Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 2 · Variant 2

1 Use logarithms to solve the equation 24x, giving your answer correct to 3 significant figures. 53x−1 = [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Introduce logarithms to both sides and use *M1 power law Obtain (3 x − 1)log5 = 4 x log2 or equivalent A1 Allow A1 for poor use of brackets if recovered later Solve linear equation for x DM1 dep *M Obtain 0.783 A1 Allow 3 sf or better 4

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2 It is given that x satisfies the equation x 1 4. Find the possible values of + = x 4 x . [4] + − −4 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Solve 3-term quadratic equation or a pair of M1 For M1, must square both sides linear equations when attempting a quadratic equation Obtain x = − 5 and x = 3 A1 Substitute (at least) one value of x (less than 4) M1 into x + 4 − x − 4 , showing correct evaluation of modulus and producing only one answer in each case Obtain –8 and 6 and no others A1 4

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3 The equation of a curve is y tan 12x 3 sin 12x. The curve has a stationary point M in the interval = + x Find the coordinates of M, giving each coordinate correct to 3 significant figures. [6] 0 < < 20. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Differentiate to obtain form M1 If a factor of 0.5 is missed, can still k1 sec 2 12 x + k 2 cos 12 x get 5/6, penalise at first A1 Obtain 12 sec 2 12 x + 23 cos 12 x A1 Equate first derivative to zero and produce *M1 cos 3 12 x = k 3 Use correct process to find one value of x DM1 Dep on *M, allow for obtaining 1.609….. , 92.2 o or 268 o Obtain x = 4.67 A1 Allow x = 4.67 or better for A1 Obtain y = 1.12 A1 Allow y = 1.12 from x = 4.66 but nothing else 6

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5 y 0, 9 P Q x O The diagram shows the curve y and a straight line. The curve crosses the y-axis at the point P. The straight line crosses the y-axis= 4e−2xat the point 0, 9 and its gradient is equal to the gradient of the curve at P. The straight line meets the curve at two points, one of which is Q as shown. (i) Show that the x-coordinate of Q satisfies the equation x 9 [6] 8 = −12e−2x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use an iterative formula based on the equation in part (i) to find the x-coordinate of Q correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) Obtain derivative of the form ke − 2 x *M1 Condone k = 4 for M1 State or imply gradient of curve at P is –8 A1 Form equation of straight line through (0, 9) *DM1 dep on *M with negative gradient Obtain y = − 8 x + 9 or equivalent A1 Equate equation of curve and equation of DM1 dep on both *M straight line Rearrange to confirm x = 89 − 12 e − 2 x A1 6 5(ii) Use iterative process correctly at least once M1 Obtain final answer 1.07 A1 Show sufficient iterations to 5 sf to justify A1 answer or show sign change in interval (1.065,1.075) 6

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1 406 (a) Find the exact value of sin x 4 sin x 6 cosx dx. [5] Ó 0 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ a 6 (b) Given that dx ln 49, find the value of the positive constant a. [5] 3x 2 = Ô0 + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Obtain 2 − 2cos2x as part of integrand B1 Obtain 3sin 2x as part of integrand B1 Allow second B1 for writing Integrate to obtain form M1  1 2  , M1 6sin x cos x d x = 6  sin x  k1 x + k 2 sin 2 x + k 3 cos2 x ∫  2  may then be implied by subsequent work Obtain 2 x − sin2 x − 32 cos2 x or A1 2 x − sin 2 x + 3sin 2 x Apply limits to obtain 12 π+ 12 A1 5 6(b) B1 6 Integrate to obtain 2ln(3 x + 2) Allow ln ( 3 x + 2 ) for B1 3 Use at least one relevant logarithm property *M1 3a + 2 (3a + 2) 2 A1 Obtain = 7 or = 49 or 2 4 equivalent without ln Solve relevant equation to find a DM1 Dep on *M1, allow for 49 = ( 3a + 2 ) 2 OE or correct working involving ( 3a + 2 ) Obtain a = 4 only A1 5

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