Cambridge A Level Mathematics 9709 — 2008 May/June Paper 1 · Variant 1

1 In the triangle ABC, AB = 12 cm, angle BAC = 60◦and angle ACB = 45◦. Find the exact length of BC. [3]

Mark scheme: 3 1 sin 60 = and sin 45 = B1 Both of these correct

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2 (i) Show that the equation 2 tan2 θ cos θ = 3 can be written in the form 2 cos2 θ + 3 cos θ −2 = 0. [2] (ii) Hence solve the equation 2 tan2 θ cos θ = 3, for 0◦≤θ ≤360◦. [3]

Mark scheme: 2 2 A1 Co m st be in s rd form

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8 Functions f and g are defined by f : x →4x −2k for x ∈ , where k is a constant, 9 g : x → for x ∈ , x ≠2. 2 −x (i) Find the values of k for which the equation fg(x) = x has two equal roots. [4] (ii) Determine the roots of the equation fg(x) = x for the values of k found in part (i). [3]

Mark scheme: (ii) x 2 + 8 x + 16 = 0 , x 2 − 16 x + 64 = 0 M1 Substituting one of the values of k. x = −4 or x = 8. A1 A1 [3] d y k 9 = d x x 3 x −2 (i) Integrating y = − k (+ c) B1 ok unsimplified − 2 k Sub (1,18) 18 = + c M1 Substitutes once (even if without + c) 2 k Sub (4,3) 3 = + c M1 2nd substitution and solution of simultaneous 32 equations for k and c → k = 32, c = 2 A1 co [4]  16  (ii) Area = − + 2 x from 1 to 1.6 B1 B1 co  x  → [−10 + 3.2] − [−16 + 2] = 7.2 M1 A1 Use of limits in an integral. co. [4] 10 OA = 2i + j + 2k, OB = 3i − 2j + pk (i) (2i + j + 2k) . (3i − 2j + pk) = 0 M1 For x1x2+y1y2+z1z2 (in (i) or (ii)) → 6 − 2 + 2p = 0 → p = −2 A1 co [2] (ii) (2i + j + 2k) . (3i − 2j + 6k) nb Part (ii) gains 4 marks if (i) missing. → 6 − 2 + 12 allow for ± this A1 co (M1 here if (i) not done)

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