Polar coordinates

5(a) θ=0 B1 B1 Initial line with correct position and shape of curve. SCB1 if correct shape with wrong starting position. Pole position must be clear. acot1π=2 3a=2 B1 Substitutes θ= 1π. AG. 6 6 3 5(b) 1 1 6 π 4cot2 ( 1π−θ ) dθ M1 Uses 1r2dθ with correct limits. 2 2 3 0 ( ) ( ) 2 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. ( ) 2 3−1π−1 3 = 4 3−1π A1 OE, must be exact. 6 3 3 3 4 5(c) cosπcosθ+sinπsinθ M1 Uses the identities for cos(A−B) and sin(A−B). Or r =2 3 3 sinπcosθ−cosπsinθ identity for tan (A – B). 3 3 x2 + y2 =2 rcosθ+ 3rsinθ M1 Applies r = x2 + y2,x=rcosθ and y=rsinθ. 3rcosθ−rsinθ ( ) ( ) 2 x+ y 3 = x 3−y x2 + y2 A1 AG. 3 © UCLES 2021 Page 10 of 14

5(a) θ=0 B1 B1 Initial line with correct position and shape of curve. SCB1 if correct shape with wrong starting position. Pole position must be clear. acot1π=2 3a=2 B1 Substitutes θ= 1π. AG. 6 6 3 5(b) 1 1 6 π 4cot2 ( 1π−θ ) dθ M1 Uses 1r2dθ with correct limits. 2 2 3 0 ( ) ( ) 2 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. ( ) 2 3−1π−1 3 = 4 3−1π A1 OE, must be exact. 6 3 3 3 4 5(c) cosπcosθ+sinπsinθ M1 Uses the identities for cos(A−B) and sin(A−B). Or r =2 3 3 sinπcosθ−cosπsinθ identity for tan (A – B). 3 3 x2 + y2 =2 rcosθ+ 3rsinθ M1 Applies r = x2 + y2,x=rcosθ and y=rsinθ. 3rcosθ−rsinθ ( ) ( ) 2 x+ y 3 = x 3−y x2 + y2 A1 AG. 3 © UCLES 2021 Page 10 of 14

5(a) B1 Approximately correct curve passing through the pole, O, with correct domain. B1 r strictly increasing over the domain 0 to π . 2 B1 Correct form at O. Approx tangential to initial line at O. [O] [θ = 0] 3 © UCLES 2021 Page 11 of 16 5(b) 1π 1 1 2 M1 Uses 1r2 dθ with correct limits. 12  −  dθ 2 2 0 π−θ π 1π 1 2 1 M1 Expands. 12 − + dθ 2 0 ( π−θ)2 π ( π−θ) π2 1 1 2 θ  1 2 π M1 A1 Integrates all terms to obtain correct form. + ln(π−θ)+ 2  π−θ π π2  0 12 2 π 1 1 2  1 3 2 1 M1 Substitute limits into an expression of the correct  + ln + −  + lnπ  =  + ln  form and simplify the log terms. 2π π 2 2π π π  22π π 2 3−4ln2 A1 AG = 4π 6 Question Answer Marks Guidance

6(a) 2 ( c(c)−(1+s)s )=0 M1 dr Finds and sets equal to 0. dθ 1−s−2s2 =0 A1 sinθ=−1, sinθ= 1 leading to θ=1π M1 A1 Solves quadratic in sinθ. 2 6 ( ) ( ) 3 3, 1π A1 Allow 2.60, 1π 3sf 2 6 6 5 6(b) B1 π Correct shape. Tangential to θ= at pole, r strictly 2 π increasing to then decreasing. 6 B1 Correct position. All in first quadrant. θ=0 2 © UCLES 2021 Page 11 of 14 6(c) 1 1 2 π 4c2(1+s)2 dθ=2 1 2 π c2 +2sc2 +s2c2 dθ M1 A1 Uses 1 2 r2 dθ with correct limits. 2 0 0 2 1 2 π 1 ( cos2θ+1 )+2sinθcos2θ+1sin22θ dθ M1 Uses double angle formulae on first and last terms. 2 4 0 1π A1 Obtains integrable form. 2 cos2θ+1+4sinθcos2θ+ 1 −1cos4θ dθ 4 4 0 5θ+1sin2θ−4cos3θ− 1 sin4θ 1 2 π M1 Integrates. 4 2 3 16  0 5π+4 A1 8 3 6 Question Answer Marks Guidance

5(a) B1 Correct symmetrical shape, closed loop. 1 5(b) Line l parallel to initial line and correct side of pole. B1 2=3sinθ+2sin2θ M1 Forms quadratic in sinθ. Or in r r=3+ 4 r sinθ= 1 M1 Solves for sinθ. 2 ( ) 4,1π A1 SC1 For finding both angles correctly 6 ( ) 4,5π A1 6 5 © UCLES 2021 Page 10 of 16 5(c) 2×1 π 6 ( 3+2sinθ)2 dθ M1 Finds the part of the required area enclosed by the 2 −π curved outer edge and two line segments from the pole. 2 Limits must be correct.  π 6 9+12sinθ+2 ( 1−cos2θ) dθ M1 Uses double angle formula and integrates. −π 2 [ 11θ−12cosθ−sin2θ ] 6 π = 22π−13 3 A1 A1 −π 3 2 2 22π−13 3+(4cosπ)×2 M1 Adds area of triangle. 3 2 6 22π− 5 3 A1 3 2 6 © UCLES 2021 Page 11 of 16

6(a) 2 ( c(c)−(1+s)s )=0 M1 dr Finds and sets equal to 0. dθ 1−s−2s2 =0 A1 sinθ=−1, sinθ= 1 leading to θ=1π M1 A1 Solves quadratic in sinθ. 2 6 ( ) ( ) 3 3, 1π A1 Allow 2.60, 1π 3sf 2 6 6 5 6(b) B1 π Correct shape. Tangential to θ= at pole, r strictly 2 π increasing to then decreasing. 6 B1 Correct position. All in first quadrant. θ=0 2 © UCLES 2021 Page 11 of 14 6(c) 1 1 2 π 4c2(1+s)2 dθ=2 1 2 π c2 +2sc2 +s2c2 dθ M1 A1 Uses 1 2 r2 dθ with correct limits. 2 0 0 2 1 2 π 1 ( cos2θ+1 )+2sinθcos2θ+1sin22θ dθ M1 Uses double angle formulae on first and last terms. 2 4 0 1π A1 Obtains integrable form. 2 cos2θ+1+4sinθcos2θ+ 1 −1cos4θ dθ 4 4 0 5θ+1sin2θ−4cos3θ− 1 sin4θ 1 2 π M1 Integrates. 4 2 3 16  0 5π+4 A1 8 3 6 Question Answer Marks Guidance

6(a) B1 Correct domain, r strictly increasing. B1 Correct gradient when 0 and 2. Maximum distance of C from the pole is 1. B1 1 4 Must be exact. π 2 3 6(b) 1 2 tan11 d M1 Uses 1r2d with correct limits. 2 2 2 0 1tan11 2  2  d M1 A1 Applies integration by parts. 2 2  0 02 4 1tan11ln(2 4) 2 A1 2 2  0 11ln81ln4 11ln2 A1 Must be exact. 4 2 2 4 2 5 © UCLES 2022 Page 12 of 15 6(c) x  tan111 2 cos B1 Uses xrcos 2 dx   tan111 2sincos  tan111 2  2 4 1 0 M1 A1 Sets derivative equal to zero. d 2 2  2 4  tan11sincos0 A1 AG 2  0.62 4  tan10.3sin0.6cos0.60.108 and B1 Shows sign change.  0.72 4  tan10.35sin0.7cos0.70.209 5 © UCLES 2022 Page 13 of 15

6(a) B1 Correct domain, r strictly increasing. B1 Correct gradient when 0 and 2. Maximum distance of C from the pole is 1. B1 1 4 Must be exact. π 2 3 6(b) 1 2 tan11 d M1 Uses 1r2d with correct limits. 2 2 2 0 1tan11 2  2  d M1 A1 Applies integration by parts. 2 2  0 02 4 1tan11ln(2 4) 2 A1 2 2  0 11ln81ln4 11ln2 A1 Must be exact. 4 2 2 4 2 5 © UCLES 2022 Page 12 of 15 6(c) x  tan111 2 cos B1 Uses xrcos 2 dx   tan111 2sincos  tan111 2  2 4 1 0 M1 A1 Sets derivative equal to zero. d 2 2  2 4  tan11sincos0 A1 AG 2  0.62 4  tan10.3sin0.6cos0.60.108 and B1 Shows sign change.  0.72 4  tan10.35sin0.7cos0.70.209 5 © UCLES 2022 Page 13 of 15

6(a) xrcos yrsin x2  y2 r2 B1 SOI r21sincosr21 1sin2a M1 Eliminates both xand yand uses the double angle 2 formula. 2a A1 AG r2  2sin2 3 6(b) B1 Polar graph with curve in correct domain. Graph needs gradient 0and r0 for all  in the domain. B1 π r strictly decreasing such that r at  is greater than 4 half r at 0. Concave graph. π Gradient at 0 and must not be vertical or 4 horizontal, respectively. 2 © UCLES 2022 Page 14 of 17 6(c) 1π B1 AG 1π 4 1tan2 R 14 r2 da d 2 0   2  1tan2  2tan 0 dt M1 Applies given substitution. t tan leading to sec2 d dt M1 Applies sec21tan2 sec21t2 d  1 1 A1 R 1a dt 2  t2 t1 0 t2 t1(t 1)2  3 B1 Completes the square. 2 4 1 M1 A1 Applies formula. R 1a   1 dt  1 atan1  2 t 1  1 2  (t 1)2  3 3  3 3  0 0 2 4 πa A1 6 3 8 © UCLES 2022 Page 15 of 17

6(a) x2 + y2 =r2 x=rcos y=rsin B1 Used. r4 =36r2( cos2−sin2 ) =36r2cos2 M1 Substitutes and applies cos2=cos2−sin2 . r2 =36cos2 A1 AG. 3 6(b) B1 Closed curve. Correct position and symmetrical about initial line. B1 Single correct loop. 6 B1 States maximum distance or labels sketch. 3 © UCLES 2022 Page 12 of 15 6(c) 18  4 cos2d=9sin2 4 M1 Forms 1 2 r2 d − 4 − 4 18 A1 2 6(d) 1 B1 y=6cos22sin cos 1 2 2cos−cos−1 2 2sin2sin=0 M1 A1 Sets dy =0. d 2tan2 M1 Applies suitable trigonometric identity. cos2cos−sin2sin=0 leading to 1= 1−tan2 =1π A1 6 3 2 A1 2 6 © UCLES 2022 Page 13 of 15

5(a) B1 Correct position and domain and r strictly increasing. B1 Decreasing positive gradient. =0 (a,0) (2a,1π) B1 Can be labelled on their sketch. 4 3 5(b) y=(2a)sin (1π ) M1 Uses y=rsin 4 a 2 A1 2 © UCLES 2022 Page 10 of 15 5(c) 1a2 1 4 π sec4d M1 Forms 1 2 r2d with correct limits. 2 0 =1a2 1 4 π sec2 ( tan2+1 ) d M1 Applies sec2=tan2+1 to obtain an 2 0 integrable form. 1π A1 = 1a21tan3+tan4 2 3  0 = 2a2 A1 3 4 5(d) x2 + y2 =r2 x=rcos M1 Substitutes to eliminate either r or . r2cos2=ar leading to x2 =a x2+y2 leading tox4 =a2( x2 +y2) M1 Completes substitution to eliminate both r and  and makes y or y2 the subject. A1 AEF y= a−2x4−x2 =x a−2x2−1 3 © UCLES 2022 Page 11 of 15

6(a) x2 + y2 =r2 x=rcos y=rsin B1 Used. r4 =36r2( cos2−sin2 ) =36r2cos2 M1 Substitutes and applies cos2=cos2−sin2 . r2 =36cos2 A1 AG. 3 6(b) B1 Closed curve. Correct position and symmetrical about initial line. B1 Single correct loop. 6 B1 States maximum distance or labels sketch. 3 © UCLES 2022 Page 12 of 15 6(c) 18  4 cos2d=9sin2 4 M1 Forms 1 2 r2 d − 4 − 4 18 A1 2 6(d) 1 B1 y=6cos22sin cos 1 2 2cos−cos−1 2 2sin2sin=0 M1 A1 Sets dy =0. d 2tan2 M1 Applies suitable trigonometric identity. cos2cos−sin2sin=0 leading to 1= 1−tan2 =1π A1 6 3 2 A1 2 6 © UCLES 2022 Page 13 of 15

5(a) 0 B1* Correct shape and domain, polar graph with r strictly decreasing but condone if not strictly decreasing close to π. DB1 Fully correct including shape at 0 correct shape at π correct l,0 B1 Identified as point furthest from the pole and given as coordinates. 3 5(b)  π 1 M1 Uses correct formula with correct limits. 1 d 2 2 1 0 π M1 A1 1 1 2   tan1  0 Integrates 2 1 . 1tan1π0.631 A1 2 4 © UCLES 2023 Page 10 of 16 5(c) sin B1 Uses yrsin y 2 1 dy  2 1 1 2cos  2 1 1 2sin M1 A1 Sets derivative equal to zero.  0 d 2 1  1 A1 AG. 0 cot10    1  B1 Shows sign change (1sf or better). 1.1 cot1.110.02…  1.1  1  and 1.2 cot1.210.209…  1.2 5 Question Answer Marks Guidance

5(a) 0 B1* Correct shape and domain, polar graph with r strictly decreasing but condone if not strictly decreasing close to π. DB1 Fully correct including shape at 0 correct shape at π correct l,0 B1 Identified as point furthest from the pole and given as coordinates. 3 5(b)  π 1 M1 Uses correct formula with correct limits. 1 d 2 2 1 0 π M1 A1 1 1 2   tan1  0 Integrates 2 1 . 1tan1π0.631 A1 2 4 © UCLES 2023 Page 10 of 16 5(c) sin B1 Uses yrsin y 2 1 dy  2 1 1 2cos  2 1 1 2sin M1 A1 Sets derivative equal to zero.  0 d 2 1  1 A1 AG. 0 cot10    1  B1 Shows sign change (1sf or better). 1.1 cot1.110.02…  1.1  1  and 1.2 cot1.210.209…  1.2 5 Question Answer Marks Guidance

5(a) r2 cos2sin2  a B1 Uses xrcos and/or yrsin. r2cos2a M1 Applies relevant double angle formulae. r2 asec2 A1 AG. 3 © UCLES 2023 Page 10 of 16 5(b) B1* Initial line drawn. Correct domain and position, r strictly increasing. dB1 Also sloping to right, concave on opposite side to pole, correct gradient when 0 and  π/4 . O a B1 3 5(c) 1a1 1 2  sec2d M1 Uses 1 2 r2d with correct limits. 2 0 1alntan 1π1 1 2 π or 1alntan2sec21 1 2 π M1 A1 Integrates. 4  4  0 4   0 1aln 3 1aln3 A1 4 8 4 © UCLES 2023 Page 11 of 16

6(a) x2−x+1+y2 =1r2−rcos+1 =1 B1 Uses x2 + y2 =r2 and x=rcos. 4 4 4 4 r(r−cos)=0 M1 Factorises. r 0r =cos A1 AG. 3 © UCLES 2023 Page 11 of 15 6(b) sin2=cos2sincos=cos M1 Sets r values equal and uses sin2=2sincos. cos0sin= 1 A1 cos0 must be recognised. 2 ( 1 3,1π ) A1 2 6 3 6(c) B1 Initial line drawn and one curve correct. B1 Other curve correct. B1 Intersection marked in correct position and both curves labelled. P o =0 3 © UCLES 2023 Page 12 of 15 6(d) 1 1 6  sin22d+1 1 2  cos2d M1 Uses 1 2 r2d with correct limits. 2 0 2 1 6 1 1 M1 Integratessin22 using identity. 16 sin22d=16 1−cos4d 2 4 0 0 1 A1 = 1−1sin46   4 4 0 1 1 M1 Integrates cos2 using identity. 12 cos2d= 12 1+cos2d 2 1 4 1 6 6 1 A1 = 1+ 1sin22   4 2 1 6 1 ( 1−1 3 ) +1 ( 1−1−1 3 ) =1 ( −3 3 ) A1 4 6 8 4 2 6 4 8 4 6 Question Answer Marks Guidance

6(a) =0 B1 Initial line drawn. Correct shape, r strictly decreasing. B1 Correct shape at extremities. 1−e−1 2 π B1 May be seen on their diagram. 3 6(b) 1 1 2 π( e−−e−1 2 π )2 d M1 Uses correct formula with correct 2 limits. 0 1 1 2 π e−2−2e−−1 2 π +e−πd A1 2 0 1π M1 A1 Integrates. 1−1e−2+2e−−1 2 π +e−π2 2 2  0 1 ( −1e−π +2e−π + 1πe−π + 1−2e−1 2 π ) = 3e−π + 1πe−π −e−1 2 π + 1 A1 2 2 2 2 4 4 4 5 © UCLES 2023 Page 12 of 15 6(c) y= ( e−−e−1 2 π ) sin B1 Uses y=rsin dy = ( e−−e−1 2 π ) cos+sin ( −e−) =0 M1 A1 Sets derivative equal to zero. d  −e−  A1 AG.    1  1+ tan=01−e−1 2 π −tan=0 2 e−−e−1 2 π  1−e0.56−1 2 π −tan0.56=0.00912 and1−e0.57−1 2 π −tan0.57=−0.00856 B1 Shows sign change. 5 Question Answer Marks Guidance

6(a) x2−x+1+y2 =1r2−rcos+1 =1 B1 Uses x2 + y2 =r2 and x=rcos. 4 4 4 4 r(r−cos)=0 M1 Factorises. r 0r =cos A1 AG. 3 © UCLES 2023 Page 11 of 15 6(b) sin2=cos2sincos=cos M1 Sets r values equal and uses sin2=2sincos. cos0sin= 1 A1 cos0 must be recognised. 2 ( 1 3,1π ) A1 2 6 3 6(c) B1 Initial line drawn and one curve correct. B1 Other curve correct. B1 Intersection marked in correct position and both curves labelled. P o =0 3 © UCLES 2023 Page 12 of 15 6(d) 1 1 6  sin22d+1 1 2  cos2d M1 Uses 1 2 r2d with correct limits. 2 0 2 1 6 1 1 M1 Integratessin22 using identity. 16 sin22d=16 1−cos4d 2 4 0 0 1 A1 = 1−1sin46   4 4 0 1 1 M1 Integrates cos2 using identity. 12 cos2d= 12 1+cos2d 2 1 4 1 6 6 1 A1 = 1+ 1sin22   4 2 1 6 1 ( 1−1 3 ) +1 ( 1−1−1 3 ) =1 ( −3 3 ) A1 4 6 8 4 2 6 4 8 4 6 Question Answer Marks Guidance

7(a) B1 Correct shape. B1 Section 1ππ correct. 2 0   B1 May be seen on their diagram. πtan1π, 0 (1.99, 0) 3 © Cambridge University Press & Assessment 2024 Page 14 of 16 7(b) π 0 M1 Uses correct formula. A 1 πtan1πd or A1 utan1udu 2 2  0 π or A 1 utan1udu 2 0 11u2 tan1u π 1 π u2 du M1 A1 Integrates by parts. 22  0 4 0 1u2 π u2 π1u2 1 π 1 M1 Rearranges integral into form which can be  du du  1 du integrated. 0 1u2 0 1u2 0 1u2 π A1 utan1u   0 . A 11u2 tan1u π 1utan1u π A1 22  0 4  0 A 1π2tan1π1  πtan1π   1  π2 1  tan1π1π2.65 A1 4 4 4 4 7 © Cambridge University Press & Assessment 2024 Page 15 of 16 7(c) y πtan1πsin B1 Uses yrsin π  1π21 tan1π M1 A1 Finds derivative. dy  πtan1πcos sin d 2 πtan1π π A1 dy 0,π2πtan1πcot tan1π0 Puts 0 and forms equation. AG. 1π2 d π1.2 B1 Shows sign change. 2π1.2tan1π1.2cot1.2 tan1π1.20.151 1π1.22 π1.3 and 2π1.3tan1π1.3cot1.3 tan1π1.30.395 1π1.32 5 © Cambridge University Press & Assessment 2024 Page 16 of 16

7(a) B1 Correct shape. B1 Section 1ππ correct. 2 0   B1 May be seen on their diagram. πtan1π, 0 (1.99, 0) 3 © Cambridge University Press & Assessment 2024 Page 14 of 16 7(b) π 0 M1 Uses correct formula. A 1 πtan1πd or A1 utan1udu 2 2  0 π or A 1 utan1udu 2 0 11u2 tan1u π 1 π u2 du M1 A1 Integrates by parts. 22  0 4 0 1u2 π u2 π1u2 1 π 1 M1 Rearranges integral into form which can be  du du  1 du integrated. 0 1u2 0 1u2 0 1u2 π A1 utan1u   0 . A 11u2 tan1u π 1utan1u π A1 22  0 4  0 A 1π2tan1π1  πtan1π   1  π2 1  tan1π1π2.65 A1 4 4 4 4 7 © Cambridge University Press & Assessment 2024 Page 15 of 16 7(c) y πtan1πsin B1 Uses yrsin π  1π21 tan1π M1 A1 Finds derivative. dy  πtan1πcos sin d 2 πtan1π π A1 dy 0,π2πtan1πcot tan1π0 Puts 0 and forms equation. AG. 1π2 d π1.2 B1 Shows sign change. 2π1.2tan1π1.2cot1.2 tan1π1.20.151 1π1.22 π1.3 and 2π1.3tan1π1.3cot1.3 tan1π1.30.395 1π1.32 5 © Cambridge University Press & Assessment 2024 Page 16 of 16

7(a) B1 One loop correct with initial line. B1 Second loop correct with correct form at the pole. 0 1π B1 May be seen on their diagram. 2 3 7(b) r5 2(rsin)(rcos)2 M1 Use of sin22sincos, and xrcos or yrsin. r5 2yx2 A1  x2  y25 2 2x2y A1 AEF 3 © Cambridge University Press & Assessment 2024 Page 14 of 16 7(c) 1 π sin(2)cosd M1 Applies 1r2 d. 2 2 0 1 M1 Attempt to integrate in a valid way. sin2cos d May apply sin(2)2sincos . 2 1 A1 Correct answer (there are alternative forms).  sincos2d cos3  [c] 3 π A1 CAO 1cos3  2  3  0 3 4 7(d) dr  1sin2cos1 22cos2cossin2sin *M1 A1 Differentiates with respect to. d 2 2cos2cossin2sin6cos34cos dM1 Correct use of relevant identities to express in terms of a single trig function. 6cos34cos0cos2 2 dM1 A1 Sets derivative equal to 0 and solves to find 3 1 2 sin2 , cos2 , 3 3 one of 1 tan2 = or =0.6155 2 A1 4 4 3 r   0.877 3 3 9 © Cambridge University Press & Assessment 2024 Page 15 of 16 7(d) Alternative method for question 7(d) 2r dr 2  2sin2coscos3  *M1 A1 Differentiates [RHS] with respect to. d 2  2  1cos2  coscos3  6cos34cos dM1 Correct use of relevant identities in terms of a single trig function 6cos34cos0cos2 2 dM1 A1 Sets derivative equal to 0 and solves to find 3 1 2 sin2 , cos2 , 3 3 one of 1 tan2 = or =0.6155 2 A1 4 4 3 r   0.877 3 3 9 6 © Cambridge University Press & Assessment 2024 Page 16 of 16

5(a) r4 =6r2sincos M1 Substitutes x=rcos, y=rsin and applies sin2=2sincos. r2 =3sin2 A1 AG. 2 5(b) B1 Correct position and symmetrical about = 1π. 4 B1 Single correct loop. =0 3 B1 States maximum distance or labels sketch. ( ) ( ) Allow 3,1π but not 1π, 3 . Allow 3sf. 4 4 3 5(c) 3 π 2sin2d= 3  −1cos2  π 2 M1 Forms 1 2 r2 d. (Allow with wrong limits.) 2 0 2 2 0 3 A1 2 2 © Cambridge University Press & Assessment 2024 Page 10 of 15 5(d) 1 1 B1 y=32sin22sin sin 1 2 2cos+sin−1 2 2cos2sin=0 M1 A1 Sets dy =0. d 2tan M1 Applies suitable trigonometric identity. sin2cos+cos2sin=0tan2=−tan =−tan 1−tan2 Accept sin3=0. = 1π A1 3 5 A1 AEF. 34 =1.40 3 22 6 Question Answer Marks Guidance

7(a) r =a ( x + y) r2 =ax+ay M1 Uses x=rcos and y =rsin to r r eliminate  x2 −ax+ y2 −ay =0 M1 A1 OE Obtains Cartesian equation. (using r2 = x2 + y2.) (x−a)2 +(y−a)2 = a2  Centre (a,a) and radius a B1 2 2 2 2 2 2 4 © Cambridge University Press & Assessment 2024 Page 13 of 16 7(b) B1 Circle B1 Correct position, passing throughO. O =0 a 2 B1 States maximum distance or labels sketch. 3 7(c) cos+sin−=0 M1 cos1.25+sin1.25−1.25=0.01 cos1.26+sin1.26−1.26=−0.002 A1 Shows sign change. 2 © Cambridge University Press & Assessment 2024 Page 14 of 16 7(d) a2   2d+a2 4 3π (cos+sin)2 d M1 Uses area formula on both curves. 2 2 0  A1 Forms area of smaller region enclosed by C and C with correct limits. 1 2  3π M1 Applies relevant identities for the circle a2  2d+ a2 4 1+2sincosd 2 2 integral to produce integrable form. 0   3π A1 For correct integration of the circle part. a2 13 +a2 +sin24 2 3  0 2    a23 + a2 ( 3π+ 1 −−sin2 ) = a2 ( 3 + 3π−+ 1cos2 ) A1 Integrates spiral and substitutes correct 6 2 4 2 2 3 4 2 limits. AG. a2 π− a2 ( 3 + 3π−+ 1cos2 ) M1 Forms area of larger region enclosed by 2 2 3 4 2 C and C . 1 2 a2 ( −3 + 1 π+− 1cos2 ) A1 2 3 4 2 © Cambridge University Press & Assessment 2024 Page 15 of 16 7(d) Alternative method for question 7(d) a2   (cos+sin)2 d−a2   2d M1 Uses area formula on both curves. 2 −1π 2 0 4 A1 Forms area of larger region enclosed by C and C with correct limits. 1 2   M1 Applies relevant identities for the circle a2  1+sin2 d−a2  2d 2 −1π 2 0 integral to produce integrable form. 4 a2− 1cos2 − a2 13  A1 For correct integration of the circle part. 2 2 −π 4 2 3  0 a2 (− 1cos2+ 1 π)− a23 = a2 ( −3 + 1 π+− 1cos2 ) A1 Integrates spiral and substitutes correct 2 2 4 6 2 3 4 2 limits. Answer not given. a2 π− a2 ( −3 + 1π+− 1cos2 ) M1 Forms area of smaller region enclosed by 2 2 3 4 2 C and C . 1 2 a2 ( 3 + 3π−+ 1cos2 ) A1 AG 2 3 4 2 7 © Cambridge University Press & Assessment 2024 Page 16 of 16

5(a) r4 =6r2sincos M1 Substitutes x=rcos, y=rsin and applies sin2=2sincos. r2 =3sin2 A1 AG. 2 5(b) B1 Correct position and symmetrical about = 1π. 4 B1 Single correct loop. =0 3 B1 States maximum distance or labels sketch. ( ) ( ) Allow 3,1π but not 1π, 3 . Allow 3sf. 4 4 3 5(c) 3 π 2sin2d= 3  −1cos2  π 2 M1 Forms 1 2 r2 d. (Allow with wrong limits.) 2 0 2 2 0 3 A1 2 2 © Cambridge University Press & Assessment 2024 Page 10 of 15 5(d) 1 1 B1 y=32sin22sin sin 1 2 2cos+sin−1 2 2cos2sin=0 M1 A1 Sets dy =0. d 2tan M1 Applies suitable trigonometric identity. sin2cos+cos2sin=0tan2=−tan =−tan 1−tan2 Accept sin3=0. = 1π A1 3 5 A1 AEF. 34 =1.40 3 22 6 Question Answer Marks Guidance

5(a) B1 Spiral shape on [0, 2π], with correct orientation, starting at pole and with one other point of intersection with the initial line. =0 B1 r strictly increasing, correct gradient at points of intersection with the initial line. 2 5(b) 1 2π 2e 1 4 d B1 1 2π r2d 2 2 0 0 = 142e 1 4  2π −1 2π 8e 1 4 d M1 A1 Integrates by parts. 2  2 0 0 = 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. 2      0 0 0 0 = ( 8π2 −32π+64 ) e 1 2 π −64 A1 6 © Cambridge University Press & Assessment 2025 Page 13 of 18 5(c) 1 B1 Uses y=rsin y=e8 sin dy =e 1 8 cos+sin ( 1e 1 8 +e 1 8  ) M1 A1 Differentiates. d 8 e 1 8 0cos+ (1+1 ) sin=0 A1 AG. 8 5cos5+ (5+1 ) sin5=−0.1399 B1 Shows sign change. 8 and 5.05cos5.05+ (5.05+1 ) sin5.05=0.1336 8 5 Question Answer Marks Guidance

5(a) B1 Spiral shape on [0, 2π], with correct orientation, starting at pole and with one other point of intersection with the initial line. =0 B1 r strictly increasing, correct gradient at points of intersection with the initial line. 2 5(b) 1 2π 2e 1 4 d B1 1 2π r2d 2 2 0 0 = 142e 1 4  2π −1 2π 8e 1 4 d M1 A1 Integrates by parts. 2  2 0 0 = 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. 2      0 0 0 0 = ( 8π2 −32π+64 ) e 1 2 π −64 A1 6 © Cambridge University Press & Assessment 2025 Page 13 of 18 5(c) 1 B1 Uses y=rsin y=e8 sin dy =e 1 8 cos+sin ( 1e 1 8 +e 1 8  ) M1 A1 Differentiates. d 8 e 1 8 0cos+ (1+1 ) sin=0 A1 AG. 8 5cos5+ (5+1 ) sin5=−0.1399 B1 Shows sign change. 8 and 5.05cos5.05+ (5.05+1 ) sin5.05=0.1336 8 5 Question Answer Marks Guidance

7(a) −esinsin+esincos2=0 M1 dr Finds and sets equal to 0. d cos2−sin=0 A1 1−sin2−sin=0 M1 Makes it a quadratic in sin and attempts to solve 1( ) A1 Correct solution. Condone 0.666 sin= −1+ 5 2 (1.208,0.666) A1 Must be to 3 decimal places 5 7(b) x=e1 2 sincos2 3  M1 Uses x=rcos dx dx =−3e 1 2 sincos 1 2sin+ 1cos 5 2e 1 2 sin=0 Finds and sets equal to 0. d 2 2 d −3sin+cos2=0 A1 −3sin+1−sin2=0 M1 Makes it a quadratic in sin and attempts to solve. 1( ) A1 Correct solution. Condone 0.308 sin= −3+ 13 2 (1.136,0.308) A1 If already penalised for rounding to 3 decimal places in part (a) then allow to 3 s.f. 5 © Cambridge University Press & Assessment 2025 Page 17 of 19 7(c) B1 Closed loop passing through O and one other point on the initial line. Does not go into 2nd and 3rd quadrants. More in the 1st than 4th quadrant. B1 −1π0.666correct. Tangent needs to 2 have positive gradient at the intersection (1,0). B1 0.666 1πcorrect. 2  Graph should approach the line = 2 O =0 3 © Cambridge University Press & Assessment 2025 Page 18 of 19 7(d) π B1 Condone absence of dθ but must include the 12 esincosd 2 −π limits. 2 π M1 Recognises integral. = 1esin2 2  −π 2 = 1 ( e−e−1) A1 Must be exact. ISW after correct answer seen. 2 3 © Cambridge University Press & Assessment 2025 Page 19 of 19

6(a) B1 Spiral shape on [0, 2π], with correct orientation, starting at pole and with one other point of intersection with the initial line. =0 B1 r increasing, correct gradient at points of intersection with the initial line. Maximum distance of C from the pole is a. B1 3 6(b) 1a2 2π tan2(1 ) d M1 Uses 1r2d with correct limits. 2 8 2 0 = 1a2 2π sec2(1 ) −1d= 1a28tan (1 ) − 2π M1 A1 Uses tan2 A=sec2 A−1. 2 0 8 2  8  0 = 1a2(8−2π)=a2(4−π) A1 Accept 0.858a2 if M1 A1 earned. 2 4 © Cambridge University Press & Assessment 2025 Page 14 of 18 6(c) y=atan (1 ) sin B1 Uses y=rsin seen or implied. 8 dy =a ( tan(1)cos+1sinsec2(1)) =0 M1 A1 Sets derivative equal to zero. d 8 8 8 cos (1 ) sin (1 ) cos+1sin=0sin (1 ) cos+1sin=0 M1 Uses sin2A=2sinAcosAin their derivative. 8 8 8 4 4 4sin (1 ) cos+sin=0 A1 AG. 4 4sin (1(4.95) ) cos4.95+sin4.95=−0.0822 and B1 Shows sign change. 4 4sin (1(5) ) cos5+sin5=0.118 4 6 Question Answer Marks Guidance

6(a) B1* Initial line drawn, pole clear and one loop in the first quadrant. DB1 Symmetrical with correct shape at extremities. = 1π. B1 States the equation of the line of 6 symmetry. 3 6(b) 1 1 3 π sin23d M1 Uses 1 2 r2d with correct limits. 2 0 =1 1 3 π 1−cos6d=1−1sin6 1 3 π M1 A1 Applies double angle formula correctly 4 0 4  6  0 and integrates. = 1 π A1 CWO 12 4 6(c) y=3sin2−4sin4 B1 Uses y=rsin. 6sincos−16sin3cos=0 M1* Differentiates correct function and sets equal to 0. Might see completing the square. sincos06−16sin2=0 A1 sin2=3 DM1 Solves quadratic in sin. 8 y= 9 . A1 16 5 © Cambridge University Press & Assessment 2025 Page 14 of 17 6(d) r4 =rsin ( 3r2 −4r2sin2 ) M1 3y y3 Uses y=rsin r = −4 r r3 ( x2 + y2)2 = y ( 3x2 −y2) M1 Uses r2 =x2 + y2 A1 Rearranges to a form with no fractions. 3 Question Answer Marks Guidance

5(a) B1 Initial line drawn and correct shape. Maximum distance of C from the pole is 1. B1 2 5(b) 1 1 8 π tan2d M1 Uses 1 2 r2d with correct limits. 2 0 = 1 1 8 πsin2 d=−1lncos21 8 π M1 A1 Integrates a correct function to get 2 0 cos2 4 0 −1lncos2 condoning the lack of 2 1 limits and factor of for M1A1. 2 =−1ln 2 =1ln2 A1 OE, must be exact. ISW 4 2 8 4 © Cambridge University Press & Assessment 2025 Page 13 of 19 5(c) 2tan M1 Uses relevant identity for tan2. r2 = 1−tan2 2 (y) M1 For the M mark uses either 2xy x2 + y2 = x = y 1− y2 x2 −y2 r= x2 +y2 or tan= . x2 x A1 y Use both r= x2 +y2 and tan= x correctly. ( x2 + y2)( x2 −y2) =2xy A1 At least one intermediate step before the answer given. x4 −y4 =2xy Needs to be in this form as answer given. x4 −2xy− y4 =0. 4 5(d) 1( cos1π )( sin1π ) −1ln2 M1 Area of triangle minus their (b). 2 8 8 8 1 2−1ln2 A1 By double angle formula or otherwise. 8 8 2 Question Answer Marks Guidance

6(a) B1* Initial line drawn, pole clear and one loop in the first quadrant. DB1 Symmetrical with correct shape at extremities. = 1π. B1 States the equation of the line of 6 symmetry. 3 6(b) 1 1 3 π sin23d M1 Uses 1 2 r2d with correct limits. 2 0 =1 1 3 π 1−cos6d=1−1sin6 1 3 π M1 A1 Applies double angle formula correctly 4 0 4  6  0 and integrates. = 1 π A1 CWO 12 4 6(c) y=3sin2−4sin4 B1 Uses y=rsin. 6sincos−16sin3cos=0 M1* Differentiates correct function and sets equal to 0. Might see completing the square. sincos06−16sin2=0 A1 sin2=3 DM1 Solves quadratic in sin. 8 y= 9 . A1 16 5 © Cambridge University Press & Assessment 2025 Page 14 of 17 6(d) r4 =rsin ( 3r2 −4r2sin2 ) M1 3y y3 Uses y=rsin r = −4 r r3 ( x2 + y2)2 = y ( 3x2 −y2) M1 Uses r2 =x2 + y2 A1 Rearranges to a form with no fractions. 3 Question Answer Marks Guidance

6(a) B1* Initial line drawn and (1,0) joined to O by a single arc lying in both of the first two quadrants and no others. DB1 First B1 awarded and also correct shape at extremities and r decreasing. 2 6(b) 1 π cos2 1d M1 Uses 1r2d with correct limits. 2 2 2 0 = 1 π cos+1d= 1sin+π M1 A1 Applies given identity and integrates. 4 0 4 0 = 1π A1 4 4 © Cambridge University Press & Assessment 2025 Page 16 of 20 6(c) y=cos1sin B1 Uses y=rsin. 2 cos1cos−1sinsin1=0 M1* Differentiates. 2 2 2 A1 Correct and equal to zero. cos1 ( 2cos2 1−1 ) −sin2 1cos1=0 DM1 2 2 2 2 cos102cos21−1−sin21=0 2 2 2 2cos2 1−1− ( 1−cos2 1 ) =03cos2 1−2=0 Forms an equation in one trig ratio using correct 2 2 2 formulae. (May be in terms of sin or tan) 2 A1 1 cos2 1= Solving equation to find one of sin2 1= 2 3 2 3 1 2 2 1 tan2 1= sin= cos= 2 2 3 3 y=2cos2 1sin1=2 (2) 1−2 = 4 1 A1 Accept 4 3. 2 2 3 3 3 3 9 6 Question Answer Marks Guidance

7(a) EITHER © Cambridge University Press & Assessment 2026 Page 21 of 23 y 1 t a a n 2 12 s i n   =  +  (B1 Uses y r s i n  = d d y 1 t a a n 2 12 c o s 12 s i n 1 t a a n 2 12 ( 2 ( 1 a s t e a c n 2 2 2 ) )2          =  +  −  +  − + *M1 Differentiates using product rule d d y ( a ( 1 t a n 2 ) ) 12 c o s 12 1 1 t a n a t a 2 n 2 32 ( 2 a s e c 2 2 ) s i n          = + − + +  − A1 or quotient rule. Must be using correct function. c o s s e 1 c 2 2 t a n s i 2 n c o s c o s 2 ( s i s n i n 2 c o s 2 ) 0           − + = − + = DM1 A1 Puts derivative = 0 and rearranges. AG. Intermediate steps must be shown. c o s ( 0 .5 ) − c o s ( 2  0 .5 ) ( s i n s i n ( 2 (  0 .5 0 .5 )) + c o s ( 2  0 .5 ) ) = 0 .2 3 5 and c o s ( 0 .6 ) − c o s ( 2  0 .6 ) ( s i n s i n ( 2 (  0 .6 0 .6 )) + c o s ( 2  0 .6 ) ) = − 0 .3 7 9 B1) Shows sign change. Values calculated to at least one decimal place. OR Alternative method for 7(a) y 2 1 t a a n 2 s i n 2   =  +  (B1 Uses y r s i n  = −2 *M1 Differentiates using product rule  dy  1   1   2y  =   2asincos−asin2   sec22  d 1+tan2 1+tan2 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(a) © Cambridge University Press & Assessment 2026 Page 22 of 23 2 y d d y ( ( 1 t a n 2 ) 2 a s i n c o s 2 a s i n 2 s e c 2 2 ) 1 t 1 a n 2 2          = + −  +  A1 or quotient rule. Must be using correct function. sec22sin sin DM1 Puts derivative = 0 and rearranges. cos− =cos− =0 1+tan2 cos2(sin2+cos2) A1 AG. Intermediate steps must be shown. sin(0.5) cos(0.5)− =0.235 and cos(20.5)( sin(20.5)+cos(20.5)) c o s ( 0 .6 ) − c o s ( 2  0 .6 ) ( s i n s i n ( 2 (  0 .6 0 .6 )) + c o s ( 2  0 .6 ) ) = − 0 .3 7 9 B1) Shows sign change. Values calculated to at least one decimal place. 6 7(b) B1 Initial line and pole clear. Curve in first quadrant through ( a , 0 ) and O. Condone pole not labelled but must be clearly identifiable. ( a , 0 ) not needed. B1 Correct shape (gradient at end points and approximately symmetrical) Maximum point should lie between 40% and 60% of width of the arc. 2 O ( a , 0 ) Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 23 of 23 12 14 0 π r 2 d a2 14 0 π 1 t 1 a n 2 d     =  + B1 Using correct formula and correct limits. v t a n 2 d d v 2 s e c 2 2    =  = M1 Applies given substitution. Condone d d v s e c 2 2   =  1 4 π 1   1 1 M1 Clearly applies a d= a dv 2 1+tan2 4 1+v sec22 0 0 s e c 2 2 1 t a n 2 2   = + Allow sec22=1+v2written explicitly. a4 0 1 1 v 1 t a 1 n 2 2 d v a4 0 ( 1 v ) 1( 1 v 2 ) d v  =   + + =   + + A1 Some working must be shown, including calculations supporting change of limits. AG. ( 1 + v ) 1( 1 + v 2 ) = 12  v 1 + 1 − v 2 v − + 1 1  *M1 Partial fractions of correct form. = 12  v 1 + 1 − v 2 v + 1 + v 2 1 + 1  A1 and separates − v v 2 + + 1 1 . a8  ln ( v + 1 ) − 12 ln ( v 2 + 1 ) + t a n − 1 v   0 DM1 Applies all three formulae from MF19 to a three-term expression from above. Follow through their constants from line above. A1 Correct.  A1 The four previous marks must have been   v+1   aln +tan−1v = 1 πa awarded and logarithms must be 8    v2 +1    16 combined for this mark to be awarded. 0 9 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

7(a) EITHER © Cambridge University Press & Assessment 2026 Page 21 of 23 y 1 t a a n 2 12 s i n   =  +  (B1 Uses y r s i n  = d d y 1 t a a n 2 12 c o s 12 s i n 1 t a a n 2 12 ( 2 ( 1 a s t e a c n 2 2 2 ) )2          =  +  −  +  − + *M1 Differentiates using product rule d d y ( a ( 1 t a n 2 ) ) 12 c o s 12 1 1 t a n a t a 2 n 2 32 ( 2 a s e c 2 2 ) s i n          = + − + +  − A1 or quotient rule. Must be using correct function. c o s s e 1 c 2 2 t a n s i 2 n c o s c o s 2 ( s i s n i n 2 c o s 2 ) 0           − + = − + = DM1 A1 Puts derivative = 0 and rearranges. AG. Intermediate steps must be shown. c o s ( 0 .5 ) − c o s ( 2  0 .5 ) ( s i n s i n ( 2 (  0 .5 0 .5 )) + c o s ( 2  0 .5 ) ) = 0 .2 3 5 and c o s ( 0 .6 ) − c o s ( 2  0 .6 ) ( s i n s i n ( 2 (  0 .6 0 .6 )) + c o s ( 2  0 .6 ) ) = − 0 .3 7 9 B1) Shows sign change. Values calculated to at least one decimal place. OR Alternative method for 7(a) y 2 1 t a a n 2 s i n 2   =  +  (B1 Uses y r s i n  = −2 *M1 Differentiates using product rule  dy  1   1   2y  =   2asincos−asin2   sec22  d 1+tan2 1+tan2 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(a) © Cambridge University Press & Assessment 2026 Page 22 of 23 2 y d d y ( ( 1 t a n 2 ) 2 a s i n c o s 2 a s i n 2 s e c 2 2 ) 1 t 1 a n 2 2          = + −  +  A1 or quotient rule. Must be using correct function. sec22sin sin DM1 Puts derivative = 0 and rearranges. cos− =cos− =0 1+tan2 cos2(sin2+cos2) A1 AG. Intermediate steps must be shown. sin(0.5) cos(0.5)− =0.235 and cos(20.5)( sin(20.5)+cos(20.5)) c o s ( 0 .6 ) − c o s ( 2  0 .6 ) ( s i n s i n ( 2 (  0 .6 0 .6 )) + c o s ( 2  0 .6 ) ) = − 0 .3 7 9 B1) Shows sign change. Values calculated to at least one decimal place. 6 7(b) B1 Initial line and pole clear. Curve in first quadrant through ( a , 0 ) and O. Condone pole not labelled but must be clearly identifiable. ( a , 0 ) not needed. B1 Correct shape (gradient at end points and approximately symmetrical) Maximum point should lie between 40% and 60% of width of the arc. 2 O ( a , 0 ) Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 23 of 23 12 14 0 π r 2 d a2 14 0 π 1 t 1 a n 2 d     =  + B1 Using correct formula and correct limits. v t a n 2 d d v 2 s e c 2 2    =  = M1 Applies given substitution. Condone d d v s e c 2 2   =  1 4 π 1   1 1 M1 Clearly applies a d= a dv 2 1+tan2 4 1+v sec22 0 0 s e c 2 2 1 t a n 2 2   = + Allow sec22=1+v2written explicitly. a4 0 1 1 v 1 t a 1 n 2 2 d v a4 0 ( 1 v ) 1( 1 v 2 ) d v  =   + + =   + + A1 Some working must be shown, including calculations supporting change of limits. AG. ( 1 + v ) 1( 1 + v 2 ) = 12  v 1 + 1 − v 2 v − + 1 1  *M1 Partial fractions of correct form. = 12  v 1 + 1 − v 2 v + 1 + v 2 1 + 1  A1 and separates − v v 2 + + 1 1 . a8  ln ( v + 1 ) − 12 ln ( v 2 + 1 ) + t a n − 1 v   0 DM1 Applies all three formulae from MF19 to a three-term expression from above. Follow through their constants from line above. A1 Correct.  A1 The four previous marks must have been   v+1   aln +tan−1v = 1 πa awarded and logarithms must be 8    v2 +1    16 combined for this mark to be awarded. 0 9 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) B1 Initial line drawn and correct parabolic shape (not vertical at furthest point), starts from the pole. [Maximum distance of C from the pole is] © Cambridge University Press & Assessment 2026 Page 14 of 20 2 ( π3 ) 12 . B1 Accept AWRT 2.05 2 5(b) 12 13 0 π s e c 2 d     M1 Uses 12 r 2 d   with correct limits. Condone omission of d. 12  t a n  130 π 12 13 0 π t a n d     = −  M1 Integrates sec2by parts. A1 Allow omission of the half and limits for these marks. 12  t a n l n s e c  13 0 π    = − A1 OE. Must have the half and the limits for this mark. ( ) A1 OE, must be exact. ISW = 1 1π 3−ln2 2 3 5 0  = Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(c) r2cos2= x2 = M1 Uses © Cambridge University Press & Assessment 2026 Page 15 of 20 x r c o s  = . x 2 = t a n − 1 y x M1 y Applies tan= x y tan ( x2) = x y= xtan ( x2) A1 AG. Must see an intermediary step in the working before the answer given 3 5(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. 16 π 3 − 12 ( 13 π 3 − l n 2 ) M1 Area of triangle minus their answer to (b) (their answer in (b) can be a decimal). 12 l n 2 A1 OE 3 Question Answer Marks Guidance

7(a) ( 2 ) 32 ( c o s ) ( s i n )   r = r r M1 Uses at least two of x 2 + y 2 = r 2 , c o s  x = r and s in  y = r . 2 ( 12 s i n 2 ) 0  r r − = M1 Uses correct identity. s i n 2 2 s i n c o s    = y r 0r = 1sin2 A1 AG. 2 3 x Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) (1sin2)2 = 1 2 (−1π)2 +5 4 Or © Cambridge University Press & Assessment 2026 Page 23 of 25 12 s i n 2 ( 14 1 π ) 2 5 12   =  − +  M1 Sets r values equal. 1sin22 ( (−1π)2 +5 ) =1sin22 ( (−1π)2 +5 ) =4 A1 AG. 4 4 4 Intermediate working must be shown. We need to see e.g. ¼ or 22 . s i n 2 ( 2  0 .5 ) ( ( 0 .5 − 14 π ) 2 + 5 ) − 4 = − 0 .4 s i n 2 ( 2  0 .6 ) ( ( 0 .6 − 14 π ) 2 + 5 ) − 4 = 0 .4 B1 Shows sign change, calculates to at least 1dp. –0.401958833 0.373343553 No conclusion required. Allow 3.598 < 4, 4.3733>4 3 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) B1 Initial line drawn and petal shape for © Cambridge University Press & Assessment 2026 Page 24 of 25 C 1 in the first quadrant, starting and ending at the pole. Ignore graph in other quadrants for this B1. B1 Convex arc shape for C 2 with the pole on the concave side. Ignore graph in other quadrants for this B1. B1 Line of symmetry labelled on the diagram or equation given and sketch symmetrical about 14 π .  = Accept labelled as 14 π . Both curves labelled. Graph only in first quadrant. B1 C Symmetrical, rounded end, 1 apparently tangential to 0  = and π 2  = at the pole. C Symmetrical, angle between curve 2 and positive direction of =0 not more than π 2 . A line drawn vertically from the start point of the arc would not intersect the petal. Graph only in first quadrant. 4 o C 14 2 π  C 1 = 0  = Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(d)   sin22d= 1  1−cos4d= 1−1sin4  M1 Integrates   0 2 0 2 4 0 © Cambridge University Press & Assessment 2026 Page 25 of 25 s i n 2 2  using correct identity. Allow M1 if at most one numerical or sign error in the integration step. A1 (Seen somewhere, condone incorrect/missing limits.) 1π 1π M1 4 1 1  −1π4 Integrates  d= tan−1  4   (−1π)2 +5 5  5   4  ( 14 1 π ) 2 5  − + . For M1 must have t a n 1 ( 14 π )  p − k − where p and k are numerical. A1 A1 for fully correct. (Seen somewhere, condone incorrect/missing limits.) 14 0 s i n 2 2 d 14 π ( 14 1 π ) 2 5 d        +  − + M1 Forms area of R with correct limits. There are other possibilities. 18 ( 14 s i n 4 ) 15 t a n 1 ( 15 ( 14 π ) )    = − − − − A1 AG. Use of limits seen and all previous marks must have been awarded. =1(−1sin4)+1 5tan−1( 1 5(1π−)) 8 4 5 5 4 6 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/