Cambridge A Level Mathematics 9709 — 2017 May/June Paper 2 · Variant 1

1 Given that 5x 34y, use logarithms to show that y mx and find the value of the constant m correct = = to 3 significant figures. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Take logarithms of both sides and apply power law to both sides M1 Allow log5 4log3 = y for M1 A1 Rearrange to the form ln5 4ln3 = y x or equivalent A1 Obtain 0.366 = m A1 Total: 3 State or imply non-modulus inequality ( ) ⩽ ( )

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2 Solve the inequality 4 3 . [4] −x ≤ −2x ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 4 x − 2 3 2x − or corresponding equation, pair of linear equations or linear inequalities M1 Attempt solution of 3-term quadratic equation, of two linear equations or of two linear inequalities M1 Obtain critical values 1 − and 7 3 A1 SR Allow B1 for x ⩽–1 only or x ⩾7 3 only if first M1 is not given State answer x ⩽ –1, x ⩾ 7 3 A1 Do not accept 7 3 ⩽x ⩽–1 or –1⩾x ⩾7 3 for A1 Total: 4

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a 3 Given that 4e Ó0 12x+3 dx = 835, find the value of the constant a correct to 3 significant figures. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Integrate to obtain form 1 2 3 + x ke where k is constant not equal to 4 M1 Obtain correct 1 2 3 8 + x e A1 Allow unsimplified for A1 Obtain 1 2 3 3 8 8 835 + − = a e e or equivalent A1 Carry out correct process to find a from equation of form 1 2 3 + = a ke c M1 Obtain 3.65 A1 If 3.65 seen with no actual attempt at integration, award B1 if it is thought that trial and improvement with calculator has been used. Total: 5

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4 The sequence of values given by the iterative formula 2x2 n xn 9 + + , xn+1 = 1 xn 2 + with x1 2, converges to = !. (i) Find the value of correct to 2 decimal places, giving the result of each iteration to 4 decimal ! places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Determine the exact value of [3] !. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Use iteration correctly at least once M1 Obtain final answer 2.08 A1 Show sufficient iterations to 4 dp to justify answer or show sign change in interval ( ) 2.075, 2.085 A1 Total: 3 4(ii) State or clearly imply equation ( ) 2 2 2 9 1 + + = + x x x x or same equation using α B1 Carry out relevant simplification M1 Obtain 3 9 A1 3

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6 y x O 160 The diagram shows the curve y tan 2x for 0 The shaded region is bounded by the curve = ≤x ≤160. and the lines x 1 and y 0. = 60 = (i) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region, giving your answer correct to 3 significant figures. 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(ii) Find the exact volume of the solid formed when the shaded region is rotated completely about the x-axis. 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Mark scheme: 6(i) State or imply correct y-values 0, 1 2 6 6 tan , tan π π B1 degree mode when working out π tan 6 etc. this gives 0.00915 and 0.0183. Allow B1. Use correct formula, or equivalent, with 1 12 h π = and y–values M1 Must be convinced they have considered 3 values for y for M1 Obtain 0.378 A1 Total: 3 Question Answer Marks Guidance 6(ii) State or imply ( ) 2 sec 2 1 d x x π − ∫ B1 Integrate to obtain 1 2 tan2 k x k x + , any non-zero constants including π or not M1 Obtain 1 2 tan2x x − or ( ) 1 2 tan2x x π − A1 Obtain ( ) 1 1 2 6 3 π π − or equivalent A1 Total: 4

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7 The parametric equations of a curve are x t3 6t 1, y t4 4t2 5. = + + = −2t3 + −12t + dy dy (i) Find and use division to show that can be written in the form at b, where a and b are dx dx + constants to be found. 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(ii) The straight line x 9 0 is the normal to the curve at the point P. Find the coordinates −2y + = of P. 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Mark scheme: 7(i) Differentiate x and y and form d d y x M1 Obtain 3 2 2 4 6 8 12 3 6 t t t t − + − + A1 First 2 marks may be implied by an attempt at division Carry out division at least as far as kt or equivalent M1 For M1, it must be division by a quadratic factor. Allow attempt at factorisation with same conditions as for division Obtain 4 3 t A1 Obtain 4 3 2 t − with complete division shown and no errors seen A1 Total: 5 Question Answer Marks Guidance 7(ii) State or imply gradient of straight line is 1 2 B1 Allow B1 if 1 9 2 2 y x = + is seen Attempt value of t from their d d y x = their negative reciprocal of gradient of line M1 Obtain 0 t = and hence ( ) 1,5 A1 Total: 3

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8 y P x O M The diagram shows the curve with equation y 3x2 ln 16x . = The curve crosses the x-axis at the point P and has a minimum point M. (i) Find the gradient of the curve at the point P. 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(ii) Find the exact coordinates of the point M. 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Mark scheme: 8(i) Apply product rule to find first derivative *M1 Obtain ( ) 1 6 6 1n 3 x x x + or equivalent A1 Allow unsimplified for A1 Identify 6 x = at P B1 Substitute their value of x at P into attempt at first derivative DM1 dep *M Obtain 18 A1 Total: 5 Question Answer Marks Guidance 8(ii) Equate their first derivative to zero and attempt solution of equation of form ( ) 1 6 1n 0 k x m + = *M1 Obtain x–coordinate of form 2 1 a a e DM1 dep *M Obtain 1 2 6e x − = or exact equivalent A1 Substitute exact x–value in the form 2 1 a a e and attempt simplification to remove ln M1 Obtain 1 54e− − or exact equivalent A1 Total: 5

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