Cambridge A Level Mathematics 9709 — 2018 May/June Paper 2 · Variant 2

2 A curve has equation y 3 ln 2x 9 ln x. = + −2 (i) Find the x-coordinate of the stationary point. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Determine whether the stationary point is a maximum or minimum point. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) Differentiate to obtain form 1 2 2 9 − + k k x x M1 Obtain correct 6 2 2 9 − + x x A1 Equate first derivative to zero and attempt solution to ... = x M1 Dependent on previous M1 Obtain 9 = x A1 4 2(ii) Use appropriate method for determining nature of stationary point M1 Second derivative or gradient or value of y Conclude minimum with no errors seen A1 2

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3 (i) Find the quotient when x4 8x2 13 −2x3 + −12x + is divided by x2 6 and show that the remainder is 1. 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(ii) Show that the equation x4 8x2 12 0 −2x3 + −12x + = has no real roots. 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Mark scheme: 3(i) Carry out division and reach at least partial quotient of form x kx M1 Obtain quotient 2 2 2 − + x x A1 Obtain remainder 1 A1 AG; necessary detail needed and all correct 3 Question Answer Marks Guidance 3(ii) State equation as 2 2 ( 6)( 2 2) 0 + − + = x x x B1 FT Following their 3-term quotient from part (i) Calculate discriminant of 3-term quadratic or equivalent M1 Obtain 4 − and state no root, also referring to no root from 2 6 + x factor A1 AG; necessary detail needed 3

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5 A curve has equation y3 sin 2x 4y 8. + = Find the equation of the tangent to the curve at the point where it crosses the y-axis. 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Mark scheme: 5 Use product rule to differentiate first term obtaining form 2 3 1 2 d sin 2 cos2 d + y k y x k y x x M1 Obtain correct 2 3 d 3 sin 2 2 cos2 d + y y x y x x A1 State 2 3 d d 3 sin 2 2 cos2 4 0 d d + + = y y y x y x x x A1 Identify 0, 2 = = x y as relevant point B1 Find equation of tangent through (0, 2) with numerical gradient M1 Dependent on previous M1 Obtain 4 2 = − + y x or equivalent A1 6

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a 1 2x 26 It is given that 1 e dx 10, where a is a positive constant. Ó 0 + = P Q 15 (i) Show that a 2 ln −a1 . 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(ii) Use the equation in part (i) to show by calculation that 1.5 a 1.6. 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(iii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. 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Mark scheme: 6(i) Rewrite integrand as 1 2 1 2e e + + x B1 Integrate to obtain form 1 2 1 2 e e + + x x x k k M1 Obtain 1 2 4e e + + x x x A1 Use limits to obtain 1 2 4e e 5 10 + + − = a a a A1 Rearrange as far as 1 2e ... = a including use of 1 1 1 2 2 2 4e e e (4 e ) + = + a a a a M1 Confirm 1 2 15 2ln 4 e   − =   +   a a a A1 AG; necessary detail needed 6 6(ii) Consider sign of 1 2 15 2ln 4 e   − −   +   a a a for 1.5 and 1.6 or equivalent M1 Obtain 0.08... − and 0.06… or equivalents and justify conclusion A1 2 6(iii) Use iterative process correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval (1.555,1.565) A1 3

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