Cambridge A Level Mathematics 9709 — 2010 May/June Paper 1 · Variant 2

2 y a y = x x O 1 3 a The diagram shows part of the curve y = x, where a is a positive constant. Given that the volume obtained when the shaded region is rotated through 360◦about the x-axis is 24π, find the value of a. [4]

Mark scheme: 1 (i) 3( 2 sin x − cos x ) = 2 (sin x − 3 cos x ) M1 E di ll i f

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4 y L1 C (–1, 3) L2 A B (3, 1) x O In the diagram, A is the point (−1, 3) and B is the point (3, 1). The line L1 passes through A and is parallel to OB. The line L2 passes through B and is perpendicular to AB. The lines L1 and L2 meet at C. Find the coordinates of C. [6]

Mark scheme: 3 f : x a 4 x − 2 x 2 ,

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5 Relative to an origin O, the position vectors of the points A and B are given by −2 4 −−→ −−→ OA = 3 and OB = 1 . 1 ! p ! −−→ −−→ (i) Find the value of p for which OA is perpendicular to OB. [2] −−→ (ii) Find the values of p for which the magnitude of AB is 7. [4]

Mark scheme: g : x a 5 x + 3 . (i) Turning point at x = 1. M1 Calculus or completing the square etc. Range is Y 2. A1 Condone < instead of Y. [2] (ii) gf(x) = 5( 4 x −x2 2 ) + 3 B1 For putting f into g. = k and use of b 2 − 4 ac M1 Setting to k, using b 2 − 4 ac → k = 13 A1 co [3] 4 Gradient of L1 is 13 . Equation of L1 is y − 3 = 13 ( x + )1 M1 A1 M1 for equation for his m. A1 co. Gradient of AB is − 12 . Perp = 2. M1 Use of m1 m2 = −1 Equation of L2 is y − 1 = 2 ( x − 3) . A1 co

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6 (i) Find the first 3 terms in the expansion of (1 + ax)5 in ascending powers of x. [2] (ii) Given that there is no term in x in the expansion of (1 −2x)(1 + ax)5, find the value of the constant a. [2] (iii) For this value of a, find the coefficient of x2 in the expansion of (1 −2x)(1 + ax)5. [3]

Mark scheme: → 6 s − 3c = 2 s − 6 c → 4 s = −3c M1 Expanding, collecting, use of t = s ÷ c 3 A1 Answer given. All correct. → tan x = − 4 [2] (ii) x = 180 – 36.9 = 143.1° or B1 co x = 360 – 36.9 = 323.1° B1√ For 180 + first answer. [2] a 2 y = x  a 2  − a2  M1 π. For using correct formula with dx = (π )   Volume = π ∫  x B1 only For correct integration of x–2    x 2  2πa 2 Use of limits 1 to 3 → M1 Must be using y2 or πy2. 3 Equates to 24π → a = 6 A1 Co, allow ±6. [4] 2

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