Continuous random variables

3(a)  9 2 x1.5 dx B1 f ( x )= 2 x, correct expression as integrand. 81 81 0 2 2 M1 Integrate, FT only on their PDF expression. × x2.5 81 5 12 A1 =2.4 5 3 3(b) 9 M1  2 x2 dx−( their a )2 81 0 2 93 6 A1 × −2.42 = 81 3 25 2 3(c) 1 M1 G(y) = y6 81 2 M1 g(y) = y5 27 Fully correct including ‘for 0  y  39 and 0 otherwise’ A1 Condone 2.08 3 © UCLES 2021 Page 8 of 12

3(a)  9 2 x1.5 dx B1 f ( x )= 2 x, correct expression as integrand. 81 81 0 2 2 M1 Integrate, FT only on their PDF expression. × x2.5 81 5 12 A1 =2.4 5 3 3(b) 9 M1  2 x2 dx−( their a )2 81 0 2 93 6 A1 × −2.42 = 81 3 25 2 3(c) 1 M1 G(y) = y6 81 2 M1 g(y) = y5 27 Fully correct including ‘for 0  y  39 and 0 otherwise’ A1 Condone 2.08 3 © UCLES 2021 Page 8 of 12

6(a) 1 M1 Integrate both parts (all powers of x increase by 1), allow M1 if x 0x<1,  no constant term. 8 F(x)=  A1 Both parts correct, with domains. 1  1  1   8x− x2  − 1  x  8. Any equivalent form, for example: 2 x− 1 x2 − 1 . 28 2  7 7 56 7 F(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1 Alternative method for question 6(a) 1 M1 Integrate both parts (all powers of x increase by 1), allow M1 if x 0x<1,  no constant term. 8 F(x)=  A1 Both parts correct, with domains.   1− 5 1 6 ( 8−x )2 1  x  8. Any equivalent form, for example: 1 ( 16x−x2 −8 ) . 56 F(x)= 0 for x < 0 and F(x) = 1 for x > 8. A1 3 6(b) 5 1  1  1 5 *M1 Correct method and attempt to simplify, ft their part (a). F(a) = leading to 8a− a2  − = 7 28 2  7 7 a2 −16a+48=0 DM1 Obtain quadratic and solve. [ ] a=4 a=12 a = 4 A1 Correct single answer, WWW. 3 © UCLES 2021 Page 11 of 13 6(c) 1 M1 Substitute x = y3 into both parts of their F(x) from part (a). y3 0  y <1,  A1 Correct functions in terms of y, any equivalent form. 8 G(y)=  1 ( ) 1  16y3 − y6 − 1 y 2. 56 7 0  y < 1 and 1  y  2seen in CDF or PDF A1 Condone  instead of <. 3 M1 Differentiate their G(y), all powers of y decrease by 1. y2 0  y< 1,  A1 Must include ‘otherwise’. 8   3 ( ) g(y)=  8y2 − y5 1  y<2, 28   0 otherwise.   5 © UCLES 2021 Page 12 of 13 6(c) Alternative method for question 6(c) 1 1 −2 M1 y= x3, dy= x 3dx, dx=3y2dy 3 0x<1, f ( x ) dx= 1 dx= 3 y2dy M1 8 8 g ( y )= 3 y2 A1 8 1x8, f ( x ) dx= 1 ( 8−x ) dx= 1 ( 8−y3 ) 3y2dy A1 28 28 g ( y )= 3 ( 8y2 −y5 ) 28 0  y < 1 and 1  y  2 seen in CDF or PDF A1 5 © UCLES 2021 Page 13 of 13

2(a) ìï1+x -1≤x≤0, M1 Differentiation attempted. ï ïï f(x)=í1-x 0<x≤1, A1 All correct, including 0 otherwise. ï ï ï0 otherwise. ïî 2 2(b)  1 1 1  1 7 1 M1 Can do by integration with correct limits. P − X = F −F − = −        2 2 2  2 8 8 3 A1 4 2 2(c) 0 1 M1 x2 ( 1+x ) dx +x2 ( 1−x ) dx −1 0 1 1  1 1  x3 + x4 + x3 − x4     3 4  3 4  1 1 1 A1 + = 12 12 6 2 © UCLES 2021 Page 7 of 14 2(d) 0 1 2 M1 Complete method with their mean squared explicit. x4 ( 1+x ) dx +x4 ( 1−x ) dx−   1  6 −1 0 1 1  1 1  1 x5 + x6 + x5 − x6 −     5 6  5 6  36 1 1 7 A1 − = 15 36 180 2 Question Answer Marks Guidance

3(a) 1  1  2  1  *M1 Correct expression integrated and equated to 1. Total prob = 1, so   a+ x  dx+  2a− x  dx=1 Powers of x correct.  5   5  0 1  1   1  ax+ x2 + 2ax− x2 = 1      10   10  1 DM1 Limits used and attempt to solve. 3a− =1 5 2 A1 Solve to give a = 5 3 3(b)  1 1( 2+x ) x2 dx+ 2 1( 4−x ) x2 dx M1 C orrect expressions integrated, FT their a. 5 5 0 1 12 1  14 1  x3+ x4 + x3− x4     53 4  53 4  13 A1 11 67 + 10 60 60 2 © UCLES 2021 Page 8 of 13 3(c) ìï 0 x<0 M1 Integration of their PDF. ï ï ï 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. F(x)=í ï ï ï ï ï 1 5 çè æ ç ç4x- 1 2 x2-1 ÷ ø ö ÷ ÷ 1x2, ï ï ï 1 x>2. ïî =0 ( x<0 ) and =1 (x>2) A1 First and last parts correct and all domains correct, with no gaps. 3 © UCLES 2021 Page 9 of 13

2(a) ìï1+x -1≤x≤0, M1 Differentiation attempted. ï ïï f(x)=í1-x 0<x≤1, A1 All correct, including 0 otherwise. ï ï ï0 otherwise. ïî 2 2(b)  1 1 1  1 7 1 M1 Can do by integration with correct limits. P − X = F −F − = −        2 2 2  2 8 8 3 A1 4 2 2(c) 0 1 M1 x2 ( 1+x ) dx +x2 ( 1−x ) dx −1 0 1 1  1 1  x3 + x4 + x3 − x4     3 4  3 4  1 1 1 A1 + = 12 12 6 2 © UCLES 2021 Page 7 of 14 2(d) 0 1 2 M1 Complete method with their mean squared explicit. x4 ( 1+x ) dx +x4 ( 1−x ) dx−   1  6 −1 0 1 1  1 1  1 x5 + x6 + x5 − x6 −     5 6  5 6  36 1 1 7 A1 − = 15 36 180 2 Question Answer Marks Guidance

3(a) 1  1  B1 Integrate and equate to 1, no errors seen. k2x2  x3k 6x x2 1   3  2   8  40 k 8 3618122  k 1    3  3 3 k  40 1 3(b) 2 6 M1* Attempt at E(X 2), correct limits, integrated. E(X 2) = k  4x3 x4 dxk  6x2 x3 dx = 0 2  1   1  k x4  x5 k 2x3 x4 = k9.6967.92      5   4  Var(X) = E(X2) – E2(X) = their (7.92)2.52 depM1 Use of correct formula, with their E(X2). 1.67 A1 3 © UCLES 2022 Page 8 of 13 3(c) 2 m 1 M1* Correct expression and attempt to integrate. kx4xdxk6x 2 0 2 2 2 B1 May be implied kx4xdx 5 0 2  1  1 depM1 First integral and 1 may be seen combined as 1 . k  6x x2   2 10 5  2  2 1 1 Obtain quadratic equation. 6m m2 122 2 10k 68 m2 12m 0 3 2 A1 Single answer. m6 30 3 2 m6 30 2.35 3 4 Question Answer Marks Guidance

3(a) 1  1  B1 Integrate and equate to 1, no errors seen. k2x2  x3k 6x x2 1   3  2   8  40 k 8 3618122  k 1    3  3 3 k  40 1 3(b) 2 6 M1* Attempt at E(X 2), correct limits, integrated. E(X 2) = k  4x3 x4 dxk  6x2 x3 dx = 0 2  1   1  k x4  x5 k 2x3 x4 = k9.6967.92      5   4  Var(X) = E(X2) – E2(X) = their (7.92)2.52 depM1 Use of correct formula, with their E(X2). 1.67 A1 3 © UCLES 2022 Page 8 of 13 3(c) 2 m 1 M1* Correct expression and attempt to integrate. kx4xdxk6x 2 0 2 2 2 B1 May be implied kx4xdx 5 0 2  1  1 depM1 First integral and 1 may be seen combined as 1 . k  6x x2   2 10 5  2  2 1 1 Obtain quadratic equation. 6m m2 122 2 10k 68 m2 12m 0 3 2 A1 Single answer. m6 30 3 2 m6 30 2.35 3 4 Question Answer Marks Guidance

4(a) E( X) x 3  1 1   dx   3 3 x0.5 x1.5 dx   M1 8 x2   8    1 32  A1 x1.5 2x0.5   83  3 2 2  1  A1 2 3  2  1 3 ( = 1.37)   8 3 3  2 3 © UCLES 2022 Page 9 of 13 4(b) 3 1 M1 Integrate. Fx x  8 x 3 1  M1 Change variable throughout in their CDF. Gy  y   8 y    gy 3  y0.5  y1.5 , 1y9 A1 Differentiate to find g(y). 16 0 otherwise A1 1y9 and 0 otherwise, dependent on second M1. 4 4(c) 3 1 M1* Correct method. Fp0.4 so p 0.4 8 p 15p2 16p150 M1 dep Obtain quadratic and solve. 5 5p33p50, p 3 25 A1 40th percentile of Y is 9 3 © UCLES 2022 Page 10 of 13

5(a) 1 3 M1 F(U) = 0.75, so 1− (12−U)2 = 144 4 (12−U)2 =36, 1 2−U =6, A1 U =6 only 2 © UCLES 2022 Page 10 of 12 5(b) 1 B1 Differentiate given PDF. [f(x) =] (12−x) 72 E(X2) = 1  2 1 ( 12x2 −x3) dx = 1   4x3− 1 x4   [= 24] M1 Integrate x2 f (x) with their f (x) . 72 72 4  0 E ( X4) = 1  2 1 ( 12x4 −x5) dx= 1   12 x5 − 1 x6   [= 1382.4] M1 Integrate x4 f (x) with their f (x) . 72 72 5 6  0 Var ( X2) =E ( X4) −E2( X2) [= 1382.4−242] M1 Using correct formula with numerical values possibly unsimplified. 1382.4−576=806(.4) A1 5 5(c)  0 y0, M1 Change the variable: 1− 1 ( 12− y2)2  144 G(y)=   1− 1 ( 12− y2)2 0 y 12, Allow domain unchanged at this stage. 144   1 y  12. 1 ( 12y− y3) M1 Differentiate their G(y) . 36   1 ( 12y− y3) 0 y 12 A1 Fully correct. g(y)=36   0 otherwise © UCLES 2022 Page 11 of 12

4(a) 1 2 B1 Integration and k + 3k/2 = 1 oe seen. k  3 2 kdx+kxdx= kx+ x2 , k+ k =1, k =   2  2 5 0 1 1 4(b)  kx 0 x1, M1 1 1  CDF or integration with x2 + seen. F(x)=k 1 5 5 x2 + 1 x 2  2 5 k 1 3 M1 Method for UQ UQ: x2 + = , 2 5 4  3 11 1 A1 UQ x2 +1=5 , x2 = x= 11    4 4 2 1 5 B1 LQ LQ: kx= , x= 4 8 1 5 A1 CAO IQR = 11− =1.03(3) 2 8 5 © UCLES 2022 Page 9 of 13 4(c) 1 2 k  k  17 M1 With correct limits. E(X) = kx dx+ kx2 dx= x2 + x3 =     2  3  15 0 1 1 2 k  k  49 M1 With correct limits. E(X2) = kx2 dx+kx3 dx= x3 + x4 =     3  4  30 0 1 2 M1 Using correct formula with numerical values. 49 17 Var(X) = −   30 15 157 A1 (=0.349) 450 4 Question Answer Marks Guidance

5(a) 1 3 M1 F(U) = 0.75, so 1− (12−U)2 = 144 4 (12−U)2 =36, 1 2−U =6, A1 U =6 only 2 © UCLES 2022 Page 10 of 12 5(b) 1 B1 Differentiate given PDF. [f(x) =] (12−x) 72 E(X2) = 1  2 1 ( 12x2 −x3) dx = 1   4x3− 1 x4   [= 24] M1 Integrate x2 f (x) with their f (x) . 72 72 4  0 E ( X4) = 1  2 1 ( 12x4 −x5) dx= 1   12 x5 − 1 x6   [= 1382.4] M1 Integrate x4 f (x) with their f (x) . 72 72 5 6  0 Var ( X2) =E ( X4) −E2( X2) [= 1382.4−242] M1 Using correct formula with numerical values possibly unsimplified. 1382.4−576=806(.4) A1 5 5(c)  0 y0, M1 Change the variable: 1− 1 ( 12− y2)2  144 G(y)=   1− 1 ( 12− y2)2 0 y 12, Allow domain unchanged at this stage. 144   1 y  12. 1 ( 12y− y3) M1 Differentiate their G(y) . 36   1 ( 12y− y3) 0 y 12 A1 Fully correct. g(y)=36   0 otherwise © UCLES 2022 Page 11 of 12

6(a) 3  1 x  1 x  M1 + c not required. F(x) 2e2 8e 2 c 28    3  1 x  1 x  9 A1 F(x) 2e2 8e 2  28  14   Correct ranges including 0x2ln3 associated with their F(x) B1 No gaps in range. F(x) = 0 for x0 and 1 for x2ln3 3 6(b) 3  8  M1 y substituted into their F. G(y) 2y 6 28 y  1 y3 B1 Seen anywhere, condone 1yeln3. 3  8  A1 Correct expression, not containing logs. g(y) 2  28 y2  3 6(c) 3  8  3 M1 3 2y 6 Their G(y)  28 y  10 10 5y2 8y200 M1 Obtain from G(y) and solve quadratic equation to find y. y1.35 A1 Single correct answer. 3 © UCLES 2023 Page 11 of 12 6(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. 17.8 A1 89 5 2 © UCLES 2023 Page 12 of 12

6(a) 3  1 x  1 x  M1 + c not required. F(x) 2e2 8e 2 c 28    3  1 x  1 x  9 A1 F(x) 2e2 8e 2  28  14   Correct ranges including 0x2ln3 associated with their F(x) B1 No gaps in range. F(x) = 0 for x0 and 1 for x2ln3 3 6(b) 3  8  M1 y substituted into their F. G(y) 2y 6 28 y  1 y3 B1 Seen anywhere, condone 1yeln3. 3  8  A1 Correct expression, not containing logs. g(y) 2  28 y2  3 6(c) 3  8  3 M1 3 2y 6 Their G(y)  28 y  10 10 5y2 8y200 M1 Obtain from G(y) and solve quadratic equation to find y. y1.35 A1 Single correct answer. 3 © UCLES 2023 Page 11 of 12 6(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. 17.8 A1 89 5 2 © UCLES 2023 Page 12 of 12

1(a) 13 2 1 M1 Integrate f(x), + c not required, at least one power correct. F(x)  x3 3x3c 6 2   1 2 1 1 1 A1 AEF x3  x3  4 2 4 Correct ranges including 1x27 associated with their F(x) A1 No gaps. F(x) = 0 for x 1 and F(x) = 1 for x 27 3 1(b) 1 M1 1 G(y)  y2 2y1  Using y  x3 in their F(x). 4 1 A1 Correct, AEF, 0 otherwise not required. g(y) y1 2 For 1 y3 B1 Seen anywhere, correct variable. 3 1(c) 1 1 M1 Equate their G(y) to ½ and solve to find y.  y2 2y1   , y12 2 OE 4 2 y  1  2 A1 Only. 2 © UCLES 2023 Page 6 of 13

4(a) 2 B1 AG 1 mx2 1 P(X ⩽ 2) = , so   = Reasoning required. 3  2  3 0 4 1 1 m   = , m= 2 3 6 6 6 M1 Attempt to integrate and correct use of correct limits to  k  2  k  1 2   x2 +c  dx= 3 so   − x +cx   =4c+ 3 k = 3 or k+12c=2 form linear equation in k and c. 2 2 k 1 k M1 Matching at x=2. Also, 2m= +c or = +c or 3k+12c=4 4 3 4 1 A1 Solve, k =1, c= 12 4 © UCLES 2023 Page 9 of 13 4(b) L B1 1 1 LQ:  xdx= , L= 3 6 4 0 U M1 Or find and use CDF. May be in terms of m. 1  k  3 UQ: +  +cdx= 3 x2  4 2 k k 5 M1 Attempt at integral and correct use of correct limits. − +Uc+ −2c= Simplify to quadratic equation, may be in terms of k and U 2 12 c. U2 −U −12=0 U =4 A1 IQR = 4− 3 A1 FT FT their UQ – their LQ, exact values only. 5 © UCLES 2023 Page 10 of 13

4(a) 6 B1 AG. 1 1 1 UQ = 4 so cdx= : c6−4= , c= Some indication of reasoning required. 4 4 8 4 4 M1 Integrate and use correct limits to form equation  1 ( 4ax−bx3) dx= 3 , 1   2ax2 − b x4   = 3 , 32a−64b=96 in a and b. 128 4 128 4  4 0 Equate at x=4: a−4b=1 M1 Solve: a=5, b=1 A1 4 4(b) m M1* Attempt at integral and use of correct limits.  1 ( 20x−x3) dx= 1 Could be in terms of a, b, c 128 2 0 Or attempt at relevant part of CDF and set OR 1 F(m)= . 1  b  5 1 F(x)=  2ax2 − x4  = x2 − x4 2 128 4  64 512 Could be in terms of a, b, c. m4 −40m2 +256=0 M1 Simplify to quartic. Could be in terms of a, b, c. m=2 2 A1 Positive answer only. 3 © UCLES 2023 Page 10 of 13 4(c) ( ) 1 4 3 7 6 1 1 M1 Their PDF multiplied by x with correct limits. E X = 20x2 −x2 dx+ x2dx 128 8 Could be in terms of a, b, c. 0 4 4 6 M1 Correct use of correct limits in their integral. 1  5 2 9 1  3 8x2 − x2 + x2 Could be in terms of a, b, c. 128 9  12  May be implied by correct final answer. 0 4 1  1  1 ( ) 4 1 A1 CAO  28 − 210  + 6 6−8 = + 6 =1.67 1.67 following (first) M0 SC B2 128 9  12 9 2 3 © UCLES 2023 Page 11 of 13

4(a) 2 B1 AG 1 mx2 1 P(X ⩽ 2) = , so   = Reasoning required. 3  2  3 0 4 1 1 m   = , m= 2 3 6 6 6 M1 Attempt to integrate and correct use of correct limits to  k  2  k  1 2   x2 +c  dx= 3 so   − x +cx   =4c+ 3 k = 3 or k+12c=2 form linear equation in k and c. 2 2 k 1 k M1 Matching at x=2. Also, 2m= +c or = +c or 3k+12c=4 4 3 4 1 A1 Solve, k =1, c= 12 4 © UCLES 2023 Page 9 of 13 4(b) L B1 1 1 LQ:  xdx= , L= 3 6 4 0 U M1 Or find and use CDF. May be in terms of m. 1  k  3 UQ: +  +cdx= 3 x2  4 2 k k 5 M1 Attempt at integral and correct use of correct limits. − +Uc+ −2c= Simplify to quadratic equation, may be in terms of k and U 2 12 c. U2 −U −12=0 U =4 A1 IQR = 4− 3 A1 FT FT their UQ – their LQ, exact values only. 5 © UCLES 2023 Page 10 of 13

7(a)   x 4x2 0x2, B1 fx4   0 otherwise. (E(X) =)  2 1 4x2 x4 dx 14x3 1x516 ( = 1.067) 04 43 5  15 (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. 04 45 9  0  45  Var( X)EX  E  X 2 used M1 Correct formula used following attempt at integration of both terms. 16 32 2 2 0.0553 A1 2 1 0 1 2 2 5  15 45   4 © Cambridge University Press & Assessment 2024 Page 14 of 15 7(b) F(x) = 1  8x2 x4 0x2 B1 16 G(y) = 1  8y y2 [0 y4] M1 Change variable in their F(x). 16 1 M1 Differentiate their G(y) . gy 82y 0 y4 16  1 A1 Complete and correct.  82y 0y4, gy16   0 otherwise. 4 7(c) 1  8mm2  1 M1 Equate their G(y) to 1 2 . 16 2 [m2 8m80,] m42 2 A1 Single answer in an exact form. 2 © Cambridge University Press & Assessment 2024 Page 15 of 15

7(a)   x 4x2 0x2, B1 fx4   0 otherwise. (E(X) =)  2 1 4x2 x4 dx 14x3 1x516 ( = 1.067) 04 43 5  15 (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. 04 45 9  0  45  Var( X)EX  E  X 2 used M1 Correct formula used following attempt at integration of both terms. 16 32 2 2 0.0553 A1 2 1 0 1 2 2 5  15 45   4 © Cambridge University Press & Assessment 2024 Page 14 of 15 7(b) F(x) = 1  8x2 x4 0x2 B1 16 G(y) = 1  8y y2 [0 y4] M1 Change variable in their F(x). 16 1 M1 Differentiate their G(y) . gy 82y 0 y4 16  1 A1 Complete and correct.  82y 0y4, gy16   0 otherwise. 4 7(c) 1  8mm2  1 M1 Equate their G(y) to 1 2 . 16 2 [m2 8m80,] m42 2 A1 Single answer in an exact form. 2 © Cambridge University Press & Assessment 2024 Page 15 of 15

5(a) x2 M1 Attempt at differentiating both parts of CDF to obtain linear expressions. 6 May be implied by first A1. 8x 12 Sketch A1 Correct shape (Two line segments touch the x-axis and meet at a point above the x- axis). A1 Correct shape with 2 and 8 and at least one of 4 or 1/3 correctly labelled. 3 © Cambridge University Press & Assessment 2024 Page 12 of 15 5(b) 4 8 B1 FT FT their PDF, with correct limits.  1 xx2dx 1 x8xdx 26 412 1 1 x3 x2  1 4x2 1 x3 M1 Attempt at integration with their PDF (not CDF) no limits needed.     6 3 12 3 May be implied by correct answer. 14 A1 Accept 4.67 or any equivalent fraction. 3 3 5(c) LQ: 1 x22  1, x 2 3 B1 Allow 3.73. 12 4 8x2 3 B1 Allow 5.55. UQ: 1  , x8 6 24 4 IQR = UQ – LQ, with values found for both M1 quartiles A1 Exact answer required. 6 6 3 4 © Cambridge University Press & Assessment 2024 Page 13 of 15

4(a) Correct sketch. B1 Label 1 and 5 on x-axis, ignore labels, if any, on y-axis. 1 4(b) kx4 k B1 For either part correct. with limits 0 and 1 gives 4 4  x2  k5x−  with limits 1 and 5 gives 8k 2   k 4 B1 AG +8k =1, k = 4 33 2 4(c)  M1 Integrate both parts. 0 x0   A1 Middle two parts correct AEF.  1 x4 0≤x1  A1 0 and 1 correct. 33 F(x)=  4  5x− 1 x2 − 17 1≤x≤5   33 2 4    1 x5.   3 © Cambridge University Press & Assessment 2024 Page 8 of 11 4(d) 1 m 1 M1* Or integrate m to 5.  kx3dx+ k(5−x)dx= Or use F(x) = 0.5. 0 1 2 4 1 1 1 1 4 25 1  1 M1 Must be a quadratic in m. +5m− m2 −5+ = or −5m+ m2 =     334 2 2 2 33 2 2  2 4  1 17 1 or 5m− m2 − =   33 2 4  2 4m2 −40m+67=0 *DM1 1( ) A1 m= 10− 33 =2.13 2 4 Question Answer Marks Guidance

4(a)  0 x2, M1 Integrate 1 ( x3−3x2 +3x )+c oe.  63   1 F(X)= ( (x−13−1 ) 2≤x≤5, A1 2 63   Correct F(x) (with c =− evaluated).  63  1 x5.   A1 0 and 1 correct, inequalities correct. 3 4(b) G(y) = 1 ( y0.75 −1 ) M1 Use transformation correctly on function . May see 63 1 ( y0.25 +1 )3 − 1 ( y0.25 +1 )2 + 1 ( y0.25 +1 ) − 2 63 21 21 63  1 M1 Differentiate and change variable in domain. y−0.25 1≤y≤256,  84 g(y)= A1 All correct and simplified.  0 otherwise.  3 4(c) 1 ( y0.75 −1 ) =0.5 M1 Equate their G(y) to 0.5 and attempt to solve. 63 y = 104 A1 AWRT 104. 2 © Cambridge University Press & Assessment 2024 Page 9 of 13 4(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.   84 84 84 7  1 1 111 A1 5461 49 2 © Cambridge University Press & Assessment 2024 Page 10 of 13

4(a) Correct sketch. B1 Label 1 and 5 on x-axis, ignore labels, if any, on y-axis. 1 4(b) kx4 k B1 For either part correct. with limits 0 and 1 gives 4 4  x2  k5x−  with limits 1 and 5 gives 8k 2   k 4 B1 AG +8k =1, k = 4 33 2 4(c)  M1 Integrate both parts. 0 x0   A1 Middle two parts correct AEF.  1 x4 0≤x1  A1 0 and 1 correct. 33 F(x)=  4  5x− 1 x2 − 17 1≤x≤5   33 2 4    1 x5.   3 © Cambridge University Press & Assessment 2024 Page 8 of 11 4(d) 1 m 1 M1* Or integrate m to 5.  kx3dx+ k(5−x)dx= Or use F(x) = 0.5. 0 1 2 4 1 1 1 1 4 25 1  1 M1 Must be a quadratic in m. +5m− m2 −5+ = or −5m+ m2 =     334 2 2 2 33 2 2  2 4  1 17 1 or 5m− m2 − =   33 2 4  2 4m2 −40m+67=0 *DM1 1( ) A1 m= 10− 33 =2.13 2 4 Question Answer Marks Guidance

2(a) 5a+ 3a=1, leading to a= 2 . B1 AG, shown convincingly. 2 13 Uses area = 1, b and c must be eliminated. Attempt to solve 2 independent equations in b and c or 3 independent equations in M1 May see any of the following: a, b and c. 3b−39c= 3a, a=b−5c, b−8c=0, 2 2 a −c= , 5a+3b−39c=1. 5−8 2 OR equation of line through (5,a)and (8,0). b=16, c= 2 A1 Correct values imply M1 A1. 39 39 3 2(b) E(X)= 5 2 xdx+ 8 x(16 − 2 x)dx M1 Attempt to integrate xf(x)with their b and c,   0 13 5 39 39 limits seen. May be in terms of a, b, c. 5 8 M1 Correct integrated expression with their b and 1 x2 +8 x2 − 2 x3 13  0 39 117  5 their c, including limits. May be in terms of a, b, c. 43 =3.31 A1 AWRT 3.31. 13 3 2(c) m M1 Uses 0m5 and area = 0.5, may be in terms 2 m=0.5 OR  2 dx =0.5 13 0 13 of a. m=13 =3.25 A1 4 2 © Cambridge University Press & Assessment 2025 Page 10 of 16 2(d) 2 x, 16 x− 1 x2(+k) B1 FT Correct expressions for integrating both pieces 13 39 39 of PDF, seen anywhere, using their b and c. k =−25 B1 FT FT their b and their c to obtain k . 39 0 y0 M1 Both parts with y substituted (condone  2 y 0≤y≤25 missing k ), ignore limits. G(y)= 13 1 3 6 9 y − 3 1 9 y− 3 2 9 5 25 y≤64 A1 FT FT their b, their c and their k .  1 y64 Fully defined including correct domain in terms of y. Domain must cover all reals, may overlap at boundaries. 4 Question Answer Marks Guidance

2(a) 5a+ 3a=1, leading to a= 2 . B1 AG, shown convincingly. 2 13 Uses area = 1, b and c must be eliminated. Attempt to solve 2 independent equations in b and c or 3 independent equations in M1 May see any of the following: a, b and c. 3b−39c= 3a, a=b−5c, b−8c=0, 2 2 a −c= , 5a+3b−39c=1. 5−8 2 OR equation of line through (5,a)and (8,0). b=16, c= 2 A1 Correct values imply M1 A1. 39 39 3 2(b) E(X)= 5 2 xdx+ 8 x (16 − 2 x ) dx M1 Attempt to integrate xf(x)with their b and c,   0 13 5 39 39 limits seen. May be in terms of a, b, c. 5 8 M1 Correct integrated expression with their b and 1 x2 +8 x2 − 2 x3 13  0 39 117  5 their c, including limits. May be in terms of a, b, c. 43 =3.31 A1 AWRT 3.31. 13 3 2(c) m M1 Uses 0m5 and area = 0.5, may be in terms 2 m=0.5 OR  2 dx =0.5 13 0 13 of a. m=13 =3.25 A1 4 2 © Cambridge University Press & Assessment 2025 Page 10 of 15 2(d) 2 x, 16 x− 1 x2(+k) B1 FT Correct expressions for integrating both pieces 13 39 39 of PDF, seen anywhere, using their b and c. k =−25 B1 FT FT their b and their c to obtain k . 39 0 y0 M1 Both parts with y substituted (condone  2 y 0≤y≤25 missing k ), ignore limits. G(y)= 13 1 3 6 9 y − 3 1 9 y− 3 2 9 5 25 y≤64 A1 FT FT their b, their c and their k .  1 y64 Fully defined including correct domain in terms of y. Domain must cover all reals, may overlap at boundaries. 4 Question Answer Marks Guidance

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 0 1  2  0  2  1  attempt to integrate and use correct limits. 1k+(64k−32k)− ( 8k−1k ) =1 2 2 25k=1, leading to k = 1 . A1 AG, shown convincingly. No errors seen. 25 Alternative method for question 3(a) 11k+177k =1 M1 Linear equation in terms of k formed by 2 2 considering areas. 25k=1, leading to k = 1 . A1 AG, shown convincingly. No errors seen. 25 2 3(b) 1 m *M1 Use of definite integrals and attempt at  kx dx+ k(8−x)dx= 1 2 integration to form equation with 0.5 and their k. 0 1 Could be in terms of k. m2−16m+39=0 or DM1 Quadratic equation in m. m=3 A1 Must reject second solution if seen. CAO Alternative method for question 3(b) 1(8−m)k(8−m)= 1 *M1 Use areas to form equation with 0.5 and their k. 2 2 Could be in terms of k. (8−m)2 =25 DM1 Quadratic equation in m. m=3 A1 Must reject second solution if seen. CAO 3 © Cambridge University Press & Assessment 2025 Page 11 of 16 3(c)   1 x2 [0≤x1] M1 Limits not required. Sight of expressions for F(x)= 50 0≤x1 and 1≤x≤8 sufficient. 2 8 5 x− 5 1 0 x2 − 2 7 5 [1≤x≤8] [F(x)=0 for x0 and F(x)=1 for x8] B1 For − 7 . 25   1 y6 [0≤y1] M1 Both parts with y substituted in their F(x). G(y)= 50 Limits not required. Sight of expressions for 2 8 5 y3 − 5 1 0 y6 − 2 7 5 [1≤y≤2] 0≤ y1 and 1≤ y≤2 sufficient. [G(y)=0 for x0 and G(y)=1 for y2] 3 y5 0≤y1 M1 Differentiation of their G(y).  25 Limits not required. Sight of expressions for g(y)=24 y2 − 3 y5 1≤y≤2 25 25 0≤ y1 and 1≤ y≤2 sufficient.  0 otherwise.  A1 Fully correct with correct limits. Alternative method for question 3(c) Let p(y)= 3 y , then q(y)=p−1(y)= y3. M1 Find inverse function. q ' (y)=3y2. A1 Differentiate inverse function. May see 3x 2 3 .   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. g(y)= 25 or g(x)= 25 2 1 5 (8− y3)(3y2) [1≤y≤2] 2 1 5 (8−x)(3x 2 3) [1≤x≤8] M Sig 1 h f t o o r f o e n x e p e r x es p s r i e o s n s s io f n o . r L 0 im ≤ it y s  n 1 ot a r n e d q u 1 ir ≤ ed y . ≤ 2 (g(y)=0 otherwise) (g(x)=0 otherwise) or 0≤x1 and 1≤x≤8 sufficient. 3 y5 0≤y1 A1 Fully correct with correct limits.  25 g(y)=24 y2 − 3 y5 1≤y≤2 25 25  0 otherwise.  5 © Cambridge University Press & Assessment 2025 Page 12 of 16

4(a)  1 kxdx+  2 kx2 dx=1  1kx2 1 +1kx3 2 =1  M1 Linear equation in terms of k formed following attempt to 0 1  2  0 3  1  integrate (all powers correct) and use of correct limits. 1k+1k(8−1) =1 May consider area of triangle (11k ) for first region. 2 3 2 17k =1, k = 6 A1 AG, shown convincingly. No errors seen. 6 17 2 4(b) 0 x0 M1 Attempt to integrate both parts, all powers correct, condone  missing c. 3 x2 0≤x1 F(x)= 17 Sight of expressions for 0≤x1 and 1≤x≤2 sufficient. 2 x3 + 1 1≤x≤2 17 17  1 x2 A1 Both parts correct with c= 1 1 7 . Sight of expressions for 0≤x1 and 1≤x≤2 sufficient. A1 All correct, fully defined including domain. Domain must cover all reals, may overlap at boundaries. 3 4(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. 2 17 17 2 m3 =15, m= 3 15 =1.55 A1 FT FT their non-zero c, to obtain m with 1≤m≤2. 4 4 2 4(d) 1 2 M1 Correct method and attempt to integrate both parts. All powers    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. 0 1 6 ( 1+2−1) = 15 =0.882 A1 CWO, AWRT 0.882. 17 2 17 2 © Cambridge University Press & Assessment 2025 Page 12 of 15

5(a) 4 M1 Equation in terms of k formed following attempt to integrate with 1   2 x 3 2   + 1   2 x  9 =1 correct limits. 163  k 4 0 1 2 (8−0)+ (3−2)=1 24 k OR 1 2 + =1 3 k 2 2 A1 AG, no errors seen. = , k =3 k 3 2 5(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. 16 3 2 Equation may be in terms of k. 0 4 OR 1 9 x−1 2 dx=0.5 3 m 1+ 2 ( m−2 ) = 1 DM1 Integrate and form an equation in m . 3 3 2 OR ( ) 2 3− m = 1 3 2 m= 81 [=5.0625] A1 Accept AWRT 5.06. 16 3 © Cambridge University Press & Assessment 2025 Page 12 of 16 5(c)   1 x 3 2 0≤x4 M1 Attempt to integrate both parts, limits not required. F(x)= 24 May be seen in part 5(a).  2x 1 2 −1 4≤x≤9 3 B1 For constant –1 obtained in expression for 4≤x≤9. May be seen in part 5(a). 1 y3 0≤y2 M1 For changing to y. G(y)=24 Limits not required.  2 y−1 2≤y≤3 3 1 y2 0≤y2 M1 For differentiation, limits not required. 8 g(y)=2 2≤y≤3 3 A1 Fully correct with correct domain covering all reals.  0 otherwise  Alternative method for question 5(c) Using chain rule (or inverse function): M1 M1 for changing to y. G(y)=F ( y2) so g(y)= d F ( y2) =2yf ( y2) . dy A1  2y ( 1 y2 ) 0≤y2 M2 Use of g(y)=2yf ( y2) twice, limits not required.  16  M1 for one expression. g(y)=   1 2y  2≤y≤3   3 y2     1 y2 0≤y2 A1 Fully correct with correct domain covering all reals. 8 g(y)=2 2≤y≤3 3  0 otherwise  5 © Cambridge University Press & Assessment 2025 Page 13 of 16

4(a) a=−1 B1 5 b= 7 B1 25 2 4(b) 1 1≤x4, M1 Attempt at differentiating both parts. 5  f(x)=1 x 4≤x≤6, 25 A1 All correct, fully defined including domain.   0 otherwise. Domain must cover all reals, must not overlap at boundaries. 2 4(c) E ( X2) = 1 4 x2dx+ 1  6 x3dx=1 x3 4 + 1 x4 6 M1 Attempt to integrate x2f(x) with correct limits. 5 1 25 4 15  1 100  4 Can be implied by answer of 73 or 14.6. 5 Var(X)=73−(529)2 M1 Variance formula used with their E ( X2) . 5 150 2.16 A1 3 4(d) 10th percentile: 1x−1 = 1 M1 Award M1 for a correct method to find at least one of the 10th or 90th 5 5 10 percentile. 90th percentile: 1 x2+ 7 = 9 50 25 10 FT their a and b. 10th percentile = 1.5 A1 90th percentile = 31 A1 AWRT 5.57. 3 © Cambridge University Press & Assessment 2025 Page 12 of 15

5(a) 4 M1 Equation in terms of k formed following attempt to integrate with 1   2 x 3 2   + 1   2 x  9 =1 correct limits. 163  k 4 0 1 2 (8−0)+ (3−2)=1 24 k OR 1 2 + =1 3 k 2 2 A1 AG, no errors seen. = , k =3 k 3 2 5(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. 16 3 2 Equation may be in terms of k. 0 4 OR 1 9 x−1 2 dx=0.5 3 m 1+ 2 ( m−2 ) = 1 DM1 Integrate and form an equation in m . 3 3 2 OR ( ) 2 3− m = 1 3 2 m= 81 [=5.0625] A1 Accept AWRT 5.06. 16 3 © Cambridge University Press & Assessment 2025 Page 12 of 16 5(c)   1 x 3 2 0≤x4 M1 Attempt to integrate both parts, limits not required. F(x)= 24 May be seen in part 5(a).  2x 1 2 −1 4≤x≤9 3 B1 For constant –1 obtained in expression for 4≤x≤9. May be seen in part 5(a). 1 y3 0≤y2 M1 For changing to y. G(y)=24 Limits not required.  2 y−1 2≤y≤3 3 1 y2 0≤y2 M1 For differentiation, limits not required. 8 g(y)=2 2≤y≤3 3 A1 Fully correct with correct domain covering all reals.  0 otherwise  Alternative method for question 5(c) Using chain rule (or inverse function): M1 M1 for changing to y. G(y)=F ( y2) so g(y)= d F ( y2) =2yf ( y2) . dy A1  2y ( 1 y2 ) 0≤y2 M2 Use of g(y)=2yf ( y2) twice, limits not required.  16  M1 for one expression. g(y)=   1 2y  2≤y≤3   3 y2     1 y2 0≤y2 A1 Fully correct with correct domain covering all reals. 8 g(y)=2 2≤y≤3 3  0 otherwise  5 © Cambridge University Press & Assessment 2025 Page 13 of 16

2(a) 0 x0 M1 Attempt to integrate both parts, condone missing c , ignore  domains.  2x2 + 4x 0≤x1 ( F(x)= )  9 9 1(x−2)3 +1 1≤x≤2 A1 Both parts correct with correct c , ignore domains. 3  1 x2 B1 Correct domains. Must cover all reals, may overlap at boundaries. Must have 0 for x0 and 1 for x2. 3 2(b) 2m2 +4m=1 M1 Identifies correct interval for their piecewise F(x) to form an 9 9 2 equation of the form equation F(m)= 1 OE. 2 ( ) A1 OE, must clearly state one answer. m= 1 −2+ 13 2 2 2(c) 0 y0 M1 Substitute x= y2 into both parts of their F(x), ignore domains.  ( G(y)= )    1 9 2 ( y y 4 2 + − 9 4 2 y ) 2 3 + 1 1 0 ≤ ≤ y y ≤ 1 2 A1 Correct functions in terms of y, AEF. 3  B1 Correct domain in terms of y. Domain may overlap at boundaries. 1 y 2 Condone missing 0 for y0 but must have 1 for y 2 . 3 © Cambridge University Press & Assessment 2025 Page 10 of 17 2(d) G ( 1 ( −2+ 13 )) =0.379 M1 Evaluates G(their m). 2 0.3790.5 so the median of Y is greater than the median of X . A1 Alternative method for question 2(d) −2+ 13 M1 Expresses m in terms of m . m2 = y x y 2 m2 =m 1 and hence m m . A1 Uses the fact squaring a number between 0 and 1 makes it smaller. y x y x Alternative method for question 2(d) Solves 2y4+4y2 =1. M1 Identifies correct interval for their piecewise G(y) to form 9 9 2 equation of the form G(y)= 1 OE. 2 m =0.8960.803=m A1 0.803 may be seen in part 2(b). y x ( ) ( ) A0 for 1 −2+ 13 0.896 without 1 −2+ 13 evaluated. 2 2 2 © Cambridge University Press & Assessment 2025 Page 11 of 17

4(a) © Cambridge University Press & Assessment 2026 Page 13 of 17 F ( 3 ) = 11 2  3 2 − 1 10 8  3 3 = 0 .5 [Since F ( 3 ) = 0 .5 B1 AG B0 if obtained using , the median of X is 3.] a = − 13 72 . May solve − 1 10 8 x 3 + 11 2 x 2 = 12 but must reject 3  3 3 as they lie outside [0, 3]. 1 4(b) − 13 2  3 2 + 71 6  3 + a = 0 .5 , a = − 13 72 B1 AG, convincingly obtained. F ( b ) = 1  − 13 2 b 2 + 71 6 b − 13 72 = 1 OR F ( b ) − F ( 3 ) = 12   − 13 2 x 2 + 71 6 x ( + a )  b 3 = 0 .5  − 13 2 b 2 + 71 6 b − 33 32 = 12 M1 Use F ( b ) = 1 or F ( b ) − F ( 3 ) = 0 .5 to form quadratic in b. May see b 2 − 1 4 b + 4 9 = 0 OE. b = 7 A1 3 4(c) f ( x ) =  − − 0 2 1 1 x + 3 6 6 7 1 x + 1 6 1 6 x 0 3 o ≤  t h x x e r ≤ ≤ w 3 , 7 , i s e . B1 Both parts differentiated correctly, ignore domain. B1FT FT their b. Fully defined including correct domain, may be in terms of b. Domain must cover all reals, may overlap at boundaries. 2 Trace ID: PC-6B4604D1D5 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/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(d) © Cambridge University Press & Assessment 2026 Page 14 of 17  3 0 1 x  − 1 3 6 x 2 + 1 6 x  d x +  7 3 1 x  − 1 1 6 x + 1 7 6  d x B1FT Correct expression, FT their f ( x ) and limits, may be in terms of b.  16 x − 17 2 x 2  3 0 +  − 11 6 x + 71 6 ln x  7 3 M1 Attempt to integrate both parts of their f ( x )  1 x . Must include natural logarithm term and limits stated (may use b). Implied by AWRT 0.496. 18 + 71 6 ln ( 73 )  = 0 .4 9 6  A1 AWRT 0.496 3 Trace ID: PC-6B4604D1D5 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/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) © Cambridge University Press & Assessment 2026 Page 13 of 17 F ( 3 ) = 11 2  3 2 − 1 10 8  3 3 = 0 .5 [Since F ( 3 ) = 0 .5 B1 AG B0 if obtained using , the median of X is 3.] a = − 13 72 . May solve − 1 10 8 x 3 + 11 2 x 2 = 12 but must reject 3  3 3 as they lie outside [0, 3]. 1 4(b) − 13 2  3 2 + 71 6  3 + a = 0 .5 , a = − 13 72 B1 AG, convincingly obtained. F ( b ) = 1  − 13 2 b 2 + 71 6 b − 13 72 = 1 M1 Use F(b)=1 or OR b F(b)−F(3)= 1 − 1 x2 + 7 x(+a) =0.5− 1 b2 + 7 b−33 = 1 2  32 16  3 32 16 32 2 F ( b ) − F ( 3 ) = 0 .5 to form quadratic in b. May see b 2 − 1 4 b + 4 9 = 0 OE. b = 7 A1 3 4(c) f ( x ) =  − − 0 2 1 1 x + 3 6 6 7 1 x + 1 6 1 6 x 0 3 o ≤  t h x x e r ≤ ≤ w 3 , 7 , i s e . B1 Both parts differentiated correctly, ignore domain. B1FT FT their b. Fully defined including correct domain, may be in terms of b. Domain must cover all reals, may overlap at boundaries. 2 Trace ID: PC-E48FEEF9A8 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/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 4(d)  3 1 1 1   7 1 1 7  B1FT Correct expression, FT their   − x2 + x  dx+  − x+  dx  x 36 6   x 16 16 0 3 © Cambridge University Press & Assessment 2026 Page 14 of 17 f ( x ) and limits, may be in terms of b.  16 x − 17 2 x 2  3 0 +  − 11 6 x + 71 6 ln x  7 3 M1 Attempt to integrate both parts of their f ( x )  1 x . Must include natural logarithm term and limits stated (may use b). Implied by AWRT 0.496. 18 + 71 6 ln ( 73 )  = 0 .4 9 6  A1 AWRT 0.496 3 Trace ID: PC-E48FEEF9A8 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/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) F ( L ) = 0 .2 5 , s o 1 1 2 ( x 2 + 4 x + 4 ) = 14 M1 May be implied by correct answer. L = 3 − 2 A1 2 Trace ID: PC-1E62B2C2D5 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/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) © Cambridge University Press & Assessment 2026 Page 15 of 15  f ( x ) =    16 15[ 0 x x + 2 + 13 11 5   o − 0 t 2 ≤ h ≤ x e r w x ≤ i  s 2 e 0  ]  B1 Both expressions seen.  E ( X 2 ) = =    − 0 12 4 2 2 x 4 x ( + 16 19 x x + 3  13 0 − ) 2 d + x +   12 5 2 0 x x 5 2 + ( 15 14 5 x x 2 3 +  2 0 ) 11 5 d x M1 Attempt to integrate their x 2 f ( x ) . For reference: 29 + 32 22 85 = 42 25 = 1 .6 8  E ( X 4 ) = =  0  −  1  3 6 x 2 x 4 6 ( + 16 x 11 5 + x 5 13  ) 0 − d 2 x + +  2  0 13 5 x x 4 7 ( + 15 2 x 17 5 x + 5 11 5 2  0 ) d x M1 Attempt to integrate their x 4 f ( x ) . For reference: 16 + 2144 = 6992 =4.439365 45 525 1575 Var ( X2) = 6992− (42)2 M1 Using correct formula with their values.   1575 25 1.62 A1 5 6(c) P ( X 2  1 ) = P ( − 1  X  1 ) = F ( 1 ) − F ( − 1 )  = 71 5 − 11 2  M1 For either F(1) or F(–1) evaluated. Could be implied by 7 , 1 , 2 or 1 seen (NFWW). 15 12 15 4 26 30 A1 AWRT 0.383 2 Trace ID: PC-1E62B2C2D5 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/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) [Continuous so] a4=b(−2)2  a=b B1 May be awarded for finding b directly (having already found a). © Cambridge University Press & Assessment 2026 Page 15 of 18 12 ( 5 − 1 )  4 a = 0 .7 5 o r  5 1 a ( x − 1 ) d x = 0 .7 5  8 a = 0 .7 5 M1 Use of area of triangle = 0.75 to form an equation in a. May see an attempt at integral leading to OR  7 b(x−7)2 dx=0.25  8b=0.25 3 5 OR  5 a(x−1)dx+  7 a(x−7)2 dx=1  8a+8a=1 3 1 5 12 a ( 4 2 − 0 2 ) = 34 . Some working may be present on sketch. OR Use of area of upper tail = 0.25 to form an equation in b. OR Use of total area = 1 to form an equation in a (or in b). a = 33 2 OR b = 33 2 A1 AG, convincingly obtained, no errors seen. 3 5(b) 0 x1 M1 Attempt to integrate both parts, all powers  correct, condone missing c, ignore domain. 3 (x−1)2 1≤x≤5   F(x)=   64 1 (x−7)3 +1 5≤x≤7 A1 Both parts fully correct. Ignore domain. 32  For reference: 1 x7 36 4 ( x − 1 ) 2 = 36 4 x 2 − 33 2 x + 36 4 , 13 2 ( x − 7 ) 3 + 1 = 13 2 x 3 − 23 12 x 2 + 1 4 7 3 2 x − 3 1 1 3 2 . B1 Correct domains. Must cover all reals, may overlap at boundaries. Must have 0 for x  1 and 1 for x  7 . 3 Trace ID: PC-1430F78DC7 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/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(c) © Cambridge University Press & Assessment 2026 Page 16 of 18  G ( y ) =   0 − 1 1 ( 13 2 3 − 6 4 3 ( − 9 y − 3 ) ) y 2 y 3 5 y  ≤ ≤  3 y y 9 ≤ ≤ 5 9 *M1 Expression for F ( 1 0 − y ) seen. DM1 Use of 1 − F ( 1 0 − y ) in both parts. A1 Both parts correct, ignore domain. May be unsimplified. ISW if seen correct in working. For reference: − 13 2 ( 3 − y ) 3 = 13 2 ( y 3 − 9 y 2 + 2 7 y − 2 7 ) , 1 − 36 4 ( 9 − y ) 2 = 16 4 ( − 3 y 2 + 5 4 y − 1 7 9 ) . B1 Correct domain must be in terms of y. Domain may overlap at boundaries. Must have 0 for y  3 and 1 for y  9 . 4 Trace ID: PC-1430F78DC7 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/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/