Cambridge A Level Mathematics 9709 — 2019 May/June Paper 1 · Variant 2

@2 A5 1 Find the coefficient of x in the expansion of . [3] x −3x ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 For 5 2 3   −     x x term in x is 10 or 5C3 or 5C2 × 2 2       x × (−3x)³ or 5 3 2 2 5.4.3 3 3! 2     −         x x or ( ) 2 5 2 5.4 2 3 2! 3   −     x x B2,1 without x’s. Can be seen in an expansion. −1080 identified B1 Allow −1080x Allow if expansion stops at this term. Allow from expanding brackets. 3

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dy 3 A curve is such that x3 . The point P 2, 9 lies on the curve. dx = −4x2 (i) A point moves on the curve in such a way that the x-coordinate is decreasing at a constant rate of 0.05 units per second. Find the rate of change of the y-coordinate when the point is at P. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the equation of the curve. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) d d d d d d = × y y x t x t = 7 × – 0.05 M1 −0.35 (units/s) or Decreasing at a rate of (+) 0.35 A1 Ignore notation and omission of units 2 3(ii) ( ) 4 4 4 = + x y x (+c) oe B1 Accept unsimplified Uses (2, 9) in an integral to find c. M1 The power of at least one term increase by 1. c = 3 or ( ) 4 4 4 y x x = + + 3 oe A1 A0 if candidate continues to a final equation that is a straight line. 3

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4 Angle x is such that sin x a b and cosx a where a and b are constants. = + = −b, (i) Show that a2 b2 has a constant value for all values of x. [3] + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) In the case where tan x 2, express a in terms of b. [2] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) 2 2 2 2 2 , 2 + + − + a ab b a ab b B1 sin²x + cos²x = 1 used → ( ) ( ) 2 2 1 + + − = a b a b M1 Appropriate use of sin²x + cos²x = 1 with ( ) 2 + a b and ( ) 2 − a b a² + b² = ½ A1 No evidence of ±2ab, scores 2/3 Alternative method for question 4(i) 2a = (s+c) & 2b = (s−c) or a = ½(s+c) & b = ½(s−c) B1 a²+b² = ( ) ( ) 2 2 1 1 4 4 + + − s c s c = ½(s²+c²) M1 Appropriate use of sin²x + cos²x = 1 a² + b² = ½ A1 Method using only 2 2 (sin ) and ( cos ) − − x b a x scores 0/3. 3 SC B1 for assuming θ is acute giving 1 5 = + a b or 2 5 −b Question Answer Marks Guidance 4(ii) sin tan cos = x x x → 2 + = − a b a b M1 Use of sin tan cos = x x x to form an equation in a and b only a = 3b A1 2

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5 C 1 rad A O r B The diagram shows a semicircle with diameter AB, centre O and radius r. The point C lies on the circumference and angle AOC radians. The perimeter of sector BOC is twice the perimeter of = 1 sector AOC. Find the value of correct to 2 significant figures. [5] 1 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 Perimeter of AOC = 2r + rθ B1 Angle COB = π – θ B1 Could be on the diagram. Condone 180 – θ. Perimeter of BOC = 2r + r(π – θ) B1 FT on angle COB if of form ( ) π θ − k , k 0. > (2r +) πr – rθ = 2((2r) + rθ) (2 + π – θ = 4 + 2θ → 2 3 π θ − = ) M1 Sets up equation using ( ) π θ − r k and ×2 on correct side. Condone any omissions of OA, OB and/or OC. θ = 0.38 A1 Equivalent answer in degrees scores A0. 5

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3x 6 The equation of a curve is y 3 cos 2x and the equation of a line is 2y 5. = + = 0 (i) State the smallest and largest values of y for both the curve and the line for 0 [3] ≤x ≤20. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 3x (ii) Sketch, on the same diagram, the graphs of y 3 cos 2x and 2y 5 for 0 [3] = + = ≤x ≤20. 0 (iii) State the number of solutions of the equation 6 cos 2x 5 for 0 [1] = −3x ≤x ≤20. 0 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) 3, –3 B1 Accept ± 3 −½ B1 2½ B1 3 Condone misuse of inequality signs. 6(ii) Only mark the curve from 0 → 2π. If the x axis is not labelled assume that 0 → 2π is the range shown. Labels on axes are not required. 2 complete oscillations of a cosine curve starting with a maximum at (0,a), a൐0 B1 Fully correct curve which must appear to level off at 0 and/or 2π. B1 Line starting on positive y axis and finishing below the x axis at 2π. Must be straight. B1 3 6(iii) 4 B1 1

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8 The position vectors of points A and B, relative to an origin O, are given by ` a ` a −−→OA 6 and −−→OB 3k , = −2 = −6 −3 where k is a constant. (i) Find the value of k for which angle AOB is [2] 90Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the values of k for which the lengths of OA and OB are equal. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The point C is such that −−→AC 2−−→CB. = (iii) In the case where k 4, find the unit vector in the direction of −−→OC. [4] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(i) 6×3+-2×k+-6×-3 = 0 (18 – 2k + 18 = 0) M1 Could be JJJG AO . JJJG OB , JJJG AO . JJJG BO or JJJG OA . JJJG BO k = 18 A1 Alternative method for question 8(i) 76 + 18 + k2 = 18 + (k + 2)2 M1 Use of Pythagoras with appropriate lengths. k = 18 A1 2 8(ii) 36 + 4 + 36 = 9 +k² + 9 M1 Use of modulus leading to an equation and solve to k= or k2 = k = ±√58 or ±7.62 A1 Accept exact or decimal answers. Allow decimals to greater accuracy. 2 Question Answer Marks Guidance 8(iii) JJJG AB = 3 6 3 −           → 2 4 2 −     =       JJJG AC then + JJJG JJJG OA AC M1 Complete method using JJJG AC = ± ⅔ JJJG AB And then + JJJG JJJG OA their AC = JJJG OC 4 2 4         −   A1 ÷ ( ) ( ) ( ) 2 2 2 4 2 4 + + − their their their M1 Divides by modulus of their JJJG OC = 4 1 2 6 4         −   or 1 6 (4i + 2j – 4k) A1 Alternative method for question 8(iii) Let JJJG OC =           p q r → 6 2 & 6 −     = +     +   JJJG JJJG p AC q CB r = 3 4 3 −     −     −−   p q r M1 Correct method. Equates coefficients leading to values for p, q, r p – 6 = 2(3 – p); q+2 = 2(4 – q); r + 6 = 2 (–3 – r) →p=4, q=2 & r= – 4 A1 ÷ ( ) ( ) ( ) 2 2 2 4 2 4 + + − their their their M1 Divides by modulus of their JJJG OC = 4 1 2 6 4         −   or 1 6 (4i + 2j – 4k) A1 Question Answer Marks Guidance 8(iii) Alternative method for question 8(iii) JJJG CB = − JJJG JJJG OB OC ( ) 2 ∴ − JJJG JJJG OB OC = JJJG OC – JJJG OA →2 JJJG OB + JJJG OA = 3 ∴ JJJG OC 3 JJJG OC = 12 6 12         −   M1 Correct method. Gets to a numerical expression for k JJJG OC from & JJJG JJJG OA OB . JJJG OC = 4 2 4         −   A1 ÷ ( ) ( ) ( ) 2 2 2 4 2 4 + + − their their their M1 Divides by modulus of their JJJG OC = 4 1 2 6 4         −   or 1 6 (4i + 2j – 4k) A1 4

More questions on Vectors

9 The curve C1 has equation y x2 7. The curve C2 has equation y2 4x k, where k is a = −4x + = + constant. The tangent to C1 at the point where x 3 is also the tangent to C2 at the point P. Find the = value of k and the coordinates of P. 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Mark scheme: 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. 2x – 2= 4 + x k (→ 4 ² 12 4 0 − + − = x x k ) *M1 Forms an equation in one variable using tangent & C2 Use of ² 4 − b ac = 0 on a 3 term quadratic set to 0. *DM1 Uses ‘discriminant = 0’ 144 = 16(4 – k) → k = − 5 A1 4 ² 12 4 0 − + − = x x k → 4 ² 12 9 0 − + = x x DM1 Uses k to form a 3 term quadratic in x x = 3 2 1 2       or , y = 1(or – 1). A1 Condone ‘correct’ extra solution. Alternative method for question 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. For C2: 1 2 (4 ) − = + dy A x k dx *M1 Finds dy dx for C2 in the form 1 2 (4 ) − + A x k Question Answer Marks Guidance 9 At P: ‘their 2’ = 1 2 (4 ) " − + A x k → ( 1 4 1 4 − = + = k x or x k ) *DM1 Equating ‘their 2’ to ‘their dy dx ’ and simplify to form a linear equation linking 4x + k and a constant. ( ) 2 2 2 4 − = + x x k → ( ) ( ) 2 2 2 2 1 4 8 3 0 − = → − + = x x x DM1 Using their 2 2 = − y x , y2 = 4x + k and their 4 1 + = x k (but not =0) to form a 3 term quadratic in x. 3 1 2 2   =     x or and from ( ) 5 1 k or = − − A1 Needs correct values for x and k. from y2 = 4x + k, y = 1(or – 1). A1 Condone ‘correct’ extra solution. Alternative method for question 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. For C2: 1 2 (4 ) − = + dy A x k dx *M1 Finds dy dx for C2 in the form 1 2 (4 ) − + A x k At P: ‘their 2’ = 1 2 (4 ) " − + A x k → ( 1 4 1 4 − = + = k x or x k ) *DM1 Equating ‘their 2’ to ‘their dy dx ’ and simplify to form a linear equation linking 4x + k and a constant. From 4x + k = 1 and y2 = 4x + k → y2 = 1 DM1 Using their 4x + k = 1 (but not =0) and C2 to form y2 = a constant Question Answer Marks Guidance 9 y = 1(or – 1) and 3 1 2 2   =     x or A1 Needs correct values for y and x. From 4 1 x k + = , k = –5 ( or – 1) A1 Condone ‘correct’ extra solution 8

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10 (a) In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is a. (i) Show that the common difference of the progression is 3a.1 [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Given that the tenth term is 36 more than the fourth term, find the value of a. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term. 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Mark scheme: 10(a)(i) S10 = S15 – S10 or S10 = S(11 to 15) M1 5(2a + 9d) oe B1 7.5(2a + 14d) – 5(2a + 9d) or 5[ 2 (a + 10d) + (a+14d)] oe A1 d = 3 a AG A1 Correct answer from convincing working 4 Condone starting with d = 3 a and evaluating both summations as 25a. 10(a)(ii) ( ) ( ) 9 36 3 + = + + a d a d M1 Correct use of ( )1 + − a n d twice and addition of ±36 a = 18 A1 2 Correct answer www scores 2/2 Question Answer Marks Guidance 10(b) S∞ = 9 ×S4; ( ) 4 1 9 1 1 − = − − a r a r r or 9(a + ar+ ar2+ar3) B1 May have 12 in place of a. 9(1 – rn) = 1 where n = 3,4 or 5 M1 Correctly deals with a and correctly eliminates ‘1 – r’ r4 = 8 9 oe A1 (5th term =) 10⅔ or 10.7 A1 4 Final answer of 10.6 suggests premature approximation – award 3/4 www.

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11 y 9 M y = 4x + 1 + 4x + 1 x O 9 The diagram shows part of the curve y 4x 1 and the minimum point M. = + + 4x 1 + dy (i) Find expressions for and y dx. [6] dx Ó ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the coordinates of M. 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The shaded region is bounded by the curve, the y-axis and the line through M parallel to the x-axis. (iii) Find, showing all necessary working, the area of the shaded region. 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Mark scheme: 11(i) ( ) 1 2 d 1 4 1 d 2 −   = +     y x x [× 4] ( ) 3 2 9 4 1 2 −   − +     x [× 4] B1B1B1 ( ) ( ) 3 3 2 2 18 8 16 4 1 4 1 4 1   −   −   +   + +   x or x x x SC If no other marks awarded award B1 for both powers of (4x +1) correct. ∫ydx = ( ) 3 2 4 1 3 2     +       x [÷ 4] + ( ) 1 2 9 4 1 1 2     +       x [÷ 4] (+C) B1B1B1 B1 B1 for each, without ÷ 4. B1 for ÷4 twice. + C not required. ( ) ( ) 3 4 1 9 4 1 ( ) 6 2 x x C   +   + + +       SC If no other marks awarded , B1 for both powers of (4x +1) correct. 6 11(ii) d 0 d = y x → 2 4 1 + x − ( ) 3 2 18 4 1 + x = 0 M1 Sets their d d y x to 0 (and attempts to solve 4x + 1 = 9 or (4x + 1)2 = 81 A1 Must be from correct differential. x = 2, y = 6 or M is (2, 6) only. A1 Both values required. Must be from correct differential. 3 Question Answer Marks Guidance 11(iii) Realises area is ∫y dx and attempt to use their 2 and sight of 0. *M1 Needs to use their integral and to see ‘their 2’ substituted. Uses limits 0 to 2 correctly → [4.5 + 13.5] – [ 1 6 + 4.5] ( = 13⅓ ) DM1 Uses both 0 and ‘their 2’ and subtracts. Condone wrong way round. (Area =) 1⅓ or 1.33 A1 Must be from a correct differential and integral. 3 13⅓ or 1⅓ with little or no working scores M1DM0A0.

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