Cambridge A Level Mathematics 9709 — 2018 Feb/March Paper 3 · Variant 2

4 The variables x and y satisfy the equation yn Ax3, where n and A are constants. It is given that = y 2.58 when x 1.20, and y 9.49 when x 2.51. = = = = (i) Explain why the graph of ln y against ln x is a straight line. 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(ii) Find the values of n and A, giving your answers correct to 2 decimal places. 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Mark scheme: 4(i) State or imply n ln y = ln A + 3 ln x B1 State that the graph of ln y against ln x has an equation which is linear in ln y and ln B1 x, or has equation of the form nY = ln A + 3X, where Y = ln y and X = ln x, and is thus a straight line. 2 4(ii) Substitute x- and y-values in n ln y = ln A + 3 ln x or in the given equation and solve M1 for one of the constants Obtain a correct constant, e.g. n = 1.70 A1 Solve for a second constant M1 Obtain the other constant, e.g. A = 2.90 A1 4

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5 The parametric equations of a curve are x 2t sin 2t, y 1 cos 2t, = + = −2 for t 1 −120 < < 20. dy (i) Show that 2 tan t. 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(ii) Hence find the x-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures. 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Mark scheme: 5(i) State correct derivative of x or y with respect to t B1 d y d y d x M1 Use = ÷ d x d t d t d y 4sin 2t A1 Obtain = , or equivalent d x 2 + 2cos2t Use double angle formulae throughout M1 Obtain the given answer correctly AG A1 5 5(ii) −1 1  B1 State or imply t = tan  −   4  Obtain answer x = – 0.961 B1 2

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7 (i) By sketching suitable graphs, show that the equation e2x 6 has exactly one real root. [2] = + e−x (ii) Verify by calculation that this root lies between 0.5 and 1. 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(iii) Show that if a sequence of values given by the iterative formula 3 + xn+1 = 1 ln 1 6exn converges, then it converges to the root of the equation in part (i). 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(iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 7(i) Sketch a relevant graph, e.g. y = 2e x B1 Sketch a second relevant graph, e.g. y = 6 + e− x , and justify the given statement B1 2 7(ii) Calculate the value of a relevant expression or values of a pair of relevant M1 expressions at x = 0.5 and x = 1 Complete the argument correctly with correct calculated values A1 2 7(iii) 1 x B1 State a suitable equation, e.g. x = ln 1 + 6e ( ) 3 Rearrange this as 2e x = 6 + e− x , or commence working vice versa B1 2 7(iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.928 A1 Show sufficient iterations to 5 d.p. to justify 0.928 to 3 d.p., or show there is a sign A1 change in the interval (0.9275, 0.9285) 3

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9 The complex number 1 2i is denoted by u. + (i) It is given that u is a root of the equation 2x3 4x k 0, where k is a constant. −x2 + + = (a) Showing all working and without using a calculator, find the value of k. 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(b) Showing all working and without using a calculator, find the other two roots of this equation. 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(ii) On an Argand diagram sketch the locus of points representing complex numbers z satisfying the equation z 1. Determine the least value of arg z for points on this locus. Give your answer −u = in radians correct to 2 decimal places. 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Mark scheme: 9(i)(a) Substitute x = 1 + 2i in the equation and attempt expansions of x 2 and 3x M1 Use i 2 =−1 correctly at least once and solve for k M1 Obtain answer k = 15 A1 3 9(i)(b) State answer 1 – 2i B1 Carry out a complete method for finding a quadratic factor with zeros 1 + 2i and M1 1 – 2i Obtain x 2 − 2 x + 5 A1 3 A1 Obtain root − , or equivalent, via division or inspection 2 4 9(ii) Show a circle with centre 1 + 2i B1 Show a circle with radius 1 B1 Carry out a complete method for calculating the least value of arg z M1 Obtain answer 0.64 A1 4

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10 The line l has equation r 4i 3j i 2j . The plane p has equation 2x 4. = + −k + - + −2k −3y −z = (i) Find the position vector of the point of intersection of l and p. 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(ii) Find the acute angle between l and p. 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(iii) A second plane q is parallel to l, perpendicular to p and contains the point with position vector 4j Find the equation of q, giving your answer in the form ax by cz d. 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Mark scheme: 10(i) Express general point of l in component form, e.g. B1 r = ( 4 + µ) i + ( 3 + 2µ) j + ( −−1 2µ) k , or equivalent NB: Calling the vector a + µb, the B1 is earned by a correct reduction of the sum to a single vector or by expressing the substitution as a distributed sum a.n + µb.n Substitute in given equation of p and solve for µ M1 Obtain final answer 2i – j + 3k from µ=− 2 A1 3 10(ii) Using the correct process, evaluate the scalar product of a direction vector for l and M1 a normal for p Using the correct process for the moduli, divide the scalar product by the product of M1 the moduli and find the inverse sine or cosine of the result Obtain answer 10.3° (or 0.179 radians) A1 3 10(iii) EITHER: State a + 2b – 2c = 0 or 2a – 3b – c = 0 (B1 Obtain two relevant equations and solve for one M1 ratio, e.g. a : b Obtain a : b : c = 8 : 3 : 7, or equivalent A1 Substitute a, b, c and given point and evaluate d M1 Obtain answer 8x + 3y + 7z = 5 A1) OR1: Attempt to calculate vector product of relevant (M1 vectors, e.g. (2i – 3j – k)×(i + 2j – 2k) Obtain two correct components of the product A1 Obtain correct product, e.g. 8i + 3j +7k A1 Use the product and the given point to find d M1 Obtain answer 8x + 3y + 7z = 5, or equivalent A1) OR2: Attempt to form a 2-parameter equation with (M1 relevant vectors State a correct equation, e.g. A1 r = 4j – k + λ(i + 2j – 2k) + µ(2i – 3j – k) State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 State answer 8x + 3y + 7z = 5, or equivalent A1) 5

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