Mathematics 9709/13 October/November 2025: question paper and mark scheme

Mark scheme

1(a) 6 6 6 5 x 6 4 x 2 6 3 x 3 M1 OE 2 ± 2 ± 2 ± 2 A correct unsimplified expansion of at least four terms. Condone 0 1 2 2 2 3 2 ± sign errors for this mark. Condone omission of brackets if x 6.5 x 6.5.4 x 2 3 recovered. or 26 1 ± 6 ± ± This mark can be implied by sight of 64 + 96 x + 60 x 2 + 20 x3 . 4 2! 4 3! 4 64 − 96 x + 60 x 2 − 20 x3 A1 Three correct simplified terms. A1 Four correct simplified terms. Accept these terms given in a list. 3 1(b) (3 × their ( −20)) + ( −1× their 60) + ( 2 × their 64) M1 Selecting the correct products using their expansion. 8 A1 Accept 8 x 3 . 2

Mark scheme

2(a) 1 1 M1 1 5 x − 3 = tan − π or − tan π [ = −1] Condone 5 x − 3 = tan π and 5x − 3 = −0.0137 4 4 4 2 A1 x= or 0.4 5 2 2(b) 5 − 5sin 2 θ = 4sin θ + 4 *M1 Replacing cos 2 θ with 1 − sin 2θ . 5sin 2 θ + 4sinθ − 1[= 0] (5sin θ −1)(sin θ + 1) = 0 DM1 Collecting terms to form a three-term quadratic. Condone only ± errors. 1 A1 sin θ = and − 1 5 θ = 2.94, 4.71 A1 AWRT 3 337 Accept π in place of 4.71 and 0.936π or π in place 2 360 of 2.94. 3 Condone repetition of π but no other values within the range. 2 π Ignore answers outside the range θ 2π. 2 4

Mark scheme

3(a) 72.030004 B1 CAO. Not AWRT. 1 3(b) 30.3888 B1 CAO. Not AWRT. 1 3(c) 30[.0] B1 CAO. 30 may be accompanied by ‘around’, ‘approximately’ etc. 1

Mark scheme

4(a) 20 + ( 30 − 1) × ( their − 5 ) M1 Correct expression for u30 using their calculated value of d. −125 A1 2 4(b) k k −5 B1 OE r= and r = 20 k k k −5 For two correct expressions for r. Can be implied by = 20 k OE. k 2 − 20k + 100[= 0] M1 Forming a 3-term quadratic. Condone only ± errors. [ r =] 12 A1 May be seen in the S∞ formula. 20 A1 S∞ = = 40 1 − 12 Alternative Method for Question 4(b) 20r = k and 20r 2 = k − 5 B1 Stating two correct relationships between k and r. This can be implied by 20r 2 = 20r − 5 OE. 20r 2 − 20r + 5[= 0] M1 Forming a three-term quadratic in r. Condone only ± errors. [ r =] 12 A1 May be seen in the S∞ formula. 20 A1 S∞ = = 40 1 − 12 4

Mark scheme

5(a) 1 2 3 B1 θ sin θ = Use of a correct expression for either or θ . 2 4 2 1 2 or cos θ = 2 4 ( ) = − 1 2 42 + 42 − 4 3 or cos θ = 2× 4× 4 2 2π B1 Correct answer can imply B1B1. [θ =] Condone inconsistent use of θ . 3 2 5(b) 1 2 2π 1 4π 1 2π *M1 Use of a correct formula with their θ or 2π − their θ ,θ > 0 . Triangle × 4 × sin or × 42 × or × 42 × 2 3 2 3 2 3 and/or minor sector may be seen in the formula for finding the minor segment. 1 2 4π 1 2 2π DM1 Addition of relevant sector area with 2π − their θ , where θ > 0, ×4 × + × 4 × sin 2 3 2 3 and triangle area with their θ , where θ > 0. 1 2π 1 2 2π Subtraction of triangle area, with their θ ,θ > 0 , from relevant Or 16π − × 42 × − × 4 × sin sector area, with their θ ,θ > 0, and subtract from circle area. 2 3 2 3 32π A1 OE. Exact equivalents only. +4 3 SC B2 following M0DM0 for 40.4 AWRT. 3 3

Mark scheme

6(a) {Stretch} {[scale] factor 2} {[parallel to] in/on y[-axis] or x-axis B2,1,0 B2 for all three {} elements correct. B1 for two {} elements invariant} correct. 3 B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation} or {[+]3 in/on [the] x[direction]} and correct. −4 Do not condone use of ‘Right’ and ‘Down’ in place of x and y {– 4 in/on [the] y [direction]} directions. 3 0 Note: The vectors can be seen as and . 0 −4 Correct transformations in the correct order without any extra B1 Combined translations (condone transformations accompanied by transformations. Down and right can be condoned for this mark if 0 the candidate’s intention is clear. a vector) must come after the stretch or, if separate, the −4 must come after the stretch. Alternative Method for Question 6(a) 3 B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation} or {[+]3 in/on [the] x [direction]} and correct. −2 Do not condone use of ‘Right’ and ‘Down’ in place of x and y {-2 in/on [the] y [direction]} directions. 3 0 Note: The vectors can be seen as and . 0 −2 {Stretch} {[scale]factor 2} {[parallel to] in/on y-[axis] or x axis B2,1,0 invariant OE} Numerically correct transformations in the correct order without B1 Combined translations (condone transformations accompanied by any extra transformations. Down and right can be condoned for 0 this mark if the candidate’s intention is clear. a vector) must come before the stretch or, if separate, the −2 must come before the stretch. 5 6(b) a=2 B1 Individual values are considered the final answer and NOT those in a ( x + b ) + c, unless individual values are not stated. 3 b = −3 B1 c = −4 B1 3

Mark scheme

7 2 1 1 2 *M1 OE y = ax − 3 + 2 y − 2 ( ax − 3) = 2 or ax − 3 = 1 For setting ‘y =’ and rearranging. y− 2 Condone only ± errors. 4 DM1 For making ax the subject. ax = +3 2 y −1 Condone only ± errors. 4 3 2 3 A1 OE [ y = or g −1 ( x ) = ] + or + a ( 2 x − 1) a 1 a Note: x and y may be interchanged from the start. a x − 2 Clear replacement of a with 6 in g −1 and, if necessary, followed by A1 This mark is dependent on getting the correct inverse. simplification to arrive at exactly the same expressions. Alternative Method for Question 7 ax + 1 *M1 OE y = 2 y ( ax − 3) = ax + 1 For setting ‘y =’ and rearranging. 2 ( ax − 3) Condone only ± errors. ax ( 2 y − 1) = 6 y + 1 DM1 For making ax ( 2 y − 1) the subject, condone only ± errors. 1 A1 OE 3x + 6x + 1 2 Note: x and y may be interchanged from the start. y = or g −1 ( x ) = a ( 2 x − 1) or 1 a x − 2 Clear replacement of a with 6 in g −1 and, if necessary, followed by A1 This mark is dependent on getting the correct inverse. simplification to arrive at exactly the same expressions. 4

Mark scheme

8(a) Centre is ( 2, 4 ) B1 [(Radius)2 =] ( 6 − 2) + (1 − 4) 2 2 M1 Use of a correct method to find ( radius )2 or radius. 1 2 This can be implied by r = 5 . ( 6 − ( −2)) + (1 − 7 )2 [= 25] 2 or 2 ( x − 2 ) 2 + ( y − 4 ) 2 = ( 5 )2 A1 OE Alternative Method for Question 8(a) y −1 y −7 B1 Correct expression using either ( 6,1) or ( −2,7 ) and ( x, y ) . Either or x−6 x+2 y −1 y − 7 M1 Product of their gradients = −1. × = −1 x−6 x+2 x2 + y 2 − 4x − 8 y − 5 = 0 A1 OE Simplification to get the correct five terms. 3 8(b) [Gradient of normal through centre =] 2 B1 SOI y − 4 = ( their gradient )( x − 2 ) *M1 Correct form of a line equation with their normal gradient i.e. not 1 Or 4 = ( their 2 )( 2 ) + c c = [0] − 2 ( x − 2)2 + ( 2x − 4)2 = 25 DM1 OE Using their 2x and their circle to form an equation in x. 5 ( x − 2 )2 = 25 ( x − 2 )2 = 5 A1 OE Accept x2 − 4x − 1= 0 . [ x =] 2 ± 5 A1 OE Exact values only. Alternative Method for Question 8(b) dy dy B1 OE 2x + 2 y − 4−8 = 0 Implicit differentiation of the correct circle equation. dx dx 1 1 *M1 dy 1 2x + 2 y − − 4 − 8 − = 0 Replacing with − in their differential which must contain 2 2 dx 2 dy two terms in . Expect y = 2 x . dx ( x − 2)2 + ( 2x − 4)2 = 25 DM1 OE Using their 2x and their circle to form an equation in x. 5 ( x − 2 )2 = 25 ( x − 2 )2 = 5 A1 OE Accept x2 − 4x − 1[= 0]. [ x =] 2 ± 5 A1 OE Exact values only. 8(b) Alternative Method 2 for Question 8(b) 1 B1 This mark can be implied by a correct differential. ( y = 25 − ( x − 2 ) + 4 or − x 2 + 4 x + 21 2 + 4 2 ) 1 *M1 Differentiating their expression for y. Use of the chain rule must dy 1 ( 2 − ) = 25 − ( x − 2 ) 2 × ( 4 − 2 x ) dx 2 be clear. 1 DM1 dy 1 − 1 1 ( 2 − ) = 25 − ( x − 2 ) 2 × ( 4 − 2 x ) 25 − ( x − 2 ) = ( 2 x − 4 ) 2 2 2 2 Equating their dx to − . 2 5x 2 − 20 x − 5 = 0 A1 OE [ x =] 2 ± 5 A1 OE Exact values only. Alternative Method 3 for Question 8(b) ( x − 2 )2 + ( ( − 12 x + c ) − 4 ) = 25 B1 Replacing y with − 12 x + c in a correct circle equation. 2 M1 Forming a three-term quadratic in x. 5 2 4 x − cx + ( c − 4 ) − 21 = 0 or 2 5 2 4 ( ) x − cx + c 2 − 8c − 5 = 0 Condone only ± errors. Using "b2 − 4ac " = 0 with their values from the quadratic in x DM1 Expect c 2 − 4 × 5 × c − 4 2 − 21 = 0. 4 ( ) 40 ± 1600 + 400 10 ± 5 5 A1 OE [ c =] = 8 2 [ x =] 2 ± 5 A1 OE Exact values only. 5

Mark scheme

9 1 4 M1 Equating the curve and the line to form a 3-term quadratic. x + = 4.5 x 2 − 9 x + 8 = 0 2 x x =1, 8 A1 SOI 567 A1 FT FT their limits from the quadratic, both limits must be positive. π × 4.52 × ( their 8 − their 1) = π or 445.3 4 *M1 For use of y dx . 2 2 [ π] x + [dx] 1 4 2 x A1 OE [ π] x2 + 4 + 2 dx = [ π ] 1 16 1 3 16 x + 4x − First A1 for two correct terms, second A1 for all three correct. 4 x 12 x May be unsimplified. May include +C. A1 DM1 Clear use of their limits in their integrated expression. At least one [ π ] 218 143 1015 + = [ π] correct power required. Both limits must be positive, or if correct 3 12 12 218 143 limits are used, then ± + is the minimum required. 3 12 343 A1 Condone a negative answer changed to a positive. Volume of solid = π A0 for omission of π. 6 SC B1: following DM0. A0 for 179.594… Alternative Method for Question 9 1 4 M1 Equating the curve and the line to form a 3-term quadratic x + = 4.5 x 2 − 9 x + 8 = 0 2 x x =1, 8 A1 SOI 4 *M1 For use of ( y − y ). 2 2 2 1 [Volume =] [ π ] 4.52 − x + [ dx ] 1 2 2 x Condone ( y1 − y2 ) for this *M1. 2 1 16 A1 OE ± [ π] 4.52 x − 4 x − x3 + 12 x A1 for each element of the integral. A1 May be unsimplified. May include +C. A1 ‘Correct’ terms from ( y − y )2 score A0A0A0 XP. 1 2 DM1 Clear use of their limits in their integrated expression. At least one [ π ] 268 193 − correct power is required. 3 6 Both limits must be positive, or if correct limits are used, then 268 193 ± − is the minimum required. 3 6 343 A1 Condone a negative answer changed to a positive. Volume of solid = π A0 for omission of π. 6 SC B1: following DM0. A0 for 179.594… 8

Mark scheme

10(a)(i) 2( x + 1)2 +8 B1 B1 for each {} element. B1 B1 3 10(a)(ii) Range or f ( x ) or y 8 B1 FT OE. FT their value of c. x 8 scores B0. 1 10(b) p ( 5m) + 4 ( 5m) − 5[= 0] and p ( −9m) + 4 ( −9m) − 5[= 0] *M1 Allow one error for this mark if the intention is clear. 2 2 5 − 20m 2 *DM1 Eliminate p by substitution or equating two expressions for p. 81 m + 4 ( −9m ) − 5[ = 0] 2 (5m) 36m + 5 or 25 m 2 + 4 ( 5m ) − 5 [ = 0 ] ( −9m )2 5 − 20m 36m + 5 or 2 = (5m) ( −9m ) 2 2520m = 280 DM1 OE Simplify as far as a linear equation or a quadratic where m = 0 is ignored or dismissed. m = 19 , p = 9 A1, A1 Alternative Method for Question 10(b) p ( 5m) + 4 ( 5m) − 5[= 0] and p ( −9m) + 4 ( −9m) − 5[= 0] *M1 Allow one error for this mark if the intention is clear. 2 2 p ( 5m) + 4 (5m) − 5 = p ( −9m) + 4 ( −9m) − 5 *DM1 Equating their two expressions. 2 2 1 45 DM1 Simplify as far as a quadratic where m = 0 is ignored or dismissed 2 56 pm − 56m = 0 m = p 5 = p or 45m = 5 and substitute for m or p. p = 9, m = 19 A1, A1 10(b) Alternative Method 2 for Question 10(b) −4 + 42 − 4 ( p )( −5) −4 − 42 − 4 ( p )( −5 ) *M1 Equating 5m and − 9m to the two solutions from the quadratic = 5m and = −9m formula. 2p 2p Allow one error for this mark if the intention is clear. −4 + 42 − 4 ( p )( −5 ) −4 − 42 − 4 ( p )( −5 ) *DM1 Attempt to solve simultaneously by equating their expressions for = m, or dividing one by the other and equating to − 95 . (5 × 2) p ( −9 × 2 ) p 16 + 20 p = 196 DM1 OE Simplify as far as a linear expression – no square root. p = 9, m = 19 A1, A1 Alternative Method 3 for Question 10(b) ( x − 5m)( x + 9m) x2 + 4mx − 45m2 [= 0] *M1 Recognising the relationship between 5m and − 9m and the quadratic equation. Or α + β = −4m and αβ = −45m 2 4 5 *DM1 4 5 4m = , −45m 2 = − Using x 2 + x − [ = 0]. Comparing coefficients or using p p p p b c α +β =− and αβ = . a a 1 DM1 Forming an equation in one unknown. m= − 45m 2 = −5m p 1 or p = 5 p 2 = 45 p m m = 19 , p = 9 A1, A1 5

Mark scheme

11(a) B1 B1 for each {} element. dy −8 × 3 6 = 2 + dx ( 3 x − 8 ) ( x − 1)2 B1 −8 × 3 *M1 Using their differentiated expression, which must contain 6 −45 + 2 = (3x − 8)−2 and ( x − 1)−2 , and x = 3 . This may be seen in their line ( 3.3 − 8 ) ( 3 − 1) 2 2 equation. −45 *DM1 Correct form of a line equation with their gradient, but not the y − 5 = their ( x − 3) negative reciprocal, and ( 3, 5 ) . 2 −45 145 or 5 = their ( 3) + c c = 2 2 −45 DM1 Replacing y with −8x and collecting terms. −8 x − 5 = 2 ( x − 3) 29x = 145 x = 5, y = −40 A1 Accept ( 5, − 40 ) . 6 11(b)(i) −24 6 *M1 dy = 0 ⇒ 24 ( x − 1) = 6 ( 3x − 8) , which must contain ( 3x − 8)−2 and ( x − 1)−2 to 0, 2 2 + Equate their ( 3x − 8) ( x − 1) 2 2 dx and clear of fractions. 4 ( x − 1)2 = ( 3x − 8 )2 Only condone ± errors. 5 x 2 − 40 x + 60 [= 0] or ±2 ( x − 1) = 3 x − 8 DM1 OE Forming a three-term quadratic. Condone only ± errors. Or taking square roots. For this method, the ± must be present. 2, 6 A1 Both values. 3 11(b)(ii) d2 y B1FT dy −24 × −2 × 3 6 × −2 Correct differentials of the elements of their , which must = + 3 dx ( 3x − 8) ( x − 1) 2 3 dx B1FT −2 −2 contain ( 3x − 8) and ( x − 1) . d 2 y their 144 their 12 M1 d2 y At x = 2, = − Replacing x with their 2 and their 6 in their , which must dx 2 ( 6 − 8)3 ( 2 − 1)3 dx 2 −3 −3 d 2 y their 144 their 12 contain ( 3x − 8) and ( x − 1) . and at x = 6, = − Correct final answers. dx 2 (18 − 8)3 ( 6 − 1)3 d2 y A1 WWW At x = 2, = −30 < 0, therefore max[imum] Correct values, or working, and > 0 and < 0 are required. dx 2 d 2 y 48 6 At x = 6, = > 0, therefore min[imum] dx 2 103 125 4