Cambridge A Level Mathematics 9709 — 2017 May/June Paper 1 · Variant 2

2 The point A has coordinates −2, 6 . The equation of the perpendicular bisector of the line AB is 2y = 3x + 5. (i) Find the equation of AB. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the coordinates of B. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Gradient = 1.5 Gradient of perpendicular = −⅔ B1 Equation of AB is ( ) 6 2 x − = − + y ⅔ Or 3 2 14 y x + = oe M1 A1 Correct use of straight line equation with a changed gradient and (− 2, 6), the (− (− 2)) must be resolved for the A1 ISW. Using = + y mx c gets A1 as soon as c is evaluated. Total: 3 2(ii) Simultaneous equations → Midpoint (1, 4) M1 Attempt at solution of simultaneous equations as far as x =, or y =. Use of midpoint or vectors → B (4, 2) M1A1 Any valid method leading to x, or to y. Total: 3

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1 2 1 −sin 1 3 (i) Prove the identity −tan 1  [3] cos 1 1 + sin 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A2 1 1(ii) Hence solve the equation −tan 1 = 2, for 0Å ≤1 ≤360Å. [3] cos 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) LHS = 2 1 s c   −     c M1 Eliminates tan by replacing with sin cos leading to a function of sin and/or cos only. = ( ) 2 1 1 ² − − s s M1 Uses s² + c² = 1 leading to a function of sin only. = ( )( ) ( )( ) 1 1 1 1 − − − + s s s s =1 sin 1 sin θ θ − + A1 AG. Must show use of factors for A1. Total: 3 3(ii) Uses part (i) → 2 – 2s = 1 + s → s = ⅓ M1 Uses part (i) to obtain s = k θ = 19.5º or 160.5° A1A1 FT FT from error in 19.5° Allow 0.340ᶜ (0.3398ᶜ) & 2.80(2) or 0.108πᶜ & 0.892πᶜ for A1 only. Extra answers in the range lose the second A1 if gained for 160.5°. Total: 3 θ θ etc

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4 O r cm 21 rad A B r cm D C The diagram shows a circle with radius r cm and centre O. Points A and B lie on the circle and ABCD is a rectangle. Angle AOB = 21 radians and AD = r cm. (i) Express the perimeter of the shaded region in terms of r and 1. 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(ii) In the case where r = 5 and 1 = 160, find the area of the shaded region. 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Mark scheme: 4(i) (AB) = 2rsinθ (or 2 2 2θ − r cos or 2 sin 2 π θ   −     rsin ) B1 Allow unsimplifed throughout eg r + r, 2 2 (Arc AB) = 2rθ B1 (P =) 2r + 2rθ + 2rsinθ (or 2 2 2 2 or sin 2 θ θ π θ −   −     rsin r cos ) B1 Total: 3 Question Answer Marks Guidance 4(ii) Area sector AOB = ( ½ r² 2θ) 25 or1 3.1 6 π B1 Use of segment formula gives 2.26 B1B1 Area triangle AOB = (½×2rsinθ×rcosθ or ½ × r2 sin2θ) 25 3 or1 0.8 4 B1 Area rectangle ABCD = (r × 2rsinθ) 25 B1 (Area =) Either 25 – (25π/6 – 25√3/4) or 22.7 B1 Correct final answer gets B4. Total: 4 =

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12 5 A curve has equation y = 3 + 2 −x. (i) Find the equation of the tangent to the curve at the point where the curve crosses the x-axis. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) A point moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.04 units per second. Find the rate of change of the y-coordinate when x = 4. 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Mark scheme: 5(i) Crosses x-axis at (6, 0) B1 6 x is sufficient. d d y x = (0 +) −12 (2 – x)−2 × (−1) B2,1,0 −1 for each incorrect term of the three or addition of + C. Tangent ( ) 6 y x = − ¾ or 4 3 18 y x = − M1 A1 Must use dy/dx, x= their 6 but not x = 0 (which gives m = 3), and correct form of line equation. Using = + y mx c gets A1 as soon as c is evaluated. Total: 5 5(ii) If x = 4, dy/dx = 3 d 3 0.04 0.12 d y t = × = M1 A1FT M1 for (“their m” from d d y x and x = 4) × 0.04. Be aware: use of x = 0 gives the correct answer but gets M0. Total: 2

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7 (a) The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20 000. 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(b) A geometric progression has a first term of 6 and a sum to infinity of 18. A new geometric progression is formed by squaring each of the terms of the original progression. Find the sum to infinity of the new progression. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) (Sn =) ( ) 32 1 8 2  + −    n n and 20000 M1 M1 correct formula used with d from 16 24 d A1 A1 for correct expression linked to 20000. → n² + 3n – 5000 (<,=,> 0) DM1 Simplification to a three term quadratic. → (n = 69.2) → 70 terms needed. A1 Condone use of 20001 throughout. Correct answer from trial and improvement gets 4/4. Total: 4 Question Answer Marks Guidance 7(b) a = 6, 18 1 = − a r → r = ⅔ M1A1 Correct S∞ formula used to find r. New progression a = 36, r = 4 9 oe M1 Obtain new values for a and r by any valid method. New S∞ = 36 4 1 9 − → 64.8 or 324 5 oe A1 (Be aware that r =−⅔ leads to 64.8 but can only score M marks) Total: 4 uuur uuur uuur uuur

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8 Relative to an origin O, the position vectors of three points A, B and C are given by −−→ −−→ −−→ OA = 3i + pj −2pk, OB = 6i + p + 4 j + 3k and OC = p −1 i + 2j + qk, where p and q are constants. (i) In the case where p = 2, use a scalar product to find angle AOB. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−→ −−→ (ii) In the case where AB is parallel to OC, find the values of p and q. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(i) Uses scalar product correctly: 3 × 6 + 2 × 6 + (−4) × 3 = 18 M1 Use of dot product with or OA AO & or OB BO only. uuuur OA = 29 , uuuur OB = 9 M1 Correct method for any one of uuur OA , uuur AO , uuur OB or . uuur BO 29 × 9 × cos AOB = 18 M1 All linked correctly. → AOB = 68.2° or 1.19ᶜ A1 Multiples of π are acceptable (e.g. 0.379πᶜ) Total: 4 8(ii) = uuur AB 3i + 4j +(3+2p)k *M1 For use of − uuur uuur OB OA , allow with p = 2 Comparing “j” DM1 For comparing, uuur OC must contain p & q. Can be implied by = uuur AB 2 uuur OC . → p = 2½ and q = 4 A1 A1 Accuracy marks only available if uuur AB is correct. Total: 4

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9 The equation of a curve is y = 8 x −2x. (i) Find the coordinates of the stationary point of the curve. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ d2y (ii) Find an expression for and hence, or otherwise, determine the nature of the stationary point. dx2 [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the values of x at which the line y = 6 meets the curve. 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(iv) State the set of values of k for which the line y = k does not meet the curve. 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Mark scheme: 9(i) ½ d 4 2 d − = − y x x B1 Accept unsimplified. = 0 when x = 2 x = 4, y = 8 B1B1 Total: 3 9(ii) 3 2 d² 2 d ² − = − y x x B1FT FT providing –ve power of x d² 1 d ² 4   = −     y x → Maximum B1 Correct d² d ² y x and x=4 in (i) are required. Followed by“< 0 or negative” is sufficient” but d² d ² y x must be correct if evaluated. Total: 2 9(iii) EITHER: Recognises a quadratic in x (M1 Eg x =u → 2 ² 8 6 0 − + = u u 1 and 3 as solutions to this equation A1 → x = 9, x = 1. A1) Question Answer Marks Guidance OR: Rearranges then squares (M1 x needs to be isolated before squaring both sides. → 2 10 9 0 − + = x x oe A1 → x = 9, x = 1. A1) Both correct by trial and improvement gets 3/3 Total: 3 9(iv) k > 8 B1 Total: 1

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