Cambridge A Level Mathematics 9709 — 2011 May/June Paper 1 · Variant 2

cos θ 1 5 (i) Prove the identity . [3] tan sin θ ≡1 + θ(1 −sin θ) cos θ (ii) Hence solve the equation 4, for [3] tan = 0◦≤θ ≤360◦. θ(1 −sin θ)

Mark scheme: cos θ cos θ 5 (i) ≡ M1 Use of t = s ÷ c tan θ 1( − sin θ ) sin θ 1( − sin θ ) 1 − sin 2 θ = M1 Replaces cos²θ with 1 − sin²θ to form sin θ 1( − sin θ ) f(sinθ). 1+ sin θ 1 = = + 1 A1 AG. Ensure all ok. Must show difference of sin θ sin θ [3] 2 squares. cos θ 1 (ii) = 4 → + 1 = 4 M1 Linking up to obtain sinθ = k. tan θ 1( − sin θ ) sin θ → sinθ = ⅓ → θ = 19.5º, 160.5º A1 A1√ co. √ 180º − 1st answer providing there are [3] no other solutions in the range 0º to 360º. x + 3

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3 6 The function f is defined by f : x x x 2. + 2x →x ∈>, ≠1 −1, (i) Show that x. [3] ff(x) = (ii) Hence, or otherwise, obtain an expression for f [2] −1(x).

Mark scheme: x + 3 6 (i) f(x) = 2 x − 1 x + 3 + 3 B1 Replacing “x” twice - must be correct 7x 2 x − 1 ff(x) = = = x M1 Correct algebra – clearing (2x − 1) 2( x + 3) 7 A1 AG – all correct. − 1 2 x − 1 [3] x + 3 (ii) y = 2 x − 1 → 2 xy − y = x + 3 M1 Attempt to make x the subject and complete → x ( 2 y − )1 = y + 3 method x + 3 → f –1(x) = A1 co 2 x − 1 [2] or since ff(x) = x, x + 3 f –1(x) = f(x) = (M1, A1) 2 x − 1

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7 The line L1 passes through the points A and B The line L2 is parallel to L1 and passes through the origin. The point C lies on L2 (2,such5) that AC(10,is9).perpendicular to L2. Find (i) the coordinates of C, [5] (ii) the distance AC. [2]

Mark scheme: 7 (i) (2, 5) to (10, 9) gradient = ½ B1 co Equation of L2 y = 12 x . B1√ √ on gradient of L1 Gradient of perpendicular = − 2 M1 Use of m1m2 = −1 Eqn of Perp y − 5 = − 2 ( x − 2 ) M1 Correct form of line eqn Sim Eqns → C(3.6, 1.8) A1 co [5] (ii) d² = 1.6² + 3.2² → d = 3.58 M1 Correct method for AC A1 co (accept with √5 in answer) [2] GCE AS/A LEVEL – May/June 2011 9709 12 8 (i) BA. BC or AB .CB B1 Correct two vectors for angle ABC.    

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