Mathematics 9709/12 May/June 2025: question paper and mark scheme

Mark scheme

1 {Stretch} {factor 2} {‘parallel to y-axis’ or ‘in y-direction’ or ‘vertically’} B2,1,0 B2 for 3 correct components. B1 for 2. 0 B2,1,0 B2 for 3 correct components. B1 for 2. {Translation} or −14 {‘parallel to the y-axis’ or ‘in the −14 y-direction’ or ‘vertically’.} Alternative Method for Question 1 0 B2,1,0 B2 for 3 correct components. B1 for 2. {Translation by} or −7 {‘parallel to the y-axis’ or ‘in the −7 y-direction’ or ‘vertically’} {Stretch} {factor 2} {‘parallel to y-axis’ or ‘in y-direction’ or ‘vertically’} B2,1,0 B2 for 3 correct components. B1 for 2. 4

Mark scheme

2 −y − 4 2 *M1 OE 2 y + 5 y = 24 For eliminating x or y. 2 Condone sign errors in the rearrangements. 24 − 5 y 2 ) or 2 + y + 4 [ = 0] 2y or 2 x ( −4 − 2 x ) + 5 ( −4 − 2 x ) = 24 2 4 y 2 − 4 y = 24 or 16 x 2 + 72 x + 56 [= 0] DM1 OE For simplifying to a 3-term quadratic; terms do not all have to be on the same side. Condone sign errors in the expansions or in the collecting of terms. 7 B1 OE y = −2, y = 3 or x = −1, x = − 2 Not for a correct ( x, y ) pair. Allow from a correct quadratic (ignore any working seen). B1 OE ( −1, − 2 ) , − , 3 7 Allow from a correct quadratic (ignore any working seen). 2 7 Condone x = −1 and y = −2 and x = − and y = 3. 2 4

Mark scheme

3 3 B1 May be seen in a list. 5 5 2 5−3 4 or or 10 × px 3 2 ( ) × x Allow with or without x’s. p Condone missing brackets only if recovered later. 64 B1 This term must now be identified if given as part of a list. 10 p 2 × Allow with or without x’s. p3 Their integer k 7 M1 640 x7 x ] = 1280[ x 7 Condone e.g. = 1280 if x 7 then disappears in p p 3 4 subsequent work. Must be from a × px ( ) 2 5−3 × x . p 1 A1 OE p= 1 2 5 5 5 −1 4 ( SC B1 for or or 5 × px 2 4 1 )× p = 4. px 4

Mark scheme

4(a) dy 3 2 1 *M1 For attempt at differentiation; at least one correct term dx = 2 ax − 12 needed. Condone poor notation throughout. DM1 dy dy dy dx 3 1 = × = a × 9 2 − 12 × 5 For correct use of chain rule with 5, x = 9 and their . dt dx dt 2 dx Condone missing brackets and allow errors in their working. dy 9 45 45a − 120 A1 OE simplified form. dt = 5 2 a − 12 or 2 a − 60 or 22.5a − 60 or 2 15 or ( 3a − 8) 2 3 4(b) 1 M1 dy 3 1 2 For setting their 2 term with at least one term correct a × − 12 = 0 dx 2 4 = 0 and substituting x = 0.25. Condone missing brackets. dy Allow a restart for if 2 terms seen and at least one term dx correct. a = 16 A1 2

Mark scheme

5(a) [Greatest] 7, [least] −1 B1B1 B2 for answers of −1 and 7 only. B1 for one correct value. Ignore incorrect identification/inequality. 2 5(b) 8 y B2,1,0 Ignore any graph outside domain ( 0, 2π ) . 7 B1 for two complete cycles, one from 0 to approximately 6 π and the other finishing at approximately 2π. Starting at 5 their greatest value and initially decreasing. 4 Condone straight lines and minimum above the x-axis or 3 joining points with straight lines. 2 1 x B2 for correct curve; condone any incorrect x-axis −π/2 π/2 π 3π/2 2π intercepts. Graph must start to level off at both 0 and 2π. −1 Ignore any y-labels, but the curve should be more above −2 the x-axis than below. 2 5(c) 3 [solutions] B1 Ignore any graphs drawn 1

Mark scheme

6 6 − 3x = 3 x-coordinate of point of intersection = 1 B1 18 1 1 B1 B1 1 B1 for 𝑘(5𝑥 + 4)2 (5 x + 4) 2 5 0 18 18 18 M1 Area = × 3 − × 2 = 5 5 5 1 M1 Line crosses x-axis at x = 2, so triangle area = × 3 × 1 2 51 A1 Total area = 10 6

Mark scheme

7(a) sin θ M1 The first two marks can be applied in the reverse order For appropriate use of tan θ = or tanθ cosθ = sinθ at least once but as soon as an error occurs, no further marks are cosθ awarded. Candidates can work from LHS to RHS or RHS For appropriate use of sin2 θ + cos2 θ = 1 or 1 + tan2 θ = sec2θ M1 to LHS for full marks. If they work on both sides simultaneously, maximum of M1M1 only if a common correct expression is reached. Condone missing brackets. If the numerator and denominator are worked on separately, they need to be brought together for the second mark. Fully correct proof A1 WWW Condone missing brackets if recovered. 3 7(b) tan θ + 7 5 *B1 OE = tan θ (tan θ + 7) = 5(tan 2 θ − 3) Using part (a) and eliminating fractions. tan θ − 3 tan θ 2 4tan 2 θ − 7tanθ − 15 = 0 ( 4 tan θ + 5) ( tan θ − 3) = 0 DM1 Factorising or other accepted method for solving their 3-term quadratic in tan θ . Condone errors made in forming 3-term quadratic. tanθ = −1.25 and tan θ = 3 B1 OE Independent of factorisation of the quadratic. And no extra solutions. Allow e.g. x = provided x = tanθ is seen. Can be implied by a correct final answer. θ = 71.6° and θ = 128.7° B1 AWRT No others in range 0 θ 180 °. Ignore any answers outside this range. Maximum 3/4 if incorrect factorisation or formula. 4

Mark scheme

8(a) ( 2, −2 ) B1 Condone x = 2 and y = −2 if seen together. (12, −2 ) B1 Condone x = 12 and y = −2 if seen together. ( 7, − 4 ) B1 3 If B0 B0 for the first two marks, then SC B1 available for x = 2 and x = 12. 8(b) 1 5 5 θ 5 θ 2 M1 Or other correct method for an isosceles triangle using tan θ = θ = 2 tan −1 or sin = or cos = 2 2 2 2 29 2 29 their A and B of the form ( , −2 ) and C of the form 2 or θ = π − 2 × tan −1 or π − 2 × 0.381 ( , −d ) , where − d < −2. 5 5 Note: tan θ = 0 / 2 unless θ is then doubled. Using 29 + 29 − 100 −21 2 or 102 = 29 + 29 − 2 × 29 × 29cosθ cos θ = = 2 × 29 × 29 29 r = 29 scores 0/2. [θ =] 2.38 A1 AWRT Final answer of 136.4° scores max 1/2. Ignore degree symbol if present. Alternative Method for Question 8(b) −2 2 M1 m2 − m1 − Use of tan θ = , where m1 and m2 are the gradients 20 1 + m2 m1 tan θ = 5 5 = − −2 2 21 of their AC and BC, where A, B and C are of the required 1+ × 5 5 form. [θ =] 2.38 A1 AWRT Ignore degree symbol if present. 2 8(c) [Length AC = Length BC = 52 + 22 ] = 29 [=5.385…] B1 Sight of 29. SOI. Condone 5.4. Could be found in parts (a) or (b), but do not award unless it is seen in part (c). 1 M1 Use of correct sector formula with their identified radius [Area of large sector =] × 29 × 29 × ( 2π − their θ ) [ = 56.587..] 2 (e.g. 29) and their ( 2π − θ ) or their θ . 1 May be embedded as part of the segment formula. or [Area of small sector =] × 29 × 29 × their θ [= 34.518…] 2 1 1 M1 Use of correct triangle formula with their identified radius Area of triangle = × 29 × 29 × sin θ or × 10 × 2 [ = 10] (e.g. 29) and their θ . 2 2 May be embedded as part of the segment formula. [Segment area = ] 66.6 A1 AWRT [Area of circle – smaller segment = π × 29 − 34.52 + 10] 4

Mark scheme

9(a) dy −24 x −2 12 B1 May be unsimplified. dx = −2 = x 2 [ +c ] 12 M1 dy + c = 0 [ c = −3] Substituting x = −2 into their of the form kx−2 + c, and ( −2 ) 2 dx setting = 0. Condone missing bracket. dy 12 A1 −24 −2 dx = x 2 − 3 or 12x − 3 must be simplified to 12, but accept c = −3 as the −2 final answer. 3 9(b) 12 2 B1 May be done by inspection or calculator, but must be x 2 − 3 = 0 x = 4 x = ±2 x = 2 WWW. Condone ±2. d 2 y 24 B1 WWW. Or other valid method. 2 = − or − 3 or < 0 maximum Ignore any working for x = −2. dx ( 2 ) 3 2 9(c) B1FT May be unsimplified. FT their nonzero c. [ y =] − 12 x−1 + their cx = − 12 − 3x [ + d ] x 12 M1 Substituting ( −2, 19 ) into their integrated expression 19 = − − 3 × ( −2 ) + d [ d = 7] −2 (defined by having at least one correct power) and including +d. 12 A1 Accept d = 7 as final answer if y = ... seen previously. y=− − 3x + 7 x Allow ' f ( x ) = ' in place of ' y = '. 3 9(d) 12 9 12 3 *M1 dy 9 2 − 3 = − 2 = x = ... x2 = 16 x = 4, y = −8 Setting their of the form kx −2 + c = − , and correctly x 4 x 4 dx 4 solving for x from their equation where x2 > 0. 4 B1 Gradient of normal = 9 y +8 4 DM1 For using their stated positive x and their stated y and a = ( x − 4) 9 changed gradient to correctly form a line equation; 4 9 condone − or . 9 4 4 x − 9 y − 88 = 0 A1 OE in correct form. 4

Mark scheme

10(a)(i) ( ) 8k − k 2 = k 2 − 4k or 2k 2 = 4k + 8k or 4k + 2 k 2 − 4k = 8k M1 OE Forming an equation in k only, clearly using the first 3 terms of an AP. [2k ( k − 6 ) = 0 ⇒] k = 6 [or k = 0] A1 Condone extra ‘solution’ k = 0. Note: can be done by inspection or with no working. 2 / 2 2 10(a)(ii) d = 8k − k 2 or 2k or k 2 − 4k = 12 and a [ = 4k ] = 24 *M1 Using a correct method to find a and d from their k. SOI. May have been found in (a)(i) but must be used in (ii). 20 20 DM1 Use of correct sum formula with n = 20 and their a and d. S20 = ( 2 × 24 + 19 × 12 ) or = ( 24 + 252 ) 2 2 S20 = 2760 A1 3 10(b) ar 3 = 36 and ar 5 = 6 B1 SOI WWW 1 or r 2 = 6 6 1 6 *M1 Using a correct method to find r. Condone ±. r= = or 36 6 6 36 3 1296 *M1 Using a correct method to find a. Condone ±. a= = 36 × 6 , 216 6 or 3 1 6 6 216 6 DM1 Using the correct formula with their a and their r < 1 . S∞ = 1 1− 6 1296 7776 A1 OE in the required form. [S∞ =] or 6 −1 216 − 6 r = 0.408 a = 529 S∞ = 894 scores 4/5. 5

Mark scheme

11(a) {( x + 2 ) }−2 B1B1 B1 for each correct { }. 2 Allow a = 2, b = −2. If contradictory, give preference to the expression. 2 11(b)(i) y = ( x + 2) − 2 y + 2 = ( x + 2) 2 2 *M1 Equating y or f −1 ( x ) or f −1 or f ( x ) to their completed square form and first step. x and y may be interchanged at this stage. Condone ± errors. x = [±] y + 2 − 2 DM1 Condone ± errors during simplification. [f −1 ( x ) =] − x + 2 − 2 A1 Do not condone x = or f ( x ) = . 3 11(b)(ii) ( ) gf ( x ) = − their completed square form − 4 or − x 2 + 4 x + 2 − 4 *M1 Using fg ( x ) = {( − x − 4 ) + 2}2 − 2 scores 0/4. gf ( x ) = − ( x + 2 ) − 2 A1 2 x = [±] − y − 2 − 2 DM1 Finding x from their completed square form, which must contain a −( x + k ) 2 term. Condone ± errors only during simplification. x = − y + 2 − 2 is DM0 (Square rooting then ÷ or × − 1 ) x and y may be interchanged at this stage. [( gf ) −1 ( x ) =] − − x − 2 − 2 A1 Do not condone x = . Alternative Method for Question 11(b)(ii) ( gf )−1 = f −1g−1 *M1 SOI Allow with their f −1 and their g −1 . Using g −1f −1 ( x ) scores 0/4. g −1 ( x ) = − x − 4 A1 ( gf )−1 ( x ) = − − x − 4 + 2 − 2 DM1 Allow with their f −1 and their g −1 . [( gf ) −1 ( x ) =] − − x − 2 − 2 A1 Do not condone x = 4