Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 2 · Variant 2

2 1 (i) Find dx. [2] Ô 4x −1 7 2 (ii) Hence find dx, expressing your answer in the form ln a, where a is an integer. [3] 4x Ô1 −1

Mark scheme: 1 (i) State indefinite integral of the form k ln (4x – 1), where k = 2, 4, or ½ M1 State correct integral ½ ln (4x – 1) A1 [2] (ii) Substitute limits correctly M1 Use law for the logarithm of a power or a quotient M1 Obtain ln 3 correctly A1 [3]

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e3x−12 The curve y has one stationary point. Find the coordinates of this stationary point. [5] 2x =

Mark scheme: 2 Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate (numerator) of derivative to zero and solve for x DM1 Obtain x = 13 A1 Obtain y = 32 A1 [5]

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3 Solve the equation 2 cosec 10, giving all solutions in the interval [6] cot21 −5 1 = 0Å ≤1 ≤360Å.

Mark scheme: 3 Use trig identity correctly to obtain a quadratic in cosec θ or sin θ M1 Solve the quadratic correctly M1 Obtain sin θ = ¼ or − ⅔ A1 Obtain one correct answer A1 Carry out correct method for second answer from either root DM1 Obtain remaining 3 answers from 14.5, 165.5, 221.8, 318.2 and no others in the range A1 [Ignore answers outside the given range] [6]

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6 (a) Find sin x x 2 dx. [4] Ó −cos (b) (i) Use the trapezium rule with 2 intervals to estimate the value of 1 20 cosec x dx, 1 Ó 40 giving your answer correct to 3 decimal places. [3] (ii) Using a sketch of the graph of y cosec x for 0 x explain whether the trapezium rule gives an under-estimate or an=over-estimate of< the≤1true20, value of the integral in part (i). [2]

Mark scheme: 6 (a) Expand brackets and use sin2 x + cos2 x = 1 M1 Obtain 1 – sin 2x A1 Integrate and obtain term of form ±k cos 2x, where k = ½, 1 or 2 M1 cos 2 x State correct integral x + ( + c ) A1 [4] 2 (b) (i) State or imply correct ordinates 1.4142…, 1.0823…, 1 B1 π Use correct formula, or equivalent, correctly with h = and three ordinates M1 8 Obtain answer 0.899 with no errors seen A1 [3] (ii) Make a recognisable sketch of y = cosec x for 0 < x Y 12 π B1 Justify statement that the trapezium rule gives an over-estimate B1 [2] GCE AS LEVEL – October/November 2013 9709 22

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7 y R x O a The diagram shows part of the curve y 8x 12ex. The shaded region R is bounded by the curve and = + 1 by the lines x 0, y 0 and x a, where a is positive. The area of R is equal to 2. = = = (i) Find an equation satisfied by a, and show that the equation can be written in the form O@2 A a . −ea 8 = O@2 A (ii) Verify by calculation that the equation a has a root between 0.2 and 0.3. [2] −ea 8 = _P2 Q (iii) Use the iterative formula to determine this root correct to 2 decimal places. −ean 8 an+1 = Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) Integrate to obtain terms 4x2 and 12 xe B1 + B1 Substitute limits correctly M1 2 1 a 1 1 Obtain correct equation in any form 4 a + e − = A1 2 2 2 Rearrange to given answer correctly A1 [5] 2 − ea (ii) Consider sign of − a , or equivalent M1 8 Complete the argument correctly with appropriate calculations A1 [2] (f (0.2 ) = .0112, f (0.3) = −.0015) (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.29 A1 Show sufficient iterations to justify its accuracy to 2 d.p. B1 0x = 0.2 0x = 0.25 0x = 0.3 0.3120 0.2992 0.2851 0.2815 0.2853 0.2894 0.2905 0.2894 0.2879 or show there is a sign change in the interval (0.285, 0.295) [3]

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