Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 2 · Variant 1

1 Use logarithms to solve the equation 5x+3 = 7x−1, giving the answer correct to 3 significant figures. [4]

Mark scheme: 1 Introduce logarithms and use power law twice M1* Obtain ( x + 3) log 5 = ( x − )1 log 7 or equivalent A1 Solve linear equation for x M1 dep Obtain 20.1 A1 [4]

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2 A curve has equation 3x 1 y + . = x −5 Find the coordinates of the points on the curve at which the gradient is [5] −4.

Mark scheme: 2 Use quotient rule or, after adjustment, product rule M1* 3 x − 15 − 3 x − 1 Obtain or equivalent A1 ( x − 52) Equate first derivative to –4 and solve for x M1 dep Obtain x-coordinates 3 and 7 or one correct pair of coordinates A1 Obtain y-coordinates –5 and 11 respectively or other correct pair of coordinates A1 [5]

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4 (i) By sketching a suitable pair of graphs, show that the equation 4 ln x = −12x has exactly one real root, [2] !. (ii) Verify by calculation that 4.5 5.0. [2] < ! < (iii) Use the iterative formula xn+1 = 8 −2 ln xn to find ! correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 4 (i) Make a recognisable sketch of y = ln x B1 Draw straight line with negative gradient crossing positive y-axis and justify one real root B1 [2] 1 (ii) Consider sign of ln x + x − 4 at 4.5 and 5.0 or equivalent M1 2 Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 4.84 A1 Show sufficient iterations to justify accuracy to 2 d.p. or show sign change in interval (4.835, 4.845) A1 [3]

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5 (a) Find tan2x sin 2x dx. [3] Ó + 1 (b) Find the exact value of dx. [4] Ó 0 3e1−2x

Mark scheme: 5 (a) Use tan 2 x = sec 2 x − 1 B1 Obtain integral of form p tan x + qx + r cos 2 x M1 1 Obtain tan x − x − cos 2 x + c A1 [3] 2 (b) Obtain integral of form k1e− 2 x M1* 3 − 2 x Obtain − 1e A1 2 Apply both limits the correct way round M1 dep 3 −1 3 Obtain − e + e or exact equivalent A1 [4] 2 2

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7 y A D B x O C The parametric equations of a curve are x 6 sin2t, y 2 sin 2t 3 cos 2t, = = + for 0 The curve crosses the x-axis at points B and D and the stationary points are A and C, as ≤t < 0. shown in the diagram. dy 2 (i) Show that cot 2t [5] 3 −1. dx = (ii) Find the values of t at A and C, giving each answer correct to 3 decimal places. [3] (iii) Find the value of the gradient of the curve at B. [3]

Mark scheme: dx 7 (i) Obtain 12 sin t cos t or equivalent for B1 dt dy Obtain 4 cos 2t − 6 sin 2t or equivalent for B1 dt dy Obtain expression for in terms of t M1 dx Use 2 sin t cos t = sin 2t A1 dy 2 Confirm given answer = cot 2t − 1 with no errors seen A1 [5] dx 3 2 (ii) State or imply tan 2t = B1 3 Obtain t = .0 294 B1 Obtain t = .1 865 B1 [3] (iii) Attempt solution of 2 sin 2t + 3 cos 2t = 0 at least as far as tan t2 = K M1 3 Obtain tan 2t = − or equivalent A1 2 13 Substitute to obtain − A1 [3] 9

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