Cambridge A Level Mathematics 9709 — 2020 May/June Paper 2 · Variant 1

3 A curve has parametric equations x et y 3e2t 1. = −2e−t, = + Find the equation of the tangent to the curve at the point for which t 0. [5] = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 State 2 d d e 2e , 6e d d − = + = t t t x y t t B1 Use d d d / d d d = y y x x t t either in terms of t or after substitution of 0 = t *M1 Obtain gradient of tangent is 2 A1 Attempt equation of tangent with numerical gradient and coordinates DM1 Obtain 2 6 = + y x or equivalent A1 5

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4 (a) Sketch, on the same diagram, the graphs of y 3x 2a and y 3x , where a is a positive constant. = + = −4a Give the coordinates of the points where each graph meets the axes. [3] (b) Find the coordinates of the point of intersection of the two graphs. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Deduce the solution of the inequality 3x 2a 3x . [1] + < −4a ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Draw two V-shaped graphs with one vertex on negative x-axis and one vertex on positive x-axis M1 Draw correct graphs related correctly to each other A1 State correct coordinates 2 4 , 2 , , 4 3 3 − a a a a A1 3 4(b) Solve linear equation with signs of 3x different or solve non-modulus equation 2 2 (3 2 ) (3 4 ) + = − x a x a M1 Obtain 1 3 = x a A1 Obtain 3 = y a A1 3 Question Answer Marks 4(c) State 1 3 < x a (FT from part (b)) B1FT 1

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5 y M x O The diagram shows part of the curve with equation y x3 cos 2x. The curve has a maximum at the point M. = 3 (a) Show that the x-coordinate of M satisfies the equation x 1.5x2 cot 2x. [3] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between 0.59 and 0.60. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Differentiate using the product rule to obtain 2 3 cos2 sin 2 − ax x bx x M1 Obtain 2 3 3 cos2 2 sin 2 − x x x x A1 Equate first derivative to zero and confirm 3 2 1.5 cot 2 = x x x AG A1 3 5(b) Consider sign of 3 2 1.5 cot 2 − x x x or equivalent for 0.59 and 0.60 M1 Obtain 0.009... − and 0.005... or equivalents and justify conclusion A1 2 5(c) Use iteration correctly at least once M1 Obtain final answer 0.596 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.5955, 0.5965] A1 3

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7 (a) Find the quotient when 9x3 1 is divided by 3x 2 , and show that the remainder is 9. −6x2 −20x + + [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 6 9x3 1 (b) Hence find dx, giving the answer in the form a ln b where a and b are −6x2 −20x + 3x 2 + Ô1 + integers. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the exact root of the equation 9e9y 0. [4] −6e6y −20e3y −8 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Carry out division at least as far as 2 3 + x kx M1 Obtain quotient 2 3 4 4 − − x x A1 Confirm remainder is 9 AG A1 3 7(b) Integrate to obtain at least 3 1k x and 2 ln(3 2) + k x terms *M1 Obtain 3 2 2 4 3ln(3 2) − − + + x x x x (FT from quotient in part (a)) A1FT Apply limits correctly DM1 Apply appropriate logarithm properties correctly M1 Obtain 125 ln64 + A1 5 7(c) State or imply 3 2 2 9 6 20 8 (3 2)(3 4 4) − − − = + − − x x x x x x (FT from quotient in part (a)) B1FT Attempt to solve cubic eqn to find positive value of x (or of 3e y ) M1 Use logarithms to solve equation of form 3e = y k where 0 > k M1 Obtain 1 ln 2 3 or exact equivalent A1 4

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