(x−2)+4 M1 Oblique asymptote - must show attempt to\ny=\nfind constant term.\n4x+4\ny = 14 x − 12 A1 Must be a linear equation.\n3\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) dy 4x2 +8x−12\n= =0\ndx (4x+4)2\n© Cambridge University Press & Assessment 2026 Page 18 of 23\n4 x 2 + 8 x − 1 2 = 0\nM1\nSets\nd\nd\ny\nx\n= 0 .\n(1,1), (−3,−7) A1 A1 for x-coordinates.\n4 4\nA1 A1 for y-coordinates correctly linked to x-\ncoordinates.\nAllow A1 for one completely correct\npoint.\n3\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c) B1 FT Axes and asymptotes labelled.\nFollow through their two asymptotes from\n(a) for this mark only.\nB1 Left branch correct and also all\napproaches to the asymptotes.\nB1 Right branch correct including a smooth\nminimum in the first quadrant.\n3\n© Cambridge University Press & Assessment 2026 Page 19 of 23\ny = 14 x − 12\nx = − 1\ny\nx\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(d) B1 Must be fully correct, including vertical\nasymptote.\n1\n\n6(e)\n© Cambridge University Press & Assessment 2026 Page 20 of 23\nx 2\n4\n−\nx\nx\n+\n+\n4\n2\n= − 2 or\nx 2\n4\n−\nx\nx\n+\n+\n4\n2\n= 2\nx 2 + 7 x + 1 0 = 0 x 2 − 9 x − 6 = 0\nM2 Finds critical points, award M1 for each\ncase. Must be using equations.\nx=−5,−2 x = 12\n(\n9 − 1 0 5\n)\n, x = 12\n(\n9 + 1 0 5\n) A1 All four correct, exact answers.\n− 5  x  − 2 12\n(\n9 − 1 0 5\n)\n x  12\n(\n9 + 1 0 5\n)\ny\nA1 CAO\n4\nx\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-12-q07",
      "question_id": "9231-2026-mj-12-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "12",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 17,
      "status": "available",
      "reason": null,
      "text": "7(a) EITHER\n© Cambridge University Press & Assessment 2026 Page 21 of 23\ny\n1 t\na\na n 2\n12\ns i n \n\n=\n\n+\n\n(B1 Uses y r s i n  =\nd\nd\ny\n1 t\na\na n 2\n12\nc o s 12 s i n\n1 t\na\na n 2\n12 ( 2\n( 1\na s\nt\ne\na\nc\nn\n2\n2\n2\n)\n)2 \n \n   \n \n=\n\n+\n\n−\n\n+\n\n−\n+\n*M1 Differentiates using product rule\nd\nd\ny\n( a ( 1 t a n 2 ) ) 12 c o s 12\n1\n1\nt a n\na\nt a\n2\nn 2\n32\n( 2 a s e c 2 2 ) s i n    \n\n \n \n=\n+ − +\n+\n\n− A1 or quotient rule. Must be using correct\nfunction.\nc o s\ns e\n1\nc 2 2\nt a n\ns i\n2\nn\nc o s\nc o s 2 ( s i\ns\nn\ni n\n2 c o s 2 )\n0\n  \n \n   \n −\n+\n= −\n+\n=\nDM1 A1 Puts derivative = 0 and rearranges.\nAG. Intermediate steps must be shown.\nc o s ( 0 .5 ) −\nc o s ( 2  0 .5 ) ( s i n\ns i n\n( 2\n(\n\n0 .5\n0 .5\n))\n+ c o s ( 2  0 .5 ) )\n= 0 .2 3 5 and\nc o s ( 0 .6 ) −\nc o s ( 2  0 .6 ) ( s i n\ns i n\n( 2\n(\n\n0 .6\n0 .6\n))\n+ c o s ( 2  0 .6 ) )\n= − 0 .3 7 9\nB1) Shows sign change. Values calculated to\nat least one decimal place.\nOR\nAlternative method for 7(a)\ny 2\n1 t\na\na n 2\ns i n 2 \n\n=\n\n+\n (B1 Uses y r s i n  =\n−2 *M1 Differentiates using product rule\n dy  1   1 \n\n2y\n\n=\n \n2asincos−asin2\n \nsec22\n d 1+tan2 1+tan2\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(a)\n© Cambridge University Press & Assessment 2026 Page 22 of 23\n2 y d\nd\ny ( ( 1 t a n 2 ) 2 a s i n c o s 2 a s i n 2 s e c 2 2 )\n1 t\n1\na n 2\n2\n    \n \n  = + − \n+\n\nA1 or quotient rule. Must be using correct\nfunction.\nsec22sin sin DM1 Puts derivative = 0 and rearranges.\ncos− =cos− =0\n1+tan2 cos2(sin2+cos2)\nA1 AG. Intermediate steps must be shown.\nsin(0.5)\ncos(0.5)− =0.235 and\ncos(20.5)( sin(20.5)+cos(20.5))\nc o s ( 0 .6 ) −\nc o s ( 2  0 .6 ) ( s i n\ns i n\n( 2\n(\n\n0 .6\n0 .6\n))\n+ c o s ( 2  0 .6 ) )\n= − 0 .3 7 9\nB1)\nShows sign change. Values calculated to\nat least one decimal place.\n6\n\n7(b) B1 Initial line and pole clear. Curve in first\nquadrant through\n(\na , 0\n)\nand O.\nCondone pole not labelled but must be\nclearly identifiable.\n(\na , 0\n)\nnot needed.\nB1 Correct shape (gradient at end points and\napproximately symmetrical) Maximum\npoint should lie between 40% and 60% of\nwidth of the arc.\n2\nO ( a , 0 )\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 23 of 23\n12 14\n0\nπ r 2 d a2 14\n0\nπ\n1 t\n1\na n 2\nd  \n\n = \n+\nB1 Using correct formula and correct limits.\nv t a n 2\nd\nd\nv\n2 s e c 2 2  \n\n=  =\nM1 Applies given substitution. Condone\nd\nd\nv\ns e c 2 2 \n\n=\n 1 4 π 1   1 1 M1 Clearly applies\na d= a dv\n2 1+tan2 4 1+v sec22\n0 0\ns e c 2 2 1 t a n 2 2   = +\nAllow sec22=1+v2written explicitly.\na4\n0 1\n1\nv 1 t a\n1\nn 2 2\nd v a4\n0\n( 1 v )\n1(\n1 v 2 )\nd v\n\n=\n \n+ +\n=\n\n\n+ +\nA1 Some working must be shown, including\ncalculations supporting change of limits.\nAG.\n( 1 + v )\n1(\n1 + v 2 )\n= 12\n\nv\n1\n+ 1\n−\nv\n2 v\n−\n+\n1\n1\n *M1 Partial fractions of correct form.\n= 12\n\nv\n1\n+ 1\n−\nv 2\nv\n+ 1\n+\nv 2\n1\n+ 1\n A1\nand separates\n−\nv\nv\n2\n+\n+\n1\n1\n.\na8  ln ( v + 1 ) − 12 ln ( v 2 + 1 ) + t a n − 1 v \n\n0\nDM1 Applies all three formulae from MF19 to\na three-term expression from above.\nFollow through their constants from line\nabove.\nA1 Correct.\n A1 The four previous marks must have been\n  v+1  \naln +tan−1v = 1 πa awarded and logarithms must be\n8    v2 +1    16 combined for this mark to be awarded.\n0\n9\nTrace ID: PC-F469E54369\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-15-polar-coordinates/answers.html",
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    {
      "id": "9231-2026-mj-13-q01",
      "question_id": "9231-2026-mj-13-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 1,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a)\n© Cambridge University Press & Assessment 2026 Page 9 of 20\nb ( ) 5    = − + + = − B1\n− 1 9 = 5 2 − 2 c M1 Uses formula for sum of squares.\nc = 2 2 A1\nEITHER:\n1 1 = 2− 2d\n(M1\nUses 1 1 1 .\n  \n  \n\n− + − + − =\n+ +\nUsing correct coefficients.\nd = − 2 A1FT) Follow through on their c.\nOR:\nAlternative method for the last 2 marks\nS\n−\n1\n2\n1\n1\n−\n9\nd\n5\n−\n=\nS\n5\n1\n1\n(\n9\n+\n5 )\n+\n2\n+\n5\n2\n(\nS\n2\n5\n0\n2\n)\n+\n( 3\n−\nd S\n− 1\n) + 1 1\n2 2 ( 3 )\n=\nd\n0\n= 0\n(M1 Uses S\n2\n− 5 S\n1\n+ 2 2 S\n0\n+ d S\n− 1\n= 0 and S\n0\n= 3\nd = − 2 A1FT) Follow through on their c.\nAvailable marks 5\n\n1(b) S +bS +cS +3d =0 M1 Uses original equation or formula for sum of\n3 2 1\ncubes, allow one arithmetic error.\nAllow FT of their b, c, and d.\nS\n3\n= − b ( − 1 9 ) − c ( 5 ) − 3 ( − 2 ) = − 1 9 9 A1FT Allow FT of their b, c, and d.\n2\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-13-q02",
      "question_id": "9231-2026-mj-13-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 2,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) Shear, rotation. B1 Both words.\nCorrect order. B1 Two transformations in the correct order with at\nleast one correct and not both the same\ntransformation.\n[x-axis fixed], with\n© Cambridge University Press & Assessment 2026 Page 10 of 20\n( 0 , 1 ) mapped to ( 2 , 1 ) . B1 For the description of the shear. Ignore any\ncomment relating to scale or shear factor\n14 π [anticlockwise] about the origin. B1 For rotation. Allow 45°\n4\n\n2(b)\nM = 12 2\n 1\n1\n1\n3\n B1 OE and must have a scalar multiple of 1 2.\n2\n12 2\n 1\n1\n1\n3\n  x\ny\n\n= 12 2\n x\nx\n+\n+ 3\ny\ny\n B1FT\nTransforms\n x\ny\n\nto\n X\nY\n\n. Follow through for\ntheir M\n12 2 ( x + 3 m x ) = m2 2 ( x + m x ) M1 Uses y=mx and Y = m X .\nx + 3 m x = m ( x + m x )  1 + 3 m = m + m 2  m 2 − 2 m − 1 = 0 A1 OE.\ny =\n(\n1 + 2\n)\nx and y =\n(\n1 − 2\n)\nx\nA1 If ambiguity over invariant points instead of\ninvariant lines, then M0 but SCB1 for the\ncorrect equations.\n5\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    },
    {
      "id": "9231-2026-mj-13-q03",
      "question_id": "9231-2026-mj-13-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3  3 −3 −1 1  B1\nP−1 =−1   = 2 2 \n6 −7 5   7 −5 \n6 6\n© Cambridge University Press & Assessment 2026 Page 11 of 20\nP D P − 1 =\n 5\n7\n3\n3\n  1\n0\n0\n3\n  −\n76\n12\n−\n12\n56\n\n=\n 5\n7\n9\n9\n  −\n76\n12\n−\n12\n56\n M1 Base case, multiplies P D or\nD P − 1 =\n −\n72\n12\n−\n12\n52\n\n.\nP D P − 1 =\n 8\n7\n−\n−\n5\n4\n\n= A\nA1 Reference to base case\nAssume that it is true for n = k , so Ak =PDkP−1 . B1 States inductive hypothesis.\nThen Ak+1 =AkA=PDkP−1PDP−1 =PDk+1P−1 M1A1 Multiplies A k with A and substitutes\nA k = P D k P − 1 .\nSo A n = P D n P − 1 is true for all positive [integers] n [by induction]. A1 States conclusion.\n7\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-17-proof-by-induction/answers.html",
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    },
    {
      "id": "9231-2026-mj-13-q04",
      "question_id": "9231-2026-mj-13-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 4,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a) 16r4 +32r3+24r2 +8r+1− ( 16r4 −32r3+24r2 −8r+1 ) =64r3+16r M1 Some evidence of expansion or evidence of\nusing difference of two squares.\nAllow one error overall.\nA1 Correct answer with like terms collected.\n© Cambridge University Press & Assessment 2026 Page 12 of 20\nr =\nr =\n...\nr =\n\n6\n\n4\n4 1 3\n4 2 5\n( n 2 n\nn\n 3 r +\nr = 1\n− 1\n− 3\n+\n1 6\n4\n4\n) 1\nn\nr\n=\n4\n1\n−\nr\n(\n=\n2\n\nn\n(\n−\n2\n1\nn\n)\n+\n4\n1 ) 4 − 1\nM1 Write enough correct rows of terms so that all\nthat are left in the answer appear plus one pair\nthat cancel.\nA1 All necessary rows of terms to demonstrate the\ncancellation. Either two at the beginning and one\nat the end or one at the beginning and two at the\nend.\n 6 4 r\nn\n= 1\nr 3 = 1 6 n 4 + 3 2 n 3 + 2 4 n 2 + 8 n − 8 n ( n + 1 )\nM1\nApplies r\nn\n= 1\nr = 12 n ( n + 1 ) .\nn n A1 AG. Must see at least one intermediate step of\n64r3 =16n2( n2 +2n+1 ) r3 =1n2(n+1)2\nworking before the answer given.\n4\nr=1 r=1\n6\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) EITHER:\n© Cambridge University Press & Assessment 2026 Page 13 of 20\n S\n2 n\n =\n2 n\nr\n= 1\nr 3 − 2 r\nn\n= 1\n( 2 r ) 3\n(*M1 Relates with sum of cubes.\n S\n2 n\n = 14 ( 2 n ) 2 ( 2 n + 1 ) 2 − 1 64 n 2 ( n + 1 ) 2 DM1 Applies r n\n= 1\nr 3 = 14 n 2 ( n + 1 ) 2 .\nS\n2 n\n= n 2 ( ( 2 n + 1 ) 2 − 4 ( n + 1 ) 2 ) = n 2 ( − 4 n − 3 ) A1) Accept − n 2 ( 4 n + 3 )\nOR:\nAlternative method\n S\n2 n\n = ( 1 3 + 3 3 + ... + ( 2 n − 1 ) 3 ) − ( 2 3 + 4 3 + ... + ( 2 n ) 3 ) (*M1 Relates with difference of cubes\n S\n2 n\n = − 1 2 r\nn\n= 1\nr 2 + 6 r\nn\n= 1\nr − n\n S\n2 n\n =\n− 1\n6\n2 n\n( n + 1 ) ( 2 n + 1 ) +\n6\n2\nn ( n + 1 ) − n\nDM1 Applies\nr\nn\n= 1\nr 2 =\nn\n6\n( n + 1 ) ( 2 n + 1 ) a n d r\nn\n= 1\nr =\nn\n2\n( n + 1 )\nS =−4n3−6n2−2n+3n2+3n−n=n2(−4n−3) A1) Accept −n2(4n+3)\n2n\nAvailable marks 3\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-13-summation-of-series/answers.html",
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    {
      "id": "9231-2026-mj-13-q05",
      "question_id": "9231-2026-mj-13-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 5,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a) B1 Initial line drawn and correct parabolic shape\n(not vertical at furthest point), starts from the\npole.\n[Maximum distance of C from the pole is]\n© Cambridge University Press & Assessment 2026 Page 14 of 20\n2 ( π3 ) 12 . B1 Accept AWRT 2.05\n2\n\n5(b) 12 13\n0\nπ s e c 2 d     M1 Uses 12 r 2 d   with correct limits.\nCondone omission of d.\n12  t a n  130 π 12 13\n0\nπ t a n d     = −  M1 Integrates sec2by parts.\nA1 Allow omission of the half and limits for these\nmarks.\n12  t a n l n s e c  13\n0\nπ    = − A1 OE. Must have the half and the limits for this\nmark.\n( ) A1 OE, must be exact. ISW\n= 1 1π 3−ln2\n2 3\n5\n0  =\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c) r2cos2= x2 = M1 Uses\n© Cambridge University Press & Assessment 2026 Page 15 of 20\nx r c o s  = .\nx 2 = t a n − 1\ny\nx\nM1 y\nApplies tan=\nx\ny\ntan\n( x2)\n=\nx\ny= xtan\n( x2) A1 AG. Must see an intermediary step in the\nworking before the answer given\n3\n\n5(d) 12 ( 2 ( π3 ) 12 c o s 13 π ) ( 2 ( π3 ) 12 s i n 13 π ) = 16 π 3 B1 Area of triangle.\n16 π 3 − 12\n(\n13 π 3 − l n 2\n) M1 Area of triangle minus their answer to (b) (their\nanswer in (b) can be a decimal).\n12 l n 2 A1 OE\n3\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-13-q06",
      "question_id": "9231-2026-mj-13-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "6(a) x2 +x+1=0 has no real roots as discriminant is [1 2 − 4 = ] − 3 B1 The discriminant needs be evaluated, compared and this is less than zero, hence\nand concluded.\nshown.\nOr shows that the expression is always positive\n(i.e. completing the square) and concludes.\ny =0 B1 Horizontal asymptote.\n2\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) dy (x2 +x+1)(1)−(x+1)(2x+1) M1\nFinds\n=\ndx ( x2 +x+1 )2\n© Cambridge University Press & Assessment 2026 Page 16 of 20\nd\nd\ny\nx\n. Allow one arithmetic error for this\nmark.\nx 2 + 2 x = 0 M1 Sets equal to 0, forms equation and attempts to\nsolve.\n(0,1) and (−2,−1) A1 Condone clearly paired x and y.\n3\n3\n\n6(c) B1 Axes and approach to asymptote.\nB1 Correct shape and position, and maximum\napparently on the axis\n( 0 , 1 ) and ( − 1 , 0 ) B1 States coordinates of just the two intersections\nwith axes, may be seen on their diagram.\n3\n\n6(d) B1FT FT provided sketch in part (c) has both positive\nand negative y values\nB1 Correct shape at (−1,0)and must have a correct\ngraph in part (c)\n2\ny\nx\nx\ny\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(e)\n© Cambridge University Press & Assessment 2026 Page 17 of 20\nx 2\nx\n+\n+\nx\n1\n+ 1\n=\n1\n3\nor\nx 2\nx\n+\n+\nx\n1\n+ 1\n\n1\n3\nM1 Writing down the equation or inequality\nx 2\nx\n+\n+\nx\n1\n+ 1\n= −\n1\n3\nor\nx 2\nx\n+\n+\nx\n1\n+ 1\n −\n1\n3\nM1 Writing down the equation or inequality\nEITHER:\nx  1 − 3\n(A1 Writing both inequalities ISW\n, x 1+ 3\nx  − 2 A1) ISW\nOR:\nAlternative for the final 2 marks\nx  − 2 , − 2  x  1 − 3 , x  1 + 3 (A2,0) All three regions correct.\nAvailable marks 4\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-13-q07",
      "question_id": "9231-2026-mj-13-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "13",
      "question_number": 7,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) i j k  1  M1 A1 Finds normal to the plane . For the method\n  mark allow one arithmetic error in calculating\n1 −1 0 = 1\n  the cross product but must be using the correct\n \n−3 1 −1  −2  direction vectors.\n© Cambridge University Press & Assessment 2026 Page 18 of 20\n( 0 ) + ( 1 ) − 2 ( 2 ) = − 3  x + y − 2 z = − 3 M1 Substitutes point using their cross product but\nmust be from using two correct direction\nvectors.\nA1 Answer must be in the form a x + b y + c z = d .\n4\n\n7(b)\n1\n6\n3\n1\n1\n2\n6\n1\n2\n6\n5\n3\n6\n6\n\n\n\n\n− −\n\n−\n\n\n +\n+\n+\n\n=\n−\n*M1 A1 Forms perpendicular distance of P from plane\n in terms of .\nFor only the method mark allow follow through\nof their d from part (a).\n3 6 6  −  DM1 Forms and solves linear inequality in .\n04 A1\n4\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 19 of 20\nO Q\n0\n1\n2\n, O R\n1\n2\n3\nQ R\n2\n1 3\n2\n \n   \n  \n   =  \n\n\n\n  \n=\n −\n−\n\n =\n\n−\n−\n−\n−\n M1 Allow one arithmetic error in calculating one\nelement.\n\nA1 FindsQR or direction of common\nperpendicular.\n\n 1−3  0 M1 Uses that dot product of QR with line direction\n    \n −    1  =0 is zero, or, alternatively, QRis a multiple of the\n   \n2−−2 2    \ncommon perpendicular\n 1\n2\n− 1\n\n, the multiple is not\n1.\n5 4   + = A1 Deduces one equation.\n2\n1 3\n2\n1\n3\n1\n0 1 1 5\n\n   \n \n\n−\n−\n−\n−\n\n\n −\n−\n\n=  + =\nM1 Deduces second equation for their\nQ R  \n−1 −3 M1 A1 Solves for and substitutes into\n7  6   \n= OR= 7 = 1 7\n18  18  18    29     29  \n18\n 1\n − \n6  \n  1\nQR= −\n 3\n \n1\n \n \n 6 \nO R  \nand\nQ R  \nor\n 1\n−\n  6\n0  \n13      1\n= OQ= 13 1 and QR= − 18 18   3\n \n2    \n1\n \n \n 6 \nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c) −3  1  B1 FT OE. FT using their common perpendicular.\n    Answer must be in equation form.\nr = 1 7 +k 2\n18   \n   \n29 −1\n   \n8\n© Cambridge University Press & Assessment 2026 Page 20 of 20\nTrace ID: PC-39A0636DD7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-14-q01",
      "question_id": "9231-2026-mj-14-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 1,
      "topic": "Summation of series",
      "topic_slug": "9231-topic-13-summation-of-series",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) 1n(n+1)(2n+1)−n(n+1)+n M1 Substitutes all three correct formulae\n6 (two from MF19).\n© Cambridge University Press & Assessment 2026 Page 9 of 25\n13 n 3 − 12 n 2 + 16 n A1 Must be in this form.\n2\n\n1(b) EITHER:\n1 1 1 1\n− = −\nr2 −2r+1 r2 +2r+1 (r−1)2 (r+1)2\n1 1 1 1 1 1 1 1 1 1 1 1 (M1 Writes enough correct sections of terms\n− + − + − + − ++ − + −\nso that three of the four sections which\n12 32 22 42 32 52 42 62 (n−2)2 n2 (n−1)2 (n+1)2\nappear in the final answer are seen.\nA1 All four correct sections correct and at\nleast one cancellation pair included.\nA1 At least five complete n- value terms,\neither first three and last two, or first\ntwo and last three.\n5\n4\n−\n1\nn 2\n−\n( n\n1\n+ 1 ) 2\nB1) OE WWW\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b) OR:\nAlternative method for 1(b)\n© Cambridge University Press & Assessment 2026 Page 10 of 25\n\n1\n2 1\n+\n1\n2 2\n+\n1\n3 2\n+\n1\n4 2\n+  +\n( n\n1\n− 2 ) 2\n+\n( n\n1\n− 1 ) 2\n\n−\n\n1\n3 2\n+\n1\n4 2\n+\n1\n5 2\n+\n1\n6 2\n+  +\n1\nn 2\n+\n( n\n1\n+ 1 ) 2\n (M1 Writes enough correct sections of terms\nso that three of the four sections which\nappear in the final answer are seen.\nA1 All four correct sections correct and at\nleast one cancellation pair included.\nA1 At least five complete n- value terms,\neither first three and last two, or first\ntwo and last three. WWW\n5 1 1 B1) OE WWW\n− −\n4 n2 (n+1)2\n4\n\n1(c) 5 B1 FT FT their 1(b) of similar form.\n4\n1\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-13-summation-of-series/answers.html",
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    },
    {
      "id": "9231-2026-mj-14-q02",
      "question_id": "9231-2026-mj-14-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 2,
      "topic": "Roots of polynomial equations",
      "topic_slug": "9231-topic-11-roots-of-polynomial-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) EITHER: (B1\ny=x−3x= y−1\n3\nCorrect substitution seen in the\n2y− 3 4 +3y−1+y−1 3 +1=0 equation.\nYou may see other forms including\n1 4\n2+3y3 + y+ y3 =0\n4 1\ny3 +3y3 =−y−2\ny(y+3)3 =−(y+2)3 M1 Obtains an equation not involving roots\nor fractional powers from a correct\nsubstitution.\n© Cambridge University Press & Assessment 2026 Page 11 of 25\ny ( y 3 + 9 y 2 + 2 7 y + 2 7 ) = − ( y 3 + 6 y 2 + 1 2 y + 8 ) M1 Expands powers of brackets.\ny 4 + 1 0 y 3 + 3 3 y 2 + 3 9 y + 8 = 0 A1 Collects like terms. Answer must be an\nequation.\n−3 +−3 +−3 +−3 =−10 B1FT) FT their answer equation.\nOR:\nAlternative method for 2(a)\n3 x 3 + 1 = − 2 x 4 − x\n( 3 x 3 + 1 ) 3 = − ( 2 x 4 + x ) 3\n(M1 Cube to make sure powers of x divide\nby 3\n2 7 x 9 + 2 7 x 6 + 9 x 3 + 1 = − 8 x 1 2 − 1 2 x 9 − 6 x 6 − x 3 M1 Expands powers of brackets.\n27y−3 +27y−2 +9y−1+1=−8y−4 −12y−3−6y−2 − y−1 B1 Correct substitution seen in the\nequation.\ny4 +10y3 +33y2 +39y+8=0 A1 Collects like terms. Answer must be an\nequation\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a)\n© Cambridge University Press & Assessment 2026 Page 12 of 25\n3 3 3 3 1 0     − + − + − + − = − B1FT) FT their answer equation.\n5\n\n2(b) 1\n6\n1\n6\n1\n6\n1\n6\n1 0 2 2 ( 3 3 )\n   \n+ + + = −\nM1\nRelates\n1\n6\n1\n6\n1\n6\n1\n6    \n+ + + to\ncoefficients.\n3 4 A1 FT FT their answer equation from 2(a) as\nlong as it is a quartic equation.\n2\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-11-roots-of-polynomial-equations/answers.html",
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    },
    {
      "id": "9231-2026-mj-14-q03",
      "question_id": "9231-2026-mj-14-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 3,
      "topic": "Proof by induction",
      "topic_slug": "9231-topic-17-proof-by-induction",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "3(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 25\nu\n1\n 1 [given] B1 Base case seen.\nAssume that u 1 for some positive integer k. B1 States inductive hypothesis.\nk\nAccept ‘Let u\nk\n 1 be true’ or ‘Suppose\nu\nk\n 1 .’\nEITHER: (M1 Considers u −1, and puts over\nk+1\nu 2 +2u +2 u 2 +2u +2−(u +4) u 2 +u −2 common denominator. Numerator\nThen u −1= k k −1= k k k = k k\nk+1 u +4 u +4 u +4 need not be simplified for this mark.\nk k k\n(u −1)(u +2)\n= k k 0.\nu +4\nk\nOr\n( u\nk\n+\n( u\n12\nk\n) 2\n+\n−\n4\n14) − 2\n 0\nA1 Factorises\nOr completes the square.\nSince u 1 [by induction,] u 1 for all positive [integers] n. A1) Must have previous M1 A1 and also\nk n\nexplicitly connect the assumption u\nk\n 1\nwith their statement. No wrong\nworking seen.\nOR:\nAlternative for 3(a) M1 A1 A1marks\nThen u\nk + 1\n− 1 =\nu\nk\n2 +\nu\nk\n2 u\n+\nk\n4\n+ 2\n− 1 =\nu\nk\n2 + 2 u\nk\nu\n+\nk\n2\n+\n−\n4\n( u\nk\n+ 4 )\n=\nu\nk\n2\nu\n+\nk\nu\n+\nk\n4\n− 2 (M1 Considers u\nk + 1\n− 1 , and puts over\ncommon denominator. Numerator\nneed not be simplified for this mark.\nu\nk\n2  1 AND u\nk\n 1 so u 2 +u −21+1−2=0 A1 Both must be seen.\nk k\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(a) Since\n© Cambridge University Press & Assessment 2026 Page 14 of 25\nu\nk\n 1 [by induction,] u 1 for all positive [integers] n. A1) Must have previous M1 A1and also\nn\nstate u\nk\n 1 either in the conclusion or\nin the argument for the M1 A1. No\nwrong working seen. Allow for all\npositive integers [n].\n5\n\n3(b) EITHER:\nu\nn + 1\n− u\nn\n=\nu\nn\n2 + 2 u\nn\n+\nu\n2\nn\n−\n+\nu\n4\nn\n( u\nn\n+ 4 )\n=\n2 (1\nu\nn\n−\n+\nu\nn4\n)\n(M1 Considers u −u , or u −u , and\nn+1 n n n+1\nputs over common denominator.\nAllow one numerical or sign error\nfor M1.\nA1\nSo u\nn\n 1  u\nn + 1\n− u\nn\n 0\n u\nn + 1\n u\nn\nA1) Uses u\nn\n 1 .\nStates result.\nOR:\nAlternative for 3(b)\nu\nn + 1\n− u\nn\n= − 2 +\nu\n1\nn\n0\n+ 4\n(M1 10\nu −u =u −2+ −u\nn+1 n n u +4 n\nn\nu\nn\n 1 , u\nn\n+ 4  5 \nu\n1\nn\n0\n+ 4\n 2\nA1\n− 2 +\nu\n1\nn\n0\n+ 4\n 0 u −u 0\nn+1 n\n u\nn + 1\n u\nn\nA1) States result.\n3\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-17-proof-by-induction/answers.html",
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    },
    {
      "id": "9231-2026-mj-14-q04",
      "question_id": "9231-2026-mj-14-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 4,
      "topic": "Matrices (Further Pure 1)",
      "topic_slug": "9231-topic-14-matrices-further-pure-1",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "4(a) 1 k 3 k 3 1 M1 Finds determinant and\n−k − =−1−k−k(−3−3k)\n1 −1 3 −1 3 1\nA1 forms correct quadratic equation.\n=3k2 +2k−1=0\n© Cambridge University Press & Assessment 2026 Page 15 of 25\nk = 13 , − 1 A1 No method need be seen for solving\nquadratic equation.\n3\n\n4(b) 1 k −11 0  1 0  M1 A1 Multiplies two matrices correctly.\n0 1 0   3 1 k   Could see\n  3 1 k  0 −1  =  0 −1 \n1 0 1\n \n4 k+1 −2\n\n3 1 −11 0  1 0   0\n1\n1\n0\n0\n1\n\n\n3\n0\n+\n2\nk\n−\n−\n−\nk\n1\n1\n\n.\n=\n k +\n2\n3\n−\n−\nk\n1\n− 1\n A1 AG. Some intermediate working must\nbe shown.\nFor first pairing of matrices must see 4\n- 2\nFor second pairing must see 0 + 2\n3\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(c)\n© Cambridge University Press & Assessment 2026 Page 16 of 25\n 1\n2\n−\n1\n1   x\ny\n\n=\n\n2\nx\nx\n−\n+\ny\ny\n B1FT\nTransforms\n x\ny\n\nto\n X\nY\n\n.\nAllow FT if left in terms of k.\n2 x + m x + c = m ( x − m x − c ) + c M1 A1 Uses y = m x + c and Y =mX +c.\nM1 A0 if c = 0 throughout.\n( m 2 + 2 ) x + m c = 0 A1 c must be present here.\nm2 +2=0 no real values of m. A1 Allow if c missing above.\nAccept ‘no solutions’.\n5\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-14-matrices-further-pure-1/answers.html",
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    {
      "id": "9231-2026-mj-14-q05",
      "question_id": "9231-2026-mj-14-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 5,
      "topic": "Vectors",
      "topic_slug": "9231-topic-16-vectors",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "5(a)\nA B  =  − 2  \ni −\nj − 2\nk A C = i − j − k B C = 3 i + k\nB1 Finds direction vectors of two lines in\nthe plane.\ni j k −1 M1 Finds normal to the plane ABC by\n  taking cross product of two of their\n−2 −1 −2 = −4\n  vectors from above. Allow one\n \n1 −1 −1  3  numerical or sign error in cross\nproduct for M1.\nA1FT Allow FT their vectors from above.\n− ( 4 ) − 4 ( 3 ) + 3 ( 3 ) = − 7  − x − 4 y + 3 z = − 7 M1 Substitutes a correct point.\nA1 CAO\n5\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)\n© Cambridge University Press & Assessment 2026 Page 17 of 25\n2\n1\n3\n1\n4\n3\n1 4 2 6 c o s c o s\n1 4\n3\n2 6\n \n \n\n −\n−\n\n=  =\nM1 Uses scalar product of\nO D  \nand their\nnormal. For M1 the scalar product and\nlengths must be evaluated but need not\nbe simplified.\nA1FT Lengths and scalar product simplified\ncorrectly.\nAcute angle between O D and Π is 90−=9.05 A1 Mark final answer. Accept 9 .0  9 .1  or\n0.16 rad. But not 9\n3\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c) EITHER: (B1\n2 5 −3\n\n     \nCD= 1 − 2 = −1\n     \n     \n3 2  1 \ni j k −3 *M1 Find common perpendicular of their\n  CD and AB. Again, at most one error in\n−2 −1 −2 = 8\n  cross product.\n \n−3 −1 1 −1\nA1 Correct answer\n© Cambridge University Press & Assessment 2026 Page 18 of 25\n1\n7 4\n −\n1\n1\n1 \n\n −\n8\n−\n3\n1\n\n=\n1 0\n7 4\n= 1 .1 6\nDM1 Uses formula for shortest distance. May\nuse any vector joining a point of AB to\na point of CD. For M1 the scalar\nproduct and length must be evaluated\nbut need not be simplified.\nA1) CAO Surd or decimal.\nOR:\nAlternative method for 5(c)\n4 −2\n   \nr= 3 + −1\n   \n   \n3 −2\nr\n5\n2\n2 1\n3\n1  =\n \n+\n −\n−\n (B1 Write down a general point on each of\nAB and CD.\n−1−2+3 M1 Finds vector joining them.\n \n1−+\n \n \n 1−2− \nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c)\n© Cambridge University Press & Assessment 2026 Page 19 of 25\n1 9 5 0   − + − =\n3 5 1 1 0   + − =\nM1 Scalar product of each line direction\nwith this vector is zero.\nBoth equations needed but can be\nunsimplified.\n1\n3\n3\n7\n =\n1\n3\n6\n7\n =\nA1 Solves the equations.\n −\n−\n43\n1 5\n3 7\n0753\n7\n\nhas magnitude\n1 0\n7 4\n= 1 .1 6\nA1)\nCAO Surd or decimal.\n5\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-14-q06",
      "question_id": "9231-2026-mj-14-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 6,
      "topic": "Rational functions and graphs",
      "topic_slug": "9231-topic-12-rational-functions-and-graphs",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "6(a) x = 0 , x=1 B1 Both vertical asymptotes.\ny = 1 B1 Horizontal asymptote and no extra\nasymptotes.\n2\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b)\n© Cambridge University Press & Assessment 2026 Page 20 of 25\nd\nd\ny\nx\n=\n− 2\n( 2\nx\nx\n2\n2\n−\n−\n4\n2\nx\nx\n+\n) 2\n2 *M1\nFinds\nd\nd\ny\nx\n. A method which clearly\ndemonstrates the use of quotient rule,\nproduct rule or rearrangement must be\nseen.\nx 2 + 2 x − 1 = 0 DM1 Sets equal to 0 and forms quadratic\nequation.\n( − 2 .4 , 0 .9 ) , ( 0 .4 , − 1 .9 ) A1 A1 Accept ( − 1 − 2 , − 12 + 2 ) ,\n(\n− 1 + 2 , − 12 − 2\n)\nA1 A0 for correct x values but y values\nmissing or incorrect.\n4\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c) B1FT Axes and asymptotes labelled. Accept\nlabelled by numerical intercepts with\nasymptotes.\nAllow FT from (a) for this B mark\nonly.\nB1 LHS correct shape and position,\ncrossing horizonal asymptote.\nThe minimum should be indicated by\nthe shape of the curve and the\nasymptotic approach. Do not expect\naccurate scale.\nB1 Both branches for\n© Cambridge University Press & Assessment 2026 Page 21 of 25\nx  0 correct.\n3\nx = 1\nx\ny\ny=1\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(d) B1FT Must be a single curve from LHS to\nRHS of the page, with no breaks.\nFollow through the max / min for their\ngraph for (c) as long as it has three\nbranches. Numerical values need not be\nconsidered for this mark.\nB1 Correct shape and intersections with\naxis and relationship with asymptotes.\n© Cambridge University Press & Assessment 2026 Page 22 of 25\n− 1  x  0 B1\nx  1 B1\n4\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-12-rational-functions-and-graphs/answers.html",
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    {
      "id": "9231-2026-mj-14-q07",
      "question_id": "9231-2026-mj-14-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 1,
      "variant": "14",
      "question_number": 7,
      "topic": "Polar coordinates",
      "topic_slug": "9231-topic-15-polar-coordinates",
      "marks": 16,
      "status": "available",
      "reason": null,
      "text": "7(a) ( 2 ) 32 ( c o s ) ( s i n )   r = r r M1 Uses at least two of\nx 2 + y 2 = r 2 , c o s  x = r and s in  y = r\n.\n2 ( 12 s i n 2 ) 0  r r − = M1 Uses correct identity.\ns i n 2 2 s i n c o s    =\ny\nr 0r = 1sin2 A1 AG.\n2\n3\nx\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b) (1sin2)2\n=\n1\n2 (−1π)2\n+5\n4\nOr\n© Cambridge University Press & Assessment 2026 Page 23 of 25\n12 s i n 2\n( 14\n1\nπ ) 2 5\n12\n\n\n=\n\n− +\n\nM1 Sets r values equal.\n1sin22 ( (−1π)2 +5 ) =1sin22 ( (−1π)2 +5 ) =4 A1 AG.\n4 4 4 Intermediate working must be shown.\nWe need to see e.g. ¼ or 22\n. s i n 2 ( 2  0 .5 ) ( ( 0 .5 − 14 π ) 2 + 5 ) − 4 = − 0 .4\ns i n 2 ( 2  0 .6 ) ( ( 0 .6 − 14 π ) 2 + 5 ) − 4 = 0 .4\nB1 Shows sign change, calculates to at\nleast 1dp.\n–0.401958833\n0.373343553\nNo conclusion required.\nAllow 3.598 < 4, 4.3733>4\n3\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c) B1 Initial line drawn and petal shape for\n© Cambridge University Press & Assessment 2026 Page 24 of 25\nC\n1\nin the first quadrant, starting and\nending at the pole. Ignore graph in\nother quadrants for this B1.\nB1 Convex arc shape for C\n2\nwith the pole\non the concave side. Ignore graph in\nother quadrants for this B1.\nB1 Line of symmetry labelled on the\ndiagram\nor equation given and sketch\nsymmetrical about 14 π .  = Accept\nlabelled as 14 π . Both curves labelled.\nGraph only in first quadrant.\nB1 C Symmetrical, rounded end, 1\napparently tangential to 0  = and\nπ\n2\n = at the pole.\nC Symmetrical, angle between curve\n2\nand positive direction of =0 not\nmore than\nπ\n2\n.\nA line drawn vertically from the start\npoint of the arc would not intersect the\npetal.\nGraph only in first quadrant.\n4\no\nC\n14\n2\nπ \nC\n1\n=\n0  =\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(d)   sin22d= 1  1−cos4d= 1−1sin4  M1 Integrates\n \n0 2 0 2 4 0\n© Cambridge University Press & Assessment 2026 Page 25 of 25\ns i n 2 2  using correct\nidentity. Allow M1 if at most one\nnumerical or sign error in the\nintegration step.\nA1 (Seen somewhere, condone\nincorrect/missing limits.)\n1π 1π M1\n4 1 1  −1π4 Integrates\n d= tan−1  4   (−1π)2 +5 5  5 \n 4 \n( 14\n1\nπ ) 2 5  − +\n. For M1 must\nhave t a n 1 ( 14 π )  p − k − where p and k\nare numerical.\nA1 A1 for fully correct.\n(Seen somewhere, condone\nincorrect/missing limits.)\n14\n0\ns i n 2 2 d\n14 π\n( 14\n1\nπ ) 2 5\nd\n\n  \n\n\n +\n\n− +\nM1 Forms area of R with correct limits.\nThere are other possibilities.\n18 ( 14 s i n 4 ) 15 t a n 1 ( 15 ( 14 π ) )    = − − − − A1 AG. Use of limits seen and all previous\nmarks must have been awarded.\n=1(−1sin4)+1 5tan−1( 1 5(1π−))\n8 4 5 5 4\n6\nTrace ID: PC-4F45C0CD5B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-15-polar-coordinates/answers.html",
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    {
      "id": "9231-2026-mj-21-q01",
      "question_id": "9231-2026-mj-21-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 1,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1\n© Cambridge University Press & Assessment 2026 Page 9 of 18\nf '( x ) = ( 1 + ( x + 1 ) 2 ) − 12 B1 Finds first derivative. Accept\nx + 1 +\n1\n( x + 1 ) 2 + 1\n\n1 +\n2 (\n2\nx\nx\n+\n+\n1\n2\n2 ) + 1\n\nor implicit\ny 'c o s h y = 1 .\nf ''( x ) = − ( 1 + ( x + 1 ) 2 ) − 32 ( x + 1 ) B1 Finds second derivative. Allow not simplified or\nimplicit\ny ''c o s h y + ( y )' 2 s i n h y = 0 .\nsinh−11=ln ( 1+ 2 ) B1\n1\nf '(0)=\n2\nf ''( 0 ) = −\n2\n1\n2\nM1 Evaluates their derivatives at zero only. Must have\nattempted both derivatives.\ns i n h − 1 ( x + 1 ) = ln ( 1 + 2 ) + 1\n2\nx −\n4\n1\n2\nx 2 A1 CWO.\n5\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-21.pdf?download=1",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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    },
    {
      "id": "9231-2026-mj-21-q02",
      "question_id": "9231-2026-mj-21-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) I = 1 (1+t)n sinπtdt=−1(1+t)ncosπt 1 + n 1 (1+t)n−1cosπtdt *M1 A1 Applies integration by parts. Allow limits missing.\nn 0  π  0 π 0 M1 for correct trigonometric functions and powers\nof (1+t).\n© Cambridge University Press & Assessment 2026 Page 10 of 18\nI\nn\n= 1π ( 2 n + 1 ) + nπ  1π (1 + t ) n − 1 s i n π t  1\n0\n− nπ  1\n0\n1π ( n − 1 ) ( 1 + t ) n − 2 s i n π t d t DM1 Applies integration by parts again. M1 for correct\ntrigonometric functions and powers of (1 + t ) .\nI\nn\n= 1π ( 2 n + 1 ) − 12\nπ\nn ( n − 1 ) I\nn − 2\n π 2 I\nn\n= π ( 2 n + 1 ) − n ( n − 1 ) I\nn − 2\nA1 AG. Must have seen limits and correct use of\nbrackets.\nCWO.\n4\n\n2(b)\nI\n0\n= \n1\n0\ns i n π t d t =  − 1π c o s π t \n1\n0\n= 2π\nB1 Finds\n.\nI\n0\n.\nπ 2 I\n2\n= π ( 2 2 + 1 ) − 2 ( 2 − 1 ) I\n0\n = 5 π − 4π  M1 Uses reduction formula from (a).\nI\n2\n= 5π − 43\nπ\nA1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-21.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-21-q03",
      "question_id": "9231-2026-mj-21-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "3(a) u = c o s h x  d u = s i n h x d x B1\n s i n h 2 x d u =  ( u 2 − 1 ) d u M1 A1 Applies correct identity s i n h 2 u = c o s h 2 u − 1 to find\nintegral in terms of u.\n= 13 u 3 − u ( + C ) =1cosh3x−coshx (+C) A1\n3\n4\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b)\n© Cambridge University Press & Assessment 2026 Page 11 of 18\ne  c o th x d x = e ln sin h x = s i n h x M1 A1 Finds integrating factor.\nd\nd x\n( y s i n h x ) = s i n h 3 x\nM1 Correct form on LHS and attempt to integrate RHS.\ny s i n h x = 13 c o s h 3 x − c o s h x + C *M1 A1 Integrates RHS using their part (a). For M1, their\nanswer to (a) must be of the correct form\nacosh3x+bcoshx with a,bnon-zero.\n0=1(5)3\n−5+CC=10\nDM1 Substitutes initial conditions. M0 if expression\n3 3 3 81 incorrectly manipulated before substitution. E.g. not\nall terms divided by s i n h x .\ny = c o s e c h x ( 13 c o s h 3 x − c o s h x + 18 01 ) A1 OE.\n7\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-26-differential-equations/answers.html",
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    {
      "id": "9231-2026-mj-21-q04",
      "question_id": "9231-2026-mj-21-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 4,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) d\nd x\n( ( x − y ) 5 ) = 5 ( x − y ) 4 ( 1 − y )' B1 Differentiates ( x − y ) 5 correctly.\nd\nd x\n( ln ( x y ) ) =\nx y '+\nx y\ny\n= y ' y − 1 + x − 1\nB1 Differentiates ln(xy) correctly.\n5 ( 1 − y )' − 12 y '− 1 = 0  y ' = 81\n1\nB1 Substitutes ( − 1 , − 2 ) into\n5 ( x − y ) 4 ( 1 − y )' + y ' y − 1 + x − 1 = 0 , AG. CWO.\nMight see y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 .\n3\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) EITHER:\n© Cambridge University Press & Assessment 2026 Page 12 of 18\n− 5 ( x − y ) 4 y ''+ 2 0 ( 1 − y )' 2 ( x − y ) 3\n(B1 B1 Differentiates 5 ( x − y ) 4 ( 1 − y )' .\n− ( y ') 2 y − 2 + y − 1 y ''− x − 2 = 0 B1\nDifferentiates y ' y − 1 + x − 1 . Might see\nx y '+\nx y\ny\ndifferentiated using product or quotient rules.\nxy(xy''+2y')−(xy'+ y)2\nE.g. .\n(xy)2\nOr − ( x y ) − 2 ( x y '+ y ) 2 + ( x y ) − 1 ( x y ''+ 2 y )' .\n− 5 (1 ) 4 y ''+ 2 0 ( 1 − 81\n1\n) 2 ( 1 ) 3 − 14 ( 81\n1\n) 2 − 12 y ''− 1 = 0 M1 Substitutes ( − 1 , − 2 ) and y'= 8 .\n11\n− 1 12 y ''+ 4 3\n1 2 1\n= 0  y '' =\n1\n83 63\n1\nA1)\nOR:\nAlternative method for question 4(b)\ny (' − y − 2 y '− 2 0 ( x − y ) 3 ( 1 − y )' ) + y ( '' y − 1 − 5 ( x − y ) 4 ) (B1 B1 Differentiates both sides of\ny (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 .\n= − 2 0 ( x − y ) 3 ( 1 − y )' + x − 2 B1\n( 81\n1\n) ( − 14  81\n1\n− 2 0  31\n1\n) + y ( '' − 12 − 5 ) = − 2 0  31\n1\n+ 1 M1 Substitutes ( − 1 , − 2 ) and y'= 8 .\n11\n− 1 12 y ''+ 4 3\n1 2 1\n= 0  y '' =\n1\n83 63\n1\nA1)\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) OR:\nSecond alternative method for question 4(b)\n( 5(x−y)4 −y−1 )( 20(x−y)3(1−y')−x−2 ) (B1 B1 Numerator of derivative of\n5(x− y)4 +x−1\n− ( 5(x−y)4 +x−1 )( 20(x− y)3(1− y')+ y−2y' ) y'= 5(x− y)4 − y−1 .\n© Cambridge University Press & Assessment 2026 Page 13 of 18\n( 5 ( x − y ) 4 − y − 1 ) 2 B1 Denominator of derivative of\ny ' =\n5\n5\n(\n(\nx\nx\n−\n−\ny\ny\n)\n)\n4\n4\n+\n−\nx\ny\n−\n−\n1\n1\n. Can be implied by ( 1 12 ) 2 .\ny '' =\n( 5 + 12 ) ( 2 0  31\n1\n− 1\n( 5\n) −\n+\n(\n12\n5\n)\n−\n2\n1 ) ( 2 0  31\n1\n+ 21\n1\n) M1 Substitutes ( − 1 , − 2 ) and y'= 8 into their\n11\nexpression.\ny '' =\n1\n83 63\n1\nA1)\nAvailable marks 5\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-23-differentiation/answers.html",
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    {
      "id": "9231-2026-mj-21-q05",
      "question_id": "9231-2026-mj-21-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 14 of 18\n 1\n0\n( 11 10 x 3 + 11\n0\n) d x  ( 1n ) ( 11 10 ( 1n ) 3 + 11\n0\n) + ( 1n ) ( 11 10 ( 2n ) 3 + 11\n0\n) +  + ( 1n ) ( 11 10 ( nn ) 3 + 11\n0\n) M1 A1 Forms the sum of the areas of the rectangles. M1 for\nsum of n complete terms. A1 for at least 3 correct\nterms including first and last. May be written in\nsummation form.\n=\n1\n10 1n\n4\nr\nn\n= 1\nr 3 + 11\n0\n=\n1 1 n 2\n4\n(\n0\nn\nn\n+\n4\n1 ) 2\n+\n1\n1\n0\nM1 Applies formula for r3 from MF19.\n= 14 10 ( 1 + 1n ) 2 + 11\n0\nA1 AG. CWO.\n4\n\n5(b)  1\n0\n( 11 10 x 3 + 11\n0\n) d x \n1\n10\nn\n+ ( 1n ) ( 11 10 ( 1n ) 3 + 11\n0\n) + ( 1n ) ( 11 10 ( 2n ) 3 + 11\n0\n) +  + ( 1n ) ( 11 10 ( n −n 1 ) 3 + 11\n0\n) M1 A1 Forms the sum of the areas of appropriate\nrectangles. M1 for sum of n complete terms. A1 for\nat least 3 correct terms including first and last.\n=\n1\n10\nn\n+\n1\n10 1n\n4\nnr −\n=\n1\n1\nr 3 + n − 1\n1 0 n\n=\n1 1 ( n\n4\n−\n0\n1\nn\n)\n4\n2 n 2\n+ 11\n0\nM1 Applies formulae from MF19. Allow\nU n + 1 10 n − 1 2 1 0 n\n= 14 10 ( 1 − 1n ) 2 + 11\n0\nA1 OE. e.g. 14 10 ( 1 + 1n ) 2 + 11\n0\n− 1 1\n1 0 n\n.\n4\n\n5(c) U\nn\n− L\nn\n= 14 10 ( 1 + 1n ) 2 − 14 10 ( 1 − 1n ) 2 M1\nTheir U\nn\n− L\nn\nc\nsimplifies to .\nn\nAccept 14 10 ( 1 ) 2 + 11\n0\n− 14 10 ( 1 ) 2 − 11\n0\n.\n11 A1 AG. Their\n→0 as n→\n10n\nL\nn\nmust be correct.\n2\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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        14
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_21.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-21.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-21-q06",
      "question_id": "9231-2026-mj-21-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6\n© Cambridge University Press & Assessment 2026 Page 15 of 18\n5 m 2 + 4 m + 1 = 0  m = − 25  15 i M1 Auxiliary equation. M1 for correct type of roots\nabi.\n y =  e − 25 x ( A c o s 15 x + B s i n 15 x ) A1 Complementary function. Allow ‘ y = ’ missing.\ny = p x 2 + q x + r  y ' = 2 p x + q  y '' = 2 p B1 Particular integral and its derivatives.\np = 1 8 p + q = 7 10p+4q+r=1 M1 Substitutes and equates coefficients.\nq = − 1 r = − 5 A1\ny = e − 25 x ( A c o s 15 x + B s i n 15 x ) + x 2 − x − 5 A1 General solution. Must have ‘y=’.\nFT on CF.\ny ' = e − 25 x ( − 15 A s i n 15 x + 15 B c o s 15 x ) − 25 e − 25 x ( A c o s 15 x + B s i n 15 x ) + 2 x − 1 M1 Differentiates. General solution must be of the\ncorrect form. Must use product rule.\n0 = A − 5 0= 1B−2 A−1\n5 5\n A = 5 , B = 1 5 DM1 A1 Uses initial conditions. A1 is both A and B correct.\ny = e − 25 x ( 5 c o s 15 x + 1 5 s i n 15 x ) + x 2 − x − 5 A1 Must have ‘ y = ’. CWO.\n10\nQuestion Answer Marks Guidance",
      "source_pages": [
        15
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-21.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2026-mj-21-q07",
      "question_id": "9231-2026-mj-21-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) z −\n1 −\nz n\nz\n+ 1\nor\nz n +\nz\n1\n−\n−\n1\nz\nor\nz ( z\nz\nn\n−\n−\n1\n1 ) B1\n1\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)\n© Cambridge University Press & Assessment 2026 Page 16 of 18\ne 2i7 π B1 States one root of z 7 = 1 , z  1 .\nei2\n7\nkπ\nk = 2 , , 6\nB1 Five other roots. No repeats or incorrect roots, z = 1\nmust be omitted.\n2\n\n7(c) EITHER:\n1\n1 1\ne 1\ne\ni\n1 i\n\n\nm\n\n=\nz m =\nz\n− z\n=\n−\n− +\n− +\n(*M1 A1 Applies sum to infinity and substitutes for z. Expect\ne\ne i\ne i\n\n −\n.\n( 1\ne 1\ne\ni\n1\n(\ni\n1\n) ( 1\ne 1\ne\ni\n1\n)\ni ) 1 e\ne\n1 i\n1 i\ne\ne\n1 i\n2\ne 2\n  \n   \n=\n−\n− +\n− +\n− −\n−\n−\n− −\n=\n− − −\n− +\n−\n−\n− +\n−\n+ −\nDM1 Rationalises denominator. Expect\ne\ne i\ne i\ne\ne\ne\ne\ni\ni e 2 e\n1 e(\ne i\ni\ne\n1\ni ) 1\n  \n    −\n −\n−\n−\n−\n\n=\n−\n+\n+\n−\n− +\n.\ne 1\n1\n( c o\n2\ns\ne 1 c\ni\no\ns\ns\ni n )\ne\ne\n2\n2  \n\n=\n−\n− −\n+\n+\n−\n−\n− e ( c\ne\no\n2\ns\n2 e\ni s\nc o\ni\ns\nn )\n1\n1  \n\n\n=\n−\n+\n+\n−  DM1 A1 Applies ei+e−i=2cos\nR e\n1\n1\ne\n2\n1\ne\nc o\n1\ns\nc o s\ne 2\ne 2 e 2\ne c\n2\no\ne\ns\nc o s\n1\n1\n \n \n\nm\n\n=\nz m\n\n=\n−\n−\n−\n−\n+\n−\n−\n=\n−\n−\n+\nDM1 A1) Takes the real part, AG.\nOR:\nAlternative method for question 7(c)\n1\n1 e\ne i\ne i ( e\nc o s\nc o s\ni) s i n\ni s i n\n  \n  \nm\n\n=\nz m =\nz\n− z\n=\n−\n=\n−\n+\n−\n(*M1 A1 Applies sum to infinity and substitutes for z.\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 17 of 18\n( e\ne\nc o s\nc o\nc o s\ns\ni s\n)\nc\ni\no\nn\ni s i\n2 s\nn\n( e\n( e\ni s i n\n( e\nc o s\nc o s\nc o s\nc o s\n))\n2 )\n+\n+\ni\ni\ni\ns\ns\ns\ns\ni n\ni n\ni n\ni n 2\n( e c o s ) s i n 2\n   \n   \n      \n \n=\n−\n+\n−\n−\n+\n −\n−\n−\n+\n+\n\n− −\nDM1 Rationalises denominator.\ne c\ne\no\n2\ns\n2 e\n1\nc o\ni\ns\ne s i n\n1\n \n\n=\n−\n− +\n+\nDM1 A1 Applies s i n 2 c o s 2 1   + = .\n   ecos−1 DM1 A1) Takes the real part, AG.\nRe zm =\n  e2 −2ecos+1\nm=1\nAvailable marks 7\nQuestion Answer Marks Guidance",
      "source_pages": [
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        17
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_21.pdf",
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      "html": "9231-topic-25-complex-numbers/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-21-q08",
      "question_id": "9231-2026-mj-21-q08",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "21",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) −1 −2 1 M1 A1 Finds determinant in terms of a and b.\na 1 −1 =4a+ab−b−2\n0 b 2\na =\nb\nb\n+\n+\n2\n4\nA1\n[since b−4]\n3\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b)\n© Cambridge University Press & Assessment 2026 Page 18 of 18\n( 1 ) ( 1 ) ( 2 ) 3 2 2 2 0       + − − = − − + = B1 Finds characteristic equation.\nA 4 − 2 A 3 − A 2 + 2 A = 0 M1 Replaces  with A and multiplies through by A .\nA4 =2 ( 2A2 +A−2I ) +A2 −2A=5A2 −4I M1 A1 Substitutes A 3 = 2 A 2 + A − 2 I . CWO\n4\n\n8(c) Eigenvalues of A are − 1 , 1 and 2 . B1 Upper diagonal matrix or characteristic equation.\n1 : 0\n0 0\n2 1\n3\n0\n0\n6 1\n0\n0\n = −\ni\n−\nj k\n=\n − \n\n  M1 A1 Uses vector product (or equations) to find\ncorresponding eigenvectors. M0 if all three vectors\nare zero vectors.\n1 :\n0\n2\n0\n2 1\n1\n2\n2\n0\n~\n1\n0\n1  = −\ni\n−\nj k\n−\n=\n\n−\n \n−\n i j k  3   1  A1 A1\n   \n=2: −3 −2 1 = −3 ~ −1\n   \n   \n0 −1 −1  3   1 \nThus P =\n 1\n0\n0\n1\n−\n0\n1\n1\n−\n1\n1\n\nand D =\n −\n0\n0\n1 0\n1\n0 1\n0\n0\n2 8\n M1 A1 Or correctly matched permutations of columns.\nA column of zeros or repeated column in P is M0.\n7\nTrace ID: PC-80D328BA4A\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
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      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2026-mj-22-q01",
      "question_id": "9231-2026-mj-22-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 1,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1\n© Cambridge University Press & Assessment 2026 Page 9 of 18\nf '( x ) = ( 1 + ( x + 1 ) 2 ) − 12 B1 Finds first derivative. Accept\nx + 1 +\n1\n( x + 1 ) 2 + 1\n\n1 +\n2 (\n2\nx\nx\n+\n+\n1\n2\n2 ) + 1\n\nor implicit\ny 'c o s h y = 1 .\nf ''( x ) = − ( 1 + ( x + 1 ) 2 ) − 32 ( x + 1 ) B1 Finds second derivative. Allow not simplified or\nimplicit\ny ''c o s h y + ( y )' 2 s i n h y = 0 .\nsinh−11=ln ( 1+ 2 ) B1\n1\nf '(0)=\n2\nf ''( 0 ) = −\n2\n1\n2\nM1 Evaluates their derivatives at zero only. Must have\nattempted both derivatives.\ns i n h − 1 ( x + 1 ) = ln ( 1 + 2 ) + 1\n2\nx −\n4\n1\n2\nx 2 A1 CWO.\n5\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-22.pdf?download=1",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s26_ms_22-p09.png"
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    },
    {
      "id": "9231-2026-mj-22-q02",
      "question_id": "9231-2026-mj-22-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) I = 1 (1+t)n sinπtdt=−1(1+t)ncosπt 1 + n 1 (1+t)n−1cosπtdt *M1 A1 Applies integration by parts. Allow limits missing.\nn 0  π  0 π 0 M1 for correct trigonometric functions and powers\nof (1+t).\n© Cambridge University Press & Assessment 2026 Page 10 of 18\nI\nn\n= 1π ( 2 n + 1 ) + nπ  1π (1 + t ) n − 1 s i n π t  1\n0\n− nπ  1\n0\n1π ( n − 1 ) ( 1 + t ) n − 2 s i n π t d t DM1 Applies integration by parts again. M1 for correct\ntrigonometric functions and powers of (1 + t ) .\nI\nn\n= 1π ( 2 n + 1 ) − 12\nπ\nn ( n − 1 ) I\nn − 2\n π 2 I\nn\n= π ( 2 n + 1 ) − n ( n − 1 ) I\nn − 2\nA1 AG. Must have seen limits and correct use of\nbrackets.\nCWO.\n4\n\n2(b)\nI\n0\n= \n1\n0\ns i n π t d t =  − 1π c o s π t \n1\n0\n= 2π\nB1 Finds\n.\nI\n0\n.\nπ 2 I\n2\n= π ( 2 2 + 1 ) − 2 ( 2 − 1 ) I\n0\n = 5 π − 4π  M1 Uses reduction formula from (a).\nI\n2\n= 5π − 43\nπ\nA1\n3\nQuestion Answer Marks Guidance",
      "source_pages": [
        10
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-22.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2026-mj-22-q03",
      "question_id": "9231-2026-mj-22-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "3(a) u = c o s h x  d u = s i n h x d x B1\n s i n h 2 x d u =  ( u 2 − 1 ) d u M1 A1 Applies correct identity s i n h 2 u = c o s h 2 u − 1 to find\nintegral in terms of u.\n= 13 u 3 − u ( + C ) =1cosh3x−coshx (+C) A1\n3\n4\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b)\n© Cambridge University Press & Assessment 2026 Page 11 of 18\ne  c o th x d x = e ln sin h x = s i n h x M1 A1 Finds integrating factor.\nd\nd x\n( y s i n h x ) = s i n h 3 x\nM1 Correct form on LHS and attempt to integrate RHS.\ny s i n h x = 13 c o s h 3 x − c o s h x + C *M1 A1 Integrates RHS using their part (a). For M1, their\nanswer to (a) must be of the correct form\nacosh3x+bcoshx with a,bnon-zero.\n0=1(5)3\n−5+CC=10\nDM1 Substitutes initial conditions. M0 if expression\n3 3 3 81 incorrectly manipulated before substitution. E.g. not\nall terms divided by s i n h x .\ny = c o s e c h x ( 13 c o s h 3 x − c o s h x + 18 01 ) A1 OE.\n7\nQuestion Answer Marks Guidance",
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      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2026-mj-22-q04",
      "question_id": "9231-2026-mj-22-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 4,
      "topic": "Differentiation",
      "topic_slug": "9231-topic-23-differentiation",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) d\nd x\n( ( x − y ) 5 ) = 5 ( x − y ) 4 ( 1 − y )' B1 Differentiates ( x − y ) 5 correctly.\nd\nd x\n( ln ( x y ) ) =\nx y '+\nx y\ny\n= y ' y − 1 + x − 1\nB1 Differentiates ln(xy) correctly.\n5 ( 1 − y )' − 12 y '− 1 = 0  y ' = 81\n1\nB1 Substitutes ( − 1 , − 2 ) into\n5 ( x − y ) 4 ( 1 − y )' + y ' y − 1 + x − 1 = 0 , AG. CWO.\nMight see y (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 .\n3\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) EITHER:\n© Cambridge University Press & Assessment 2026 Page 12 of 18\n− 5 ( x − y ) 4 y ''+ 2 0 ( 1 − y )' 2 ( x − y ) 3\n(B1 B1 Differentiates 5 ( x − y ) 4 ( 1 − y )' .\n− ( y ') 2 y − 2 + y − 1 y ''− x − 2 = 0 B1\nDifferentiates y ' y − 1 + x − 1 . Might see\nx y '+\nx y\ny\ndifferentiated using product or quotient rules.\nxy(xy''+2y')−(xy'+ y)2\nE.g. .\n(xy)2\nOr − ( x y ) − 2 ( x y '+ y ) 2 + ( x y ) − 1 ( x y ''+ 2 y )' .\n− 5 (1 ) 4 y ''+ 2 0 ( 1 − 81\n1\n) 2 ( 1 ) 3 − 14 ( 81\n1\n) 2 − 12 y ''− 1 = 0 M1 Substitutes ( − 1 , − 2 ) and y'= 8 .\n11\n− 1 12 y ''+ 4 3\n1 2 1\n= 0  y '' =\n1\n83 63\n1\nA1)\nOR:\nAlternative method for question 4(b)\ny (' − y − 2 y '− 2 0 ( x − y ) 3 ( 1 − y )' ) + y ( '' y − 1 − 5 ( x − y ) 4 ) (B1 B1 Differentiates both sides of\ny (' y − 1 − 5 ( x − y ) 4 ) = − 5 ( x − y ) 4 − x − 1 .\n= − 2 0 ( x − y ) 3 ( 1 − y )' + x − 2 B1\n( 81\n1\n) ( − 14  81\n1\n− 2 0  31\n1\n) + y ( '' − 12 − 5 ) = − 2 0  31\n1\n+ 1 M1 Substitutes ( − 1 , − 2 ) and y'= 8 .\n11\n− 1 12 y ''+ 4 3\n1 2 1\n= 0  y '' =\n1\n83 63\n1\nA1)\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) OR:\nSecond alternative method for question 4(b)\n( 5(x−y)4 −y−1 )( 20(x−y)3(1−y')−x−2 ) (B1 B1 Numerator of derivative of\n5(x− y)4 +x−1\n− ( 5(x−y)4 +x−1 )( 20(x− y)3(1− y')+ y−2y' ) y'= 5(x− y)4 − y−1 .\n© Cambridge University Press & Assessment 2026 Page 13 of 18\n( 5 ( x − y ) 4 − y − 1 ) 2 B1 Denominator of derivative of\ny ' =\n5\n5\n(\n(\nx\nx\n−\n−\ny\ny\n)\n)\n4\n4\n+\n−\nx\ny\n−\n−\n1\n1\n. Can be implied by ( 1 12 ) 2 .\ny '' =\n( 5 + 12 ) ( 2 0  31\n1\n− 1\n( 5\n) −\n+\n(\n12\n5\n)\n−\n2\n1 ) ( 2 0  31\n1\n+ 21\n1\n) M1 Substitutes ( − 1 , − 2 ) and y'= 8 into their\n11\nexpression.\ny '' =\n1\n83 63\n1\nA1)\nAvailable marks 5\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        11,
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        13
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-22.pdf?download=1",
      "html": "9231-topic-23-differentiation/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-22-q05",
      "question_id": "9231-2026-mj-22-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 14 of 18\n 1\n0\n( 11 10 x 3 + 11\n0\n) d x  ( 1n ) ( 11 10 ( 1n ) 3 + 11\n0\n) + ( 1n ) ( 11 10 ( 2n ) 3 + 11\n0\n) +  + ( 1n ) ( 11 10 ( nn ) 3 + 11\n0\n) M1 A1 Forms the sum of the areas of the rectangles. M1 for\nsum of n complete terms. A1 for at least 3 correct\nterms including first and last. May be written in\nsummation form.\n=\n1\n10 1n\n4\nr\nn\n= 1\nr 3 + 11\n0\n=\n1 1 n 2\n4\n(\n0\nn\nn\n+\n4\n1 ) 2\n+\n1\n1\n0\nM1 Applies formula for r3 from MF19.\n= 14 10 ( 1 + 1n ) 2 + 11\n0\nA1 AG. CWO.\n4\n\n5(b)  1\n0\n( 11 10 x 3 + 11\n0\n) d x \n1\n10\nn\n+ ( 1n ) ( 11 10 ( 1n ) 3 + 11\n0\n) + ( 1n ) ( 11 10 ( 2n ) 3 + 11\n0\n) +  + ( 1n ) ( 11 10 ( n −n 1 ) 3 + 11\n0\n) M1 A1 Forms the sum of the areas of appropriate\nrectangles. M1 for sum of n complete terms. A1 for\nat least 3 correct terms including first and last.\n=\n1\n10\nn\n+\n1\n10 1n\n4\nnr −\n=\n1\n1\nr 3 + n − 1\n1 0 n\n=\n1 1 ( n\n4\n−\n0\n1\nn\n)\n4\n2 n 2\n+ 11\n0\nM1 Applies formulae from MF19. Allow\nU n + 1 10 n − 1 2 1 0 n\n= 14 10 ( 1 − 1n ) 2 + 11\n0\nA1 OE. e.g. 14 10 ( 1 + 1n ) 2 + 11\n0\n− 1 1\n1 0 n\n.\n4\n\n5(c) U\nn\n− L\nn\n= 14 10 ( 1 + 1n ) 2 − 14 10 ( 1 − 1n ) 2 M1\nTheir U\nn\n− L\nn\nc\nsimplifies to .\nn\nAccept 14 10 ( 1 ) 2 + 11\n0\n− 14 10 ( 1 ) 2 − 11\n0\n.\n11 A1 AG. Their\n→0 as n→\n10n\nL\nn\nmust be correct.\n2\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-22-q06",
      "question_id": "9231-2026-mj-22-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6\n© Cambridge University Press & Assessment 2026 Page 15 of 18\n5 m 2 + 4 m + 1 = 0  m = − 25  15 i M1 Auxiliary equation. M1 for correct type of roots\nabi.\n y =  e − 25 x ( A c o s 15 x + B s i n 15 x ) A1 Complementary function. Allow ‘ y = ’ missing.\ny = p x 2 + q x + r  y ' = 2 p x + q  y '' = 2 p B1 Particular integral and its derivatives.\np = 1 8 p + q = 7 10p+4q+r=1 M1 Substitutes and equates coefficients.\nq = − 1 r = − 5 A1\ny = e − 25 x ( A c o s 15 x + B s i n 15 x ) + x 2 − x − 5 A1 General solution. Must have ‘y=’.\nFT on CF.\ny ' = e − 25 x ( − 15 A s i n 15 x + 15 B c o s 15 x ) − 25 e − 25 x ( A c o s 15 x + B s i n 15 x ) + 2 x − 1 M1 Differentiates. General solution must be of the\ncorrect form. Must use product rule.\n0 = A − 5 0= 1B−2 A−1\n5 5\n A = 5 , B = 1 5 DM1 A1 Uses initial conditions. A1 is both A and B correct.\ny = e − 25 x ( 5 c o s 15 x + 1 5 s i n 15 x ) + x 2 − x − 5 A1 Must have ‘ y = ’. CWO.\n10\nQuestion Answer Marks Guidance",
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        15
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_22.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-22.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2026-mj-22-q07",
      "question_id": "9231-2026-mj-22-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a) z −\n1 −\nz n\nz\n+ 1\nor\nz n +\nz\n1\n−\n−\n1\nz\nor\nz ( z\nz\nn\n−\n−\n1\n1 ) B1\n1\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)\n© Cambridge University Press & Assessment 2026 Page 16 of 18\ne 2i7 π B1 States one root of z 7 = 1 , z  1 .\nei2\n7\nkπ\nk = 2 , , 6\nB1 Five other roots. No repeats or incorrect roots, z = 1\nmust be omitted.\n2\n\n7(c) EITHER:\n1\n1 1\ne 1\ne\ni\n1 i\n\n\nm\n\n=\nz m =\nz\n− z\n=\n−\n− +\n− +\n(*M1 A1 Applies sum to infinity and substitutes for z. Expect\ne\ne i\ne i\n\n −\n.\n( 1\ne 1\ne\ni\n1\n(\ni\n1\n) ( 1\ne 1\ne\ni\n1\n)\ni ) 1 e\ne\n1 i\n1 i\ne\ne\n1 i\n2\ne 2\n  \n   \n=\n−\n− +\n− +\n− −\n−\n−\n− −\n=\n− − −\n− +\n−\n−\n− +\n−\n+ −\nDM1 Rationalises denominator. Expect\ne\ne i\ne i\ne\ne\ne\ne\ni\ni e 2 e\n1 e(\ne i\ni\ne\n1\ni ) 1\n  \n    −\n −\n−\n−\n−\n\n=\n−\n+\n+\n−\n− +\n.\ne 1\n1\n( c o\n2\ns\ne 1 c\ni\no\ns\ns\ni n )\ne\ne\n2\n2  \n\n=\n−\n− −\n+\n+\n−\n−\n− e ( c\ne\no\n2\ns\n2 e\ni s\nc o\ni\ns\nn )\n1\n1  \n\n\n=\n−\n+\n+\n−  DM1 A1 Applies ei+e−i=2cos\nR e\n1\n1\ne\n2\n1\ne\nc o\n1\ns\nc o s\ne 2\ne 2 e 2\ne c\n2\no\ne\ns\nc o s\n1\n1\n \n \n\nm\n\n=\nz m\n\n=\n−\n−\n−\n−\n+\n−\n−\n=\n−\n−\n+\nDM1 A1) Takes the real part, AG.\nOR:\nAlternative method for question 7(c)\n1\n1 e\ne i\ne i ( e\nc o s\nc o s\ni) s i n\ni s i n\n  \n  \nm\n\n=\nz m =\nz\n− z\n=\n−\n=\n−\n+\n−\n(*M1 A1 Applies sum to infinity and substitutes for z.\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 17 of 18\n( e\ne\nc o s\nc o\nc o s\ns\ni s\n)\nc\ni\no\nn\ni s i\n2 s\nn\n( e\n( e\ni s i n\n( e\nc o s\nc o s\nc o s\nc o s\n))\n2 )\n+\n+\ni\ni\ni\ns\ns\ns\ns\ni n\ni n\ni n\ni n 2\n( e c o s ) s i n 2\n   \n   \n      \n \n=\n−\n+\n−\n−\n+\n −\n−\n−\n+\n+\n\n− −\nDM1 Rationalises denominator.\ne c\ne\no\n2\ns\n2 e\n1\nc o\ni\ns\ne s i n\n1\n \n\n=\n−\n− +\n+\nDM1 A1 Applies s i n 2 c o s 2 1   + = .\n   ecos−1 DM1 A1) Takes the real part, AG.\nRe zm =\n  e2 −2ecos+1\nm=1\nAvailable marks 7\nQuestion Answer Marks Guidance",
      "source_pages": [
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        17
      ],
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      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2026-mj-22-q08",
      "question_id": "9231-2026-mj-22-q08",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "22",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 14,
      "status": "available",
      "reason": null,
      "text": "8(a) −1 −2 1 M1 A1 Finds determinant in terms of a and b.\na 1 −1 =4a+ab−b−2\n0 b 2\na =\nb\nb\n+\n+\n2\n4\nA1\n[since b−4]\n3\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b)\n© Cambridge University Press & Assessment 2026 Page 18 of 18\n( 1 ) ( 1 ) ( 2 ) 3 2 2 2 0       + − − = − − + = B1 Finds characteristic equation.\nA 4 − 2 A 3 − A 2 + 2 A = 0 M1 Replaces  with A and multiplies through by A .\nA4 =2 ( 2A2 +A−2I ) +A2 −2A=5A2 −4I M1 A1 Substitutes A 3 = 2 A 2 + A − 2 I . CWO\n4\n\n8(c) Eigenvalues of A are − 1 , 1 and 2 . B1 Upper diagonal matrix or characteristic equation.\n1 : 0\n0 0\n2 1\n3\n0\n0\n6 1\n0\n0\n = −\ni\n−\nj k\n=\n − \n\n  M1 A1 Uses vector product (or equations) to find\ncorresponding eigenvectors. M0 if all three vectors\nare zero vectors.\n1 :\n0\n2\n0\n2 1\n1\n2\n2\n0\n~\n1\n0\n1  = −\ni\n−\nj k\n−\n=\n\n−\n \n−\n i j k  3   1  A1 A1\n   \n=2: −3 −2 1 = −3 ~ −1\n   \n   \n0 −1 −1  3   1 \nThus P =\n 1\n0\n0\n1\n−\n0\n1\n1\n−\n1\n1\n\nand D =\n −\n0\n0\n1 0\n1\n0 1\n0\n0\n2 8\n M1 A1 Or correctly matched permutations of columns.\nA column of zeros or repeated column in P is M0.\n7\nTrace ID: PC-45F835F92B\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q01",
      "question_id": "9231-2026-mj-23-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 1,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1\n© Cambridge University Press & Assessment 2026 Page 9 of 19\nw 5 = − 3 2 = 3 2 e iπ B1 Converts to exponential form.\nz\n1\n= 2 e 1i5 π M1 A1 Finds one root in required form.\nz\n2\n= 2 e − 1i5 π A1 Finds another root.\nz\n3\n= 2 e 3i5 π , z\n4\n= 2 e − 3i5 π , z\n5\n= 2 e − iπ = − 2 A1 Finds other three roots.\n5\nQuestion Answer Marks Guidance",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-25-complex-numbers/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q02",
      "question_id": "9231-2026-mj-23-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 2,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 1 + ( 12 x − 12 ) 2 = 1 + 14 x − 1 B1 2 dy\nFinds 1+   .\ndx\n2 π \n2\n0\nx 1 +\n1\n4\nx − 1 d x\nM1 Correct formula for surface area with limits.\nIntegral must be in terms of x.\n2 2 A1 Integrand simplified correctly.\n=2π x+ 1 dx or π 4x+1dx\n4\n0 0\n= 2 π\n\n2\n3\n x + 1\n4\n\n32 \n2\n0\nM1 Integrates and substitutes limits. Integrand must\n2\nπ 3 or  (4x+1) 2 be of the form\n6 \n0\n( p x + a ) n with a and p nonzero\nand consistent with previous working and n  1 .\n= 1 33 π A1 AEF.\n5\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q03",
      "question_id": "9231-2026-mj-23-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 3,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3(a) EITHER:\n© Cambridge University Press & Assessment 2026 Page 10 of 19\nd\nd x\n( ( x + 1 ) ln n ( x + 1 ) ) = n ( x + 1 ) ln n − 1\n( x\n(\n+\nx\n1\n+\n)\n1 ) + ln n ( x + 1 )\n(B1 CAO (AEF).\n ( x + 1 ) ln n ( x + 1 ) \na\n0\n= n I\nn − 1\n+ I\nn\nM1 Integrates both sides using the limits given.\nI\nn\n= ( a + 1 ) ln n ( a + 1 ) − n I\nn − 1\nA1) Substitutes limits and rearranges. AG.\nOR:\nAlternative method for question 3(a)\nI\nn\n= \na\n0\nln n ( x + 1 ) d x =  x ln n ( x + 1 ) \na\n0\n− n \na\n0\nx ln n\n(\n−\nx\n( 1\n+\nx\n1\n+\n)\n1 )\nd x\n(B1 CAO (Condone missing limits).\n\n= a l n n ( a + 1 ) − n \na\n0\nln n − 1 ( x + 1 ) d x + n \na\n0\nln n − 1\n( x\n(\n+\nx\n1\n+\n)\n1 ) \n= a l n n ( a + 1 ) − n I\nn − 1\n+  ln n ( x + 1 ) \na\n0\nM1 Expression needs to be fully integrated and in\nterms of a and I\nn − 1\n.\nI\nn\n= ( a + 1 ) ln n ( a + 1 ) − n I\nn − 1\nA1) Substitutes limits, convincing work to AG.\nAvailable marks 3\n\n3(b) I =(a+1)ln(a+1)−a B1 CAO.\n1\nI\n2\n= ( a + 1 ) ln 2 ( a + 1 ) − 2 I\n1\nM1 Applies reduction formula to find I\n2\n.\nI\n2\n= ( a + 1 ) ln 2 ( a + 1 ) − 2 ( a + 1 ) ln ( a + 1 ) + 2 a A1\n3\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        10
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q04",
      "question_id": "9231-2026-mj-23-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 4,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4\n© Cambridge University Press & Assessment 2026 Page 11 of 19\n4 9 m 2 + 1 4 m + 1 = 0  m = − 17 M1 Auxiliary equation.\nx = e − 17 t ( A t + B ) A1 Complimentary function.\nx = k e 17 t  x ' = 17 k e 17 t  x '' = 14\n9\nk e 17 t B1 Particular integral and its derivatives.\nk e 17 t + 2 k e 17 t + k e 17 t = 1 9 6 e 17 t  4 k = 1 9 6 M1 Substitutes and equates coefficients.\nk = 4 9 A1\nx = e − 17 t ( A t + B ) + 4 9 e 17 t A1 General solution.\nx ' = A e − 17 t − 17 e − 17 t ( A t + B ) + 7 e 17 t M1* General solution must be of correct form for any\nfurther marks.\n7 = B + 4 9 0 = A − 17 B + 7  A = − 1 3 , B = − 4 2 dM1 A1 Substitutes initial conditions to find A and B.\nx = e − 17 t ( − 1 3 t − 4 2 ) + 4 9 e 17 t A1\n10\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        11
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q05",
      "question_id": "9231-2026-mj-23-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 5,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 12 of 19\ns e c h t =\ne t\n2\n+ e − t\nt a n h t =\ne\ne\nt\nt\n−\n+\ne\ne\n−\n−\nt\nt\nB1\n1 −\n\ne t\n2\n+ e − t\n 2\n=\n( e t +\n( t e\ne\n+\n− t\ne\n)\n−\n2\nt\n−\n2 )\n4\n=\ne 2 t +\n( e\n2\nt\n+\n+\n− 2 t e\n2 ) − t e\n− 4\n=\n(\n(\ne\ne\nt\nt\n−\n+\ne\ne\n−\n−\nt\nt\n)\n)\n2\n2\nM1 A1 Expands, gets to e 2 t +\n( e\n2\nt\n+\n+\n− 2 t e\n2 ) − t e\n− 4 for M1, AG.\nVariable used must be consistent throughout. A0\nif a mixture of variables is used.\n3\n\n5(b) d\nd\ny\nt\n= t s i n h t + c o s h t ,\nd\nd\nx\nt\n= t c o s h t + s i n h t\nB1 CAO\nd\nd\ny\nx\n=\nd\nd\ny\nt\n\nd\nd\nt\nx\n=\nt\nt\ns\nc\ni n\no\nh\ns h\nt\nt\n+\n+\nc\ns\no\ni\ns\nn\nh\nh\nt\nt\nM1 Uses chain rule.\nt\nt\ns\nc\ni n\no\nh\ns h\nt\nt\n+\n+\nc\ns\no\ni\ns\nn\nh\nh\nt\nt\n\ns\ns\ne\ne\nc\nc\nh\nh\nt\nt\n=\nt t a\nt +\nn h\nt a\nt\nn\n+\nh\n1\nt\nA1 Divides numerator and denominator by c o s h t ,\nAG.\n3\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c)\n© Cambridge University Press & Assessment 2026 Page 13 of 19\nd\nd t\n d\nd\ny\nx\n\n=\n( t + t a n h t ) ( t s e c h 2 t +\n(\nt a\nt\nn\n+\nh\nt\n) t\na n\n−\nh\n( t\n2 ) t\nt a n h t + 1 ) ( 1 + s e c h 2 t ) *M1 A1 Applies quotient rule.\n t 2 s e c h 2 t + t t a n h t + t s e c h 2 t t a n h t +\n( t\nt a\n+\nn\nt\nh\na\n2 t\nn h\n−\nt )\nt\n2\nt a n h t − t t a n h t s e c h 2 t − 1 − s e c h 2 t \n t 2 s e c h 2 t +\n( t\nt a\n+\nn\nt\nh\na\n2\nn\nt\nh\n−\nt )\ns\n2\ne c h 2 t − 1 \nt 2 s e c h 2 t + ( 1\n( t\n−\n+\ns e c h\nt a n h\n) 2 t\n2 ) t\n− s e c h 2 t − 1\nM1 Simplifies numerator to the form ( t2 −2 ) sechnt\nby applying t a n h 2 t = 1 − s e c h 2 t .\n=\n( 2 t\n( t\n−\n+\n2\nt\n) s\na n\ne\nh\nc\nt\nh\n)\n2\n2\nt A1\nd\nd\n2\nx\ny\n2\n=\n( 2 t\n( t\n−\n+\n2\nt\n) s\na n\ne\nh\nc\nt\nh\n)\n2\n2\nt\n\nt c o s h\n1\nt + s i n h t\n=\n( 2 t\n( t\n−\n+\n2\nt\n) s\na n\ne\nh\nc\nt\nh\n)\n3\n3\nt dM1 A1 dt Uses chain rule, multiplies by their for M1.\ndx\nConvincing argument to final result for A1\n6\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q06",
      "question_id": "9231-2026-mj-23-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 6,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a)\n© Cambridge University Press & Assessment 2026 Page 14 of 19\nd\nd\ny\nx\n+\n1 −\ny\nx 2\n= 1 − x\nB1 Divides through by 1 − x 2 .\ne 1 1 − x 2 d x  = e ta n h −1x o r e 12 ln  11 +− xx   M1 A1 a  dx Finds integrating factor. Allow e1−x2 for M1.\n1 1+x A1 AG. Must have obtained all previous marks.\nln  1+x\n=e2 1−x =\n1−x\n4\n\n6(b) d\nd x\n\ny\n1\n1\n+\n−\nx\nx\n\n= 1 + x\nM1 A1 Correct form on LHS and RHS.\ny 1\n1\n+\n−\nx\nx\n= 23 ( 1 + x ) 32 + C M1 A1 Integrates RHS.\n1 = 23 + C  C = 13 M1 Substitutes initial conditions.\ny 1\n1\n+\n−\nx\nx\n= 23 ( 1 + x ) 32 + 13 A1 y=2(1+x)(1−x) 1 2 +1(1+x)−1 2(1−x) 1 2 . Final 3 3\nanswer must be in terms of x and not be in\nterms of integrating factor.\n6\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        14
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2026-mj-23-q07",
      "question_id": "9231-2026-mj-23-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 7,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 12,
      "status": "available",
      "reason": null,
      "text": "7(a)\n© Cambridge University Press & Assessment 2026 Page 15 of 19\n\n1\nx\n− x 2\nd x = −\n1\n2\n\n1\nu\nd u\nM1 A1\nApplies given substitution. Allow \na\nu\nd u for\nM1.\n= − 1 − x 2 [ + C ] A1 Accept without ‘ + C ‘. Must be in terms of x .\n3\n\n7(b)  1\nn\ns i n − 1\n1\nn\n+\n1\nn\ns i n − 1\n2\nn\n+  +\n1\nn\ns i n − 1\nn\nn\n\n= r\nn\n= 1\n1\nn\ns i n − 1\n r\nn\n\n \n1\ns0 i n − 1 x d x\nM1 M1 correct comparison (no limits required)\nA1 A1 correct limits on both sides\n\n1\n0\ns i n − 1 x d x =  x s i n − 1 x \n1\n0\n−\n\n1\n0 1\nx\n− x 2\nd x\n x s i n − 1 x + 1 − x 2 \n1\n0\nM1 Integrates by parts.\nA1 CAO with correct signs\n=\n1\n2\nπ − 1\nA1 Clear substitution of limits to given result\n5\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 16 of 19\n1\nn\ns i n − 1\n1\nn\n+\n1\nn\ns i n − 1\n2\nn\n+  +\n1\nn\ns i n − 1\nn −\nn\n1\n=\nn\nr\n−\n=\n1\n1\n1\nn\ns i n − 1\n r\nn\n M1 Forms the sum of the areas of appropriate\nrectangles.\nnr −\n=\n1\n1\n1\nn\ns i n − 1\n r\nn\n\n=\nn\nr\n= 1\n1\nn\ns i n − 1\n r\nn\n\n−\n 1\nn\n  π\n2\n A1 Adjusts summation with reference to missing\nterm.\n nr −\n=\n1\n1\n1\nn\ns i n − 1\n r\nn\n\n \n1\ns0 i n − 1 x d x\n A1\nn 1 r 1 π\n sin−1   π−1+\nn n 2 2n\nr=1\nCAO\n3\n\n7(d) 12 π − 1 B1 Dependent on obtaining full marks in 7(c).\n1\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-24-integration/answers.html",
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    {
      "id": "9231-2026-mj-23-q08",
      "question_id": "9231-2026-mj-23-q08",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "23",
      "question_number": 8,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 15,
      "status": "available",
      "reason": null,
      "text": "8(a) EITHER:\n© Cambridge University Press & Assessment 2026 Page 17 of 19\n6\na\n2\n− 1\n4\nb\n2 − 1 2\n1\n− 8\n= − 1 2 a b − 9 6 a − 6 b − 1 2 0 = 0\n(M1A1 Finds determinant in terms of a and b.\na = −\n2\nb( +\nb\n2\n+\n0\n8 )\n[ b  − 8 .\nA1)\n]\nOR:\nAlternative method for 8(a)\n( 6 + 1 2 a ) x + 3 6 y = 2 5 (R1+12R2)\n( 2 + 8 a ) x + ( b + 3 2 ) y = 1 9\n(M1 Row eliminates\n(R3+8R2)\nz from two of the equations and\nforms an equation from resulting coefficients\nbeing in ratio.\n6\n2\n+\n+\n1 2\n8\na\na\n=\nb\n3\n+\n6\n3 2\nA1 Correct ratio equation\na = −\n2\nb( +\nb\n2\n+\n0\n8 )\nA1) CAO\n[b−8.]\nAvailable marks 3\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b)\n© Cambridge University Press & Assessment 2026 Page 18 of 19\n−\n6\n2\nx\n14x\n−\nx\n−\n+\n1 2\n4\n3 2\ny\ny\ny\n−\n+\n−\n1 2 z =\nz =\n8 z =\n1\n2\n3\n,\n,\n,\n − 8 ( 2 ) = 3\nB1 Must show an inconsistent system.\nTwo distinct parallel planes. B1 B1 Accept a clear sketch for parallel planes and\nthird plane intersecting both.\nOther plane not parallel.\n3\n\n8(c)  6 −12 −121 −6 1 M1 MultipliesAwith given eigenvector.\n      \n−1 4 1 0 = 0 =−6 0\n      \n      \n 2 −8 −8 1 −6 1\n−6 A1\n2\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(d)\n© Cambridge University Press & Assessment 2026 Page 19 of 19\n6\n2\n1 4\n1 2\n8 8\n1\n1\n2\n3 2 2 3 6 7 2\n\n   \n\n−\n−\n−\n−\n− −\n−\n−\n= − + + −\nM1 Characteristic equation.\n2 , 6 , 6  = − A1\n2 : 4\n1\n1\n2\n2 1\n1\n2\n1 2\n8\n4\n3\n2\n1\n =\n−\ni\n−\nj\n−\nk\n=\n\n−\n\n\n\n−\n M1 A1 Uses vector product (or equations) to find\ncorresponding eigenvectors.\n6 : 0\n1\n1 2\n2\n1\n1\n2 1\n3\n2\n1\n6\n2\n~ 1\n3\n1\n =\n−\ni\n−\n−\nj\n−\nk\n=\n −\n−\n  −\n−\n A1\nThus P =\n 3\n2\n− 1\n−\n1\n−\n3\n1\n1\n0\n1\n\nand D =\n 2\n0\n0\n0\n6\n0\n0\n0\n− 6\n M1 A1 Or correctly matched permutations of columns.\nM0A0 if a zero eigenvector is included.\n7\nTrace ID: PC-D4137E1CBF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_23.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-23.pdf?download=1",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-24-q01",
      "question_id": "9231-2026-mj-24-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 1,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1(a)\n© Cambridge University Press & Assessment 2026 Page 9 of 19\n3\n8\n8\nk\n1\n2\n7\n2\n4\n= 3\n1\n2\n2\n4\n− k\n8\n8\n2\n4\n+ 7\n8\n8\n1\n2\n= 0 − k ( 1 6 ) + 7 ( 8 )\nM1A1 Evaluates determinant. For M1 allow sign\nerrors.\nk  72 A1\nThree planes intersect at a unique point. B1\n4\n\n1(b) 3x+ 7 y+ 7z= 1, B1 OE. Obtains a clear numerical contradiction.\n2 −25x+7=1\n8x+ y+ 2z= 2, y=2−2z−8x 1 = 6 CWO. Might see 1= 2 or 1=− 1 . Accept\n25 22\n−8x+4=3 8 25\n8x+ 2y+ 4z= 3, 3 7 7 1\n2\n \ncorrect row echelon form e.g. 0 1 2 2\n 25\n \n0 0 0 *\n \nwith * non zero.\nTriangular prism. B1\n2\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        9
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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    },
    {
      "id": "9231-2026-mj-24-q02",
      "question_id": "9231-2026-mj-24-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 2,
      "topic": "Matrices (Further Pure 2)",
      "topic_slug": "9231-topic-22-matrices-further-pure-2",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "2(a) EITHER:\n© Cambridge University Press & Assessment 2026 Page 10 of 19\nd e t ( ) d e t\na\n5 1\n1\n( a ) ( 1 ) 5\n\n  \n\nA − I =\n − −\n−\n\n= − − +\n(M1A1 Finds d e t ( )  A − I\n2 ( a 2 ) ( 1 2 ) 5 0 a 7  =  − − + =  = A1\n2 8 1 2 0 6    − + =  = A1)\nOR:\nAlternative method for 2(a)\n a\n5\n−\n1\n1   x\ny\n\n= 2\n x\ny\n\n\na\n5\nx\nx\n−\n+\ny\ny\n=\n=\n2\n2\nx\ny\n ( a − 2 ) x = 5 x\n(M1A1 Uses that 2 is eigenvalue to form an equation in\na.\na = 7 A1\n7\n5 1\n1 x\ny\nx\ny\n7\n5\nx\nx\ny\ny\nx\ny\n( 7 )\n5\n1\n6\n\n  \n \n −   \n=\n \n\n−\n+\n=\n=\n − =\n−\n =\nA1) CWO.\nAvailable marks 4\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b) 7 −1x x 7x−y=2x x 1 *M1A1 M0 if both of their answers are zero vectors.\n  =2   ~  i j −1 1\n5 1 y y 5x+y=2y y 5\nAccept = ~  and\n5 −1 −5 5\n© Cambridge University Press & Assessment 2026 Page 11 of 19\ni\n1 −\nj\n1\n=\n −\n−\n1\n1\n\n~\n 1\n1\n\n.\nM1 for setting up and solving equations.\n 7\n5\n−\n1\n1   x\ny\n\n= 6\n x\ny\n\n\n7\n5\nx\nx\n−\n+\ny\ny\n=\n=\n6\n6\nx\ny\n\n x\ny\n\n~\n 1\n1\n A1\nThus P =\n 1\n5\n1\n1\n 32 0  DM1A1 Or correctly matched permutations of columns.\nand D= .\nM0 if a column of zeros appears in P. Accept\n 0 7776\n2 5\nand 6 5 in D.\n5\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        10,
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-22-matrices-further-pure-2/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-24-q03",
      "question_id": "9231-2026-mj-24-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 3,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3 EITHER:\n© Cambridge University Press & Assessment 2026 Page 12 of 19\ne  1 d x = e x\n(M1A1 Finds integrating factor. Allow sign error for M1\ne.g.\ne\n − 1 d x\n.\nd\nd x\n( y e x ) = e 32 x M1 Correct form on LHS.\nyex =e 3 2 x = 2e 3 2 x +C *M1A1 Integrates RHS. M1 for integrating RHS of the\n3\ncorrect form a e b x .\n3 = 23 + C DM1 Substitutes initial conditions.\ny = 23 e 12 x + 73 e − x A1)\nOR:\nAlternative method for 3\ny = a e 12 x  y ' = 12 a e 12 x (M1A1\n1ae 1 2 x +ae 1 2 x =e 1 2 x a=2 M1A1\n2 3\ny= 2e 1 2 x +Be−x *M1\n3\n3 = 23 + B DM1 Substitutes initial conditions.\ny= 2e 1 2 x +7e−x A1)\n3 3\nAvailable marks 7\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        12
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
      "image_paths": [
        "../answer-assets/9231_s26_ms_24-p12.png"
      ]
    },
    {
      "id": "9231-2026-mj-24-q04",
      "question_id": "9231-2026-mj-24-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 4,
      "topic": "Hyperbolic functions",
      "topic_slug": "9231-topic-21-hyperbolic-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 19\nx = 4 s i n h 2 t + 3 , y = 3 s i n h 2 t − 4 B1 Differentiates x and y with respect to t.\nx 2 + y 2 = ( 4 s i n h 2 t + 3 ) 2 + ( 3 s i n h 2 t − 4 ) 2 = 2 5 ( s i n h 2 2 t + 1 ) M1 Expands and simplifies x 2 + y 2 . Must be in\nterms of t.\n=25cosh22t A1\n5  1\n0\nc o s h 2 t d t = 52  s i n h 2 t 1\n0\n= 54 ( e 2 − e − 2 ) M1A1 Applies correct formula for arc length. M1 for\nintegrating the correct form a c o s h 2 t .\n5\n\n4(b) y\nx\n=\n3\n4\ns\ns\ni n\ni n\nh\nh\n2\n2\nt\nt\n−\n+\n4\n3\nB1 Finds first derivative.\nd 3sinh2t−4 (4sinh2t+3)(6cosh2t)−(3sinh2t−4)(8cosh2t) *M1\n = Differentiates\ndt4sinh2t+3 (4sinh2t+3)2\nd\nd\ny\nx\nwith respect to t.\nd\nd\ny\nx\nmust\nhave the correct form\na\nc\ns\ns\ni n\ni n\nh\nh\n2\n2\nt\nt\n+\n+\nb\nd\n.\n24sinh2tcosh2t+18cosh2t−24sinh2tcosh2t+32cosh2t A1 Correct numerator expanded showing that\n(4sinh2t+3)2 24sinh2tcosh2twill cancel.\n \n50cosh2t\n= \n\n(4sinh2t+3)2\n\n \nd\nd\n2\nx\ny\n2\n=\nd\nd t\n 3\n4\ns\ns\ni n\ni n\nh\nh\n2\n2\nt\nt\n−\n+\n4\n3\n\n\nd\nd\nt\nx\n=\n( 4\n5\ns\n0\ni n\nc\nh\no s\n2\nh\nt\n2\n+\nt\n3 ) 3\nDM1A1 Applies chain rule.\n5\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        13
      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-21-hyperbolic-functions/answers.html",
      "image_paths": [
        "../answer-assets/9231_s26_ms_24-p13.png"
      ]
    },
    {
      "id": "9231-2026-mj-24-q05",
      "question_id": "9231-2026-mj-24-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 5,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5\nSum\n© Cambridge University Press & Assessment 2026 Page 14 of 19\n=\n\n1\n \n+\n\n1\n 7\n+  +\n ( 4 n 2\n1\n+ 4 n + 3 )\n\n\nn\n0\n ( 4 x 2\n1\n+ 4 x + 3 )\nd x\nM1A1 Forms sum of the areas of the rectangles. M1 for\ncorrect comparison (inequality). A1 for limits.\n4x2+4x+3=(2x+1)2\n+2\nB1 Completes the square\n n\n0 ( 2 x +\n1\n1 ) 2 + 2\nd x =\n\n12 s i n h − 1\n 2 x\n\n+\n\n1   n\n0\nM1A1 Uses formula\n=\n\n12 ln\n 2 x\n\n+\n\n1\n+\n( 2 x + 1\n2\n) 2 + 2   n\n0\n=\n\n12 ln\n 2 x + 1 +  ( 4\n\nx\n\n2 + 4 x + 3 )   n\n0\nM1A1 Uses logarithmic form of s i n h − 1 .\n2n+1+ ( 4n2 +4n+3 ) 1+ M1 Inserts limits.\n= 1ln −1ln \n2    2   \n \n1+2n+ ( 4n2 +4n+3 ) A1 AG. CWO.\n= 1ln \n2  + \n \n9\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
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    },
    {
      "id": "9231-2026-mj-24-q06",
      "question_id": "9231-2026-mj-24-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 6,
      "topic": "Integration",
      "topic_slug": "9231-topic-24-integration",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) 1π 1π 1π *M1A1 Applies integration by parts. For M1, parts\nI =2 excosn xdx=excosn x2 +n2 excosn−1xsinxdx\nn −1π   −1π −1π chosen must lead to the given answer. Correct\n2 2 2\nexponents.\n© Cambridge University Press & Assessment 2026 Page 15 of 19\nI\nn\n= n   e x c o s n − 1 x s i n x  12\n−\nπ\n12 π\n−  1 π 2\n1 − π\n2\ne x ( c o s n x − ( n − 1 ) c o s n − 2 x s i n 2 x ) d x  DM1 Applies integration by parts again. Correct\nexponents.\nI\nn\n= − n I\nn\n+ n ( n − 1 ) \n1 π\n2\n1 − π\n2\ne x c o s n − 2 x s i n 2 x d x\nA1 AG. Withhold for missing x in trig functions; for\nexample,cosn−2sin2x. No limits seen is A0.\nI\nn\n= − n I\nn\n+ n ( n − 1 ) \n1 π\n2\n1 − π\n2\ne x c o s n − 2 x d x − n ( n − 1 ) \n1 π\n2\n1 − π\n2\ne x c o s n x d x\nM1 Applies sin2x=1−cos2x to given answer.\nI =−nI +n(n−1)I −n(n−1)I  ( n2 +1 ) I =n(n−1)I A1 AG.\nn n n−2 n n n−2\n6\n\n6(b) I = 1 2 π exdx=e 1 2 π −e−1 2 π. B1 Finds\n0\n−1π\n2\nI 0 .\n5 I\n2\n= 2 I\n0\nor 17I =12I . M1 Uses reduction formula from (a).\n4 2\nI = 2 ( e 1 2 π −e−1 2 π ) A1\n2 5\n17I =12I I = 24 ( e 1 2 π −e−1 2 π ) A1\n4 2 4 85\n4\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-24-integration/answers.html",
      "image_paths": [
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    },
    {
      "id": "9231-2026-mj-24-q07",
      "question_id": "9231-2026-mj-24-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 7,
      "topic": "Complex numbers",
      "topic_slug": "9231-topic-25-complex-numbers",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "7(a)\n© Cambridge University Press & Assessment 2026 Page 16 of 19\nc o s 7 R e ( c i s ) 7  = + *M1 Uses binomial theorem.\n=c7 −21s2c5+35s4c3−7s6c A1\n=c7 −21c5( 1−c2) +35c3( 1−c2)2 −7c ( 1−c2)3 M1 Uses\n=c7 −21c5( 1−c2) +35c3( 1−2c2 +c4) −7c ( 1−3c2 +3c4 −c6)\ns 2 = 1 − c 2 .\n=64c7 −112c5 +56c3−7c A1 Must have shown expansion of ( 1 − c 2 ) 3 clearly.\ns e c 7\n6 4 c 7 1 1 2 c\n1\n5 5 6 c 3 7 c\ns\ns\ne\ne\nc\nc\n7\n7\n\n\n\n =\n− + −\n\nDM1 Divides numerator and denominator by c7.\n(Must see working for this.)\nsec7 A1 AG. CWO\nsec7=\n64−112sec2+56sec4−7sec6\n6\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b)\nx7 =2 ( 7x6 −56x4 +112x2 −64 ) \nx7\n=−2\nM1A1 Relates with equation in (a). M1 for setting\n64−112x2 +56x4 −7x6\n© Cambridge University Press & Assessment 2026 Page 17 of 19\nx s e c  = and deriving sec7=a. A1 for\na = − 2 . M1 can be implied from\nx7\n=−2.\n64−112x2 +56x4 −7x6\nsec7=−2cos7=−1 M1 Solves\n2\nc o s 7 12 .  = −\nx = s e c ( 22\n1\nπ ) A1 Gives one correct solution.\nx = s e c ( q π ) , q = − 1 62\n1\n, − 1 02\n1\n, − 42\n1\n, 82\n1\n, 1 42\n1\n = 23  , 22 01 A1 Gives six other solutions. Allow different values\nof q if all seven solutions are found. Might see\nq = 22\n1\n, 82\n1\n, 1 42\n1\n, 22 01 , 22 61 , 32 21 , 32 81 .\n5\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        16,
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-25-complex-numbers/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-24-q08",
      "question_id": "9231-2026-mj-24-q08",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 2,
      "variant": "24",
      "question_number": 8,
      "topic": "Differential equations",
      "topic_slug": "9231-topic-26-differential-equations",
      "marks": 13,
      "status": "available",
      "reason": null,
      "text": "8(a)\n© Cambridge University Press & Assessment 2026 Page 18 of 19\nd\nd\nv\nx\n= 3 x y 2\nd\nd\ny\nx\n+ y 3\nB1\nd2v d2y dy dy  dy\n=3xy2 + 6xy +3y2 +3y2\ndx2 dx2 dx dx  dx\n= 3 x y 2\nd\nd\n2\nx\ny\n2\n+ 6 x y\n d\nd\ny\nx\n 2\n+ 6 y 2\nd\nd\ny\nx\n*B1 Clear use of product and chain rules.\n13\nd\nd\n2\nx\nv\n2 +\nd\nd\nv\nx + v = x y 2\nd\nd\n2\nx\ny\n2 + 2 x y\n d\nd\ny\nx\n 2\n+ 2 y 2\nd\nd\ny\nx + 3 x y 2\nd\nd\ny\nx + y 3 + x y 3 = x + 5\nDM1A1 Substitutes, AG. For M1 must see terms of\n13 d\nd\n2\nx\nv\n2\nseparately.\n4\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n8(b) 1m2+m+1=0\n3\n© Cambridge University Press & Assessment 2026 Page 19 of 19\n m = − 32  32 i  M1 Auxiliary equation. M1 is for the correct type of\nroots.\n v =  e − 32 x ( A c o s 32 x + B s i n 32 x ) A1 Complementary function. Allow ‘ v = ’ missing.\n[ v = ] a x + b  [ v ' = ] a  [ v '' = ] 0 B1 Particular integral and its derivatives.\na + a x + b = x + 5 M1 Substitutes and equates coefficients.\na = 1 , b = 4 A1\nv = x y 3 = e − 32 x ( A c o s 32 x + B s i n 32 x ) + x + 4 A1 Substitutes for y and find v in terms of x. Must\nnot see i. FT on CF. Must have “ v = ”\nor ‘ x y 3 = ’.\ny=3 x−1e−3 2 x( Acos 3x+Bsin 3x ) +1+4x−1 A1 Makes y the subject.\n2 2\n7\n\n8(c) e − 32 x → 0  y  3 1 + 4 x − 1 M1A1 M1 for stating e−kx →0 (for k0) and\nderiving 3a+bx−1 . Condone e − 32 x  0 .\n2\nTrace ID: PC-F3F6C8368E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_24.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-24.pdf?download=1",
      "html": "9231-topic-26-differential-equations/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q01",
      "question_id": "9231-2026-mj-31-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 1,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) (1x)πr2x+(9r+x)18πr3 =x ( πr2x+18πr3) M1 Moments equation with three terms, dimensionally\n2 8 correct so must be of the form\nr2x2 +(r+x)r3 =x ( r2x+r3) where\n,,,, are all non-zero constants.\nFor reference:\nShape Distance Volume\nCylinder\n© Cambridge University Press & Assessment 2026 Page 9 of 22\n12 x π r 2 x\nHemisphere 38  3 r + x 12  43 π ( 3 r ) 3 = 1 8 π r 3\nTotal x πr2(x+18r)\nB1 At least two terms correct.\nA1 All correct.\n2x2 +81r2 +72xr A1\nx = OE, for example,\n4x+72r\n12 π x 2 r 2\nπ\n+\n2 r\n8 14\nx +\nπ r\n1\n4\n8\n+\nπ r\n1\n3\n8 π r 3 x\n.\nAll individual terms must be simplified.\nISW after correct answer seen.\n4\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b) x−x *M1\ntan= For\n3r\n© Cambridge University Press & Assessment 2026 Page 10 of 22\nt a n\n3\n =\nx −\nr\nx\nor t a n\n3\n =\nx −\nr\nx\nor t a n\n3\n =\nx\nr\n− x\nor t a n\n3\n =\nx −\nr\nx\nonly (allow equivalents with x\nreplaced with r or vice versa).\n2x2 +81x2 +72x2 DM1\n−x For\n4x+72x\ntan=\n3x\nt a n\n3\n =\nx −\nx\nx\nor t a n\n3\n =\nx −\nx\nx\nonly with their x\ncorrectly substituted – their xmust be dimensionally\ncorrect and have come from an expression of the\nform\na x 2 +\nd\nb\nx\nr\n+\n2\ne\n+\nr\nc x r\nwith a, b, c, d and e being non-\nzero.\nMust be in terms of either x or r only.\n155x 79\nIf correct, may see −x= x in numerator.\n76 76\nFor reference: t a n\n7 9\n2 2 8\n =  .\n=19.1 A1 AWRT 19.1 (19.110752…) or 0.334 radians\n(0.3335455…).\n79\nAllow tan=− =19.1for full marks (but\n228\nA0 if given as a negative angle).\n3\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        9,
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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      ]
    },
    {
      "id": "9231-2026-mj-31-q02",
      "question_id": "9231-2026-mj-31-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2 (F =)71242 [=1344] B1 Correct unsimplified expression for the radial force.\nradial\nAttempt to apply N2L either horizontally or vertically to form an equation. M1 Correct number of terms, dimensionally correct,\nallow sign errors and sin/cos mix only. Values of\nsin/cos must be substituted. Allow 37 or better\n(36.8698…) for the angle between AP and the\nhorizontal and 23 or better (22.6198…) for the angle\nbetween BP and the horizontal.\n3T = 5T +7g\n5 A 13 B\n© Cambridge University Press & Assessment 2026 Page 11 of 22\n(  3 9 T\nA\n= 2 5 T\nB\n+ 4 5 5 0 ) A1 Allow 37 or better (36.8698…) for the angle between\nAP and the horizontal and 23 or better (22.6198…)\nfor the angle between BP and the horizontal, for\nexample, T\nA\ns i n ( 3 7 ) = T\nB\ns i n ( 2 3 ) + 7 g .\n45 T\nA\n+ 11 23 T\nB\n= 7  1 2  4 2 (  1 3 T\nA\n+ 1 5 T\nB\n= 2 1 8 4 0 ) A1 Allow 36 or better (36.8698…) for the angle between\nAP and the horizontal and 23 or better (22.6198…)\nfor the angle between BP and the horizontal, for\nexample, T cos(37)+T cos(23)=71242.\nA B\nT =675 and T =871 A1 AWRT 675 and 871 only.\nA B\n5\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        11
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      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q03",
      "question_id": "9231-2026-mj-31-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a)\n© Cambridge University Press & Assessment 2026 Page 12 of 22\n8 − 0 .5 v 2 = 6 .4 v\nd\nd\nv\nx\n*B1 Correct use of Newton’s second law – acceleration\nmust be v\nd\nd\nv\nx\n.\n 6.4v *M1 Separate variables correctly (from a differential\nx= dv=−6.4ln8−0.5v2 +c\n8−0.5v2\nequation of the form\nor\n 6.4v\nx=− dv=−6.4ln 0.5v2 −8 +c\n0.5v2 −8\n2\nd\nd\n   + v = v\nv\nx\n) and obtain an\nequation of the form x = k\n1\nln ( k\n2\n+ k\n3\nv 2 ) where k\n1\n, k\n2\nand k\n3\nare non-zero constants OR\nx= Aln(k +k v)+Bln(k +k v) where A and B\n1 2 3 4\nmust have the same sign. Condone lack of +c.\nx=−6.4ln8−0.5v2 +c (*) A1\nOE, for example,\nor\nx=−6.4ln 0.5v2 −8 +c (**)\n− 12 ln ( 4 − v ) − 12 ln ( 4 + v ) =\n1\nx\n2 .8\n+ c .\nCondone lack of modulus signs in (*). However, if\nsolving x = −\n\n0\n6\n.5\n.4\n2 v\nv\n− 8\nd v then modulus must be\nincluded in (**). Condone lack of +c.\nx = 0 , v = 0  c = 6 .4 ln 8 DM1 Use correct initial conditions to find c.\nNote that 0 = − 6 .4 ln ( 0 − 8 ) + c leading to a value of c\nis DM0.\n1 .3 7 m s–1 A1 WWW\nAWRT 1.37 (1.3711489…) – or exact, for example,\nv = 4 1 − e\n−\n18\n.\n5\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b) dv B1 Correct use of Newton’s second law – acceleration\n8−0.5v2 =6.4\ndt\nmust be\n© Cambridge University Press & Assessment 2026 Page 13 of 22\nd\nd\nv\nt\n.\n 12.8 4+v M1 Separate variables correctly (from a differential\nt= dv=1.6ln +c\n16−v2 4−v equation of the form 2 d\nd\n   + v = v\nt\n) and obtain an\nequation of the form\nt = A ln ( k\n1\nv + k\n2\n) + B ln ( k\n3\nv + k\n4\n) where\nA , B , k\n1\n, k\n2\n, k\n3\nand k are non-zero constants but A\n4\nand B must have different signs.\nAllow a r t a n h ( )   t = v for any , 0   where\n1 .   \n4+v\nt =1.6ln +c(*)\n4−v\nor\nt = − 1 .6 ( ln v − 4 − ln v + 4 ) + c\nA1 Allow unsimplified and\n(**)\nt = 3 .2 a r t a n h ( 14 v ) + c .\nCondone lack of +c. Condone lack of modulus signs\nin (*), but in (**) must see modulus sign on at least\nthe ln ( v − 4 ) term.\nFind c: t = 0 when v = 0  c = 0\nt = 1 .6 ln\n4\n4\n+\n−\nv\nv\n(*)\nor\nt = 1 .6 ln\n4\nv\n+\n−\nv\n4\nA1 As a minimum must state, ‘\n(**)\nt = 0 , v = 0  c = 0 ’.\nISW if re-arranged to make v the subject.\nAllow as two separate log terms.\nAllow t = 3 .2 a r t a n h ( 14 v ) .\nCondone lack of modulus sign for (*) but required in\n(**) in at least the ln ( v − 4 ) term.\nIf answer is in terms of x, for example\n4+x\nt =1.6ln , then maximum B1 M1 A0 A0\n4−x\npossible. Allow recovery for full marks.\n4\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        12,
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q04",
      "question_id": "9231-2026-mj-31-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 4,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 14 of 22\nv\n1\n= 11 23 e u and v\n2\n= 51\n3\nu\nor\nv s i n 11 23 e u  = and v c o s 51\n3\nu  =\nB1 v\n1\nis the velocity component of P perpendicular to\nthe barrier after impact and v\n2\nis velocity component\nof P parallel to the barrier after impact. May see LHS\nof v\n1\nequation as s i n  v or c o s  v and similar for\nv , but must have opposite trigonometric ratios.\n2\nAllow vsin=eusin(67) or better and\nc o s c o s ( 6 7 )  v = u  or better.\n12 m v 2 = 41\n3\n 12 m u 2   v 2 = 41\n3\nu 2  *B1 Correct equation relating kinetic energies.\nThis mark can be implied by seeing v2 = 4 u2.\n13\n( 11 23 e u ) 2 + ( 51\n3\nu ) 2 = 41\n3\nu 2 DM1 Correct use of 2 s i n 2 2 c o s 2 2   v + v = v to form an\nequation in e (and possibly u) only.\nAllow minor slips, for example, e rather than e 2 but\nother terms must have been squared.\nOne component must be either 12eu or 5 eu and the\n13 13\nother must be either 51\n3\nu or 11 23 u (or using sin(67),\ncos(67), or better).\nAllow consistent sin/cos mix only.\nB1 B1 DM1 can be implied by ( 11 23 e ) 2 + ( 51\n3\n) 2 = 41\n3\n.\ne=1 3 ONLY A1 AWRT 0.433 or better (0.4330127…).\n4\n4\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) tan= 1 312 tan= 3 3=1.03923... *B1FT FT their e provided\n4 5  5 \n© Cambridge University Press & Assessment 2026 Page 15 of 22\n0  e  1 to find an expression\nfor tan or .\nIf correct, may see an angle of 46.102… (which may\nbe seen on a diagram).\nOr for c o s 52\n6\n1 3  = or s i n 32\n6\n3 9  = .\ntan(60−)=0.247... DB1 Follow through their t a n  or  (the angle between\nP and the first barrier after the first collision). If\nexact and correct, may see t a n ( 6 0 ) 17 3  − = .\nOR for finding the angle between the second barrier\nand P before the second collision – if correct, this\nangle is 13.897…\nMay be seen on a diagram.\ns i n s i n ( 6 0 )   w = e v −\nc o s c o s ( 6 0 )   w = v −\nM1 Allow, for example, w for\n1\ns i n  w and w\n2\nfor\nc o s  w – both equations must be correct.\nTheir value for  does not need to be substituted for\nthis mark.\nAllow e or their value for e.\nCan be implied by tan=etan(60−).\ntan=etan(60−)= 1 31 3= 3  =6.12 A1 AWRT 6.12 (6.1155035…) or 6.11 WWW.\n 7 4 28 \n4\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q05",
      "question_id": "9231-2026-mj-31-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) 1m ( 2 ag )2 =mga(1−cos)+ 1mv2\n2 2\nor\n© Cambridge University Press & Assessment 2026 Page 16 of 22\nm g a 12 m ( 2 a g ) 2 m g a c o s 12 m v 2  − +  = − +\n*M1 Three term energy equation (OE).\nMust be dimensionally correct.\nAllow sign errors and sin/cos mix only.\nMass must be present.\nv2 =2agcos+2ag\n \nmv2 B1 N2L radially.\nT −mgcos=\na\nc o s ( 2 c o s 2 )   T − m g =\nm\na\n a g + a g\nDM1 Combine 2 equations (N2L radially and energy).\nTheir N2L radially can have sign errors and sin/cos\nmix only. Dimensionally correct.\n[T =]mg(3cos+2) A1 OE\n4\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b) mg(3cos+2)=0\n© Cambridge University Press & Assessment 2026 Page 17 of 22\nc o s 23    = −  *M1 Setting their dimensionally correct expression for T\nfrom 5(a) to zero (OE from a restart, so may see\nc o s 23  = being derived). Their expression for T\nmust be two terms only with only one term\ncontaining either sine or cosine.\nv2 =2ag(−2)+2ag v2 = 2ag DM1 Find an expression (possibly unsimplified) for\n3  3 \nv 2 or\nv when the string goes slack in terms of a (and\npossibly g). Their expression for v 2 must be two\nterms only with only one term containing either sine\nor cosine.\nDM1 For\nv = 2ag 1− ( −2)2 1 30ag\ny 3 3  9 \ns i n  v following through their v and their value\nof c o s .\nIf correct, might see (1.9245…)a.\nDependent on both previous method marks.\n0= ( 1 30ag )2 +2(−g)s\n9 y\n  s\ny\n= 52\n7\na  DM1 For use of v 2 = u 2 + 2 a s (or equivalent complete\nmethod) with v = 0 , a = − g and their v\ny\nfor u.\nIf correct, might see (0.1851…)a.\nDependent on all previous method marks.\n50a A1 Condone 1.85a or better (1.85185…)a or\n27\n52\n7\na g .\nDependent on full marks in part (a) (or a fully\ncorrect restart in this part).\n5\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q06",
      "question_id": "9231-2026-mj-31-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 6,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "6(a) EITHER: (B1 Correct expression for the EPE.\n15x2\n3x2\n \n22.5\n15x2 M1 Conservation of energy with exactly 3 terms (GPE\n1.5g(2.5+x)= + 11.5v2 and KE terms must be correct). Dimensionally\n22.5 2\ncorrect – allow sign errors and slips in the EPE term\n(but must be of the form\n© Cambridge University Press & Assessment 2026 Page 18 of 22\nk\n1\nx 2 ).\n37.5+15x=3x2 +0.75v2 A1) AG – at least one line of working from application of\nconservation of energy to given answer. Any error\nv2 =50+20x−4x2 =−4(x2 −5x−12.5)\nseen is A0.\nOR: (B1\n15 dv\n1.5g− x=1.5v\n2.5 dx\n dv \nv =10−4x\n \n dx \n12 v 2 = 1 0 x − 2 x 2 + c\nx = 0 , v 2 = 2  1 0  2 .5 = 5 0  c = 2 5\nM1 Integrate correctly and explicitly use the correct\ninitial condition to find c. As a minimum, must see\nx = 0 , v 2 = 5 0  c = 2 5 (OE, for example, may find\nthat the constant for their expression is 50).\n1v2 =10x−2x2+25v2 =−4(x2−5x−12.5) A1) AG\n2\nAny error seen is A0.\nAvailable marks 3\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) 6.83 only B1\nAWRT 6.83 (6.8301270…) or\n© Cambridge University Press & Assessment 2026 Page 19 of 22\n52\n(\n1 + 3\n)\n.\nIf −1.830127... seen it must be explicitly rejected.\nIf working shown, then it must be correct.\n1\n\n6(c) [Maximum occurs when] x = 2 .5 B1 The correct answer for v can imply this mark. Must\nbe clear that 2.5 is the extension at maximum speed.\n v 2 = 7 5   v = 8 .6 6 m s–1 B1 AWRT 8.66 (8.660254…) or exact, for example,\n5 3 .\n2\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
      "source_pdf": "_source-pdfs/2026-May-June/ms/9231_s26_ms_31.pdf",
      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2026-mj-31-q07",
      "question_id": "9231-2026-mj-31-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "31",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "7 EITHER:\ns =10+53 [13] and\nhor 5\n© Cambridge University Press & Assessment 2026 Page 20 of 22\n s\nv e r\n=  45  5   4 \n(*B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n4 u t s i n 12 g t 2  = − and 1 3 c o s  = u t DM1 Applying s = u t + 12 a t 2 both horizontally and\nvertically.\nAllow sin/cos mix only.\nNo sign errors, must be using a=−g.\nDependent on previous B1 mark.\nucos 4 usin−gt 3 B1 Use perpendicular condition to form a correct\n= or =−\nequation in u, t and\ngt−usin 3 ucos 4\n−3ucos=4usin−4gt\n. \n− 3\n 1 3\nt\n\n= 4\n 4 + 12t g t 2 \n− 4 g t\nM1 Eliminate two of\n2gt2 =55\n \n, ,  u t to obtain an equation in one\nvariable only – dependent on all previous marks.\nCan be implied by t = 12 1 1 = 1 .6 5 8 3 1 ...\nu c o s 21 61 1 1 , u s i n 72 12 1 1 5 3 .8     = =   =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu = 1 3 .3 A1)\nAWRT 13.3 (13.26735…) or 7 74 44 5 .\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7 OR:\nAlternative Method 1\ns =10+53 [13] and\nhor 5\n© Cambridge University Press & Assessment 2026 Page 21 of 22\n s\nv e r\n=  45  5   4 \n(*B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n4=13tan−\ng132\n( 1+tan2 )\nDB1 Correct trajectory equation in terms of u and  in\nany form.\n2u2\ntan− g13( 1+tan2 ) =−3\nu2 4\n4 2 t a n 4 1 3 (1 t a n 2 ) 3 2     u − g   + = − u \nor\n4 ucos = or\n3 u2sin2−2g4\n3\n4\ns i n\nc o s\n1\nc\n3\no s\n\n\n\n− =\nu\nu\n−\nu\ng\nM1 Differentiate general trajectory equation, set equal to\n− 34 and use their horizontal distance travelled to\nform an equation in u and  only. Allow a single\nslip in the differentiation of the trajectory equation.\nOR\nUse perpendicular condition to form a correct\nequation in u and .\n1 6 9\n2 6\n( 1\nt a n\nt a n 2\n8\n) 4 1\n4\n3\nt a\n( 1\nn\nt a n\n3\n2 )  \n \ng +\n−\n=\ng  +\n+\nM1 Attempt to combine their two equations to arrive at\nan equation in either u or  only – dependent on all\nprevious marks.\nMay see, if correct, 1 6 9\n1 6\nt a n 2 1 3 t a n 42 95 76 [ 0 ]   − − = .\nt a n 75 12   =   5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu = 1 3 .3 A1)\nAWRT 13.3 (13.26735…) or 7745 .\n44\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7 OR:\nAlternative Method 2\ns =10+53 [13] and\nhor 5\n© Cambridge University Press & Assessment 2026 Page 22 of 22\n s\nv e r\n=  45  5   4 \n(B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n2\n5ucos 3\n   =u2sin2−2104\n 4 5\n91\n6\nu 2 c o s 2 u 2 s i n 2 8 0     = − \n1 3 c o s\n5 c o\n4\ns 3\n5 0\ns\n1\ni n\n0\n \n = u \n u\n +\nu \n1 3\n2 c\n4\no\n0\ns\n( 3 c o s 4 s i n )\n\n \n\n =\nu\n+\n\nM1 M1 for at least one equation in terms of u and only.\nAllow sin/cos mix on the  terms only (no other\nerrors allowed).\nA1 For two different correct equations in u and only.\nMight see 35 54 u c o s u s i n 1 0\nu\n1\nc\n3\no s\n \n\n−  = −  .\n8 0\nc\n1\no\n3\ns 3 c o s\n4 0\n4 s i n 1\n9\n6\nc o s 2 s i n 2  \n  \n− =\n\n+\n \n−\n M1 Attempt to combine their two equations to arrive at\nan equation in either u or  only – dependent on all\nprevious marks.\nt a n 75 12   =   5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu = 1 3 .3 A1)\nAWRT 13.3 (13.26735…) or 7745 .\n44\nAvailable marks 6\nTrace ID: PC-76076FFBDF\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-31.pdf?download=1",
      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    {
      "id": "9231-2026-mj-32-q01",
      "question_id": "9231-2026-mj-32-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 1,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a)\n© Cambridge University Press & Assessment 2026 Page 9 of 22\n( 12 x )  π r 2 x + ( 98 r + x )  1 8 π r 3 = x  ( π r 2 x + 1 8 π r 3 ) M1 Moments equation with three terms, dimensionally\ncorrect so must be of the form\nr2x2 +(r+x)r3 =x ( r2x+r3) where\n,,,, are all non-zero constants.\nFor reference:\nShape Distance Volume\nCylinder 12 x π r 2 x\nHemisphere 38  3 r + x 14π(3r)3 =18πr3\n2 3\nTotal x π r 2 ( x + 1 8 r )\nB1 At least two terms correct.\nA1 All correct.\nx =\n2 x 2 +\n4\n8\nx\n1 r\n+\n2\n7\n+\n2 r\n7 2 x r A1\nOE, for example,\n12 π x 2 r 2\nπ\n+\n2 r\n8 14\nx +\nπ r\n1\n4\n8\n+\nπ r\n1\n3\n8 π r 3 x\n.\nAll individual terms must be simplified.\nISW after correct answer seen.\n4\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b)\n© Cambridge University Press & Assessment 2026 Page 10 of 22\nt a n\n3\n =\nx −\nr\nx *M1\nFor t a n\n3\n =\nx −\nr\nx\nor t a n\n3\n =\nx −\nr\nx\nor t a n\n3\n =\nx\nr\n− x\n3r\nor tan= only (allow equivalents with x\nx−x\nreplaced with r or vice versa).\nt a n\n2 2\n4\n8 1 2\n73 2\n7 2 2\n =\nx +\nx +\nx +\nx\nx\nx\n− x\nDM1\nFor t a n\n3\n =\nx −\nx\nx\nor t a n\n3\n =\nx −\nx\nx\nonly with their x\ncorrectly substituted – their xmust be\ndimensionally correct and have come from an\nexpression of the form\na x 2 +\nd\nb\nx\nr\n+\n2\ne\n+\nr\nc x r\nwith a, b, c, d\nand e being non-zero.\nMust be in terms of either x or r only.\nIf correct, may see\n1 5\n7\n5\n6\nx\n− x =\n7\n7\n9\n6\nx in numerator.\nFor reference: t a n\n7 9\n2 2 8\n =  .\n1 9 .1  =  A1 AWRT 19.1 (19.110752…) or 0.334 radians\n(0.3335455…).\nAllow t a n\n7 9\n2 2 8\n1 9 .1   = −  =  for full marks (but\nA0 if given as a negative angle).\n3\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-32.pdf?download=1",
      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2026-mj-32-q02",
      "question_id": "9231-2026-mj-32-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 2,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "2\n© Cambridge University Press & Assessment 2026 Page 11 of 22\n( F\nra d ia l\n= ) 7  1 2  4 2 [ = 1 3 4 4 ] B1 Correct unsimplified expression for the radial force.\nAttempt to apply N2L either horizontally or vertically to form an equation. M1 Correct number of terms, dimensionally correct,\nallow sign errors and sin/cos mix only. Values of\nsin/cos must be substituted. Allow 37 or better\n(36.8698…) for the angle between AP and the\nhorizontal and 23 or better (22.6198…) for the angle\nbetween BP and the horizontal.\n35 T\nA\n= 51\n3\nT\nB\n+ 7 g (  3 9 T\nA\n= 2 5 T\nB\n+ 4 5 5 0 ) A1 Allow 37 or better (36.8698…) for the angle between\nAP and the horizontal and 23 or better (22.6198…)\nfor the angle between BP and the horizontal, for\nexample, T sin(37)=T sin(23)+7g.\nA B\n45 T\nA\n+ 11 23 T\nB\n= 7  1 2  4 2 (  1 3 T\nA\n+ 1 5 T\nB\n= 2 1 8 4 0 ) A1 Allow 36 or better (36.8698…) for the angle between\nAP and the horizontal and 23 or better (22.6198…)\nfor the angle between BP and the horizontal, for\nexample, T cos(37)+T cos(23)=71242.\nA B\nT\nA\n= 6 7 5 and T =871 A1 AWRT 675 and 871 only.\nB\n5\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-32.pdf?download=1",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2026-mj-32-q03",
      "question_id": "9231-2026-mj-32-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 3,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a)\n© Cambridge University Press & Assessment 2026 Page 12 of 22\n8 − 0 .5 v 2 = 6 .4 v\nd\nd\nv\nx\n*B1 Correct use of Newton’s second law – acceleration\nmust be v\nd\nd\nv\nx\n.\nx = \n8\n6\n−\n.4 v\n0 .5 v 2\nd v = − 6 .4 ln 8 − 0 .5 v 2 + c\nor\nx = − \n0\n6\n.5\n.4\n2 v\nv\n− 8\nd v = − 6 .4 ln 0 .5 v 2 − 8 + c\n*M1 Separate variables correctly (from a differential\nequation of the form 2 d\nd\n   + v = v v\nx\n) and obtain an\nequation of the form x = k 1 ln ( k 2 + k 3 v 2 ) where k 1 , k 2\nand k\n3\nare non-zero constants OR\nx = A ln ( k\n1\n+ k\n2\nv ) + B ln ( k\n3\n+ k\n4\nv ) where A and B\nmust have the same sign. Condone lack of +c.\nx = − 6 .4 ln 8 − 0 .5 v 2 + c (*)\nor\nx = − 6 .4 ln 0 .5 v 2 − 8 + c\nA1\nOE, for example,\n(**)\n− 12 ln ( 4 − v ) − 12 ln ( 4 + v ) =\n1\nx\n2 .8\n+ c .\nCondone lack of modulus signs in (*). However, if\nsolving x = −\n\n0\n6\n.5\n.4\n2 v\nv\n− 8\nd v then modulus must be\nincluded in (**). Condone lack of +c.\nx=0,v=0c=6.4ln8 DM1 Use correct initial conditions to find c.\nNote that 0=−6.4ln(0−8)+c leading to a value of\nc is DM0.\n1 .3 7 m s–1 A1 WWW\nAWRT 1.37 (1.3711489…) – or exact, for example,\nv = 4 1 − e\n−\n18\n.\n5\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b)\n© Cambridge University Press & Assessment 2026 Page 13 of 22\n8 − 0 .5 v 2 = 6 .4\nd\nd\nv\nt\nB1 Correct use of Newton’s second law – acceleration\nmust be\nd\nd\nv\nt\n.\n 12.8 4+v M1 Separate variables correctly (from a differential\nt= dv=1.6ln +c\n16−v2 4−v equation of the form 2 d\nd\n   + v = v\nt\n) and obtain an\nequation of the form\nt = A ln ( k\n1\nv + k\n2\n) + B ln ( k\n3\nv + k\n4\n) where\nA , B , k\n1\n, k\n2\n, k\n3\nand k\n4\nare non-zero constants but A\nand B must have different signs.\nAllow a r t a n h ( )   t = v for any ,0 where\n1 .   \n4+v\nt =1.6ln +c(*)\n4−v\nor\nt = − 1 .6 ( ln v − 4 − ln v + 4 ) + c\nA1 Allow unsimplified and t=3.2artanh (1v ) +c.\n4\nCondone lack of +c. Condone lack of modulus signs\nin (*), but in (**) must see modulus sign on at least\nthe (**) ln ( v − 4 ) term.\nFind c: t = 0 when v = 0  c = 0\nt = 1 .6 ln\n4\n4\n+\n−\nv\nv\nA1 As a minimum must state, ‘t=0,\n(*)\nor\n4+v\nt =1.6ln (**)\nv−4\nv = 0  c = 0 ’.\nISW if re-arranged to make v the subject.\nAllow as two separate log terms.\nAllow t = 3 .2 a r t a n h ( 14 v ) .\nCondone lack of modulus sign for (*) but required in\n(**) in at least the ln ( v − 4 ) term.\nIf answer is in terms of x, for example\nt = 1 .6 ln\n4\n4\n+\n−\nx\nx\n, then maximum B1 M1 A0 A0\npossible. Allow recovery for full marks.\n4\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-32.pdf?download=1",
      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    },
    {
      "id": "9231-2026-mj-32-q04",
      "question_id": "9231-2026-mj-32-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 4,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 14 of 22\nv\n1\n= 11 23 e u and v\n2\n= 51\n3\nu\nor\nv s i n 11 23 e u  = and v c o s 51\n3\nu  =\nB1 v\n1\nis the velocity component of P perpendicular to\nthe barrier after impact and v is velocity component\n2\nof P parallel to the barrier after impact. May see LHS\nof v\n1\nequation as s i n  v or c o s  v and similar for\nv\n2\n, but must have opposite trigonometric ratios.\nAllow s i n s i n ( 6 7 )  v = e u  or better and\nc o s c o s ( 6 7 )  v = u  or better.\n12 m v 2 = 41\n3\n 12 m u 2   v 2 = 41\n3\nu 2  *B1 Correct equation relating kinetic energies.\nThis mark can be implied by seeing v 2 = 41\n3\nu 2 .\n( 11 23 e u ) 2 + ( 51\n3\nu ) 2 = 41\n3\nu 2 DM1 Correct use of 2 s i n 2 2 c o s 2 2   v + v = v to form an\nequation in e (and possibly u) only.\nAllow minor slips, for example, e rather than e 2 but\nother terms must have been squared.\nOne component must be either 11 23 e u or 51\n3\ne u and the\nother must be either 51\n3\nu or 11 23 u (or using sin(67),\ncos(67), or better).\nAllow consistent sin/cos mix only.\nB1 B1 DM1 can be implied by\n(12e)2 +(5)2\n= 4.\n13 13 13\ne=1 3 ONLY A1 AWRT 0.433 or better (0.4330127…).\n4\n4\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b)\n© Cambridge University Press & Assessment 2026 Page 15 of 22\nt a n 14 3 1 25 t a n 35 3 1 .0 3 9 2 3 ...   =    = =  *B1FT FT their e provided 0  e  1 to find an expression\nfor tan or .\nIf correct, may see an angle of 46.102… (which may\nbe seen on a diagram).\nOr for c o s 52\n6\n1 3  = or s i n 32\n6\n3 9  = .\nt a n ( 6 0 ) 0 .2 4 7 ...  − = DB1 Follow through their t a n  or  (the angle between\nP and the first barrier after the first collision). If\nexact and correct, may see tan(60−)=1 3.\n7\nOR for finding the angle between the second barrier\nand P before the second collision – if correct, this\nangle is 13.897…\nMay be seen on a diagram.\ns i n s i n ( 6 0 )   w = e v −\nc o s c o s ( 6 0 )   w = v −\nM1 Allow, for example, w\n1\nfor s i n  w and w\n2\nfor\nc o s  w – both equations must be correct.\nTheir value for  does not need to be substituted for\nthis mark.\nAllow e or their value for e.\nCan be implied by t a n t a n ( 6 0 )   = e − .\nt a n e t a n ( 6 0 ) 17 3 14 3 32\n8\n6 .1 2     = − =  =   =  A1 AWRT 6.12 (6.1155035…) or 6.11 WWW.\n4\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-32.pdf?download=1",
      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2026-mj-32-q05",
      "question_id": "9231-2026-mj-32-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 5,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 16 of 22\n12 m ( 2 a g ) 2 m g a (1 c o s ) 12 m v 2   = − + *M1 Three term energy equation (OE).\nMust be dimensionally correct.\nor −mga+ 1m ( 2 ag )2 =−mgacos+ 1mv2 Allow sign errors and sin/cos mix only.\n2 2 Mass must be present.\nv2 =2agcos+2ag\n \nc o s\n2\n T − m g =\nm v\na\nB1 N2L radially.\nc o s ( 2 c o s 2 )   T − m g =\nm\na\n a g + a g\nDM1 Combine 2 equations (N2L radially and energy).\nTheir N2L radially can have sign errors and sin/cos\nmix only. Dimensionally correct.\n[ ] ( 3 c o s 2 )  T = m g + A1 OE\n4\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)\n© Cambridge University Press & Assessment 2026 Page 17 of 22\n( 3 c o s 2 ) 0  m g + = c o s 23    = −  *M1 Setting their dimensionally correct expression for T\nfrom 5(a) to zero (OE from a restart, so may see\nc o s 23  = being derived). Their expression for T\nmust be two terms only with only one term\ncontaining either sine or cosine.\nv 2 = 2 a g  ( − 23 ) + 2 a g   v 2 = 23 a g  DM1 Find an expression (possibly unsimplified) for v 2 or\nv when the string goes slack in terms of a (and\npossibly g). Their expression for v2 must be two\nterms only with only one term containing either sine\nor cosine.\nv\ny\n= 23 a g  1 − ( − 23 ) 2   19 3 0 a g \nDM1 For s i n  v following through their v and their value\nof c o s .\nIf correct, might see (1.9245…)a.\nDependent on both previous method marks.\n0 = ( 19 3 0 a g ) 2 + 2 ( − g ) s\ny\n  s\ny\n= 52\n7\na  DM1 For use of v 2 = u 2 + 2 a s (or equivalent complete\nmethod) with v = 0 , a = − g and their v\ny\nfor u.\nIf correct, might see (0.1851…)a.\nDependent on all previous method marks.\n50a A1 Condone 1.85a or better (1.85185…)a or 5 ag.\n27 27\nDependent on full marks in part (a) (or a fully\ncorrect restart in this part).\n5\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-32-q06",
      "question_id": "9231-2026-mj-32-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 6,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "6(a) EITHER: (B1 Correct expression for the EPE.\n15x2\n3x2\n \n22.5\n© Cambridge University Press & Assessment 2026 Page 18 of 22\n1 .5 g ( 2 .5 + x ) =\n1\n2\n5\n\nx\n2\n2\n.5\n+ 12  1 .5  v 2\nM1 Conservation of energy with exactly 3 terms (GPE\nand KE terms must be correct). Dimensionally\ncorrect – allow sign errors and slips in the EPE term\n(but must be of the form k\n1\nx 2 ).\n3 7 .5 + 1 5 x = 3 x 2 + 0 .7 5 v 2\n v 2 = 5 0 + 2 0 x − 4 x 2 = − 4 ( x 2 − 5 x − 1 2 .5 )\nA1) AG – at least one line of working from application of\nconservation of energy to given answer. Any error\nseen is A0.\nOR:\n1 .5 g −\n1\n2\n5\n.5\nx = 1 .5 v\nd\nd\nv\nx\n\n v\nd\nd\nv\nx\n= 1 0 − 4 x\n\n(B1\n12 v 2 = 1 0 x − 2 x 2 + c\nx = 0 , v 2 = 2  1 0  2 .5 = 5 0  c = 2 5\nM1 Integrate correctly and explicitly use the correct\ninitial condition to find c. As a minimum, must see\nx = 0 , v 2 = 5 0  c = 2 5 (OE, for example, may find\nthat the constant for their expression is 50).\n12 v 2 = 1 0 x − 2 x 2 + 2 5  v 2 = − 4 ( x 2 − 5 x − 1 2 .5 ) A1) AG\nAny error seen is A0.\nAvailable marks 3\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) 6.83 only B1\nAWRT 6.83 (6.8301270…) or\n© Cambridge University Press & Assessment 2026 Page 19 of 22\n52\n(\n1 + 3\n)\n.\nIf − 1 .8 3 0 1 2 7 ... seen it must be explicitly rejected.\nIf working shown, then it must be correct.\n1\n\n6(c) [Maximum occurs when] x = 2 .5 B1 The correct answer for v can imply this mark. Must\nbe clear that 2.5 is the extension at maximum speed.\nv2 =75v=8.66 m s–1 B1 AWRT 8.66 (8.660254…) or exact, for example,\n \n5 3 .\n2\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2026-mj-32-q07",
      "question_id": "9231-2026-mj-32-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "32",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "7 EITHER:\n© Cambridge University Press & Assessment 2026 Page 20 of 22\n s\nh o r\n=  1 0 + 5  35 [  1 3 ] and  s\nv e r\n=  45  5   4 \n(*B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n4 u t s i n 12 g t 2  = − and 1 3 c o s  = u t DM1 Applying s = u t + 12 a t 2 both horizontally and\nvertically.\nAllow sin/cos mix only.\nNo sign errors, must be using a=−g.\nDependent on previous B1 mark.\nucos 4\n= or\ngt−usin 3\ns i n\nc o s\n3\n4\n\n\nu\nu\n− g t\n= −\n 3 c o s 4 s i n 4     − u = u − g t\nB1 Use perpendicular condition to form a correct\nequation in u, t and . \n− 3\n 1 3\nt\n\n= 4\n 4 + 12t g t 2 \n− 4 g t\nM1 Eliminate two of u,,t to obtain an equation in one\n2gt2 =55\n  variable only – dependent on all previous marks.\nCan be implied by t=1 11=1.65831...\n2\nu c o s 21 61 1 1 , u s i n 72 12 1 1 5 3 .8     = =   =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu=13.3 A1)\nAWRT 13.3 (13.26735…) or 7 74 44 5 .\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7 OR:\nAlternative Method 1\n© Cambridge University Press & Assessment 2026 Page 21 of 22\n s\nh o r\n=  1 0 + 5  35 [  1 3 ] and  s\nv e r\n=  45  5   4 \n(*B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n4 1 3 t a n\n2\n1 3\n2\n2 ( 1 t a n 2 )   = − g \nu\n+ DB1 Correct trajectory equation in terms of u and  in\nany form.\nt a n g\nu\n1\n2\n3 ( 1 t a n 2 ) 34   −  + = −\n4 2 t a n 4 1 3 (1 t a n 2 ) 3 2     u − g   + = − u \nor\n4\n3 2 s i n\nc\n2\no s\n2 4\n\n\n=\nu\nu\n− g \nor 3\n4\ns i n\nc o s\n1\nc\n3\no s\n\n\n\n− =\nu\nu\n−\nu\ng\nM1 Differentiate general trajectory equation, set equal to\n− 34 and use their horizontal distance travelled to\nform an equation in u and  only. Allow a single\nslip in the differentiation of the trajectory equation.\nOR\nUse perpendicular condition to form a correct\nequation in u and .\n1 6 9\n2 6\n( 1\nt a n\nt a n 2\n8\n) 4 1\n4\n3\nt a\n( 1\nn\nt a n\n3\n2 )  \n \ng +\n−\n=\ng  +\n+\nM1 Attempt to combine their two equations to arrive at\nan equation in either u or  only – dependent on all\nprevious marks.\nMay see, if correct, 1 6 9\n1 6\nt a n 2 1 3 t a n 42 95 76 [ 0 ]   − − = .\ntan= 71\n \n52\n5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7 74 44 5 .\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7 OR:\nAlternative Method 2\n© Cambridge University Press & Assessment 2026 Page 22 of 22\n s\nh o r\n=  1 0 + 5  35 [  1 3 ] and  s\nv e r\n=  45  5   4 \n(B1 For both unsimplified distances – may see these two\nvalues on a diagram.\n2\n5ucos 3\n =u2sin2−2104  \n 4 5\n91\n6\nu 2 c o s 2 u 2 s i n 2 8 0     = − \n1 3 c o s\n5 c o\n4\ns 3\n5 0\ns\n1\ni n\n0\n \n = u \n u\n +\nu \nM1 M1 for at least one equation in terms of u and \nonly.\nAllow sin/cos mix on the  terms only (no other\nerrors allowed).\n u2cos \n13= (3cos+4sin) \n40 A1 For two different correct equations in u and only.\n \n13\nMight see −35ucos=usin−10 .\n5 4 ucos\n8 0\nc\n1\no\n3\ns 3 c o s\n4 0\n4 s i n 1\n9\n6\nc o s 2 s i n 2  \n  \n− =\n\n+\n \n−\n M1 Attempt to combine their two equations to arrive at\nan equation in either u or  only – dependent on all\nprevious marks.\nt a n 75 12   =   =53.8 A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…).\nu = 1 3 .3 A1)\nAWRT 13.3 (13.26735…) or 7 74 44 5 .\nAvailable marks 6\nTrace ID: PC-852B15CDF8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2026-mj-33-q01",
      "question_id": "9231-2026-mj-33-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "1  Rcos=mg B1 Resolve vertically.\n© Cambridge University Press & Assessment 2026 Page 9 of 20\nR s i n m ( a s i n ) 2    = B1 Resolve horizontally.\nB1 for 2  R = m a (with s i n  cancelled).\nB0 for s i n 2   R = m r unless r replaced with s i n  a later in\nworking.\nB1 for s i n\n3\n5\n2   R = m a .\nSight of t a n ( s i n ) 2   m g = m a earns B1 B1.\ng ( a c o s ) 2  =\ng\nor 2 = or\na cos\nc o s\na\ng\n2\n\n\n=\nM1 OE, eliminate R and m to get an expression in 2  .\nMust be obtained from equations with sin/cos mix only.\nt a n\n13 5\ng\na 2 \n =\n13 a a a c o s  = − , c o s 23  = B1\n12 5\n13 5\ng\na 2 \n=\n3\n2\n =\ng\na\nA1\nOE, for example, 1 .2 2\ng\na\n.\n5\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-33-q02",
      "question_id": "9231-2026-mj-33-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 2,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "2 EITHER:\n© Cambridge University Press & Assessment 2026 Page 10 of 20\nm v\nd\nd\nv\nx\n= − k\n1\nx 2\n(*B1\nN2L to form a differential equation in terms of v\nd\nd\nv\nx\n.\nAccept k.\nB0 for m v\nd\nd\nv\nx\n= − k\n1\nd 2\n.\n m v d v = −\n k\nx 2\nd x leading to\nm v\n2\n2\n=\nk\nx\n+ c\n*M1 Separate variables correctly and integrate to obtain an\nequation of the form m v 2\nx\n( c )\n\n= + where 0.\nCondone omission of +c.\n 1212 m\nm\nv\nu\n2\n2\n=\n1\n4\n *B1 Use kinetic energy loss to find v when x= 1d.\nv2 =1u2 or v=1u 2\n4 2 Implied by x = 12 d and v= 1u as a boundary condition.\n2\nx=d, v=u \nm u\n2\n2\n=\nk\nd\n+ c\nDM1 Use both sets of boundary conditions correctly to obtain two\nequations in c, k and d using their\nmu2 2k\nx= 1d, v= 1u  = +c\n2 2 8 d\nm v\n2\n2\n=\nk\nx\n+ c .\nBe aware k can be negative here, k =  38 m d u 2 .\nc may already have been found, c =  78 m u 2 .\n1\n4\n k\nd\n+ c\n\n=\n2\nd\nk\n+ c implies B1M1.\n−3mdu2 7mu2 DM1 Dependent on all marks so far.\n0= 8 + Substitute\nx 8\nv = 0 .\nc = −\n7\n3\nk\nd\nk 7k\n0= −\nd 3d\n3d A1) SC B3 for correct answer if m is not present throughout.\n7\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2 OR:\nAlternative method with x measured from A.\n© Cambridge University Press & Assessment 2026 Page 11 of 20\nm v\nd\nd\nv\nx\n= − k\n( d\n1\n− x ) 2\n(*B1\nN2L to form a differential equation in terms of v\nd\nd\nv\nx\n.\nAccept k.\n m v d v = −\n\n( d\nk\n− x ) 2\nd x leading to\nm v\n2\n2\n= −\n( d\nk\n− x )\n+ c\n*M1 Separate variables correctly and integrate to obtain an\n\nequation of the form mv2 = (+c) where 0.\n(d −x)\nCondone omission of +c.\n 1212 m\nm\nv\nu\n2\n2\n=\n1\n4\n *B1 Use kinetic energy loss to find v when\nv2 =1u2 or v=1u\n4 2\nx = 12 d .\nImplied by x = 12 d and v= 1u as a boundary condition.\n2\nx = 0 , v = u\nmu2 k\n =− +c\n2 d\nx= 1d, v= 1u\n2 2\n\nm u\n8\n2\n= −\n( d −\nk\n12 d )\n+ c\nDM1 Use both sets of boundary conditions correctly to obtain two\nmv2 k\nequations in c, k and d using their =− +c.\n2 (d −x)\nBe aware k can be negative here, k =  38 m d u 2 .\nc may already have been found, c=7mu2.\n8\n3mdu2 7mu2 M1 Dependent on all marks so far.\n0= 8 + Substitute v=0.\n(d−x) 8\nx = 47 d , minimum distance = d − 47 d = 37 d A1) SC B3 for correct answer if m is not present throughout.\nAvailable marks 6\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-33-q03",
      "question_id": "9231-2026-mj-33-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 5,
      "status": "available",
      "reason": null,
      "text": "3\nAt B,\n© Cambridge University Press & Assessment 2026 Page 12 of 20\nc o s\n2\n m g =\nm v\na\nB1 N2L radially. If their equation contains R, only award this\nmark when R is set to 0.\n1mv2−1mu2 =mga(cos−cos) *M1 Energy equation with 4 terms and correct dimensions (mass\n2 2 must be present). Allow sign errors and sin/cos mix only.\nA1 Fully correct energy equation.\na g c o s ( 23 ) 2 a g c o s 2 a g ( c o s c o s )     − = − DM1 Combine their two equations and substitute u = 23 v to obtain\nan equation in, for example, c o s  and c o s  only.\nAllow sin/cos mix only in their N2L equation radially.\nMay, instead, see the following:\nc o s ( 23 ) 2 c o s 2 ( c o s c o s )     − = −\nc o s 49 c o s 2 ( c o s c o s )     − = −\nc o s : c o s 2 3 : 1 8   = A1 OE\n5\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2026-mj-33-q04",
      "question_id": "9231-2026-mj-33-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 4,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 20\n( d 3 + 1\n3\nd 2 h ( k + 1 ) ) x = 13 k h d 2 ( 12 d + 14 k h )\nM1 Dimensionally correct moments equation, which must be of\nthe form ( d3+d2h(k+1)) x =khd2(d+kh) where\n, , ,  are all non-zero constants.\nFor reference:\nVolume From AB\nCube d 3 0\nVertical pyramid 13 h d 2 0\nHorizontal pyramid 1khd2\n3\n12 d + 14 k h\nObject d 3 + 13 d 2 h ( k + 1 ) x\nB1 Either side of equation correct.\nA1 Fully correct moments equation.\nkh(2d +kh) A1 1khd2(1d +1kh)\nx = OE, for example, 3 2 4 or\n4(hk+h+3d) d3+ 1d2h(k+1)\n3\n1\n2\n2\nd\nd\nk\n+\nh +\n4 k\nk\nh\n2 h\n+\n2\n4 h\n.\nISW\n4\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b)\n© Cambridge University Press & Assessment 2026 Page 14 of 20\nx = 12 d B1FT SOI, equate their x to 12 d .\nk 2 h 2 − 2 d h − 6 d 2  = 0  M1 Rearrange to obtain a 3-term quadratic in h of the form\n2 2 2  0     k h + d h + d = . Allow, for example, an equation of\nthe form k 2 h 2 − 2 d h = 6 d 2 .\nTheir x\nkh(d+kh)\nmust be of the form , where\n(hk+h+d)\n, , , ,   are all non-zero constants\nh =\nd + d\nk\n1\n2\n+ 6 k 2 \n=\nd\nk 2\n(\n1 + 1 + 6 k 2\n) \nA1\nOE, for example, h =\n2 d + 4 d\n2\n2\nk\n+\n2\n2 4 k 2 d 2\n.\nIf seen, must reject\nd\nk 2\n(\n1 − 1 + 6 k 2\n)\n.\n3\n\n4(c)\n1 .5 d =\nd\nk 2\n(\n1 + 1 + 6 k 2\n) M1 Dependent on M1 in 4(b).\nEquating their h to 1.5d or substituting h = 1.5d into\nk 2 h 2 − 2 d h − 6 d 2 = 0 .\nk=2 only A1 If seen, k=0 and k = − 2 must be rejected.\n2\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      ],
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      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2026-mj-33-q05",
      "question_id": "9231-2026-mj-33-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 5,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 15 of 20\n35 u\nA\n= 45 v\nA\n*B1 Velocity perpendicular to line of centres.\nSubstitute values of s i n  and sin.\nOE, for example, 3 u\nA\n= 4 v\nA\n.\n1mu 2 −1m(3u )2  DB1 Calculate the percentage loss in KE of A.\n2 A 2 4 A 100=43.75(%) OE, for example,\n 1mu 2 \n 2 A \n1 74 5 .\nCondone 43.8.\n2\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b)\n© Cambridge University Press & Assessment 2026 Page 16 of 20\n72\n5\nu\nB\n= 35 v\nB\n 7 u\nB\n= 1 5 v\nB\n B1 Velocity perpendicular to line of centres.\nValues for trigonometric functions must be substituted.\nmu cos−mu cos=−mv cos+mv cos M1 Dimensionally correct momentum equation (m required).\nA B A B\nAllow sin/cos mix only.\nleading to\nValues for trigonometric functions must be substituted.\nm (4u ) −m (24u ) =−mv 3+mv 4.\n5 A 25 B A 5 B 5 Allow 37 or better (36.8698…) for , .\nAllow 16 or better (16.2602…) for .\nAllow 53 or better (53.131…) for .\nOE, for example,\nm (4u ) −m (24u ) =−m (3u ) 3+m (7 u ) 4 .\n5 A 25 B 4 A 5 15 B 5\nMay be implied by 1 5 u\nA\n= 1 6 u\nB\n, 7 v\nA\n= 1 2 v\nB\n.\n( c o s c o s ) c o s c o s     e u\nA\n+ u\nB\n= v\nA\n+ v\nB\nM1 Use Newton’s experimental law to form an equation.\nAllow sin/cos mix only.\nleading to\nValues for trigonometric functions must be substituted.\ne (4u +24u ) =3v +4v\n5 A 25 B 5 A 5 B Allow 37 or better (36.8698…) for , .\nAllow 16 or better (16.2602…) for .\nAllow 53 or better (53.131…) for .\nSigns for approach velocities must be consistent with those\nfor velocities before the collision in momentum equation.\nOE, for example, e (4u +24u ) = 9 u +28u .\n5 A 25 B 20 A 75 B\ne= 8 =0.471 A1 Substitute expressions for\n17\nu\nA\n, v\nA\n, u\nB\n, v\nB\nand simplify.\ne = − 8 8\n2 6 3\n = − 0 .3 3 5  from correct working B1M1M1A0.\n4\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      ],
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      "html": "9231-topic-36-momentum/answers.html",
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    },
    {
      "id": "9231-2026-mj-33-q06",
      "question_id": "9231-2026-mj-33-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 6,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "6(a) 1.6mgx B1 Correct use of Hooke’s law.\nT =\n10a\n© Cambridge University Press & Assessment 2026 Page 17 of 20\nT = 0 .2 m g  45  = 0 .1 6 m g  M1 Use of N2L, must be considering same mass on both sides.\nAllow sin/cos mix. Allow sign error.\nMust replace s i n  with a numerical value.\n0 .1 6 m g =\n1 .6\n1\nm\n0\ng\na\nx A1 AG, set up correct equation and find length.\nx = a\nOP =10a−a=9a.\n3\n\n6(b) 12  0 .3 m v 2 = 1 0 a  45  0 .3 m g\nOR\nv 2 = 2  1 0 a  g  45\n*M1 Energy equation before collision.\nSame mass on both sides (allow m instead of 0.3m).\nCorrect number of terms, dimensionally correct.\nSubstitute value for s i n  (may be done after m cancelled).\nAllow sin/cos mix.\nOR\nUse kinematics to find an equation containing v, a and g,\nincluding substituting value for s i n . Allow sin/cos mix.\nv = 4 a g A1 Award A1 for v2 =16ag.\n0.3m'4 ag'=(0.3m+0.2m)w DM1 Dimensionally correct momentum equation, masses must be\npresent, with their v substituted.\nw=2.4 ag A1\nOE, for example,\n1\n0\n.2\n.5\na g .\nSC B1 for correct answer if m is missing throughout.\n4\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(c) x defined as the distance from O (2 GPE terms):\n© Cambridge University Press & Assessment 2026 Page 18 of 20\n1\n2\n 0 .5 m  \" 2 .4 2 a g \" + 0 .5 m g  9 a \n4\n5\n+\n1 .6 m\n2 0\ng\na\na 2\n= 0 .5\n+\nm\n1\ng\n.6\n\nm\nx\ng\n\n( 1\n2 0\n4\n5\n0\na\na − x ) 2\nOR\nx defined as the final extension (2 GPE terms):\n1\n2\n 0 .5 m  \" 2 .4 2 a g \" + 0 .5 m g  ( x − a ) \n4\n5\n+\n1 .6 m\n2 0\ng\na\na 2\n=\n1 .6\n2\nm\n0\ng\na\nx 2\nOR\nx defined as the change in extension (one GPE term):\n1\n2\n 0 .5 m  \" 2 .4 2 a g \" + 0 .5 m g x \n4\n5\n+\n1 .6 m\n2 0\ng\na\na 2\n=\n1 .6 m g\n2\n(\n0\nx\na\n+ a ) 2\nB1 At least one correct EPE.\nM1 Dimensionally correct energy equation with 2 EPE terms,\none KE term and one/two GPE term(s) depending on how x\nis defined.\nAllow sign errors. Allow sin/cos mix.\nValues for trigonometric functions must be substituted.\nAllow slips in the EPE terms but must be of the form k\n1\nx 2 .\nDo not allow sign errors in the EPE terms.\nMass of particle needs to be present.\nSimplified equations for reference:\n3\n2\n6\n5\nm a g +\n2\n5\nm g ( 9 a − x ) +\n2 m\n2\ng\n5\na\na\n2\n=\n2 m g ( 1\n2\n0\n5\na\na\n− x ) 2\nOR\n3\n2\n6\n5\nm a g +\n2\n5\nm g ( x − a ) +\n2 m\n2\ng\n5\na\na\n2\n=\n2 m\n2\ng\n5\nx\na\n2\nOR\n3\n2\n6\n5\nm a g +\n2\n5\nm g x +\n2 m\n2\ng\n5\na\na\n2\n=\n2 m g (\n2\nx\n5\n+\na\na ) 2\nA1FT All terms correct. FT their w 2 .\n x =  3 a A1 If seen, must reject alternative solutions outside [0, 10a].\n4\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "html": "9231-topic-34-hooke-s-law/answers.html",
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    },
    {
      "id": "9231-2026-mj-33-q07",
      "question_id": "9231-2026-mj-33-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "33",
      "question_number": 7,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "7(a)\n© Cambridge University Press & Assessment 2026 Page 19 of 20\ns i n  V\ny\n= u\nP\n− g t , and so 2 5 s i n 1 0  V\ny\n=  − t .\nAt impact, V\ny\n= 0 , hence, 2 5  45 − 1 0 t = 0 , t = 2\nB1\nAG, minimum working\n.\n2 5  45 − 1 0 t = 0 OR\n2 5 \ng\n45\n=\n2\n1\n0\n0\n.\nB0 if 20−10t=0 alone with no previous working shown.\nB0 if not using g = 10.\n1\n\n7(b) h=202−1g22 *M1 Use\n2\ns = u t + 12 a t 2 to find h.\nSubstitute t = 2, u = 2 5  45 or 2 5  35 and a =  g .\nMay see h = 2 5  45  2  12 g  2 2 or h = 2 5  35  2  12 g  2 2\nor 0 = 2 0 2 − 2 g h\n20\nor h= 2.\n2\nMust be clear that h = 20 is the maximum height.\n2 5 t\nQ\n− 12 g  t\nQ\n2 = '2 0 ' DM1 Use s = u t + 12 a t 2 to find the time Q takes to get to h.\nSubstitute u = 2 5 and a =  g .\nAllow their h.\nThis line fully correct implies M1M1.\nt =1, T =2−t =1 A1 Exclude\nQ Q\ntQ = 4 .\nMust distinguish between T and tQ (for example, t = 1, T = 1\nearns A1).\n3\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(c)\n© Cambridge University Press & Assessment 2026 Page 20 of 20\nc o s c o s  1 5    v = u\nP\n=\nleading to\ns i n\nc\n1\no\n5\ns\ns i n  \n\nv = \n*M1 Use the fact that horizontal component of the velocity is\nunchanged, meaning vcos=15, to find the vertical\ncomponent of the velocity of P after the collision, V\ny\n.\nImplied by 1 5 t a n  V\ny\n= .\n0 ( s i n ) 2 2   = v − g  y\nm a x\n\n0\nc\n1\no\n5\ns\ns i n\n2\n2 \n\n=\n \n− g  y\nm a x\nDM1 Use v 2 = u 2 + 2 a s .\nSubstitute their V\ny\nand a =  g with vertical component of\nvelocity at maximum height set to 0 to find h.\ny\nm a x\n4 54 t a n 2  = A1\nGreatest height above ground 2 0 4 54 t a n 2  = + A1 OE, for example, 2 0 1 1 .2 5 t a n 2  + .\n4\nTrace ID: PC-202C151F9E\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "html": "9231-topic-31-motion-of-a-projectile/answers.html",
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    },
    {
      "id": "9231-2026-mj-34-q01",
      "question_id": "9231-2026-mj-34-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 1,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 4,
      "status": "available",
      "reason": null,
      "text": "1\n© Cambridge University Press & Assessment 2026 Page 9 of 25\nT c o s m g  = B1\nm(2 ag)2 B1 m(2 ag)2\nTsin= Do not accept Tsin= , r=asin\nasin r\nuntil both combined with r eliminated.\nc\nm\no\ng\ns\ns i n\nm ( 2\na s i\na\nn\ng ) 2\n\n \n =\nM1 Eliminating T from their dimensionally correct\nhorizontal and vertical equations.\nTheir horizontal and vertical equations can\ncontain sin/cos mix only (therefore their r must\nbe either asin or a c o s  only).\n1−cos2=4cos cos=−2+ 5 ONLY A1 WWW OE, accept AWRT 0.236 (0.2360679…).\n \nNo working is required to solve the quadratic\nequation, but if present then it must be correct.\nIf c o s 2 5  = −  , then c o s 2 5  = − − must\nbe rejected (but no reasoning is required for\nrejection).\n4\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
        9
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-34.pdf?download=1",
      "html": "9231-topic-33-circular-motion/answers.html",
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    },
    {
      "id": "9231-2026-mj-34-q02",
      "question_id": "9231-2026-mj-34-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 2,
      "topic": "Equilibrium of a rigid body",
      "topic_slug": "9231-topic-32-equilibrium-of-a-rigid-body",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) EITHER:\nTaking moments about AD:\n© Cambridge University Press & Assessment 2026 Page 10 of 25\n( a 2 − 18 a 2 ( 1 + 12 ) ) x = a 2  12 a − 18 a 2  12 ( a − 16 a  12 ) − 18 a 2  56 a\n13a2x = 1a3− 1 a2(11a ) − 5 a3  16 2 16 12 48 \nTaking moments about AB:\n( a 2 − 18 a 2 ( 1 + 12 ) ) y = a 2  12 a − 18 a 2  12  16 a − 18 a 2  56 a\n  11 36 a 2 y = 12 a 3 − 19\n6\na 3 − 54\n8\na 3 \n(M1 Dimensionally correct moments equation about\nAD with all terms present.\nAt least two correct terms.\nFor reference:\nArea From AD From AB\nABCD a2 12 a 12 a\nBFE 18 a 2  12 a − 16 a  12 16 a\nCFG 18 a 2 56 a 56 a\nShape\nBCDEF\na 2 − 18 a 2 ( 1 + 12 ) x y\nM1 Dimensionally correct moments equation about\nAB with all terms present.\nAt least two correct terms.\nA1 One correct unsimplified equation.\nx = 51\n2\na A1 AG WWW, A0 if any errors seen. Allow a\ncorrect equation followed by this answer with no\nintermediate working for this mark.\ny = 37 78 a A1) WWW but no working required.\nOE, answer must be exact.\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) OR:\nAlternative – rectangle and two triangles.\nTaking moments about AD:\n© Cambridge University Press & Assessment 2026 Page 11 of 25\n( 12 a 2 + 18 a 2 + 31\n6\na 2 ) x = 12 a 2  14 a + 18 a 2  71\n2\na + 31\n6\na 2  34 a\nTaking moments about AB:\n( 12 a 2 + 18 a 2 + 31\n6\na 2 ) y = 12 a 2  12 a + 18 a 2  13 a + 31\n6\na 2  12 a\n(M1 Dimensionally correct moments equation about\nAD with all terms present.\nAt least two correct terms.\nM1 Dimensionally correct moments equation about\nAB with all terms present.\nAt least two correct terms.\nA1 One correct unsimplified equation.\nx = 5 a A1 AG WWW, A0 if any errors seen. Allow a\n12\ncorrect equation followed by this answer with no\nintermediate working for this mark.\ny = 37 78 a A1) WWW but no working required.\nOE, answer must be exact.\nAvailable marks 5\n\n2(b) 1a− 5 a M1 1a− 5 a a− y\ntan= 2 12 For tan= 2 12 or tan= OE\na− y a− y 1a− 5 a\n2 12\nwith their y substituted. Their y must be of the\nform ka, where 0  k  1 .\nNote that use of y = 37a leads to tan=13.\n78 82\n=9.0 A1 WWW, must come from a correct (but possibly\ninexact) value of y =(0.4743...)a.\nAllow AWRT 9 (9.0085037…) or 0.157\n(0.157228…) in radians.\n2\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-32-equilibrium-of-a-rigid-body/answers.html",
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    },
    {
      "id": "9231-2026-mj-34-q03",
      "question_id": "9231-2026-mj-34-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 3,
      "topic": "Circular motion",
      "topic_slug": "9231-topic-33-circular-motion",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a)\n© Cambridge University Press & Assessment 2026 Page 12 of 25\nT\nA\n+ m g c o s 6 0 =\nm u\na\n2 M1 N2L radially at A (radius must be a) with correct\nnumber of dimensionally correct terms.\nAllow sin/cos mix and sign errors only.\nAngle must be substituted.\nm(2u)2 B1 Correct application of N2L radially at B.\nT −mg =\nB a\nv = 2 u\nmust be substituted. Radius must be a.\n12 m  ( 2 u ) 2 − 12 m u 2 = m g a (1 + c o s 6 0 ) *M1 Three-term energy equation (OE) for the motion\nat A and at B. Must be dimensionally correct.\n  3 2 u2 = 3 2 gau2 =ga  Allow sign errors and sin/cos mix only. Mass\nmust be present. Angle must be substituted.\nT\nA\n+ m g c o s 6 0 =\nm\na\n g a   T\nA\n= 12 m g \nor\nT\nB\n− m g =\nm  ( 2\na\ng a ) 2\n  T\nB\n= 5 m g \nDM1 Substitute their expression for u (which must be\nin terms of a and possibly g) into either of their\nN2L equations to obtain an equation containing\nT or T (not both). Their N2L must contain the\nA B\ncorrect number of dimensionally correct terms\n(with sin/cos mix and sign errors only).\nT\nA\n= 12 m g or T\nA\n= 5 m or T\nB\n= 5 m g or T\nB\n= 5 0 m A1 For either tension correct in terms of m (and\npossibly g) or, for example, 2 T\nA\n= 1 0 m .\nk = 1 0 A1 WWW, accept T\nB\n= 1 0 T\nA\nOE.\n6\n\n3(b) [Loss in KE =] 2mga B1 WWW, if KE terms seen then K E\nb o tto m\n= 2 m g a\nand K E\nto p\n= 0 .\n1\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    },
    {
      "id": "9231-2026-mj-34-q04",
      "question_id": "9231-2026-mj-34-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 4,
      "topic": "Momentum",
      "topic_slug": "9231-topic-36-momentum",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "4(a) Case 1:\n© Cambridge University Press & Assessment 2026 Page 13 of 25\nm u = k m v\nA\n− m v\nB\n0 .2 u = v\nA\n+ v\nB\nCase 2: m u = k m v\nA\n+ m v\nB\n0 .2 u = v\nA\n− v\nB\nM1 PCLM: must include m (but not g).\nNEL: must be 0.2 multiplied by u.\nRHS of one equation must be the sum of v\nA\nand\nv\nB\nand the RHS of the other equation must be a\ndifference of v and v . For example, allow for\nA B\ncase 2: 0 .2 u = v\nB\n− v\nA\nfor M1.\nCase 1: v\nB\n=\n 0 .2\nk\nk\n+\n−\n1\n1 \nu or v\nB\n=\n k −\n6\n5 \nv\nA\nor 6v =(k−5)v\nB A\nCase 2: v\nB\n=\n 1 −\nk\n0\n+\n.2\n1\nk \nu or v\nB\n=\n 5 −\n6\nk \nv\nA\nor 6 v\nB\n= ( 5 − k ) v\nA\nA1 Any equivalent (possibly unsimplified) correct\nexpression for v B in terms of u or a correct\nequation in v and\nA\nv\nB\n(allow with m present but\nnot u). For reference, v\nA\n=\n5 ( k\n6\n+ 1 )\nu .\nCase 1: 0 .2 k − 1  0  k  5\nCase 2: 1 − 0 .2 k  0  k  5\nA1 AG WWW, set correct expression for v\nB\n 0\n(case 1) or v\nB\n 0 (case 2) followed by k > 5.\nAccept (with equivalents for case 2), for\nexample,\n 0 .2\nk\nk\n+\n−\n1\n1 \nu  0  k  5 or\n0 .2\nk\nk\n+\n−\n1\n1\n 0  k  5 , OE with v\nB\nin terms of\nv\nA\n, for example, ( v\nB\n= )\n k −\n6\n5 \nv\nA\n 0  k  5 .\nAward the final two A1 marks for\nv\nA\n=\n1\n1\n.2\n+\nu\nk\n\n1\n1\n.2\n+\nu\nk\n− 0 .2 u  0  1 + k  6 so k  5\nwith the first A1 mark for ( v\nB\n= )\n1\n1\n.2\n+\nu\nk\n− 0 .2 u .\nA0 for any errors seen.\n3\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b)\n© Cambridge University Press & Assessment 2026 Page 14 of 25\nw\nB\n= 0 .8 v\nB\n\n= 0 .8 \n0 .2\nk\nk\n+\n−\n1\n1\nu\n B1 Multiply their expression for v\nB\n(in any form)\nby  0.8 or for quote w\nB\n=  0 .8 v\nB\nprovided it is\nclear.\nd 2d d\n+ =\nv w v\nB B A\n\n\n 0 .2\nk\nd\nk\n+\n−\n1\n1 \nu\n+\n\n0 .8 \n\n2 d\n0 .2\nk\nk\n+\n−\n1\n1 \nu\n\n=\n 1\nk\nd\n.2\n+ 1\n\nu\n\nM1 d 2d d\nFor correctly stating or applying + =\nv w v\nB B A\nwith their expressions. Must be an equation in\nterms of speeds and distances only (not time).\nFor reference, if correct, then may see\n 0 .2\nk\n1\nk\n+\n−\n1\n1 \n+\n\n0 .8 \n2\n0 .2\nk\nk\n+\n−\n1\n1 \n=\n1\nk\n1\n.2\n+ 1\nOE\nor\n0 .2\n1\nk − 1\n+\n0 .8  (\n2\n0 .2 k − 1 )\n=\n5\n6\n.\nCandidates may derive the result that v\nB\n= 3 .5 v\nA\n(possibly leading to\nk\n6\n− 5\n=\n2\n7\n) as part of their\nworking.\nk = 2 6 A1 WWW\n3\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-34.pdf?download=1",
      "html": "9231-topic-36-momentum/answers.html",
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    {
      "id": "9231-2026-mj-34-q05",
      "question_id": "9231-2026-mj-34-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 5,
      "topic": "Linear motion under a variable force",
      "topic_slug": "9231-topic-35-linear-motion-under-a-variable-force",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "5(a) The initial velocity u > 0 and the initial acceleration\n© Cambridge University Press & Assessment 2026 Page 15 of 25\nu 2 − k u 2  0 . B1 The following three parts are all required:\n• u > 0 or initial velocity > 0 or in words (for\nexample, initial velocity is positive).\nHowever, ‘velocity is > 0’ only is B0.\n• The expression u2 −ku2explicitly seen\n(OE, for example, − u 2 ( k − 1 ) not with m).\n• a < 0 OE (for example, u 2 − k u 2 < 0, ‘the\nacceleration is negative’ condone lack of\n‘initial’ as this is covered by seeing the\ncorrect expression).\n1\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b) mv dv =m ( v2 −ku2) or v dv =v2 −ku2 B1 Acceleration must be v dv .\ndx dx dx\n v\nx= dvx= 1ln v2 −ku2 +c\nv2 −ku2 2\nor\n© Cambridge University Press & Assessment 2026 Page 16 of 25\nx = − \nk u 2\nv\n− v 2\nd v  x = 12 ln k u 2 − v 2 + c\n*M1 Separate variables correctly and obtain an\nequation of the form x = A ln B v 2 + C k u 2 where\nA, B and C are non-zero constants.\nOr x = A ln ( B v + C k u ) + D ln ( E v + F k u ) where\nA, B, C, D, E and F are non-zero constants, but\nA and D must have the same sign.\nCondone lack of modulus and condone no +c.\nx\n\n=\nc\n0 ,\n=\nv\n−\n=\n12\nu\nln\n\nu\n0\n2\n=\n− k\n12\nu\nln\n2\nu 2\no\n−\nr\nk u\nc\n2\n=\n+\n−\nc\n12 ln ( k u 2 − u 2 ) o r c = − 12 ln k u 2 − u 2\nDM1 Use correct initial conditions of x = 0 and v = u\ncorrectly to find c.\nNote that c=−1ln ( u2 −ku2) is M0 (as this\n2\ngives an undefined value for c).\nISW after seeing a correct c.\nNote that these values of c are dependent on\nthe side in which the candidate added +c (so\nmay see c being the opposite sign).\n\nx =\n1\n2\nln\nk\nu\nu\n2\n2\n−\n−\nk\nv\nu\n2\n2\n, v = 0 \n\nx =\n1\n2\nln\n\nk\nk\n− 1\n 1 k A1 1 k\nor x= ln WWW, OE, for example, ln ,\n2 k−1 2 1−k\n1  k \nln  etc.\n \n2 1−k\n \n1  k \nNote: ln  is A0.\n2 1−k\n4\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c) m dv =m ( v2 −ku2) or dv =v2 −ku2 B1 Acceleration must be dv .\ndt dt dt\n© Cambridge University Press & Assessment 2026 Page 17 of 25\nt =\n\nv 2 −\n1\nk u 2\nd v =\n2 u\n1\nk\nln\nv\nv\n−\n+\nu\nu\nk\nk\n+ c\nM1 Separate variables correctly and obtain an\nequation of the form\nor\n 1 1 u k +v\nt=− dv=− ln +c \nku2 −v2 2u k u k −v\nt =\nu\nK\nk\nln\n\n\nv\nv\n\n\nu\nu\nk\nk\nwhere K is a non-zero constant (or equivalent as\ntwo separate log terms). Condone lack of\nmodulus and condone no +c.\n[When t = 0, v = u  ]\n1 1− k\nc=− ln\n2u k 1+ k\nor\nc =\n2 u\n1\nk\nln\nk\nk\n+\n−\n1\n1\nA1\nOE, for example, c = −\n2 u\n1\nk\nln\n k\nk\n−\n+\n1\n1\n\nor\n−\n2 u\n1\nk\nln\n 1\n1\n−\n+\nk\nk\n\nbut −\n2 u\n1\nk\nln\n 1\n1\n−\n+\nk\nk\n\n(or any other undefined expression) is A0.\nAward even if u appears in all terms in the\nfraction in the argument of the natural logarithm.\nNote that these values of c are dependent on\nthe side in which the candidate added +c (so\nmay see c being the opposite sign).\n\nt =\n2 u\n1\nk\nln\nv\nv\n−\n+\nu\nu\nk\nk\n−\n2 u\n1\nk\nln\n1\n1\n−\n+\nk\nk\ns o w h e n v = 0 \n\n[ t = ]\n2 u\n1\nk\nln\n k\nk\n+\n−\n1\n1\n\nA1 1− k \n1\nWWW OE, for example, − ln \n2u k 1+ k \n \n1  k −1 1  k +k\nor − ln  or ln  1 1− k \n2u k   k +1   2u k  k− k   but A0 for − ln    etc.\n2u k 1+ k \nMust have scored the previous A1 mark.\nAllow u in all terms in the fraction in the\nargument of the natural logarithm.\n4\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-35-linear-motion-under-a-variable-force/answers.html",
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    {
      "id": "9231-2026-mj-34-q06",
      "question_id": "9231-2026-mj-34-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 6,
      "topic": "Motion of a projectile",
      "topic_slug": "9231-topic-31-motion-of-a-projectile",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "6(a)\n© Cambridge University Press & Assessment 2026 Page 18 of 25\nd 12 ( g s i n ) T 2  = \nor\n1mv2 =mgdsin\n2\nv 2 2 g d s i n   =\n*B1 Throughout this part allow t instead of T for\nfull marks.\nFirst line of working must be a direct\nsubstitution into s=ut+1at2 (although u = 0\n2\nand v=(gsin)T 2gdsin=(gsin)T\ncan be implied). Therefore, if the first line of\nworking is 2 d g s i n T 2  =  then award B0. If\nworking backwards from AG this is B0.\nEnergy approach – must get to a correct\nequation in T for B1.\n2d DB1 AG, at least one line of working from\n2d =gsinT2 T =\ngsin\nor\n2gdsin 2gdsin 2d\nT = = =\ngsin g2sin2 gsin\nd 12 ( g s i n ) T 2  =  to given answer OE.\nAllow d = 12 a T 2 with one line of working\nleading to T =\n2 d\na\nand then given answer\nprovided a g s i n  = is explicitly stated.\nB1 B0 for a g s i n  = , therefore d = 12 a T 2\nT\n2 d\na g\n2\ns\nd\ni n \n = = .\nEnergy: T\n2\ng\ng d\ns i\ns\nn\ni n\ng\n2\ns\nd\ni n\n\n \n= = is B0.\n2\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) EITHER:\n© Cambridge University Press & Assessment 2026 Page 19 of 25\nd c o s u T c o s  d u T    =  =\n(*B1 Throughout this part allow t instead of T for\n2d full marks.\nor dcos=u cos Stating d=uT implies this mark.\ngsin\nh u T s i n ( 2 ) 12 g T 2   − = − −\nor\nh u\ng\n2\ns\nd\ni n\ns i n ( 2 )\n1\n2\ng\ng\n2\ns\nd\ni n\n2\n \n \n− =  − −\n \n*M1 Applying s = u t + 12 a t 2 (or other complete\nmethod) to find the vertical distance travelled\nby P. Allow s i n  for s i n ( 2 )   − .\nAllow y for the vertical height but not  d .\nAllow sign errors only.\nd s i n u\ng\n2\ns\nd\ni n\ns i n\n1\n2\ng\ng\n2\ns\nd\ni n\n2\n \n \n− =  −\n \nor\nu T s i n u T s i n\n1\n2\ng T 2   − = −\nor\nd s i n d s i n\n1\n2\ng T 2   − = −\nA1 OE, obtaining a second correct equation in two\nof d, u, and T. Can be implied by T\n4 u s i\ng\nn \n=\nor T 2\n4 d s i\ng\nn \n= .\n4 d s i\ng\nn\ng\n2\ns\nd\ni n\n2\n4 s i n\ns i\n2\nn\n\n\n \n\n=\n \n\n\n= or s i n s i n\ns i\n1\nn\n \n\n− = −\nDM1 Eliminate all given unknowns to obtain an\nequation in  only.\n=45 A1) OE, for example, 14 π or AWRT 0.785\n(0.7853981…) radians.\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) OR:\nAlternative approach 1 – considering the motion of P parallel to the slope\n© Cambridge University Press & Assessment 2026 Page 20 of 25\nd ( u c o s 2 ) T 12 ( g s i n ) T 2   = +\n(*M1 Throughout this part allow t instead of T for\nfull marks.\nApplying s = u t + 12 a t 2 parallel to the slope.\nAllow sign errors only. Must be d on the LHS.\nA1 Fully correct equation.\nd ( u c o s 2 )\ng\n2\ns\nd\ni n\n1\n2\n( g s i n )\ng\n2\ns\nd\ni n\n2\n \n \n=  +  \n \nA1 Eliminate one of T, d or u to obtain a correct\nequation in the other two variables, for example,\nT2gsin 1\n=(ucos2)T + (gsin)T2.\n2 2\n u T c o s 2 0   =  c o s 2 0  = DM1 Eliminate all given unknowns to obtain an\nequation in  only.\n4 5  =  A1) OE, for example, 14 π or AWRT 0.785\n(0.7853981…) radians.\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) OR:\nAlternative approach 2 – equation of the trajectory\n© Cambridge University Press & Assessment 2026 Page 21 of 25\nd c o s u T c o s  d u T    =  = or d c o s u\ng\n2\ns\nd\ni n\nc o s  \n\n=\n(*B1 Throughout this part allow t instead of T for\nfull marks.\nStating d=uT implies this mark. Or for another\ncorrect equation in at least two of d, u and T.\nd s i n d c o s t a n\ng\n2\n(\nu\nd\n2\nc\nc\no\no\ns\ns 2\n) 2 \n  \n\n− =  −\n*M1 Applying the equation of the trajectory.\nAllow sign errors only.\nLHS must be d s i n   .\ng d 4 u 2 s i n    =  A1 Fully correct unsimplified equation.\n4 u 2 s i n 2 2 u 2   =   2 s i n 2 1  = DM1 Eliminate all given unknowns to obtain an\nequation in  only.\n4 5  =  A1) OE, for example, 14 π or AWRT 0.785\n(0.7853981…) radians.\nAvailable marks 5\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-34-q07",
      "question_id": "9231-2026-mj-34-q07",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 3,
      "variant": "34",
      "question_number": 7,
      "topic": "Hooke's law",
      "topic_slug": "9231-topic-34-hooke-s-law",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "7(a)\n© Cambridge University Press & Assessment 2026 Page 22 of 25\nR = m g  11 23  = 1 2 0\n1 3\nm  B1 Correct expression for the normal contact force.\nAllow 23 or better (22.61986…) for angle.\nAllow R = m g c o s ( s i n − 1 ( 51\n3\n) ) .\nAngle must be substituted.\nT 11 23 m g 51\n3\nm g m 54 g  −  − =  M1 Use of N2L parallel to the slope. Must contain\nthe correct number of dimensionally correct\nrelevant terms. Allow sin/cos mix, allow sign\nerrors. Must include a component of the weight\nnot mass. Allow with T or their possibly\nincorrect T (but all other values must be\nsubstituted). Normal contact force must be either\n11 23 m g or 51\n3\nm g (OE, using an angle of 23 or\nbetter).\n6mg3a B1 Correct unsimplified expression for T using\nT = = 9mg\n8a  4  Hooke’s law.\n9mg−12mg− 5 mg=m5g A1 Allow AWRT 0.667 (but not 0.666).\n \n4 13 13 4\n= 2\n3\n4\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b) EITHER:\nMETHOD 1 (SEE SC ON PAGE 25 FOR THOSE WHO ASSUME THAT P\nCOMES TO REST AFTER THE STRING IS SLACK)\n© Cambridge University Press & Assessment 2026 Page 23 of 25\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2  = 2 78 m g a \n(*B1 1 6mg\nB0 for  3a2 unless recovered.\n2 8a\n1\n2\n\n6 m g\n8\n( 3\na\na ) 2\n=\n1\n2\n\n6 m\n8 a\ng\n ( 3 a − d ) 2 +\n1\n5\n3\n m g d +\n2\n3\n 1\n1\n2\n3\nm g d\n\nor\n1 6mg(3a)2 1 6mg 5 212 \n =  e2 + mg(3a−e)+  mg(3a−e)\n2 8a 2 8a 13 313 \nor\n1\n2\n\n6 m g\n8\n( 3\na\na ) 2\n=\n1\n2\n\n6 m g ( x\n8\n−\na\n8 a ) 2\n+\n1\n5\n3\nm g (1 1 a − x ) +\n2\n3\n 1\n1\n2\n3\nm g (1 1 a − x )\n\n*M1 Work-energy equation using their  with the\ncorrect number of dimensionally correct relevant\nterms. Allow sign errors. Allow sin/cos mix on\nthe GPE and work done by friction.\nFinal EPE must be dimensionally correct, and its\nextension X must be consistent with their GPE\nand work done by friction terms, but allow other\nslips in their EPE term.\nIf their X in EPE implies that X  0 , then M0.\nA1FT Fully correct equation.\nFT their value of .\n3 d\n8 a\n2\n−\n5\n4\nd = 0  d =\n1 0\n3\na\n[  3 a s o P m o v i n g w h e n t h e s t r i n g i s s la c k ]\nA1\nCorrectly obtaining\nor (3e+a)(e−3a)=0e=−1a\n3\nd =\n1 0\n3\na\nor e = − 13 a or\nx = 2 33 a from ( 3 x − 2 3 a ) ( x − 1 1 a ) = 0\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2 = m g d \n1\n5\n3\n+\n2\n3\n\n 1\n1\n2\n3\nm g d\n 1 6mg\nor  (3a)2 =mgd or\n2 8a\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2 = m g ( 3 a − e ) \n1\n5\n3\n+\n2\n3\n\n 1\n1\n2\n3\nm g ( 3 a − e )\n\nDM1 OE, work-energy equation with correct number\nof dimensionally correct relevant terms. Allow\nsign errors. Allow sin/cos mix on the GPE and\nOE work done by friction (but these two terms must\ncontain a component of weight). Must be using\ncorrect EPE at this stage.\n d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1) WWW 7.625a (condone 7.63a).\nFor reference,\ne=−3aOC =11a−3a−3a= 61a.\n8 8 8\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b) OR: (*B1 Correct unsimplified expression for the initial\nMETHOD 2 (Showing KE or velocity is > 0 when the string goes slack) EPE.\n1 6mg\n (3a)2 = 27mga\n2 8a  8 \n1 6mg 5 2 12 \n2\n\n8a\n(3a)2 =KE\nslack\n+3amg\n13\n+\n3\n\n13\nmg3a\n\nor\n© Cambridge University Press & Assessment 2026 Page 24 of 25\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2 = K E\nsla c k\n+ 3 m g a\n*M1 Work-energy equation using their  with the\ncorrect number of dimensionally correct relevant\nterms. Allow sign errors. Allow sin/cos mix on\nthe GPE and work done by friction (but these\ntwo terms must contain a component of weight\nand the correct distance of 3a). EPE must be of\nthe form k\n1\n( 3 a ) 2 where k\n1\nis dimensionally\ncorrect, but allow other slips. May see\nK E\nsla c k\n= 12 m v 2 .\nA1FT Fully correct equation.\nFT their value of . \nK E = 38 m g a or v 2 = 34 g a [  0 s o P i s m o v i n g w h e n t h e s t r i n g i s s la c k ] . A1 WWW, correct expression for the KE when the\nstring goes slack OE. No comment required that\nthis is positive (OE).\n1 6mg 5 2 12 \n (3a)2 =mgd +  mgd\n2 8a 13 3 13 \nor\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2 = m g d\nDM1 Work-energy equation with correct number of\ndimensionally correct relevant terms – allow\nsign errors. Allow sin/cos mix on the GPE and\nwork done by friction (but these two terms must\ncontain a component of weight). Must be using\ncorrect EPE at this stage.\n d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1) WWW 7.625a (condone 7.63a).\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n7(b) OR:\nSC FOR THOSE WHO ASSUME THAT P COMES TO REST AFTER THE\nSTRING IS SLACK\n© Cambridge University Press & Assessment 2026 Page 25 of 25\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2  = 2 78 m g a \n(*B1 1 6mg\nB0 for  3a2 unless recovered.\n2 8a\n1\n2\n\n6 m\n8 a\ng\n ( 3 a ) 2 = m g d \n1\n5\n3\n+\n2\n3\n\n 1\n1\n2\n3\nm g d\n DM1 OE, work-energy equation with correct number\nof dimensionally correct relevant terms. Allow\nsign errors. Allow sin/cos mix on the GPE and\nor\nwork done by friction (but these two terms must\n1 6mg\n (3a)2 =mgd contain a component of weight). Must be using\n2 8a the correct expression for EPE and their  from\n7(a).\n d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1 7.625a (condone 7.63a)\nMaximum of three marks for the SC.\nThe maximum number of marks available to candidates who assume that P\ncomes to rest after the string is slack with no further work is 3 marks as above.\nHowever, full marks available for a candidate who explicitly shows why P is still in\nmotion when the string goes slack. An example of such an approach is as follows.\n[Sum of work done by friction to travel 3a up the plane and the GPE is given by] M1 For sum of the two terms with sin/cos mix only\nmg3a 5 +2 (12mg3a ) =3mga and their .\n13 3 13\nA1FT FT their .\n[As] 2 78 m g a  3 m g a , [P is still in motion.] A1) Correct comparison with the EPE.\nAvailable marks 6\nTrace ID: PC-051B44EFC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-41-q01",
      "question_id": "9231-2026-mj-41-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 1,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) Data has outliers / is skewed / lacks symmetry, suggesting population / distribution may B1 Condone no explicit mention of\nnot be symmetrical. population/distribution or a definite statement\nrelating to population.\nAllow comment about tied ranks.\n1\n© Cambridge University Press & Assessment 2026 Page 7 of 17\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b) H : population median is 8. B1 Allow m, but not .\n0\nH 1 : population median is greater than 8. If m defined, must see population.\nMedian, not mean or average.\n© Cambridge University Press & Assessment 2026 Page 8 of 17\nX ~ B ( 1 5 , 0 .5 ) B1 SOI, may be implied by N ( 7 .5 , 3 .7 5 ) .\n15 15 15\nP(X ≤3)=(0.5)15 +  (0.5)15 +  (0.5)15 +  (0.5)15\n 1   2  3\nOR\n\n 3 (  0 .5\n3\n)\n.7\n−\n5\n7 .5 \nM1 Attempt at using binomial distribution with 4\nterms. May calculate P(X ≥12).\nImplied by AWRT 0.018.\nOR Attempt to use correct normal approximation,\nignoring continuity correction.\nImplied by AWRT  2 .0 7 ,  2 .3 2 o r  2.58.\n0 .0 1 7 6 A1 AWRT 0.018\nOR M1A1 for unsupported 0.018 or 2.07 from\n2.07 normal approximation.\n‘0.0176’ < 0.05, reject H / significant. M1 Compare their 0.0176 with 0.05 and appropriate\n0\nOR conclusion (may be in terms of H ).\n1\n2.07 > 1.645, reject H / significant. Their 0.0176 must come from use of binomial.\n0\nTest result can be implied by an attempt at an\nappropriate conclusion in context.\nFor normal approximation, compare their  2 .0 7\n(must come from attempt at standardising) with\n1.645 or 1.64 or 1.65, signs must be consistent.\nSufficient evidence to support the manager’s belief. A1 Correct conclusion from correct working, ignoring\nhypotheses.\nIn context. Level of uncertainty in language is\nused (for example, not ‘prove’).\nDo not accept statements such as ‘there is\ninsufficient evidence to suggest…’.\n6\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-41-q02",
      "question_id": "9231-2026-mj-41-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "2(a) H : reliability of service is independent of bus company. B1 Allow dependent, association or relationship for\n0\nH : reliability of service is not independent of bus company. not independent, for example,\n1\n“H : No association/relationship between…,\n0\nH : Association/relationship between…”.\n1\nMust refer to ‘reliability’ and ‘company’ in H .\n0\nM1 Calculate expected frequencies (must be seen).\n7 (12.25) 22 (19.11) 20 (17.64)\nAt least two correct to at least 2 significant\nfigures.\n14 (9.25) 11 (14.43) 12 (13.32)\nA1 All expected frequencies correct to at least 3\n4 (3.50) 6 (5.46) 4 (5.04)\nsignificant figures.\nM1 Combine rows (for example rows 2 and 3).\n7 (12.25) 22 (19.11) 20 (17.64)\nOR\nCombine columns (for example columns A and B)\n18 (12.75) 17 (19.89) 16 (18.36)\nOR\n29 (31.36) 20 (17.64)\n25 (23.68) 12 (13.32)\n10 (8.96) 4 (5.04)\n© Cambridge University Press & Assessment 2026 Page 9 of 17\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) Combining rows 2 and 3:\n(7−12.25)2 (22−19.11)2 (16−18.36)2\n+ ++\n12.25 19.11 18.36\n© Cambridge University Press & Assessment 2026 Page 10 of 17\n = 2 .2 5 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .1 6 1 7 6 + 0 .4 1 9 9 1 + 0 .3 0 3 3 6 \nOR\n1\n7\n2\n2\n.2 5\n+\n1\n2\n9\n2 2\n.1 1\n+  +\n1\n1\n8\n2 6\n.3 6\n− 1 0 0\nCombining columns A and B:\n(29−31.36)2 (20−17.64)2 (4−5.04)2\n+ ++\n31.36 17.64 5.04\n = 0 .1 7 7 6 0 + 0 .3 1 5 7 4 + 0 .0 7 3 5 8 + 0 .1 3 0 8 1 + 0 .1 2 0 7 1 + 0 .2 1 4 6 0 \nOR\n3\n2\n1\n2 9\n.3 6\n+\n1\n2\n7\n2 0\n.6 4\n+  +\n5\n2 4\n.0 4\n− 1 0 0\nNo rows / columns combined:\n( 7 −\n1\n1\n2\n2 .2\n.2 5\n5 ) 2\n+\n( 2 2 −\n1\n1\n9\n9\n.1\n.1\n1\n1 ) 2\n+  +\n( 4 −\n5\n5 .0\n.0 4\n4 ) 2\nM1 At least 2 expressions (not including 100) of\ncorrect form seen.\nMay see individual terms not summed.\nMay be implied by 2 contributions correct to 3\nsignificant figures seen.\nMay be implied by AWRT 5.89 or AWRT 1.03 or\nAWRT 6.73.\n[= 2.250+0.43705+0.31574+2.4392+0.81531+0.13081+0.07143+0.05341+0.21460]\nCombining rows 2 and 3: Test statistic = 5.888 A1 AWRT 5.89\nOR OR\nCombining columns A and B: Test statistic = 1.033 AWRT 1.03\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) Combining rows 2 and 3: M1 Compare their test statistic with 5.991 (or 9.488 if\n‘5.888’ < 5.991, accept H / do not reject H / not significant. no rows/columns combined) and appropriate\n0 0\nconclusion (may be in terms of H ).\nOR 1\nTest result can be implied by an attempt at an\nCombining columns A and B:\nappropriate conclusion in context ignoring\n‘1.033’ < 5.991, accept H / do not reject H / not significant.\n0 0 hypotheses.\nOR\nNo cells combined:\n‘6.73’ < 9.488, accept H / do not reject H / not significant.\n0 0\nInsufficient evidence to suggest reliability (of service) is not independent of (bus) A1 Correct conclusion from correct working ignoring\ncompany. hypotheses.\nOR In context. Must refer to both reliability and\nInsufficient evidence to suggest reliability (of service) is associated with (bus) company. company.\nOR Level of uncertainty in language is used (for\nInsufficient evidence to suggest that reliability (of service) is dependent on (bus) example not ‘prove’).\ncompany. Allow ‘Not enough evidence / no sufficient\nevidence that / to show / to conclude that …’\nDo not accept ‘No evidence…’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…’.\n8 No cells combined can earn a maximum of 5/8\n(B1 M1 A1 M0 M1 A0 M1 A0).\n\n2(b) Company A (appears less reliable). B1\nThe observed frequencies are higher than the (combined) expected frequencies for DB1 Comparison of observed and expected\ncompany A for late and cancelled, but not for company B and company C. frequencies.\n2\n© Cambridge University Press & Assessment 2026 Page 11 of 17\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    },
    {
      "id": "9231-2026-mj-41-q03",
      "question_id": "9231-2026-mj-41-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 (x−x)2 =29144−1 ( 4452) =854.7 M1 Either expression, may be embedded.\n7\nMay be implied by\n(y−y)2 =11622−1 ( 3102) =944.2\n9\n© Cambridge University Press & Assessment 2026 Page 12 of 17\n5 94 82 3 or 4 23 46 9 or 142.45 or\n118.03 seen.\ns 2p =\n'8 5 4 .7 '\n7 +\n+\n9\n'9\n−\n4\n2\n4 .2 ' M1 Correct form for pooled variance.\n128.495 A1 1 1 3 3 3\n8 8 2\n3 , SOI, may be implied by s = 1 1 .3 3 6 .\n(445−310)\n2.145 '128.5'\n(1 +1) B1 B1 for 2.145 seen.\n7 9 7 9\nM1 Correct form, must be a t-value.\nFT their s. If variances combined, must use\n( 4 47 5 − 3 19 0 )  2 .1 4 5 5 94 82 3 + 4 23 46 9 .\n1 6 .9 , 4 1 .4  A1 Accept with inequality signs or open brackets\nor  − 4 1 .4 , − 1 6 .9  from Y − X .\nCondone  4 1 .4 , 1 6 .9  .\nDo not accept 2 9 .1 5  1 2 .2 5 .\n6\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2026-mj-41-q04",
      "question_id": "9231-2026-mj-41-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 17\nF ( 3 ) = 11\n2\n 3 2 −\n1\n10\n8\n 3 3 = 0 .5\n[Since F ( 3 ) = 0 .5\nB1 AG\nB0 if obtained using\n, the median of X is 3.]\na = − 13 72 .\nMay solve −\n1\n10\n8\nx 3 + 11\n2\nx 2 = 12 but must reject\n3  3 3 as they lie outside [0, 3].\n1\n\n4(b) − 13\n2\n 3 2 + 71\n6\n 3 + a = 0 .5 , a = − 13 72 B1 AG, convincingly obtained.\nF ( b ) = 1  − 13\n2\nb 2 + 71\n6\nb − 13 72 = 1\nOR\nF ( b ) − F ( 3 ) = 12   − 13\n2\nx 2 + 71\n6\nx ( + a )  b\n3\n= 0 .5  − 13\n2\nb 2 + 71\n6\nb − 33 32 = 12\nM1 Use F ( b ) = 1 or F ( b ) − F ( 3 ) = 0 .5 to form\nquadratic in b.\nMay see b 2 − 1 4 b + 4 9 = 0 OE.\nb = 7 A1\n3\n\n4(c)\nf ( x ) =\n −\n−\n0\n2 1 1 x +\n3 6 6\n7 1 x +\n1 6 1 6\nx 0\n3\no\n≤\n\nt h\nx\nx\ne r\n≤\n≤\nw\n3 ,\n7 ,\ni s e .\nB1 Both parts differentiated correctly, ignore domain.\nB1FT FT their b.\nFully defined including correct domain, may be in\nterms of b.\nDomain must cover all reals, may overlap at\nboundaries.\n2\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(d)\n© Cambridge University Press & Assessment 2026 Page 14 of 17\n 3\n0\n1\nx\n\n−\n1\n3 6\nx 2 +\n1\n6\nx\n\nd x +\n 7\n3\n1\nx\n\n−\n1\n1\n6\nx +\n1\n7\n6\n\nd x\nB1FT Correct expression, FT their f ( x ) and limits, may\nbe in terms of b.\n 16 x − 17\n2\nx 2  3\n0\n+  − 11\n6\nx + 71\n6\nln x  7\n3\nM1 Attempt to integrate both parts of their f ( x )  1\nx\n.\nMust include natural logarithm term and limits\nstated (may use b).\nImplied by AWRT 0.496.\n18 + 71\n6\nln ( 73 )  = 0 .4 9 6  A1 AWRT 0.496\n3\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    {
      "id": "9231-2026-mj-41-q05",
      "question_id": "9231-2026-mj-41-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) H :\n0\n© Cambridge University Press & Assessment 2026 Page 15 of 17\nd\n5  = , H :\n1 d\n5  \nOR\nH :\n0 A B\n5   − = , H :\n1 A B\n5   − \nB1 If  not used, must have population means.\nAllow use of  in place of \nd\n.\nM1 Calculates differences, at most 2 errors.\nd 6 24 12 8 5 26 13 2\ns 2d =\n1\n7\n\n'1 6 9 4 '−\n'9 6\n8\n2'  \n=\n5 4\n7\n2\n= 7 7 .4 3\n M1 Must be unbiased.\ns=8.799\n t = \n'9 6\n8'7\n'\n−\n7 .4\n8\n5\n3 '\nM1 M0 if 5 omitted.\n2.25 A1 AWRT 2.25\n‘2.25’ > 1.415, reject H / significant. M1 Compare their 2.25 with 1.415 and appropriate\n0\nconclusion (may be in terms of H ). Their 2.25\n1\nmust come from an attempt at standardising.\nTest result can be implied by an attempt at an\nappropriate conclusion in context ignoring\nhypotheses.\nFor 2-sample t-test allow comparison with 1.345.\nSufficient evidence to support Margaret’s claim. A1 Correct conclusion from correct working, ignoring\nOR hypotheses.\nSufficient evidence to suggest that the course will improve a student’s score by more In context. Level of uncertainty in language is\nthan 5 marks (on average). used (for example, not ‘prove’).\nDo not accept statements such as “there is\ninsufficient evidence to suggest…”.\n7 Biased variance: max 4/7 (B1M1M0M1A0M1A0)\n2-sample: max 2/7 (B1M0M0M0A0M1A0)\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b) Population of differences are normally distributed. B1 B1 for underlying/parent distribution of\ndifferences is normal.\nB0 for population mean is normally distributed.\nB0 for (underlying) data is normal.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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    {
      "id": "9231-2026-mj-41-q06",
      "question_id": "9231-2026-mj-41-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "41",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) k(a+1)2 =1 *B1\n© Cambridge University Press & Assessment 2026 Page 16 of 17\nk a 2 + 2 a k + k = 1\nk a 2 − k = 13 *M1 Uses G  ( 1 ) = 13 to form equation in k and a.\n− k ( a + 1 ) 2 + 2 k a ( a + 1 ) = 13\n(\na\na\n2\n+\n−\n1\n1\n) 2\n=\n1\n3\nDM1 Uses their equations to eliminate k or a.\na = 2 A1 AG, convincingly obtained.\nk = 19 B1\n5\n\n6(b) EITHER:\nG ''X ( t ) =\n9\n2\nt 3\n, G ''X ( 1 ) = 2\n9\n = 2 k \n(B1FT CWO with their k.\nDifferentiates their G 'X ( t ) and evaluates at t = 1 .\nMay be in terms of k.\nVar(X)= 2+1−(1)2 = 4 B1FT) Uses their values correctly in correct formula to\n9 3 3 9 obtain positive value.\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) OR:\n© Cambridge University Press & Assessment 2026 Page 17 of 17\nE ( X 2 ) = 59  = 5 k \n(B1FT\nx –1 0 1\nP(X =x) k 4k 4k\nVar(X)=5−(1)2\n= 4\nB1FT) Uses their values correctly in correct formula to\n9 3 9 obtain positive value.\nAvailable marks 2\n\n6(c) EITHER: (B1FT SOI\n1  k3  FT their k. May be in terms of k.\nG (t)= (2t+1)6 = (2t+1)6 \nY 729t3\n\nt3\n\nCoefficient of t 0 term:\n7\n1\n2 9\n 6\n3\n\n( 2 ) 3\n\n = k 3\n 6\n3\n\n( 2 ) 3\n M1 Identifies appropriate term and attempt to\ncalculate. May be in terms of k.\nP ( Y = 0 ) = 17 62 09 A1) AWRT 0.219\nOR:\nRelevant values for ( X\n1\n, X\n2\n, X\n3\n)\n(B1 All seven cases considered, SOI.\nare (–1, 0, 1), 6 possible orderings, and (0,0,0).\n6144+(4)3 =6k(4k)2 +(4k)3 M1 Must be correct form using their k, may be in\n9 9 9 9   terms of k.\n17 62 09 A1)\nAvailable marks 3\nTrace ID: PC-6B4604D1D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-42-q01",
      "question_id": "9231-2026-mj-42-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 1,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "1(a) Data has outliers / is skewed / lacks symmetry, suggesting population / distribution may B1 Condone no explicit mention of\nnot be symmetrical. population/distribution or a definite statement\nrelating to population.\nAllow comment about tied ranks.\n1\n© Cambridge University Press & Assessment 2026 Page 7 of 17\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n1(b) H : population median is 8. B1 Allow m, but not .\n0\nH 1 : population median is greater than 8. If m defined, must see population.\nMedian, not mean or average.\n© Cambridge University Press & Assessment 2026 Page 8 of 17\nX ~ B ( 1 5 , 0 .5 ) B1 SOI, may be implied by N ( 7 .5 , 3 .7 5 ) .\n15 15 15\nP(X ≤3)=(0.5)15 +  (0.5)15 +  (0.5)15 +  (0.5)15\n 1   2  3\nOR\n\n 3 (  0 .5\n3\n)\n.7\n−\n5\n7 .5 \nM1 Attempt at using binomial distribution with 4\nterms. May calculate P(X ≥12).\nImplied by AWRT 0.018.\nOR Attempt to use correct normal approximation,\nignoring continuity correction.\nImplied by AWRT 2.07, 2.32 or 2.58.\n0 .0 1 7 6 A1 AWRT 0.018\nOR M1A1 for unsupported 0.018 or 2.07 from\n2.07 normal approximation.\n‘0.0176’ < 0.05, reject H / significant. M1 Compare their 0.0176 with 0.05 and appropriate\n0\nOR conclusion (may be in terms of H ).\n1\n2.07 > 1.645, reject H / significant. Their 0.0176 must come from use of binomial.\n0\nTest result can be implied by an attempt at an\nappropriate conclusion in context.\nFor normal approximation, compare their  2 .0 7\n(must come from attempt at standardising) with\n1.645 or 1.64 or 1.65, signs must be consistent.\nSufficient evidence to support the manager’s belief. A1 Correct conclusion from correct working, ignoring\nhypotheses.\nIn context. Level of uncertainty in language is\nused (for example, not ‘prove’).\nDo not accept statements such as ‘there is\ninsufficient evidence to suggest…’.\n6\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-42-q02",
      "question_id": "9231-2026-mj-42-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 2,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "2(a) H : reliability of service is independent of bus company. B1 Allow dependent, association or relationship for\n0\nH : reliability of service is not independent of bus company. not independent, for example,\n1\n“H : No association/relationship between…,\n0\nH : Association/relationship between…”.\n1\nMust refer to ‘reliability’ and ‘company’ in H .\n0\nM1 Calculate expected frequencies (must be seen).\n7 (12.25) 22 (19.11) 20 (17.64)\nAt least two correct to at least 2 significant\nfigures.\n14 (9.25) 11 (14.43) 12 (13.32)\nA1 All expected frequencies correct to at least 3\n4 (3.50) 6 (5.46) 4 (5.04)\nsignificant figures.\nM1 Combine rows (for example rows 2 and 3).\n7 (12.25) 22 (19.11) 20 (17.64)\nOR\nCombine columns (for example columns A and B)\n18 (12.75) 17 (19.89) 16 (18.36)\nOR\n29 (31.36) 20 (17.64)\n25 (23.68) 12 (13.32)\n10 (8.96) 4 (5.04)\n© Cambridge University Press & Assessment 2026 Page 9 of 17\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) Combining rows 2 and 3:\n(7−12.25)2 (22−19.11)2 (16−18.36)2\n+ ++\n12.25 19.11 18.36\n© Cambridge University Press & Assessment 2026 Page 10 of 17\n = 2 .2 5 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .1 6 1 7 6 + 0 .4 1 9 9 1 + 0 .3 0 3 3 6 \nOR\n1\n7\n2\n2\n.2 5\n+\n1\n2\n9\n2 2\n.1 1\n+  +\n1\n1\n8\n2 6\n.3 6\n− 1 0 0\nCombining columns A and B:\n(29−31.36)2 (20−17.64)2 (4−5.04)2\n+ ++\n31.36 17.64 5.04\n = 0 .1 7 7 6 0 + 0 .3 1 5 7 4 + 0 .0 7 3 5 8 + 0 .1 3 0 8 1 + 0 .1 2 0 7 1 + 0 .2 1 4 6 0 \nOR\n3\n2\n1\n2 9\n.3 6\n+\n1\n2\n7\n2 0\n.6 4\n+  +\n5\n2 4\n.0 4\n− 1 0 0\nNo rows / columns combined:\n(7−12.25)2 (22−19.11)2 (4−5.04)2\n+ ++\n12.25 19.11 5.04\n[ = 2 .2 5 0 + 0 .4 3 7 0 5 + 0 .3 1 5 7 4 + 2 .4 3 9 2 + 0 .8 1 5 3 1 + 0 .1 3 0 8 1 + 0 .0 7 1 4 3 + 0 .0 5 3 4 1 + 0 .2 1 4 6 0 ]\nM1 At least 2 expressions (not including 100) of\ncorrect form seen.\nMay see individual terms not summed.\nMay be implied by 2 contributions correct to 3\nsignificant figures seen.\nMay be implied by AWRT 5.89 or AWRT 1.03 or\nAWRT 6.73.\nCombining rows 2 and 3: Test statistic = 5.888 A1 AWRT 5.89\nOR OR\nCombining columns A and B: Test statistic = 1.033 AWRT 1.03\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(a) Combining rows 2 and 3: M1 Compare their test statistic with 5.991 (or 9.488 if\n‘5.888’ < 5.991, accept H / do not reject H / not significant. no rows/columns combined) and appropriate\n0 0\nconclusion (may be in terms of H ).\nOR 1\nTest result can be implied by an attempt at an\nCombining columns A and B:\nappropriate conclusion in context ignoring\n‘1.033’ < 5.991, accept H / do not reject H / not significant.\n0 0 hypotheses.\nOR\nNo cells combined:\n‘6.73’ < 9.488, accept H / do not reject H / not significant.\n0 0\nInsufficient evidence to suggest reliability (of service) is not independent of (bus) A1 Correct conclusion from correct working ignoring\ncompany. hypotheses.\nOR In context. Must refer to both reliability and\nInsufficient evidence to suggest reliability (of service) is associated with (bus) company. company.\nOR Level of uncertainty in language is used (for\nInsufficient evidence to suggest that reliability (of service) is dependent on (bus) example not ‘prove’).\ncompany. Allow ‘Not enough evidence / no sufficient\nevidence that / to show / to conclude that …’\nDo not accept ‘No evidence…’.\nDo not accept statements such as ‘there is\nsufficient evidence to suggest…’.\n8 No cells combined can earn a maximum of 5/8\n(B1 M1 A1 M0 M1 A0 M1 A0).\n\n2(b) Company A (appears less reliable). B1\nThe observed frequencies are higher than the (combined) expected frequencies for DB1 Comparison of observed and expected\ncompany A for late and cancelled, but not for company B and company C. frequencies.\n2\n© Cambridge University Press & Assessment 2026 Page 11 of 17\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-42-q03",
      "question_id": "9231-2026-mj-42-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 3,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "3 (x−x)2 =29144−1 ( 4452) =854.7\n7\n© Cambridge University Press & Assessment 2026 Page 12 of 17\n ( y − y ) 2 = 1 1 6 2 2 − 19 ( 3 1 0 2 )  = 9 4 4 .2 \nM1 Either expression, may be embedded.\nMay be implied by 5 94 82 3 or 4 23 46 9 or 142.45 or\n118.03 seen.\ns 2p =\n'8 5 4 .7 '\n7 +\n+\n9\n'9\n−\n4\n2\n4 .2 ' M1 Correct form for pooled variance.\n128.495 A1 1 1 3 3 3\n8 8 2\n3 , SOI, may be implied by s = 1 1 .3 3 6 .\n( 4 47 5 − 3 19 0 )  2 .1 4 5 '1 2 8 .5 (' 17 + 19 ) B1 B1 for 2.145 seen.\nM1 Correct form, must be a t-value.\nFT their s. If variances combined, must use\n( 4 47 5 − 3 19 0 )  2 .1 4 5 5 94 82 3 + 4 23 46 9 .\n1 6 .9 , 4 1 .4  A1 Accept with inequality signs or open brackets\nor  − 4 1 .4 , − 1 6 .9  from Y − X .\nCondone  4 1 .4 , 1 6 .9  .\nDo not accept 2 9 .1 5  1 2 .2 5 .\n6\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2026-mj-42-q04",
      "question_id": "9231-2026-mj-42-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 4,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 17\nF ( 3 ) = 11\n2\n 3 2 −\n1\n10\n8\n 3 3 = 0 .5\n[Since F ( 3 ) = 0 .5\nB1 AG\nB0 if obtained using\n, the median of X is 3.]\na = − 13 72 .\nMay solve −\n1\n10\n8\nx 3 + 11\n2\nx 2 = 12 but must reject\n3  3 3 as they lie outside [0, 3].\n1\n\n4(b) − 13\n2\n 3 2 + 71\n6\n 3 + a = 0 .5 , a = − 13 72 B1 AG, convincingly obtained.\nF ( b ) = 1  − 13\n2\nb 2 + 71\n6\nb − 13 72 = 1 M1 Use F(b)=1 or\nOR\nb F(b)−F(3)= 1 − 1 x2 + 7 x(+a) =0.5− 1 b2 + 7 b−33 = 1\n2  32 16  3 32 16 32 2\nF ( b ) − F ( 3 ) = 0 .5 to form\nquadratic in b.\nMay see b 2 − 1 4 b + 4 9 = 0 OE.\nb = 7 A1\n3\n\n4(c)\nf ( x ) =\n −\n−\n0\n2 1 1 x +\n3 6 6\n7 1 x +\n1 6 1 6\nx 0\n3\no\n≤\n\nt h\nx\nx\ne r\n≤\n≤\nw\n3 ,\n7 ,\ni s e .\nB1 Both parts differentiated correctly, ignore domain.\nB1FT FT their b.\nFully defined including correct domain, may be in\nterms of b.\nDomain must cover all reals, may overlap at\nboundaries.\n2\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(d)  3 1 1 1   7 1 1 7  B1FT Correct expression, FT their\n  − x2 + x  dx+  − x+  dx\n x 36 6   x 16 16\n0 3\n© Cambridge University Press & Assessment 2026 Page 14 of 17\nf ( x ) and limits, may\nbe in terms of b.\n 16 x − 17\n2\nx 2  3\n0\n+  − 11\n6\nx + 71\n6\nln x  7\n3\nM1 Attempt to integrate both parts of their f ( x )  1\nx\n.\nMust include natural logarithm term and limits\nstated (may use b).\nImplied by AWRT 0.496.\n18 + 71\n6\nln ( 73 )  = 0 .4 9 6  A1 AWRT 0.496\n3\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    },
    {
      "id": "9231-2026-mj-42-q05",
      "question_id": "9231-2026-mj-42-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "5(a) H :\n0\n© Cambridge University Press & Assessment 2026 Page 15 of 17\nd\n5  = , H : 5\n1 d\nOR\nH :\n0 A B\n5   − = , H :\n1 A B\n5   − \nB1 If  not used, must have population means.\nAllow use of  in place of \nd\n.\nM1 Calculates differences, at most 2 errors.\nd 6 24 12 8 5 26 13 2\ns 2d =\n1\n7\n\n'1 6 9 4 '−\n'9 6\n8\n2'  \n=\n5 4\n7\n2\n= 7 7 .4 3\n M1 Must be unbiased.\ns=8.799\n t = \n'9 6\n8'7\n'\n−\n7 .4\n8\n5\n3 '\nM1 M0 if 5 omitted.\n2.25 A1 AWRT 2.25\n‘2.25’ > 1.415, reject H / significant. M1 Compare their 2.25 with 1.415 and appropriate\n0\nconclusion (may be in terms of H ). Their 2.25\n1\nmust come from an attempt at standardising.\nTest result can be implied by an attempt at an\nappropriate conclusion in context ignoring\nhypotheses.\nFor 2-sample t-test allow comparison with 1.345.\nSufficient evidence to support Margaret’s claim. A1 Correct conclusion from correct working, ignoring\nOR hypotheses.\nSufficient evidence to suggest that the course will improve a student’s score by more In context. Level of uncertainty in language is\nthan 5 marks (on average). used (for example, not ‘prove’).\nDo not accept statements such as “there is\ninsufficient evidence to suggest…”.\n7 Biased variance: max 4/7 (B1M1M0M1A0M1A0)\n2-sample: max 2/7 (B1M0M0M0A0M1A0)\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b) Population of differences are normally distributed. B1 B1 for underlying/parent distribution of\ndifferences is normal.\nB0 for population mean is normally distributed.\nB0 for (underlying) data is normal.\n1\nQuestion Answer Marks Guidance",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-42.pdf?download=1",
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    {
      "id": "9231-2026-mj-42-q06",
      "question_id": "9231-2026-mj-42-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "42",
      "question_number": 6,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) k(a+1)2 =1 *B1\n© Cambridge University Press & Assessment 2026 Page 16 of 17\nk a 2 + 2 a k + k = 1\nk a 2 − k = 13 *M1 Uses G  ( 1 ) = 13 to form equation in k and a.\n− k ( a + 1 ) 2 + 2 k a ( a + 1 ) = 13\n(\na\na\n2\n+\n−\n1\n1\n) 2\n=\n1\n3\nDM1 Uses their equations to eliminate k or a.\na = 2 A1 AG, convincingly obtained.\nk = 19 B1\n5\n\n6(b) EITHER: (B1FT CWO with their k.\nG '' (t)= 2 , G '' (1)= 2 =2k Differentiates their\nX 9t3 X 9\nG 'X ( t ) and evaluates at t = 1 .\nMay be in terms of k.\nVar(X)= 2+1− (1)2 = 4 B1FT) Uses their values correctly in correct formula to\n9 3 3 9 obtain positive value.\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b) OR: (B1FT x –1 0 1\nE ( X2) = 5 =5k P(X =x) k 4k 4k\n9\nVar(X)=5− (1)2 = 4 B1FT) Uses their values correctly in correct formula to\n9 3 9 obtain positive value.\nAvailable marks 2\n\n6(c) EITHER: (B1FT SOI\n1  k3  FT their k. May be in terms of k.\nG (t)= (2t+1)6 = (2t+1)6 \nY 729t3\n\nt3\n\nCoefficient of\n© Cambridge University Press & Assessment 2026 Page 17 of 17\nt 0 term:\n7\n1\n2 9\n 6\n3\n\n( 2 ) 3\n\n = k 3\n 6\n3\n\n( 2 ) 3\n M1 Identifies appropriate term and attempt to\ncalculate. May be in terms of k.\nP ( Y = 0 ) = 17 62 09 A1) AWRT 0.219\nOR:\nRelevant values for ( X\n1\n, X\n2\n, X\n3\n)\n(B1 All seven cases considered, SOI.\nare (–1, 0, 1), 6 possible orderings, and (0,0,0).\n6  19  49  49 + ( 49 ) 3 \n\n= 6  k  ( 4 k ) 2 + ( 4 k ) 3  M1 Must be correct form using their k, may be in\nterms of k.\n17 62 09 A1)\nAvailable marks 3\nTrace ID: PC-E48FEEF9A8\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-43-q01",
      "question_id": "9231-2026-mj-43-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1\n© Cambridge University Press & Assessment 2026 Page 9 of 15\nx =\n1 9 2\n9\n.2 B1 OE, for example, 21.356 or 9 6 1\n4 5\n.\nAccuracy must be 3 significant figures or better.\ns 2 =\n1\n8\n\n4 1 2 7 .6 4 −\n1 9 2\n9\n.2 2  M1 Correct form with their values. May be implied by sight of\n2.887… or 2 5 9 9\n9 0 0\nfor s2 or 1.69934… for s.\nt =\n1 9 2\n9\n.2\ns\n9\n−\n2\n2 0\nM1 Correct form with their values.\n2.39 A1 AWRT 2.39 or 2.40. Accept ± here.\n'2 .3 9 '  2 .3 0 6 , reject H / significant. M1 Compare their 2.39 with 2.306 and appropriate conclusion (may\n0\nbe in terms of H ).\n1\nTest result can be implied by an attempt at an appropriate\nconclusion in context.\nTheir 2.39 must come from an attempt at standardising.\nThere is sufficient evidence to support that 2 0   . A1 Correct conclusion from correct working. No contradictions.\nLevel of uncertainty in language used (for example, not ‘prove’).\nAllow , or (population) mean.\nDo not accept statements such as ‘there is insufficient evidence to\nsuggest…’.\n6\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-43.pdf?download=1",
      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2026-mj-43-q02",
      "question_id": "9231-2026-mj-43-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 2,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "2(a) Population of differences may not be normally distributed. B1 Allow underlying/parent distribution for population.\nB0 for population mean may not be normal.\nB0 for (underlying) data may not be normal.\n1\n\n2(b) H : difference in population medians =0\n0\nH : difference in population medians\n1\n© Cambridge University Press & Assessment 2026 Page 10 of 15\n 0\nB1 Allow use of m\nd\nor m\n1\n− m\n2\nOE. If m defined, must include\nreference to population. Median not mean or average.\nDifferences: 4, 8, 15, 3, –1, –2, 5, –7, 11, 6 M1 Attempt at differences (at most 2 errors).\nSigned ranks: 4, 8, 10, 3, –1, –2, 5, –7, 9, 6 M1 Attempt at signed ranks from an attempt at differences.\n P = 4 5 , Q = 1 0  T=10 A1\nCritical value = 8 B1\n'1 0 '  '8 ' , accept H / do not reject H / not significant. M1 Compare their 10 with their 8 (their 8 may be 13, 10, 8, 5 or 3)\n0 0\nand appropriate conclusion (may be in terms of H ).\n1\nTheir 10 must come from a valid attempt at ranking.\nTest result can be implied by an attempt at an appropriate\nconclusion in context.\nInsufficient evidence to support that there is (on average) a A1 Correct conclusion from correct working ignoring hypotheses.\ndifference between the marks (awarded by the teachers). In context.\nLevel of uncertainty in language used (for example, not ‘prove’).\nDo not accept ‘No evidence…’.\nDo not accept statements such as ‘there is sufficient evidence to\nsuggest…’.\n7\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
      "source_pages": [
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      "html": "9231-topic-44-non-parametric-tests/answers.html",
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    },
    {
      "id": "9231-2026-mj-43-q03",
      "question_id": "9231-2026-mj-43-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 3,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "3(a)\n© Cambridge University Press & Assessment 2026 Page 11 of 15\nG 'X ( t ) = 0 .3 + 0 .8 t + 0 .6 t 2 M1 Attempt at differentiating G\nX\n( t ) .\nMay be implied by expression for G 'X ( 1 ) seen.\nE ( X ) = 1 .7 A1 CWO\nG ''X ( t ) = 0 .8 + 1 .2 t M1 Attempt at differentiating G 'X ( t ) .\nMay be implied by expression for G ''X ( 1 ) seen.\nV a r ( X ) = 2 + 1 .7 − 1 .7 2 M1 Use correct formula using their G ' (1) and\nX\nG ''X ( 1 ) .\n0.81 A1 CWO\n5\n\n3(b) G\nY\n( t ) = 0 .1 t 2 + 0 .3 t 5 + 0 .4 t 8 + 0 .2 t 1 1 B1 Correct coefficients in a 4-term polynomial.\nB1 Correct powers of t.\n2\nQuestion Answer Marks Guidance",
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      "source_pdf_url": "https://pastpapers.co/api/file/caie/a-level/mathematics-further-9231/2026-may-june/9231-mathematics---further-june-2026-mark-scheme-43.pdf?download=1",
      "html": "9231-topic-45-probability-generating-functions/answers.html",
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    },
    {
      "id": "9231-2026-mj-43-q04",
      "question_id": "9231-2026-mj-43-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 4,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "4(a)  2 2   1  2 7  M1 Attempt at integral with correct limits.\n dx = − = \n x3    x2  36\n1.5  1.5 \na= 7 200=38.89 A1 AG, convincingly obtained with no errors seen.\n36\n5.89 or 5.90 B1 Summing to 200 gives 5.89, integration gives 5.90.\n3\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) H : f is a good fit for the data. B1 Allow ‘satisfactory model for the data’.\n0\nH : f is not a good fit for the data. Must refer to f and data.\n1\nCombine 5th and 6th columns: (16, 9.72) B1 SOI\nOR 5.90 gives 9.73 instead of 9.72.\nCombine final two columns: (20, 16.33)\n5th and 6th columns combined:\n© Cambridge University Press & Assessment 2026 Page 12 of 15\n( 9 7 −\n1\n1\n1\n1\n1\n1\n.1\n.1\n1\n1 ) 2\n+  +\n( 1 6 −\n9\n9\n.7\n.7\n2\n2 ) 2\n+\n( 1 4 −\n1\n1\n2\n2\n.5\n.5\n0\n0 ) 2\n=1.792+0.389+1.389+2.786+4.057+0.18\nFinal 2 columns combined:\n(97−111.11)2 (10−5.89)2 (20−16.33)2\n++ +\n111.11 5.89 16.33\n = 1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 2 .8 6 8 ( 2 .8 4 9 ) + 0 .8 2 5 \nM1 At least 2 terms correct.\n972 162 142\nOE, for example, ++ + −200.\n111.11 9.72 12.50\nUse of 9.73 instead of 9.72 from using 5.90 for b gives\n1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 4 .0 4 0 + 0 .1 8 .\nOE, for example,\n1\n9\n1\n7\n1\n2\n.1 1\n+  +\n1\n5\n0\n.8\n2\n9\n+\n1\n2\n6\n2 0\n.3 3\n− 2 0 0 .\nUsing 5.90 instead of 5.89 for b gives\n1 .7 9 2 + 0 .3 8 9 + 1 .3 8 9 + 2 .7 8 6 + 2 .8 4 9 + 0 .8 2 5 .\nM1 possible if no columns combined, may be implied by answer\nin the range 10.61 to 10.63.\n5th and 6th columns combined: 10.6 A1 SC Unsupported answer of 10.6 or 10.0 or 10.1 scores B1.\nFinal two columns combined: 10.0 or 10.1 Must combine columns to earn this mark.\n'1 0 .6 '  9 .2 3 6 , reject H / significant. M1 Compare their 10.6 with 9.236 and appropriate conclusion (may\n0\nbe in terms of H ). If no evidence of columns combined, compare\n1\ntheir 10.6 with 10.64. Test result can be implied by appropriate\nconclusion that is consistent with their 10.6.\nSufficient evidence to suggest that f is not a good fit (for the data). A1 Correct conclusion from correct working, ignoring hypotheses.\nLevel of uncertainty in language used (for example, not ‘prove’).\nDo not accept statements such as ‘there is insufficient evidence to\nsuggest…’.\n6\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-43-chi-squared-tests/answers.html",
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    {
      "id": "9231-2026-mj-43-q05",
      "question_id": "9231-2026-mj-43-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 5,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 11,
      "status": "available",
      "reason": null,
      "text": "5(a)\n© Cambridge University Press & Assessment 2026 Page 13 of 15\nH\n0\n:\nx y\nH :1\nx y\n    =  B1 If  not used, must have population means.\n1  6.632 \ns x 2 = 43 1.21− 44    =4.906510−3 \n \nor s 2y =\n1\n5 1\n\n1 .6 5 −\n8 .5\n5\n6\n2\n2 \n = 4 .7 2 3 4  1 0 − 3 \nM1 Either expression, may be embedded.\nImplied by 2.02310−4.\ns2 s2 4.906510−3 4.723410−3 M1 Correct form with their sample variances.\nx + y = + For pooled s, award this M1 for correct form of pooled estimate.\n44 52 44 52\n2 .0 2 3  1 0 − 4 A1 May be implied by s = 0 .0 1 4 2 2 .\nAccept 3 significant figure accuracy or better.\nA0 for pooled s.\nz =\n6 .6 3\n4 4\n2 s\nx\n4 4\n−\n+\n8 .5\n5 2\n2 s\ny\n5 2\n6\n= − 0 .9 8 ( 0 )\nM1 Value of z obtained using correct form with their sample\nvariances. Order of numerator may be reversed.\nA1 Accept –0.977 to –0.980.\nMay see 0.980 for opposite order of numerator.\nUse of pooled s scores M0 A0.\n' − 0 .9 8 0 '  − 1 .2 8 2 , accept H\n0\n/ do not reject H\n0\n/ not significant.\nOR\nP ( Z  − 0 .9 8 0 ) = 0 .1 6 3 5  0 .1 , accept H\n0\n/ do not reject H\n0\nM1 Compare their z with\n/ not\nsignificant.\n 1 .2 8 2 (consistent signs) and appropriate\nconclusion (may be in terms of H ). 1\nAccept 0.1637.\nTest result can be implied by a conclusion that is consistent with\ntheir − 0 .9 8 0 or 0.1635.\nInsufficient evidence to support Rhiannon’s claim. A1 Correct conclusion from correct working ignoring hypotheses.\nIn context. Level of uncertainty in language used (for example,\nnot ‘prove’). Condone ‘no sufficient evidence’. Do not accept\nstatements such as ‘there is sufficient evidence to suggest…’.\n8 Use of pooled s scores max 3/8 B1 M1 M0 A0 M0 A0 M1 A0.\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(b) M1 Accept their s (may be pooled).\n6.63 8.56 s2 s2\n − 1.645 x + y Use correct formula with any z-value.\n 44 52  44 52 May subtract the means in opposite order.\nB1 1.645 SOI. Allow 1.64 or 1.65.\n© Cambridge University Press & Assessment 2026 Page 14 of 15\n[ − 0 .0 3 7 3 , 0 .0 0 9 4 7 ] A1 Accept 0.00947 to 0.0095.\nAccept with inequality signs or open brackets or\n − 9 .4 7  1 0 − 3 , 0 .0 3 7 3 \n\nfrom Y − X .\nCondone  9 .4 7  1 0 − 3 , − 0 .0 3 7 3 \n\n.\nDo not accept − 0 .0 1 3 9  0 .0 2 3 4 .\n3\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-43-q06",
      "question_id": "9231-2026-mj-43-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "43",
      "question_number": 6,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "6(a) F ( L ) = 0 .2 5 , s o 1\n1 2\n( x 2 + 4 x + 4 ) = 14 M1 May be implied by correct answer.\nL = 3 − 2 A1\n2\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(b)\n© Cambridge University Press & Assessment 2026 Page 15 of 15\n f ( x ) = \n\n\n16\n15[\n0\nx\nx\n+\n2 +\n13\n11\n5\n\n\no\n−\n0\nt\n2\n≤\nh\n≤\nx\ne r w\nx\n≤\ni\n\ns\n2\ne\n0\n\n]\n B1 Both expressions seen.\n E ( X 2 ) =\n=\n \n\n−\n0\n12\n4\n2 2 x\n4 x\n(\n+\n16\n19\nx\nx\n+\n3 \n13\n0\n−\n)\n2\nd\n+\nx +\n\n\n12\n5\n2\n0\nx\nx\n5\n2\n+\n( 15\n14\n5\nx\nx\n2\n3\n+\n\n2\n0\n) 11\n5\nd x M1 Attempt to integrate their x 2 f ( x ) .\nFor reference: 29 + 32 22 85 = 42 25 = 1 .6 8\n E ( X 4 ) =\n=\n 0 \n−\n 1\n 3 6\nx\n2\nx\n4\n6\n(\n+\n16 x\n11\n5\n+\nx 5\n13\n\n)\n0\n−\nd\n2\nx\n+\n+\n\n2 \n0\n13\n5\nx\nx\n4\n7\n(\n+\n15 2 x\n17\n5\nx\n+\n5\n11\n5\n2\n\n0\n) d x M1 Attempt to integrate their x 4 f ( x ) .\nFor reference: 16 + 2144 = 6992 =4.439365\n45 525 1575\nVar ( X2) = 6992− (42)2 M1 Using correct formula with their values.\n  1575 25\n1.62 A1\n5\n\n6(c) P ( X 2  1 ) = P ( − 1  X  1 ) = F ( 1 ) − F ( − 1 )  = 71\n5\n− 11\n2\n M1 For either F(1) or F(–1) evaluated.\nCould be implied by 7 , 1 , 2 or 1 seen (NFWW).\n15 12 15 4\n26 30 A1 AWRT 0.383\n2\nTrace ID: PC-1E62B2C2D5\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    {
      "id": "9231-2026-mj-44-q01",
      "question_id": "9231-2026-mj-44-q01",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 1,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 6,
      "status": "available",
      "reason": null,
      "text": "1 934.3 B1 Either expression, may be embedded.\ns2 = =23.956\nx 39 Accuracy must be 3 significant figures or better.\nMay be implied by\n756.2\ns2 = =12.817 y 59 17.25,\n© Cambridge University Press & Assessment 2026 Page 9 of 18\n6 94 , 0.71875, 23 32 , 18 4 6 or 0.84779 seen.\nAllow B1 for correct form of pooled estimate.\ns 2 =\n'2 3 .9\n4 0\n5 6 '\n+\n'1 2 .8\n6 0\n1 7 '[\n= 0 .8 1 2 5 ]\nM1 Correct form using their values.\nM0 for pooled estimate.\nA1 For fully correct expression SOI.\nImplied by AWRT 0.812 or 0.813 or s = 0 .9 0 1\nor fully correct interval.\n6 7\n4\n.1\n0\n−\n8 4\n6\n.1\n0\n 1 .9 6 '0 .8 1 2 5  ' = 0 .2 7 5 8  1 .7 6 6 7 \nM1 Correct form with any z-value.\nCondone t = 1.98.\nFT their s. If pooled must use s2( 1 + 1 ) .\np 40 60\nB1 1.96 seen.\n − 1 .4 9 , 2 .0 4  A1 CWO\nAccept with inequality signs or open brackets.\nCondone  2 .0 4 , − 1 .4 9  .\nDo not accept 0 .2 7 6  1 .7 6 7 .\n6 Use of pooled s scores max 3/6 B1 M0 A0 M1\nB1 A0 leading to [–1.39, 1.94].\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-42-inference-using-normal-and-t-distributions/answers.html",
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    },
    {
      "id": "9231-2026-mj-44-q02",
      "question_id": "9231-2026-mj-44-q02",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 2,
      "topic": "Probability generating functions",
      "topic_slug": "9231-topic-45-probability-generating-functions",
      "marks": 7,
      "status": "available",
      "reason": null,
      "text": "2(a) G ' (t)= 1 ( t2 +2 ) + 2t(t+1)= 1 ( 3t2 +2t+2 )\n X 6 6 6 \n© Cambridge University Press & Assessment 2026 Page 10 of 18\n E ( X ) =  76\nB1 AWRT 1.17\n G ''X ( t ) = t + 13 \n\nG ''X ( 1 ) = 43\nM1 Attempt to differentiate their G 'X ( t ) and\nevaluate at t = 1 . May be implied by an\nexpression for G '' (1) or 4 seen.\nX 3\nV a r ( X ) = 43 + 76 − ( 76 ) 2 M1 Use correct formula with their values.\n43 16  = 1 .1 4  A1 AWRT 1.14, NFWW (for example, an incorrect\nG ''X ( t ) , w i t h G ''X ( 1 ) = 43 scores max B1 M1 M1\nA0 XP).\n4\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n2(b) EITHER: (*M1 SOI, must be cubed.\nG Y (t)=    1 6 (t+1)( t2 +2 )   3    =    1 6 ( t3+t2 +2t+2 )   3  \n© Cambridge University Press & Assessment 2026 Page 11 of 18\n2\n11\n6\n( 8 + 2 4 t + 3 6 t 2 +  )  = 12\n7\n+ 19 t + 16 t 2 +   DM1 Attempt to expand to form a polynomial up to at\nleast t2. May be unsimplified. May be implied\nby a sum of coefficients or AWRT 0.315.\nISW with powers of t that are 3 or above.\nP(Y 2)= 8 + 24 + 36 = 1 + 1+ 1 =17 =0.315 A1) AWRT 0.315\n216 216 216 27 9 6 54\nOR:\nAlternative method for X\n1\n, X\n2\n, X\n3\n:\nP(all zero)= (1)3 , P( two 0s, one 1)=3 (1)2  (1)\n3 3 3\nP ( t w o 0 s , o n e 2 ) = 3  ( 13 ) 2  ( 16 ) , P ( o n e 0 , t w o 1 s ) = 3  13  ( 13 ) 2\n(M1 Consider all 4 cases.\nM1 Correct probability calculations for at least 2\ncases.\nOnly 2 cases considered can score a maximum\nof M0 M1 A0.\n15 74 A1) AWRT 0.315, CWO\nAvailable marks 3\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-44-q03",
      "question_id": "9231-2026-mj-44-q03",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 3,
      "topic": "Chi-squared tests",
      "topic_slug": "9231-topic-43-chi-squared-tests",
      "marks": 9,
      "status": "available",
      "reason": null,
      "text": "3(a) X ~N ( 236.5,1 1.62 )\n \n c = 1 2 0  P ( 2 5 0 ≤ X ≤ 2 6 0 ) =  1 2 .1 0\nB1 12.09 or 12.10 or 12.11\nAccept 12.1 for 12.10.\nd =120(1−Φ(2.026))=2.57 or [(120 – total) =] 2.56 B1 2.55, 2.56 or 2.57.\n \n2\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b) H : A normal distribution fits the data. B1 0 Condone\nH : A normal distribution does not fit the data.\n1\nOR\nH : A normal distribution is a satisfactory/suitable model for the masses (of this\n0\nspecies of frog).\nH : A normal distribution is not a satisfactory/suitable model for the masses (of this\n1\nspecies of frog).\n© Cambridge University Press & Assessment 2026 Page 12 of 18\nN ( 2 3 6 .5 ,1 1 .6 2 ) included in\nhypotheses.\nMust mention distribution/model and\ndata/masses.\nCombines expected frequencies in either first two cells or last two cells. M1 At least one cell must have an expected\nfrequency less than 5 seen. May be implied by a\ncorresponding contribution seen.\n( 1 1 −\n9\n9\n.2\n.2\n8\n8 ) 2\n+\n( 2 0 −\n2\n2\n5\n5\n.2\n.2\n2\n2 ) 2\n+  +\n( 1 0 −\n'1\n 1\n4\n4\n.6\n.6\n7\n7\n'\n)' 2\n0.3188+1.0804+0.0134+2.5404+1.4866\nOR\n1\n9\n1\n.2\n2\n8\n+\n2\n2\n5\n2 0\n.2 2\n+  +\n'1\n1\n4\n2 0\n.6 7 '\n− 1 2 0\nM1 At least 2 expressions (not including 120) of\ncorrect form seen.\nMay see individual terms not summed.\nMay be implied by 2 contributions correct to 3\ndecimal places seen.\nMay be implied by AWRT 5.43 or 5.44 or 5.45.\nFor contribution from combining last two cells\nallow AWRT 1.48 or 1.49 to 3 significant\nfigures.\nNo combining gives\nFor last two cells: 12.09 gives 1.3836, 2.57 gives\n0.3251+0.1415+1.0804+0.0134+2.5404+1.3893+0.1225\n0.1264.\n5.440 A1 AWRT 5.43 or 5.44 or 5.45\nCritical value is 5.991 B1 With both first 2 cells and last 2 cells combined.\n'5.440' < 5.991, accept H / do not reject H / not significant. M1 Compare their 5.44 with 5.991 or 7.815 or 9.488\n0 0\nor 11.07 or 12.59 and appropriate conclusion\n(may be in terms of H ).\n1\nTest result can be implied by an attempt at an\nappropriate conclusion in context ignoring\nhypotheses.\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n3(b) Insufficient evidence to suggest the masses (of this species of frog) are not normally A1 Correct conclusion from correct working (using\ndistributed. 5.99) ignoring hypotheses.\nIn context.\nOR Level of uncertainty in language used (for\nexample, not ‘prove’).\nInsufficient evidence to reject the biologist’s belief. Allow ‘Not enough evidence / no sufficient\nevidence that / to show / to conclude that …’.\nOR Do not accept ‘No evidence…’.\nDo not accept statements such as “there is\nInsufficient evidence to suggest that a normal distribution is not a suitable model / sufficient evidence to suggest…”.\ngood fit (for the data).\n7\nQuestion Answer Marks Guidance",
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    {
      "id": "9231-2026-mj-44-q04",
      "question_id": "9231-2026-mj-44-q04",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 4,
      "topic": "Non-parametric tests",
      "topic_slug": "9231-topic-44-non-parametric-tests",
      "marks": 8,
      "status": "available",
      "reason": null,
      "text": "4(a) Reduces the uncertainty/variability from choosing cyclists of different abilities. B1 Allow for an attempt at an explanation that\npairing/differences/matching leads to improved\nOR accuracy/precision/clarity/sensitivity/certainty.\nControls for differences between different cyclists.\n1\n© Cambridge University Press & Assessment 2026 Page 13 of 18\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n4(b) H : population median difference= 0 B1 Allow\n0\nH : population median difference > 0 1\n© Cambridge University Press & Assessment 2026 Page 14 of 18\nm , m\nd\n, m\na\n− m\nc\nOE but not .\nIf m defined, must see population.\nMedian not mean or average.\n3.9 3 8 –1.4 1.8 3.7 0.1 0.2 –0.3 2.9 M1 Attempt at (signed) differences. At most 2\nerrors. May be seen with table before 4(a).\n9 7 10 –4 5 8 1 2 –3 6 M1 Attempt at (signed) ranks, ranks may be\nreversed. May be seen with table before 4(a).\n[P=48] Q = 7 A1 Ignore any errors in positive ranks.\n'7' < 10, reject H / significant. M1 Compare their 7 with 10 and appropriate\n0\nconclusion (may be in terms of H ). Note that if\n1\ntheir 7 is 10, they should still reject H .\n0\nTheir 7 must come from attempt at ranks and be\nthe smaller of their P and Q. Test result can be\nimplied by an attempt at an appropriate\nconclusion in context ignoring hypotheses.\nSufficient evidence to support the club captains’ belief. A1 Correct conclusion from correct working\nignoring hypotheses.\nOR A0 for Q=7 from incorrect ranks.\nIn context. Level of uncertainty in language used\nSufficient evidence to suggest that the anticlockwise route takes longer (to complete (for example, not ‘prove’).\non average than the clockwise route). Allow ‘(Some/enough) evidence that/to\nshow/conclude that…’.\nDo not accept statements such as ‘there is\ninsufficient evidence to suggest…’.\n6\n\n4(c) The population of differences is symmetrical. B1 Allow underlying/parent distribution for\npopulation.\n1\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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    {
      "id": "9231-2026-mj-44-q05",
      "question_id": "9231-2026-mj-44-q05",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 5,
      "topic": "Continuous random variables",
      "topic_slug": "9231-topic-41-continuous-random-variables",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "5(a)\n[Continuous so]\na4=b(−2)2\n a=b\nB1 May be awarded for finding b directly (having\nalready found a).\n© Cambridge University Press & Assessment 2026 Page 15 of 18\n12 ( 5 − 1 )  4 a = 0 .7 5 o r \n5\n1\na ( x − 1 ) d x = 0 .7 5  8 a = 0 .7 5\nM1 Use of area of triangle = 0.75 to form an\nequation in a. May see an attempt at integral\nleading to\nOR\n 7 b(x−7)2 dx=0.25  8b=0.25\n3\n5\nOR\n 5 a(x−1)dx+  7 a(x−7)2 dx=1  8a+8a=1\n3\n1 5\n12 a ( 4 2 − 0 2 ) = 34 .\nSome working may be present on sketch.\nOR Use of area of upper tail = 0.25 to form an\nequation in b.\nOR Use of total area = 1 to form an equation in\na (or in b).\na = 33\n2\nOR b = 33\n2\nA1 AG, convincingly obtained, no errors seen.\n3\n\n5(b) 0 x1 M1 Attempt to integrate both parts, all powers\n correct, condone missing c, ignore domain.\n3 (x−1)2 1≤x≤5\n\n\nF(x)=\n\n 64\n1 (x−7)3 +1 5≤x≤7 A1 Both parts fully correct. Ignore domain.\n32  For reference:\n1 x7\n36\n4\n( x − 1 ) 2 = 36\n4\nx 2 − 33\n2\nx + 36\n4\n,\n13\n2\n( x − 7 ) 3 + 1 = 13\n2\nx 3 − 23 12 x 2 + 1 4 7\n3 2\nx − 3 1 1\n3 2\n.\nB1 Correct domains.\nMust cover all reals, may overlap at boundaries.\nMust have 0 for x  1 and 1 for x  7 .\n3\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n5(c)\n© Cambridge University Press & Assessment 2026 Page 16 of 18\n G ( y ) = \n 0\n−\n1\n1\n( 13\n2\n3 −\n6 4\n3\n(\n−\n9\ny\n−\n3 )\n) y 2\ny\n3\n5\ny\n\n≤\n≤\n\n3\ny\ny\n9\n≤\n≤\n5\n9\n*M1 Expression for F ( 1 0 − y ) seen.\nDM1 Use of 1 − F ( 1 0 − y ) in both parts.\nA1 Both parts correct, ignore domain. May be\nunsimplified. ISW if seen correct in working.\nFor reference:\n− 13\n2\n( 3 − y ) 3 = 13\n2\n( y 3 − 9 y 2 + 2 7 y − 2 7 ) ,\n1 − 36\n4\n( 9 − y ) 2 = 16\n4\n( − 3 y 2 + 5 4 y − 1 7 9 ) .\nB1 Correct domain must be in terms of y.\nDomain may overlap at boundaries.\nMust have 0 for y  3 and 1 for y  9 .\n4\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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      "html": "9231-topic-41-continuous-random-variables/answers.html",
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    {
      "id": "9231-2026-mj-44-q06",
      "question_id": "9231-2026-mj-44-q06",
      "subject": "9231",
      "year": 2026,
      "session": "May/June",
      "session_code": "mj",
      "paper": 4,
      "variant": "44",
      "question_number": 6,
      "topic": "Inference using normal and t-distributions",
      "topic_slug": "9231-topic-42-inference-using-normal-and-t-distributions",
      "marks": 10,
      "status": "available",
      "reason": null,
      "text": "6(a) H :  =, H:   B1 If in words, must have population means.\n0 A B 1 A B\n© Cambridge University Press & Assessment 2026 Page 17 of 18\n ( x − x ) 2 = '9 7 7 8 '−\n'2 7\n8\n0 2'\n = 6 6 5 .5 \nB1 For correct sum and sum of squares for either x\nor y.\nMay be implied, for example, by either 95.1 or\n56.75 seen.\n'225'2\n(y− y)2 ='6079'− =454.0\nM1 Correct form for both expressions using their\n9\nvalues. May be embedded.\n'665.5'+'454.0' M1 Correct form of pooled variance using their\ns2 =\np 8+9−2 values.\n2 23 30 9  = 7 4 .6 3 3  A1 AWRT 74.6, implied by 2.08.\n( t = )\n'2\n'7\n7\n8\n4\n0\n.6\n'\n3\n−\n3\n'2\n('\n2\n9\n18\n5\n+\n'\n19 )\nM1 Must be in correct form using their s.\n2 .0 8 4 A1 AWRT 2.08\n'2.084' > 1.753, reject H / significant. M1 Compare their\n0\n 2 .0 8 4 with  1 .7 5 3 (consistent\nsigns) and appropriate conclusion (may be in\nterms of H\n1\n).\nTheir  2 .0 8 4 must come from an attempt at\nstandardising.\nTest result can be implied by an attempt at an\nappropriate conclusion in context.\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/\n\n6(a) Sufficient evidence to suggest that there is a difference in the (average) percentage A1 Correct conclusion from correct working,\nof waste (that is recycled in City A and City B). ignoring hypotheses.\nIn context. Level of uncertainty in language used\n(for example, not ‘prove’).\nAllow ‘(Some/enough) evidence that/to\nshow/conclude that…’.\nDo not accept statements such as ‘there is\ninsufficient evidence to suggest…’.\n9 Combined s max 5/9\nB1B1M1M0A0M1A0M1A0\n\n6(b) Population of percentage of household waste recycled in each city are normal. B1 B1 for underlying/parent distributions are\nnormal.\nB0 for population mean normally distributed.\nB0 for (population of/underlying) data/sample\nis normal.\nAllow singular, for example, population is\nnormal.\nAllow use of A and B to denote population /\nrandom variable.\nMust not refer to differences.\n1\n© Cambridge University Press & Assessment 2026 Page 18 of 18\nTrace ID: PC-1430F78DC7\nRe-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable.\nLicensed for hosting on papacambridge.com only.\nP92a31p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026\nDownloaded from PapaCambridge - https://papacambridge.com/",
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