Cambridge A Level Mathematics 9709 — 2023 Feb/March Paper 2 · Variant 2

1 2π 11 Find the exact value of 2 tan2 2x dx. [4] Ó 0 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Express integrand as 2sec 2 12 x − 2 B1 Integrate to obtain form a tan 12 x + bx M1 where ab  0 . Obtain correct 4tan 12 x − 2 x A1 Obtain 4 − π A1 4

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4 (a) Sketch, on the same diagram, the graphs of y 2x and y 3x [2] = −11 = −3. (b) Solve the inequality 2x 3x [3] −11 < −3. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the smallest integer N satisfying the inequality 2 ln N 3 ln N [2] −11 < −3. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw approximately correct graph of y = 3 x − 3 with greater B1 Crossing x-axis between origin and vertex of first graph. gradient 2 4(b) Attempt solution of linear equation where signs of 2x and 3x are M1 different Solve −2 x + 11 = 3x − 3 to obtain x = 145 A1 OE Conclude x  145 A1 OE Alternative method for Question 4(b) Attempt solution of 3-term equation (2 x − 11) 2 = (3 x − 3) 2 to M1 Or equivalent inequality. obtain at least one value of x Obtain at least x = 145 A1 OE Conclude x  145 A1 OE 3 4(c) Attempt value of N (maybe non-integer at this stage) using M1 logarithms and their answer to part (b). Conclude with single integer 17 A1 2

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a 4 3 5 It is given that dx ln 10, where a is a constant greater than 1. 1 2x x + = Ô1 + 3/ (a) Show that a 90 1 2a [5] = + −2. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 3 significant figures. Use an initial value of 1.7 and give the result of each iteration to 5 significant figures. 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Mark scheme: 5(a) Integrate to obtain form k1 ln(1 + 2 x ) + k 2 ln x *M1 k1  0, k 2  0 . Obtain correct 2ln(1 + 2 x ) + 3ln x A1 Use limits correctly and equate to ln10 DM1 Apply relevant logarithm properties correctly and arrange as far as DM1 a 3 = ... 3 −2 A1 AG Confirm given result a = 90(1 + 2a ) with sufficient detail 5 5(b) Use iteration process correctly at least once M1 Need to see 1.6848 . Obtain final answer 1.68 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign A1 change in the interval [1.675, 1.685] 3

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6 y P M x O 4e2x 9 The diagram shows the curve with equation y . The curve has a minimum point M and + ex 2 = crosses the y-axis at the point P. + (a) Find the exact value of the gradient of the curve at P. 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(b) Find the exact coordinates of M. 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Mark scheme: 6(a) Differentiate using quotient rule M1 OE x 2 x 2 x x A1 OE (e + 2) 8e − (4e + 9) e Obtain (e x + 2) 2 Substitute x = 0 in first derivative and attempt evaluation M1 Obtain 119 A1 4 6(b) Equate first derivative to zero and attempt factorisation or M1 equivalent Solve a three-term quadratic equation in ex to obtain e x = ... M1 (2e x + 9)(2e x − 1) = 0 . Obtain x-coordinate ln 12 or − ln2 A1 Obtain y-coordinate 4 A1 4

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7 y P x O The diagram shows the curve with parametric equations x k tan t, y 3 sin 2t sin t, = = −4 for 0 t 1 It is given that k is a positive constant. The curve crosses the x-axis at the point P. < < 2π. (a) Find the value of cos t at P, giving your answer as an exact fraction. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ dy (b) Express in terms of k and cos t. [4] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Given that the normal to the curve at P has gradient 10,9 find the value of k, giving your answer as an exact fraction. 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Mark scheme: 7(a) Use identity sin2t = 2sin t cos t B1 Attempt solution of y = 0 for cost M1 Obtain cost = 23 A1 Or exact equivalent. 3 7(b) d x d y *M1 Differentiate to obtain at least one of and correct d t d t dy 6cos2t − 4cos t A1 Obtain = dx k sec 2 t d y DM1 Using correct identities. Attempt to express in terms of cost d x dy 6(2cos 2 t − 1)cos 2 t − 4cos 3 t A1 OE Obtain = dx k 4 7(c) d y *M1 Substitute value from part (a) in expression for involving k d x and cost Equate to − 109 and solve for k DM1 Obtain k = 43 A1 3

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