Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 3 · Variant 3

sin x 2 The equation of a curve is y for x Show that the gradient of the curve is positive 1 cosx, = −0 < < 0. for all x in the given interval. + [4]

Mark scheme: 2 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 1 Use Pythagoras to simplify the derivative to , or equivalent A1 1 + cos x Justify the given statement, − 1 < cosx < 1 statement, or equivalent A1 [4]

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3 Express the equation cot 1 tan as a quadratic equation in tan Hence solve this equation for 21 = + 1 1. [6] 0Å < 1 < 180Å.

Mark scheme: 3 Use the tan 2A formula to obtain an equation in tan θ only M1 Obtain a correct horizontal equation A1 Rearrange equation as a quadratic in tan θ, e.g. 3tan 2 θ+ 2tanθ−=1 0 A1 Solve for θ (usual requirements for solution of quadratic) M1 Obtain answer, e.g. 18.4° A1 Obtain second answer, e.g. 135° , and no others in the given interval A1 [6] 2

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4 x 6 Let I dx. −1 2 x x = Ô1 + 2 u (i) Using the substitution u x, show that I du. [3] −1 u 1 = = Ô1 + (ii) Hence show that I 1 ln 49. [6] = +

Mark scheme: 1 6 (i) State or imply d u = d x B1 2 x Substitute for x and dx throughout M1 Justify the change in limits and obtain the given answer A1 [3] B (ii) Convert integrand into the form A + M1* u + 1 Obtain integrand A =1, B = −2 A1 Integrate and obtain u − 2ln(u + 1) A1 + A1 Substitute limits correctly in an integral containing terms au and bln(u + 1), where ab ≠ 0 DM1 Obtain the given answer following full and correct working A1 [6] [The f.t. is on A and B.]

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7 Throughout this question the use of a calculator is not permitted. The complex number z is defined by z i. The complex conjugate of z is denoted by z*. = ï2 − ï6 (i) Find the modulus and argument of z. [2] (ii) Express each of the following in the form x iy, where x and y are real and exact: + (a) z 2z*; + z* (b) . iz [4] (iii) On a sketch of an Argand diagram with origin O, show the points A and B representing the complex numbers z* and iz respectively. Prove that angle AOB is equal to 1 [3] 60.

Mark scheme: 7 (i) State modulus 2 2 , or equivalent B1 State argument − 13π (or −60°) B1 [2] (ii) (a) State answer 3 2 + 6 i B1 (b) EITHER: Substitute for z and multiply numerator and denominator by conjugate of iz M1 Simplify the numerator to 4 3 + 4 i or the denominator to 8 A1 Obtain final answer 12 3 + 12 i A1 OR: Substitute for z, obtain two equations in x and y and solve for x or for y M1 Obtain x = 12 3 or y = 12 A1 Obtain final answer 12 3 + 12 i A1 [4] (iii) Show points A and B in relatively correct positions B1 Carry out a complete method for finding angle AOB, e.g. calculate the z ∗ argument of M1 i z Obtain the given answer A1 [3] A Bx + C

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9 y 1 x O a 20 k The diagram shows the curves y x cosx and y x, where k is a constant, for 0 x The curves = = < ≤120. touch at the point where x a. = 2 (i) Show that a satisfies the equation tan a a. [5] = @ A 2 (ii) Use the iterative formula to determine a correct to 3 decimal places. Give the tan−1 an+1 = an result of each iteration to 5 decimal places. [3] (iii) Hence find the value of k correct to 2 decimal places. [2] [Question 10 is printed on the next page.]

Mark scheme: 9 (i) Differentiate both equations and equate derivatives M1* k Obtain equation cos a − a sin a = − 2 A1 + A1 a k State a cos a = and eliminate k DM1 a Obtain the given answer showing sufficient working A1 [5] (ii) Show clearly correct use of the iterative formula at least once M1 Obtain answer 1.077 A1 Show sufficient iterations to 5 d.p. to justify 1.077 to 3 d.p., or show there is a sign change in the interval (1.0765, 1.0775) A1 [3] (iii) Use a correct method to determine k M1 Obtain answer k = 0.55 A1 [2]

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