Cambridge A Level Mathematics 9709 — 2016 Feb/March Paper 3 · Variant 2

3 The equation x5 x2 0 has one positive root. −3x3 + −4 = (i) Verify by calculation that this root lies between 1 and 2. [2] (ii) Show that the equation can be rearranged in the form O@ A 3 4 x 3x . [1] = + x2 −1 (iii) Use an iterative formula based on this rearrangement to determine the positive root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 3 (i) Consider sign of x 5 − 3 x 3 + x 2 − 4 at x = 1 and x = 2, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2] (ii) Rearrange the given quintic equation in the given form, or work vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.78 A1 Show sufficient iterations to 4 d.p. to justify 1.78 to 2 d.p., or show there is a sign change in the interval (1.775, 1.785) A1 [3]

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6 A curve has equation sin y ln x x sin y, = −2 for −120 ≤y ≤120. dy (i) Find in terms of x and y. [5] dx (ii) Hence find the exact x-coordinate of the point on the curve at which the tangent is parallel to the x-axis. [3]

Mark scheme: 6 (i) EITHER: State correct derivative of sin y with respect to x B1 Use product rule to differentiate the LHS M1 Obtain correct derivative of the LHS A1 Obtain a complete and correct derived equation in any form A1 dy Obtain a correct expression for in any form A1 dx OR: State correct derivative of sin y with respect to x B1 Rearrange the given equation as sin y = x / (ln x + 2) and attempt to differentiate both sides B1 Use quotient or product rule to differentiate the RHS M1 Obtain correct derivative of the RHS A1 dy Obtain a correct expression for in any form A1 [5] dx dy (ii) Equate to zero and obtain a horizontal equation in ln x or sin y M1 dx Solve for ln x M1 Obtain final answer x = 1/ e , or exact equivalent A1 [3]

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7 The variables x and y satisfy the differential equation dy xex+y, dx = and it is given that y 0 when x 0. = = (i) Solve the differential equation and obtain an expression for y in terms of x. [7] (ii) Explain briefly why x can only take values less than 1. [1] ` a ` a ` a

Mark scheme: 7 (i) Separate variables and attempt integration of one side M1 Obtain term − e − y A1 Integrate x ex by parts reaching x e x ±∫ e x d x M1 Obtain integral xe x − e x A1 Evaluate a constant, or use limits x = 0, y = 0 M1 Obtain correct solution in any form A1 Obtain final answer y = − ln(e x (1 − x )) , or equivalent A1 [7] (ii) Justify the given statement B1 [1]

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10 (a) Find the complex number z satisfying the equation z* 1 2iz, where z* denotes the complex conjugate of z. Give your answer in the form x iy, where+ =x and y are real. [5] + (b) (i) On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities z 1 and Im z where Im z denotes the imaginary part of z. + −3i ≤1 ≥3, [4] (ii) Determine the difference between the greatest and least values of arg z for points lying in this region. [2]

Mark scheme: 10 (a) Substitute and obtain a correct equation in x and y B1 Use 2i = −1 and equate real and imaginary parts M1 Obtain two correct equations, e.g. x + 2y +1 = 0 and y + 2x = 0 A1 Solve for x or for y M1 Obtain answer z = 13 − 23 i A1 [5] (b) (i) Show a circle with centre −+1 3 i B1 Show a circle with radius 1 B1 Show the line Im z = 3 B1 Shade the correct region B1 [4] (ii) Carry out a complete method to calculate the relevant angle M1 Obtain answer 0.588 radians (accept 33.7°) A1 [2]

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