Cambridge A Level Mathematics 9709 — 2025 May/June Paper 1 · Variant 2

1 y 6 4 y = f( x) 2 – 4 – 2 0 2 4 6 8 10 12 x – 2 – 4 – 6 – 8 y = g( x) The diagram shows the graphs with equations y = f ( x) and y = g ( x) . Describe fully a sequence of two transformations which transforms the graph of y = f ( x) to the graph of y = g ( x) . Make clear the order in which the transformations should be applied. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 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

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2 Find the coordinates of the points of intersection of the curve and the line with equations 2xy + 5y 2 = 24 and 2x + y + 4 = 0 . [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   2 y  or 2 x ( −−4 2 x ) + 5 ( −−4 2 x ) 2 = 24 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).  7  B1 OE ( −1, − 2 ) ,  − , 3  Allow from a correct quadratic (ignore any working seen).  2  7 Condone x = −1 and y = −2 and x = − and y = 3. 2 4

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2 43 The coefficient of x7 in the expansion of e px + xo is 1280. p Find the value of the constant p. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: 3 5 − 3  4  3 B1 May be seen in a list. 5 5 2 px x  Allow with or without x’s.  or  or 10  ( )  3 2  p  Condone missing brackets only if recovered later. 2 64 B1 This term must now be identified if given as part of a list. 10p  3 Allow with or without x’s. p Their integer k 7 7 M1 640 x 7   x  = 1280 x  Condone e.g. = 1280 if 7x then disappears in p p 3 5 − 3  4  2 px subsequent work. Must be from a  x  . ( )   p  1 A1 OE p = 1 5 −1  4  2 5 5 2 px     p = 4. SC B1 for  or  or 5  ( ) 4 1  px  4

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4 A point P is moving along the curve with equation y = ax 2 - 12 x in such a way that the x-coordinate of P is increasing at a constant rate of 5 units per second. (a) Find the rate at which the y-coordinate of P is changing when x = 9 . Give your answer in terms of the constant a. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that the curve has a minimum point when x = 1 , find the value of a. [2] 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 1 *M1 For attempt at differentiation; at least one correct term 3  dy = ax 2 − 12 needed.    dx  2 Condone poor notation throughout.  dy dy dx   3 12  DM1 For correct use of chain rule with 5, x = 9 and their dy . =  =  a  9 − 12   5  dx    dt dx dt  2  Condone missing brackets and allow errors in their working.  dy   9  45 45 a − 120 A1 OE simplified form. = 5  a − 12  or a − 60 or 22.5a − 60 or    dt   2  2 2 15 or ( 3a − 8 ) 2 3 4(b) 1 M1 d y 3  1  2 For setting their 2 term with at least one term correct a    − 12 = 0 d x 2  4  = 0 and substituting x = 0.25. Condone missing brackets. d y Allow a restart for if 2 terms seen and at least one term d x correct.  a = 16 A1 2

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5 The equation of a curve is y = 4 cos 2x + 3 for 0 G x G 2 r. (a) State the greatest and least possible values of y. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Sketch the curve. [2] y x O 1 3 rr r rr 2r 2 2 (c) Hence determine the number of solutions of the equation 4 cos 2 x + 3 = 2x - 1 for 0 G x G 2 r. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

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

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6 y P x O 9 The diagram shows the curve with equation y = 1 and the line y = 6 - 3x . The line and the ( 5x + 4) 2 curve intersect at the point P which has y-coordinate 3. Find the area of the shaded region. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 6 − 3x = 3  x-coordinate of point of intersection = 1 B1 1 B1 B1 1 1   B1 for 𝑘(5𝑥+ 4) 2  18(5 x + 4) 2   5  0 18  18  M1 Area =18 −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

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tan i + 7 sin i cos i + 7 cos 2 i 7 (a) Prove the identity / . [3] tan 2 i - 3 1 - 4 cos 2 i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ sin i cos i + 7 cos 2 i 5 (b) Hence solve the equation = for 0° G i G 180° . [4] 1 - 4 cos 2 i tani ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) sin M1 The first two marks can be applied in the reverse order For appropriate use of tan = or tancos= sin at least once but as soon as an error occurs, no further marks are cos awarded. Candidates can work from LHS to RHS or RHS For appropriate use of sin 2 + cos 2 = 1 or 1 + tan 2 = sec2 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+ 7 5 2 *B1 OE =  tan(tan+ 7) = 5(tan − 3) 2 Using part (a) and eliminating fractions. tan − 3 tan  4tan 2 − 7tan− 15 = 0 ( 4 tan+ 5) ( tan− 3 ) = 0 DM1 Factorising or other accepted method for solving their 3-term quadratic in tan. Condone errors made in forming 3-term quadratic. tan= −1.25 and tan= 3 B1 OE Independent of factorisation of the quadratic. And no extra solutions. Allow e.g. x = provided x = tan is seen. Can be implied by a correct final answer. = 71.6 and = 128.7 B1 AWRT No others in range 0  180 . Ignore any answers outside this range. Maximum 3/4 if incorrect factorisation or formula. 4

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8 y x O A B C The diagram shows the circle with equation x 2 + y 2 - 14x + 8y + 36 = 0 and the line y =-2 . The line intersects the circle at the points A and B. The centre of the circle is C. (a) Find the coordinates of A, B and C. 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(b) Find the angle ACB in radians. Give your answer correct to 3 significant figures. 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(c) The chord AB divides the circle into two segments. Find the area of the larger segment. 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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  −1 5    5   2 M1 Or other correct method for an isosceles triangle using   = 2tan tan   = or sin  = or cos  =   , −2 ) and C of the form  2  2  2   2  29  2  29 their A and B of the form ( −1 2 ( , − d ) , where − d −2. or  = π −2 tan or π −2 0.381 5 5 Note: tan=  0/ 2 unless  is then doubled. Using 2  29 + 29 − 100 −21  2 or 10 = 29 + 29 − 2  29  29cos  cos = =    2  29  29 29  r = 29 scores 0/2. [ =] 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 Use of tan = m2 − m1 , where m1 and m2 are the gradients 5 5  20  1 + m2 m1 tan = = − −2 2  21  of their AC and BC, where A, B and C are of the required 1 +  5 5 form. [ =] 2.38 A1 AWRT Ignore degree symbol if present. 2 8(c) B1 Sight of 29. SOI. [Length AC = Length BC = 52 + 22 ] = 29 [=5.385…] 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  ( 2− their )  = 56.587.. 2 (e.g. 29) and their ( 2− ) or their . 1 May be embedded as part of the segment formula. or [Area of small sector =]  29  29  their [= 34.518…] 2 1 1 M1 Use of correct triangle formula with their identified radius Area of triangle =  29  29  sin or  10  2  = 10  (e.g. 29) and their . 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

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10 (a) The first, second and third terms of an arithmetic progression are 4k, k2 and 8k respectively, where k is a non-zero constant. (i) Find the value of k. 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(ii) Find the sum of the first 20 terms of the progression. 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(b) The fourth and sixth terms of a geometric progression are 36 and 6 respectively. The common ratio of the progression is positive. a Find the sum to infinity of the progression. Give your answer in the form , where a, b and c b - c are integers. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 10(a)(i) 2 2 2 2 M1 OE k − 4 k = 8 k 8k − k = k − 4k or 2k = 4k + 8k or 4 k + 2 ( ) Forming an equation in k only, clearly using the first 3 terms of an AP. [2 k ( 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 2 k or k 2 − 4 k  = 12 and a  = 4 k  = 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 2 1 or r = 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

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11 (a) Express x 2 + 4 x + 2 in the form ( x + a ) 2 + b , where a and b are integers. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The functions f and g are defined as follows. f ( )x = x 2 + 4x + 2 for x G-2 g ( )x =- x - 4 for x H-2 (b) (i) Find an expression for f -1 ( )x . 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(ii) Find an expression for ( gf ) -1 ( )x . 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Mark scheme: 11(a) 2 B1B1 B1 for each correct { }. {( x + 2 ) }−2 Allow a = 2, b = −2. If contradictory, give preference to the expression. 2 11(b)(i) y = ( x + 2 ) 2 − 2  y + 2 = ( x + 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 x 2 + 4 x + 211(b)(ii)  − 4 *M1 Using fg ( x ) = {( −−x 4 ) + 2}2 − 2 scores 0/4. their completed square form − 4 or − ( )  gf ( x ) =−   x + 2 ) 2 − 2 A1  gf ( x ) =− ( 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. −1 A1 Do not condone x = . [( gf ) ( x ) =] − −−x 2 − 2 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 . −1 A1 Do not condone x = [( gf ) ( x ) =] − −−x 2 − 2 4

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