Cambridge A Level Mathematics 9709 — 2024 May/June Paper 1 · Variant 1

1 (a) Express 3y 2 - 12y - 15 in the form 3 ( y + a) 2 + b , where a and b are constants. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the exact solutions of the equation 3x 4 - 12x 2 - 15 = 0 . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 2 3 2 27   y or 2, 27 a b   B1 B1 2 1(b)   2 2 2 9 x   leading to 2 2 3 x   M1 Must be 2 x unless substitution is clear. 2 2 1 or 5 x x   M1 Allow omission of -1 if ±3 seen. 5  x A1 B1 SC if M1M1 not awarded. Ignore ± i, i, –i, √–1. Use of calculator with no working scores 0/3. Alternative method for Question 1(b) 3 x 4 – 12 x 2 – 15 = 0 leading to    2 2 3 5 1 [   x x = 0] (M1) 2 2 1 or 5 x x   (M1) Allow omission of –1 if factors seen. Factorising or other valid method. 5  x (A1) B1 SC if M1M1 not scored. Ignore ± i, i, –i, √–1. Use of calculator with no working scores 0/3. 3

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2 y 1 x r 3 O 1 r r 2 r 5 r 3r 7 r 2 2 2 2 – 1 – 2 – 3 – 4 – 5 The diagram shows two curves. One curve has equation y = sin x and the other curve has equation y = f ( x) . (a) In order to transform the curve y = sin x to the curve y = f ( x) , the curve y = sin x is first reflected in the x-axis. Describe fully a sequence of two further transformations which are required. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find f ( x) in terms of sinx. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) {Stretch}{factor 3}{ in y-direction} B2,1,0 2 out of 3 scores B1. {Translation}    0 2        B2,1,0 Accept shift. Alternative Method for Question 2(a) {Translation}  0 2 3                  (B2,1,0) 2 out of 3 scores B1. Accept shift. {Stretch}{factor 3}{ in y-direction} (B2,1,0) 4 2(b)  f { 3sin }{ 2      x x } B1 B1 No marks awarded if extra terms seen. 2

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3 The coefficient of x3 in the expansion of ( 3 + ax) 6 is 160. (a) Find the value of the constant a. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the coefficient of x3 in the expansion of ( 3 + ax) 6 ( 1 - 2x) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 3 20 27 160    a M1 Allow 6C3  33 3 160   a . Accept 540a3 with no other working for M1.  2 3  a A1 Allow 0.667 AWRT. SC B1 is a = 2 3 with no other working. 2 Question Answer Marks Guidance 3(b) Coefficient of 2 x is   2 2 15 81 540 3          their B1 FT May be in a list. 6C2 and 34 must be evaluated but may be implied by later work. Condone 540 with no working. 160 1 2 540 their M1 920  A1 Condone 3 920  x . 3

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4 The equation of a curve is y = f ( x) , where f ( x) = ( 2x - 1) 3x - 2 - 2 . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places. A(2, 4), B(2.0001, k), C(2.001, 4.00625), D(2.01, 4.06261), E(2.1, 4.63566), F(3, 11.22876) (a) Find the value of k. Give your answer correct to 5 decimal places. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The table shows the gradients of the chords AB, AC, AD and AF. Chord AB AC AD AE AF Gradient of 6.2501 6.2511 6.2608 7.2288 chord (b) Find the gradient of the chord AE. Give your answer correct to 4 decimal places. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Deduce the value of f l ( 2) using the values in the table. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) [k] = 4.00063 B1 CAO 1 4(b) [Gradient AE] = 6.3566 B1 CAO 1 4(c) Suggests that  f' 2 6.25     B1 CAO 1

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sin 2 x - cos x - 15 (a) Prove the identity / - cos x . [3] 1 + cos x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ sin 2 x - cos x - 1 1 (b) Hence solve the equation = for 0° G x G 360° . [3] 2 + 2 cos x 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 2 2 sin cos 1 1 cos cos 1 1 cos 1 cos         x x x x x x or 2 cos cos 1 cos    x x x 2 2 sin cos 1   x x . Allow use of s, c, t or omission of x throughout. =   cos 1 cos 1 cos    x x x M1 For factorising. = cos  x A1 3 5(b) 1 1 cos 2 4   x ⇒ 1 1 cos 2         x M1 120   x or 240   x A1 A1 FT FT for 360 – their answer. A1 A0 if extra solution(s) in range. SC B1 if answer in radians for both 2π 3 , 4π 3 . 3

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6 y O x 2 The function f is defined by f ( x) = 2 + 4 for x 1 0 . The diagram shows the graph of y = f ( x) . x (a) On this diagram, sketch the graph of y = f -1 ( x) . Show any relevant mirror line. [2] (b) Find an expression for f -1 ( x) . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Solve the equation f ( x) = 4.5 . [1] ............................................................................................................................................................ ............................................................................................................................................................ (d) Explain why the equation f -1 ( x) = f ( x) has no solution. [1] ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) B1 For curve in correct quadrant. B1 Fully correct including line  y x . Horizontal asymptote closer to x axis than vertical asymptote is to y axis. 2 Question Answer Marks Guidance 6(b) 2 2 4 x y   leading to   2 4 2   y x or   2 2 4   y x M1 Allow x and y swapped around.   2 2 4   y x leading to  2 4 y x   or  2 4   x y M1  1 2 f 4 x x       A1 3 6(c)  2  x B1 1 6(d) Because 1 f  is always negative and f is always positive or curves do not intersect B1 Accept other correct answers e.g. ‘f is only defined for positive values of x and f–1 is only defined for negative values of x’ or ‘domains do not overlap’ or ‘the y values cannot be the same’ or ‘the x values cannot be the same’. 1

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7 B C i rad A D 15 cm O 10 cm In the diagram, AOD and BC are two parallel straight lines. Arc AB is part of a circle with centre O and radius 15 cm . Angle BOA = i radians. Arc CD is part of a circle with centre O and radius 10 cm . r radians. Angle COD = 12 (a) Show that i = .07297 , correct to 4 decimal places. 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(b) Find the perimeter and the area of the shape ABCD. Give your answers correct to 3 significant figures. 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Mark scheme: 7(a) Angle  = 1 1 π 10 10 cos or sin 0.7297 2 15 15     B1 Condone working in degrees if converted to radians at the end. AG 1 Question Answer Marks Guidance 7(b) BC = 2 2 15 10 11.18... or 5 5        B1 Arc AB =   15 0.7297 10.9455   B1 Perimeter = their BC + their arc AB + 25 + 5π M1 Perimeter = 62.8 A1 AWRT Area sector AOB =   2 1 15 0.7297 82.09 2    B1 Area = 1 10 2  their BC + their sector AOB + 2 π 10 4  M1 Area = 217 A1 AWRT 7

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8 (a) The first three terms of an arithmetic progression are 25, 4p - 1 and 13- p , where p is a constant. Find the value of the tenth term of the progression. 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(b) The first three terms of a geometric progression are 25, 4q - 1 and 13- q , where q is a positive constant. Find the sum to infinity of the progression. 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Mark scheme: 8(a) 2 4 1 25 13    p p *M1 40 9  p A1 40 74 4 1 25 9 9                  d DM1 Using their p to find d. 10th term = 25 9 49   d A1 Alternative Method for first 3 marks of Question 8(a) 4 26   d p , 14 5   d p , p + 2d = -12 Any two (*M1) Allow unsimplified or equivalent. Solving simultaneously to find p or d (DM1) 40 9 p        , 74 9  d (A1) 4 Question Answer Marks Guidance 8(b)     2 4 1 25 13    q q ⇒   2 16 17 324 0    q q M1     4 16 81 0 q q    leading to  4 q  M1 Solve 3 term quadratic with real solutions.  3 5  r A1 Ignore 17 20  . Sum to infinity = 25 125 3 2 1 5   A1 Ignore extra solution. SC B1 if no method shown for solving quadratic. Alternative Method for Question 8(b) 25r = 4q – 1, 25r2 = 13 – q leading to 2 100r + 25r – 51 = 0 (M1)    5 3 20 17 0    r r (M1) Solve 3 term quadratic with real solutions. r = 3 5 (A1) Ignore 17 20  . Sum to infinity = 25 125 3 2 1 5   (A1) Ignore extra solution. SC B1 if no method shown for solving quadratic. 4

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9 y 1 (1, 1) O 2.4 x 1 The diagram shows part of the curve with equation y = 1 and the lines x = 2.4 and y = 1. The 3 ( 5x - 4 ) curve intersects the line y = 1 at the point (1, 1). Find the exact volume of the solid generated when the shaded region is rotated through 360° about the x-axis. 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Mark scheme: 9 Volume of cylinder = 2 7 7 π 1 π 5 5    May be done using 2.4 1 1  . This would be the only mark available if candidate integrates y. Volume under curve =   2 3 1 π d 5 4 x x   M1 No further marks available if y  . =   1 3 3 π 5 4 5 x         B1 B1 Calculator used for integration scores no further marks. = 1 3 3 3 π 8 1 π 5 5              M1 Uses limits 1, 2.4 in an integral of y2. Volume = 7 3 π π 5 5  = 4π 5 A1 SC B1 if the only error is not showing substitution. 6

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10 The equation of a circle is ( x - 3) 2 + y 2 = 18 . The line with equation y = mx + c passes through the point ( 0 , - 9) and is a tangent to the circle. Find the two possible values of m and, for each value of m, find the coordinates of the point at which the tangent touches the circle. 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Mark scheme: 10 2 2 3 18 9 x y y mx     2 2 3 9 18 x mx     M1 Finding equation of tangent and substituting into circle equation. Must be 9  mx . 2 2 2 6 9 18 81 18 x x m x mx       leading to       2 2 1 6 18 72 0 m x m x      M1 Brackets expanded and all terms collected on one side of the equation. May be implied in the discriminant. m cannot be numeric.       2 2 6 18 4 1 72 0      m m *M1 Use of 2 4  b ac . Not in quadratic formula. m cannot be numeric, c must be numeric.   2 36 216 252 0    m m 2 leading to 6 7 0 m m        DM1 Simplifies to 3 term quadratic. 1 or 7 m m   A1 Condone no method for solving quadratic shown. 1 m  leading to 2 2 24 72 0 x x    leading to 6 x  DM1 Must be correct x for their quadratic. 7 m  leading to 2 50 120 72 0 x x    leading to 6 5 x  DM1 Must be correct x for their quadratic.   6 3 6, 3 , , 5 5          A1 Question Answer Marks Guidance 10 Alternative Method 1 for first 4 marks of Question 10  2 3 1 0 9 1    m m (M1) Use of the formula for the length of a perpendicular from a point to a line.  2 3 1 0 9 1    m m = 18 (M1) Equates length of a perpendicular from a point to a line to the radius. (3 m – 9)2 = 18( 2 m + 1) (M1) Squares and clears the fraction. 9 2 m - 54 m + 81 = 0 2 leading to 6 7 0 m m        (M1) Alternative Method 2 for first 3 marks of Question 10 (3 – x )(9 + 6 x - 2 x )-1/2 = m (M1) OE Differentiates implicitly or otherwise and equates d d y x to m. ( 2 1m ) 2 x – 6(1 + 2) m x + 9(1 – 2 m )[ = 0] (M1) Brackets expanded and all terms collected on one side of the equation. May be implied in the discriminant. 36( 2 1 m )2 – 4( 2 1 m ) × 9(1 – 2 m )[ = 0] (M1) Use of 2 4 . b ac  8

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11 y O x 4 3 A function is defined by f ( x) = 3 - + 2 for x ! 0 . The graph of y = f ( x) is shown in the diagram. x x (a) Find the set of values of x for which f ( x) is decreasing. 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(b) A triangle is bounded by the y-axis, the normal to the curve at the point where x = 1 and the tangent to the curve at the point where x =- 1. Find the area of the triangle. Give your answer correct to 3 significant figures. 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Mark scheme: 11(a) 4 2 d 12 3 d   y x x x 4 2 d 12 3 0 d y x x x    leading to 4 2 3 12 0 x x   or -12 + 3x2 = 0 M1 Set = 0 or uses , ⩽ and simplifies. Must be from 4 2 d d   y A B x x x .   2 2 3 4 0 x x   leading to 2 x only A1 SC B1 for 2  x if M0 scored. 2 0 and 0 2     x x or (-2, 0) and (0, 2) or 2 2   x and x 0  B1FT Allow and/or. B1FT Allow 2 0 and / or 0 2    x x   but only B1B0 if 0 included in either or both. Allow [–2, 0) and (0, 2]. Allow B1B0 for 2 2 x   or (–2, 2). Must be from 4 2 d d   y A B x x x . 5 B marks only available if d d y x 4 2   A B x x . Question Answer Marks Guidance 11(b) [At 1] 3 and tan 9 x y m    *M1 Using their d d y x . 1 1 norm 9 9    m DM1 Equation of normal is   1 1 26 3 1 leading to 9 9 9 y x y x            A1 At 1, 1, 9    x y m M1 Equation of tangent is    1 9 1 leading to 9 8 y x y x     A1 Meet when 1 26 49 9 8 leading to 1.19512 , 9 9 41 x x x             M1 Equates their tangent and their normal. Area = 1 26 1 .19512 8 2 9 their their          M1 If 2 1   y y is used integration must be correct and substitution shown. 6.51 A1 AWRT Accept fraction wrt 6.51 8

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