Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 2 · Variant 3

1 Solve the inequality x 1 3x 5 . [4] + < +

Mark scheme: 1 Either State or imply non-modular inequality ( x + 1) 2 < (3 x + 5 ) 2 , or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values −2 and − 32 A1 State correct answer x < −2 or x > − 32 A1 Or Obtain one critical value, e.g. x = −2, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 Obtain the other critical value similarly B2 State correct answer x < −2 or x > − 32 B1 [4] 4

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2 y x O P The diagram shows the curve y x4 2x The curve cuts the positive x-axis at the point P. = + −9. (i) Verify by calculation that the x-coordinate of P lies between 1.5 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation 3O@9 A x . x = −2 (iii) Use the iterative formula _P Q 3 9 xn+1 = xn −2 to determine the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 2 (i) Consider sign of x4 + 2x – 9 at x = 1.5 and x = 1.6 M1 Complete the argument correctly with appropriate calculations A1 [2] (f (1.5 ) = −.0 9375f, (1.6 ) = .0 7536 ) (ii) Rearrange x4 + 2x – 9 = 0 to given equation or vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. B1 [3] xo = 1.5 xo = 1.55 xo = 1.6 1.5874 1.5614 1.5362 1.5424 1.5556 1.5685 1.5653 1.5520 1.5536 1.5604 1 5595 1.5561 1.5565 or show there is a sign change in the interval (1.555, 1.565) 2

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3 The equation of a curve is y 12e2x 4x. Find the exact x-coordinate of each of the stationary points of the curve and determine= the−5exnature+ of each stationary point. [6]

Mark scheme: 3 Obtain derivative e2x – 5ex + 4 B1 Equate derivative to zero and carry out recognisable solution method for a quadratic in ex M1 Obtain ex = 1 or ex = 4 A1 Obtain x = 0 and x = ln 4 A1 Use an appropriate method for determining nature of at least one stationary point M1  d 2 y 2 x x d 2 y d 2 y  = 2e − 5e , when x = ,0 = − (3), x = ln ,4 = + (12 )  dx 2 dx 2 dx 2   Conclude maximum at x = 0 and minimum at x = ln 4 (no errors seen) A1 [6] ( )

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5 The parametric equations of a curve are x cos y 4 = 21 −cos 1, = sin21, for 0 ≤1 ≤0. dy 8 cos (i) Show that [4] 1 dx 1 cos = −4 1. (ii) Find the coordinates of the point on the curve at which the gradient is [4] −4.

Mark scheme: dx dy 5 (i) State = −2 sin 2θ + sin θ or = 8 sin θ cos θ B1 d θ d θ dy d y d x Use = ÷ M1 d x d θ dθ Use sin 2θ = 2sinθ cosθ M1 Obtain given answer correctly A1 [4] (ii) Equate derivative to −4 and solve for cos θ M1 Obtain cos θ = ½ A1 Obtain x = −1 A1 Obtain y = 3 A1 [4] 2

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