Cambridge A Level Mathematics 9709 — 2012 May/June Paper 1 · Variant 2

3 The coefficient of x3 in the expansion of is 90. Find the value of the positive (a + x)5 + (2 −x)6 constant a. [5]

Mark scheme: –1 8 g = + 3 , x ≠ 0 B1 Allow if a linear denominator. x [4] +ve gradient, +ve y intercept (ii) +ve gradient, +ve y intercept y = f(x) y = x States, or shows the line y = x as a line y y = f–1(x) B1 of symmetry. B1 B1 [3] x

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1 1 5 (i) Prove the identity tan x [2] + tan x ≡ sin x cos x. 2 (ii) Solve the equation 1 3 tan x, for [4] sin x cos x = + 0◦≤x ≤180◦.

Mark scheme: 1 1 5 tan x + ≡ tan x sin x cos x sin x cos x (i) LHS = + M1 Use of tan = sin/cos twice cos x sin x sin 2 x + cos 2 x 1 M1 Use of s² + c² = 1 appropriately – = = sin x cos x sin xcos x [2] everything correct. 2 (ii) = 3 tan x + 1 sin x cos x M1 Uses part (i) to obtain eqn in tanx only 1 Uses (i) 2(tan x + ) = 3 tan x + 1 tan x → tan2 x + tan x − 2 = 0 DM1 Correct soln of quadratic eqn → tanx = 1 or −2 B1 A1 co. Must have correct quadratic co → x = 45º or 116.6º [4]

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6 A B 2.4 rad 8 cm O The diagram shows a metal plate made by removing a segment from a circle with centre O and radius 8 cm. The line AB is a chord of the circle and angle AOB 2.4 radians. Find = (i) the length of AB, [2] (ii) the perimeter of the plate, [3] (iii) the area of the plate. [3]

Mark scheme: 6 (i) cosine rule or 2×r × sin ½.(2.4) M1 Any complete valid method. → 14.9 cm A1 co [2] (ii) Perimeter = (i) + rθ M1 Uses s = rθ with 2.4, or π − 2.4, θ = 2π − 2.4, B1 or 2π − 2.4 → 46.0 cm A1 Anywhere in parts (ii) or (iii). [3] Adds 31.1 to (i) for . (iii) Area = Sector + triangle ½×8² (2π − 2.4) + ½×8²sin 2.4 M1 M1 124.3 + 21.6 → 146 cm². A1 Uses ½r²θ. Uses any valid method. [3] co

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7 (a) In an arithmetic progression, the sum of the first n terms, denoted by Sn, is given by Sn n2 8n. = + Find the first term and the common difference. [3] (b) In a geometric progression, the second term is 9 less than the first term. The sum of the second and third terms is 30. Given that all the terms of the progression are positive, find the first term. [5]

Mark scheme: 7 (a) Sn = n² + 8n. S1 = 9 → a = 9 B1 co S2 = 20 → a + d = 11 → d = 2 M1 A1 Realises that S2 is a + (a + d). co (or equating n² + 8n with Sn and comparing [3] coefficients) (b) a −ar = 9 B1 co ar + ar 2 = 30 B1 co Eliminates a → 3r 2 + 13r − 10 = 0 M1 Complete elimination of r or a or → 2 a 2 − 57 a + 81 = 0 Correct quadratic. → r = ⅔ A1 → a = 27 A1 co (condone 27 or 1.5) [5] GCE AS/A LEVEL – May/June 2012 9709 12

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8 (i) Find the angle between the vectors 3i and 2i 3j [4] −4k + −6k. The vector −−→OA has a magnitude of 15 units and is in the same direction as the vector 3i The −4k. vector −−→OB has a magnitude of 14 units and is in the same direction as the vector 2i 3j + −6k. (ii) Express −−→OA and −−→OB in terms of i, j and k. [3] (iii) Find the unit vector in the direction of −−→AB. [3] [Questions 9 and 10 are printed on the next page.]

Mark scheme: 8 (i) 3i − 4k, 2i + 3j − 6k. Dot product = 6 + 24 = 30 M1 Uses x1x2 + y1y2 + z1z2 = 25 × 49 cos θ M1 Method for modulus → angle = 31º or 0.54(1) radians. M1 A1 Links everything correctly. co [4] (ii) OA = (3i − 4k) × (15 ÷ 5) → 9i − 12k M1 A1 M mark for ×(15 ÷ 5) or ×(14 ÷ 7) OB = (2i + 3j − 6k) × (14 ÷ 7) A1 for OA → 4i + 6j − 12k. A1 A1 for OB [3] (iii) AB = b − a = −5i + 6j M1 Correct use for either AB or BA → Magnitude of 61 or 7.81 M1 Complete method for unit vector. A1 co → Unit vector of (−5i + 6j) ÷ 61 [3] 2

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9 y A y = – x2 + 8x –10 B x O The diagram shows part of the curve y 8x which passes through the points A and B. The = −x2 + −10 curve has a maximum point at A and the gradient of the line BA is 2. (i) Find the coordinates of A and B. [7] (ii) Find y dx and hence evaluate the area of the shaded region. [4] ã

Mark scheme: 9 y = − x 2 + 8 x − 10 dy (i) = −2x + 8 B1 co dx = 0 when x = 4, A is (4, 6) M1A1 Sets to 0 and attempt to solve for x. co. Equation of AB is y − 6 = 2 ( x − 4 ) M1 Correct form of equation. Sim eqns with eqn of curve M1 Eliminates x or y completely → x 2 −x6 + 8 = 0 or y 2 −y8 + 12 = 0 A1 Method for quadratic eqn = 0. → B (2, 2) A1 co (Must not be guessed from diagram) [7] 2 x 3 2 (ii) ∫ − x + 8 x − 10 dx = − 3 + 4 x − 10 x B2,1 3 terms, loses 1 for each error Uses his x limits 2 to 4 M1 Uses x limits correctly – allow ± → 9⅓ A1 co – allow ± (2 must have been [4] correctly found, not guessed) GCE AS/A LEVEL – May/June 2012 9709 12 8 10 f : x a 2 x + 5 g : x a x − 3 –1 B1 co (i) f = ½(x − 5) M1 A1 Attempt at x the subject. co but (f(x) 1 8

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