Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 3 · Variant 1

1 Find the quotient and remainder when x4 is divided by x2 2x [3] + −1. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks 1 Commence division and reach a partial quotient x 2 + kx M1 Obtain quotient x 2 − 2 x + 5 A1 Obtain remainder − 12 x + 5 A1 3

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5 The equation of a curve is 2x4 xy3 y4 10. + + = dy y3 (i) Show that + . [4] dx = −8x33xy2 4y3 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence show that there are two points on the curve at which the tangent is parallel to the x-axis and find the coordinates of these points. 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Mark scheme: 5(i) 3 2 dy 3 B1 State or imply y + 3 xy as derivative of xy dx 3 dy 4 B1 State or imply 4 y as derivative of y dx d y M1 Equate derivative of the LHS to zero and solve for d x Obtain the given answer A1 4 5(ii) Equate numerator to zero *M1 Obtain y = −2x, or equivalent A1 Obtain an equation in x or y DM1 Obtain final answer x = −1, y = 2 and x = 1, y = −2 A1 4

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6 The variables x and y satisfy the differential equation dy 4 cos2y tan x, dx = for 0 1 and x 0 when y 1 Solve this differential equation and find the value of x when ≤x < 20, = = 40. y 1 [8] = 30. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ 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Mark scheme: 6 Separate variables correctly and attempt integration of one side B1 Obtain term tan y , or equivalent B1 Obtain term of the form k lncos x , or equivalent M1 Obtain term − 4lncos x , or equivalent A1 Use x = 0 and y = 14π in solution containing a tan y and b lncos x to evaluate a M1 constant, or as limits Obtain correct solution in any form, e.g. tan y = 4lnsec x + 1 A1 Substitute y = 13πin solution containing terms a tan y and b lncos x , and use correct M1 method to find x Obtain answer x = 0.587 A1 8

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9 y R x O 2 2x 1 The diagram shows the curve y The shaded region R is enclosed by the curve, = + x2 e−1 for x ≥0. the x-axis and the lines x 0 and x 2. = = (i) Find the exact values of the x-coordinates of the stationary points of the curve. 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(ii) Show that the exact value of the area of R is 18 . [5] −42e ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(i) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and obtain a 3 term quadratic equation in x M1 Obtain answers x = 2 ± 3 A1 4 9(ii) 2 − 12 x − 12 x *M1 xe d x Integrate by parts and reach k (1 + x )e + l ∫ 2 − 12 x − 12 x A1 xe dx , or equivalent Obtain −2(1 + x )e + 4 ∫ 2 − 12 x A1 Complete the integration and obtain ( −18 − 8 x − 2 x )e , or equivalent Use limits x = 0 and x = 2 correctly, having fully integrated twice by parts DM1 Obtain the given answer A1 5

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