Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 2 · Variant 1

2 A curve has equation y 3 tan 2x1 cos 2x. = Find the gradient of the curve at the point for which x 1 [5] = 3π. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 1 1 2 1 B1 State or imply derivative of tan x is sec x or derivative of 2 2 2 cos2x is −2sin2x Attempt use of product rule to find first derivative *M1 3 2 1 1 A1 or (unsimplified) equivalent. Obtain correct sec x cos2 x − 6tan x sin2 x 2 2 2 1 DM1 Substitute π into attempt at first derivative and attempt evaluation to 3 find the gradient Obtain −4 A1 5

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10 4 3 (a) Find dx, giving your answer in the form ln a, where a is an integer. [4] 2x Ô 4 −5 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 10 (b) Find the exact value of [2] e2x−5 dx. Ó 4 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Obtain 2ln(2 x − 5) B1 Apply limits correctly M1 For integral of form k ln(2 x − 5) . Use one relevant logarithm property correctly M1 For integral of form k ln(2 x − 5) . Apply second logarithm property correctly and obtain ln25 A1 4 3(b) 1 2 x− 5 B1 Integrate to obtain e 2 1 15 1 3 B1FT or exact equivalent, FT on their ke 2 x −.5 Obtain final answer e − e 2 2 2

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4 (a) Sketch, on the same diagram, the graphs of y 3x and y 2x 7. [2] = −5 = + (b) Solve the equation 3x 2x 7. [3] −5 = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Hence solve the equation 2 3y 7, giving your answer correct to 3 significant figures. 3y+1 −5 = × + [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 2 x + 7 with smaller gradient B1 And crossing y-axis above y-intercept of modulus graph. 2 4(b) Solve 3x −=5 2 x + 7 to obtain x = 12 B1 Attempt solution of linear equation where signs of 3x and 2x are M1 3x −=5 −2 x − 7 OE. different 2 A1 Obtain x = − 5 Alternative solution for question 4(b) State or imply non-modulus equation (3 x − 5) 2 = (2 x + 7) 2 B1 Must be working with (3 x − 5) 2 = (2 x + 7) 2 Attempt solution of 3-term quadratic equation M1 2 A1 Obtain − and 12 5 3 4(c) Apply logarithms and use power law for 3y = k where k  0 or M1 Using their positive answer from part (b) correct equivalent or greater accuracy; and no other values. Obtain 2.26 A1 2

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7 The curve with equation e2x y3 y 11 has a stationary point at p, q . −18x + + = (a) Find the exact value of p. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Show that q 3 2 18 ln 3 [2] = + −q. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Show by calculation that the value of q lies between 2.5 and 3.0. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Use an iterative formula, based on the equation in (b), to find the value of q correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 3 2 d y B1 Differentiate y to obtain 3 y d x Differentiate complete equation to produce at least one term involving M1 d y using implicit differentiation. d x 2 x 2 dy dy A1 Obtain 2e − 18 + 3 y + = 0 dx dx dy 1 A1 Substitute = 0 to obtain either p = 2 ln9 or p = ln3 dx 4 7(b) Substitute value of p in original equation and rearrange as far as y 3 = ... M1 Allow in terms of ln9 . or q3 = … Obtain given result q = 3 2 + 18ln3 − q or y = 3 2 + 18ln3 − y with A1 AG sufficient detail 2 7(c) Consider sign of q − 3 2 + 18ln3 − q or equivalent for 2.5 and 3.0 M1 Obtain −0.18... and 0.34... with sufficient detail and justify A1 OE conclusion 2 7(d) Use iteration process correctly at least once M1 Obtain final answer q = 2.673 A1 Answer required to exactly 4 s.f. Show sufficient iterations to 6 sf to justify answer or show sign change A1 in the interval [2.6725, 2.6735] 3

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