Cambridge A Level Mathematics 9709 — 2021 Feb/March Paper 2 · Variant 2

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

Mark scheme: 1(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of 2 y x = + with smaller positive gradient B1 Crossing y-axis between 0 and y-intercept of first graph. 2 1(b) Solve 3 5 2 x x − = + to obtain 7 2 x = B1 Attempt solution of linear equation where signs of 3x and x are different. M1 Obtain 3 4 x = A1 Alternative method for question 1(b) State or imply non-modulus equation 2 2 (3 5) ( 2) x x − = + B1 Attempt solution of 3-term quadratic equation M1 Obtain 3 4 and 7 2 A1 3

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2 Solve the equation cot 8 for 0 [5] sec21 1 = < 1 < π. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 State 2 1 cos 8 sin cos θ θ θ × = B1 OE involving sinθ and cosθ only. Attempt use of sin 2θ identity to obtain sin 2 k θ = M1 Obtain 1 sin 2 4 θ = A1 Use correct process to find two values of θ between 0 and π M1 Allow if working in degrees. Obtain 0.126 and 1.44 A1 AWRT Alternative method for question 2 State 2 1 tan 8 tan θ θ + = B1 Attempt solution of 3-term quadratic equation to find values of tanθ M1 OE involving tanθ only. Obtain 8 60 tan 2 θ ± = A1 OE Solve tan ... θ = to find two values of θ between 0 and π M1 Allow if working in degrees. Obtain 0.126 and 1.44 A1 AWRT 5

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3 The parametric equations of a curve are x e2t cos 4t, y 3 sin 2t. = = Find the gradient of the curve at the point for which t 0. [5] = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Attempt use of product rule to find d d x t M1 Obtain 2 2 2e cos4 4e sin 4 t t t t − A1 Obtain d 6cos2 d y t t = B1 Divide to find d d y x with 0 t = substituted M1 Obtain 3 A1 CWO 5

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4 y x O 1 3 5x The shaded region is bounded by the The diagram shows part of the curve with equation y = 4x3 1. curve and the lines x 1, x 3 and y 0. + = = = dy (a) Find and hence find the x-coordinate of the maximum point. [4] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Differentiate using quotient rule (or product rule) M1* Obtain 3 3 3 2 5(4 1) 60 (4 1) x x x + − + A1 OE Equate first derivative to zero and attempt solution DM1 Obtain 1 2 x = A1 4 Question Answer Marks Guidance 4(b) Use y values 5 10 15 , , 5 33 109 or decimal equivalents B1 Use correct formula, or equivalent, with 1 h = M1 Obtain 1 20 15 1 2 33 109   + +     or equivalent and hence 0.87 A1 3 4(c) State over-estimate with reference to top of each trapezium above curve B1 Or clear equivalent. 1

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9 5 (a) Given that 2 ln x 1 ln x ln x 9 , show that x x 2. [3] + + = + = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ _ 9 (b) It is given that the equation x has a single root. x 2 = + Show by calculation that this root lies between 1.5 and 2.0. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (b), to find the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Use the power law correctly *M1 Use correct process to obtain equation with no logarithms DM1 Confirm 9 2 x x = + A1 AG; condone absence of justification for choice of positive root. 3 5(b) Consider sign of 9 2 x x − + or equivalent for 1.5 and 2 M1 Obtain 0.1... − and 0.5 or equivalents and justify conclusion A1 AG 2 Question Answer Marks Guidance 5(c) Use iteration process correctly at least once M1 Obtain final answer 1.58 A1 Final answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign change in interval [1.575, 1.585] A1 3

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6 The polynomial p x is defined by p x x3 ax b, = + + where a and b are constants. It is given that x 2 is a factor of p x and that the remainder is 5 when p x is divided by x . + −3 (a) Find the values of a and b. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence find the exact root of the equation p e2y 0. [5] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Substitute 2 x = − and equate to zero *M1 Substitute 3 x = and equate to 5 *M1 Obtain 8 2 0 a b −− + = and 27 3 5 a b + + = or equivalents A1 Solve a pair of relevant linear simultaneous equations for a or b DM1 Dependent at least one M mark. Obtain 6 a = − and 4 b = − A1 5 Question Answer Marks Guidance 6(b) Attempt division by 2 x + at least as far as 2 x kx + M1 Obtain 2 2 2 x x − − A1 Obtain (at least) the positive root 2 12 2 + or exact equivalent A1 Equate 2 e y to positive root, apply logarithms and use power law M1 Obtain 1 2 12 ln 2 2   +       or 1 ln(1 3) 2 + or exact equivalent A1 5

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7 (a) Express 5 3 cos x 5 sin x in the form R cos x , where R 0 and 0 1 [3] + −! > < ! < 2π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) As x varies, find the least possible value of 4 5 3 cos x 5 sin x, + + and determine the corresponding value of x where x [3] −π < < π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 (c) Find [3] 2 Ô 5 3 cos 5 sin d1. 31 + 31 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) State 10 R = B1 Use appropriate trigonometry to find α M1 Obtain 1 6 α = π A1 3 7(b) State 6 − B1 FT Following their value of R. Attempt to find x from their cos( ) 1 x α − = − M1 Obtain 1 6 x − π = −π and hence 5 π 6 − A1 3 7(c) State integrand of form 2 1 1 sec 3 π 6 k θ   −     *M1 Integrate to obtain form 2 1 tan 3 π 6 k θ   −     DM1 Obtain 1 1 tan 3 π 300 6 c θ   − +     A1 3

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