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

1 Given that 2y 93x, use logarithms to show that y kx and find the value of k correct to 3 significant figures. = = [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Apply logarithms to both sides and apply power law at least once M1 Rearrange to the form 3ln9 ln 2 = y x OE A1 Obtain 9.51 = k A1 3

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1 2 Find the exact coordinates of the stationary point on the curve with equation y 5xe 2x. [5] = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Differentiate using product rule to obtain 1 1 2 2 e e + x x a bx *M1 Obtain correct 1 1 2 2 5 5e e 2 + x x x OE A1 Equate first derivative to zero and solve for x DM1 Obtain x-coordinate –2 A1 Obtain y-coordinate 1 10e− − A1 5

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3 The equation of a curve is cos 3x 5 sin y 3. + = Find the gradient of the curve at the point 1 1 . [5] 9π, 6π ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Differentiate cos3x to obtain 3sin3 − x Differentiate 5sin y to obtain d 5cos d y y x B1 Obtain d 3sin3 5cos 0 d − + = y x y x OE B1 Substitute x and y values to find value of first derivative M1 Obtain 3 5 A1 5

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5 (a) Sketch, on the same diagram, the graphs of y 2x and y 3x 5. [2] = −3 = + (b) Solve the inequality 3x 5 2x . [3] + < −3 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of 3 5 = + y x B1 2 5(b) State equation 3 5 (2 3) + = − − x x or corresponding inequality B1 Attempt solution of linear equation / inequality where signs of 3x and 2x are different M1 State answer 2 5 < − x A1 Alternative method for question 5(b) Square both sides of equation / inequality and attempt solution of 3-term quadratic equation / inequality M1 Obtain (eventually) only 2 5 − A1 State answer 2 5 < − x A1 3

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a 4 7 It is given that 8x dx 10, where a is a positive constant. 2x 1 + = Ô0 + (a) Show that a 2.5 ln 2a 1 . [4] = −0.5 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using the equation in part (a), show by calculation that 1 a 2. [2] < < ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Integrate to obtain the form 2 1 2 ln(2 1) + + k x k x *M1 Obtain correct 2 2ln(2 1) 4 + + x x A1 Use limits correctly and attempt rearrangement DM1 Confirm 2.5 0.5ln(2 1) = − + a a AG A1 4 Question Answer Marks 7(b) Consider sign of 2.5 0.5ln(2 1) − − + a a or equivalent for 1 and 2 M1 Obtain 0.3... − and 0.6... or equivalents and justify conclusion A1 2 7(c) Use iteration process correctly at least once M1 Obtain final answer 1.358 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.3575, 1.3585] A1 3

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8 (a) Show that 3 sin cot 6 [2] 21 1  cos21. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the equation 3 sin cot 5 for 0 [3] 21 1 = < 1 < π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 2π 1(c) Find the exact value of 3 sin x cot 2x dx. [5] 1 Ó 4π ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Use at least one of sin 2 2sin cos θ θ θ = and cos cot sin θ θ θ = B1 Use both and conclude 2 6cos θ AG B1 2 8(b) Attempt solution of 2 5 cos 6 θ = to find at least one value M1 Obtain 0.421 A1 Obtain 2.72 A1 3 Question Answer Marks 8(c) Express integrand in form cos + a b x M1 Obtain correct integrand 3 3cos + x A1 Integrate to obtain sin + px q x *M1 Apply limits correctly DM1 Obtain 3 3 3 4 2 π + − or exact equivalent A1 5

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