Cambridge A Level Mathematics 9709 — 2010 May/June Paper 3 · Variant 2

1 Solve the equation 2x 1 5, + 2x = −1 giving your answer correct to 3 significant figures. [4]

Mark scheme: 1 EITHER: Attempt to solve for 2 x M1 Obtain 2x = 6/4, or equivalent A1 Use correct method for solving an equation of the form 2x = a, where a > 0 M1 Obtain answer x = 0.585 A1 n + 6) / 5) / ln 2 B1 OR: State an appropriate iterative formula, e.g. xn + 1 = ln((2 x Use the iterative formula correctly at least once M1 Obtain answer x = 0.585 A1 Show that the equation has no other root but 0.585 A1 [4] [For the solution 0.585 with no relevant working, award B1 and a further B1 if 0.585 is shown to be the only root.] 2 ∫

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π 2 Show that x2 sin x dx π2 [5] ã 0 = −4.

Mark scheme: 2 Integrate by parts and reach ± x 2 cos x ± ∫ 2 x cos x d x M1 Obtain − x 2 cos x + ∫ 2 x cos x dx , or equivalent A1 Complete the integration, obtaining − x 2 cos x + 2 x sin x + 2 cos x , or equivalent A1 Substitute limits correctly, having integrated twice M1 Obtain the given answer correctly A1 [5]

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3 Showing your working and without using a calculator It is given that cos a a 35, where 0◦< = < 90◦. to evaluate a, (i) find the exact value of [3] sin(a −30◦), (ii) find the exact value of tan 2a, and hence find the exact value of tan 3a. [4]

Mark scheme: 3 (i) State or imply sin a = 4/5 B1 Use sin(A – B) formula and substitute for cos a and sin a M1 1 Obtain answer ( 4 3 − )3 , or exact eqivalent A1 [3] 10 (ii) Use tan 2A formula and substitute for tan a, or use sin 2A and cos 2A formulae, substitute sin a and cos a, and divide M1 24 Obtain tan 2 a = − , or equivalent A1 7 Use tan(A + B) formula with A = 2a , B = a and substitute for tan 2a and tan a M1 44 Obtain tan 3a = − A1 [4] 117

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4 y 1 x O p 2p M sin x The diagram shows the curve y for 0 x and its minimum point M. x = < ≤2π, (i) Show that the x-coordinate of M satisfies the equation x tan x. = [4] (ii) The iterative formula π xn+1 = tan−1(xn) + can be used to determine the x-coordinate of M. Use this formula to determine the x-coordinate of M correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 4 (i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain the given answer correctly A1 [4] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 4.49 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the interval (4.485, 4.495) A1 [3]

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5 The polynomial 2x3 5x2 ax b, where a and b are constants, is denoted by It is given that is a factor of+ +and that+ when is divided by the remainderp(x).is 9. (2x + 1) p(x) p(x) (x + 2) (i) Find the values of a and b. [5] (ii) When a and b have these values, factorise completely. [3] p(x)

Mark scheme: 5 (i) Substitute x = − 12 , equate to zero and obtain a correct equation, e.g. − 1 + 5 − 1 a + b = 0 B1 4 4 2 Substitute x = −2 and equate to 9 M1 Obtain a correct equation, e.g. − 16 + 20 − 2 a + b = 9 A1 Solve for a or for b M1 Obtain a = −4 and b = −3 A1 [5] GCE AS/A LEVEL – May/June 2010 9709 32 (ii) Attempt division by 2x + 1 reaching a partial quotient of x 2 + kx M1 Obtain quadratic factor x 2 + 2 x − 3 A1 Obtain factorisation ( 2 x + 1)( x + 3)( x − )1 A1 [3] [The M1 is earned if inspection has an unknown factor of x 2 + ex + f and an equation in e and/or f, or if two coefficients with the correct moduli are stated without working.] [If linear factors are found by the factor theorem, give B1 + B1 for (x −1) and (x + 3), and then B1 for the complete factorisation.] 1 dy

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6 The equation of a curve is x ln y 2x 1. = + dy (i) Show that [4] dx = −yx2. (ii) Find the equation of the tangent to the curve at the point where y 1, giving your answer in the form ax by c 0. = [4] + + =

Mark scheme: 1 dy 6 (i) EITHER: State or imply as derivative of ln y B1 y dx x dy State correct derivative of LHS, e.g. ln y + B1 y dx dy Differentiate RHS and obtain an expression for M1 dx Obtain given answer A1 2 x + 1 OR 1: State ln y = , or equivalent, and differentiate both sides M1 x 1 dy State correct derivative of LHS, e.g. B1 y dx State correct derivative of RHS, e.g. − 1 / x 2 B1 Rearrange and obtain given answer A1 OR 2: State y = exp( 2 + 1 / x ) , or equivalent, and attempt differentiation by chain rule M1 State correct derivative of RHS, e.g. − exp(2 + 1 / x /) x 2 B1 + B1 Obtain given answer A1 [4] [The B marks are for the exponential term and its multiplier.] (ii) State or imply x = − 12 when y = 1 B1 Substitute and obtain gradient of −4 B1√ Correctly form equation of tangent M1 Obtain final answer y + 4x + 1 = 0, or equivalent A1 [4]

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7 The variables x and t are related by the differential equation dx e2t cos2x, dt = where t When t 0, x 0. ≥0. = = (i) Solve the differential equation, obtaining an expression for x in terms of t. [6] (ii) State what happens to the value of x when t becomes very large. [1] (iii) Explain why x increases as t increases. [1]

Mark scheme: 7 (i) Separate variables correctly and attempt integration of both sides B1 Obtain term tan x B1 Obtain term − 12 e−2 t B1 Evaluate a constant or use limits x = 0, t = 0 in a solution containing terms a tan x and be–2t M1 Obtain correct solution in any form, e.g. tan x = 12 − 12 e −2 t A1 Rearrange as x = tan −1 ( 12 − 12 e −2 t ) , or equivalent A1 [6] (ii) State that x approaches tan −1 ( 12 ) B1 [1] (iii) State that 1 − e − 2 t increases and so does the inverse tangent, or state that e−2 t cos 2 x is positive B1 [1] GCE AS/A LEVEL – May/June 2010 9709 32 2 2 2 

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8 The variable complex number is given by ß 1 cos 2θ i sin 2θ, ß = + + where θ takes all values in the interval 2π θ 12π. −1 < < (i) Show that the modulus of is 2 cos θ and the argument of is θ. [6] ß ß 1 (ii) Prove that the real part of ß is constant. [3]

Mark scheme: 2 2 2 8 (i) EITHER: State a correct expression for zor z , e.g. 1( + cos 2θ ) + (sin 2θ ) B1 Use double angle formulae throughout or Pythagoras M1 Obtain given answer 2cos θ correctly A1 State a correct expression for tangent of argument, e.g. (sin 2θ /(1 + cos 2θ ) B1 Use double angle formulae to express it in terms of cos θ and sin θ M1 Obtain tan θ and state that the argument is θ A1 OR: Use double angle formulae to express z in terms of cos θ and sin θ M1 Obtain a correct expression, e.g. 1 + cos 2 θ − sin 2 θ + 2i sin θ cos θ A1 Convert the expression to polar form M1 Obtain 2 cos θ (cos θ + i sin θ ) A1 State that the modulus is 2 cosθ A1 State that the argument is θ A1 [6] (ii) Substitute for z and multiply numerator and denominator by the conjugate of z, or equivalent M1 Obtain correct real denominator in any form A1 Identify and obtain real part equal to 12 A1 [3]

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9 The plane p has equation 3x 2y 13. A second plane q is perpendicular to p and has equation ax y 4, where a is a constant.+ + 4ß = + + ß = (i) Find the value of a. [3] (ii) The line with equation r j 2j meets the plane p at the point A and the plane q at the point B. Find the length= −kof+ AB.λ(i + + 2k) [6]

Mark scheme: 9 (i) State or imply a correct normal vector to either plane, e.g. 3i + 2j + 4k or ai + j + k B1 Equate scalar product of normals to zero and obtain an equation in a, e.g. 3a + 2 + 4 = 0 M1 Obtain a = −2 A1 [3] (ii) Express general point of the line in component form, e.g. (λ , 1 + 2λ , −1 + 2λ) B1 Either substitute components in the equation of p and solve for λ , or substitute components and the value of a in the equation of q and solve for λ M1* Obtain λ = 1 for point A A1 Obtain λ = 2 for point B A1 Carry out correct process for finding the length of AB M1(dep*) Obtain answer AB = 3 A1 [6] [The second M mark is dependent on both values of λ being found by correct methods.]

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