Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 2 · Variant 3

1 Solve the inequality [3] |x −2| ≥|x + 5|.

Mark scheme: 1 EITHER State or imply non-modular inequality ( x − 2 )2 ≥ ( x + 5 )2 , or corresponding equation or pair of linear equations M1 3 Obtain critical value − A1 2 3 State correct answer x ≤ − A1 2 OR State a correct linear equation for the critical value, e.g. x – 2 = – x – 5, or corresponding correct linear inequality, e.g. x – 2 ≥ – x – 5 M1 3 Obtain critical value − A1 2 3 State correct answer x ≤ − A1 [3] 2

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giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 32x−1, [4]

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2x – 1) log 3 or equivalent A1 Solve for x M1(dep*) Obtain answer x = 1.87 A1 [4]

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3 Solve the equation 2 cos 2θ 4 cos θ = −3, for [4] 0◦≤θ ≤180◦.

Mark scheme: 3 Make relevant use of the cos 2θ formula M1 Obtain a correct quadratic in cos θ A1 Solve a quadratic in cos θ M1 Obtain answer θ = 60 and no others in the range A1 [4] (Ignore answers outside the given range) dx − 2 dy 2

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4 The parametric equations of a curve are 2 x y , for t 0. t = ln(1 −2t), = < dy 1 (i) Show that . [3] −2t dx = t2 (ii) Find the exact coordinates of the only point on the curve at which the gradient is 3. [3]

Mark scheme: dx 2 dy − 24 (i) State = or = −2t B1 dt 1 − 2t dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to 3 and solve for t M1 State or imply that t = –1 c.w.o. A1 Obtain coordinates (ln 3, –2) A1 [3] GCE AS LEVEL – October/November 2012 9709 23

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5 y B (q, cos q ) C R x O A 12p The diagram shows the curve y cos x, for 0 2π. A rectangle OABC is drawn, where B is the = ≤x ≤1 point on the curve with x-coordinate θ, and A and C are on the axes, as shown. The shaded region R is bounded by the curve and by the lines x θ and y 0. = = (i) Find the area of R in terms of θ. [2] (ii) The area of the rectangle OABC is equal to the area of R. Show that 1 θ θ . −sinθ cos = [1] 1 θn (iii) Use the iterative formula −sin , with initial value θ1 0.5, to determine the value θn+1 = cos θn = of θ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 5 (i) Attempt to integrate and use limits θ and π M1 Obtain 1– sin θ A1 [2] (ii) State that area of rectangle = θcos θ, equate area of rectangle to area of R and rearrange to given equation B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.555, 0.565) B1 [3]

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6 (a) Use the trapezium rule with two intervals to estimate the value of 1 1 dx, 6 2ex ä 0 + giving your answer correct to 2 decimal places. [3] (b) Find dx. [4] (ex −2)2 ä e2x

Mark scheme: 6 (a) State or imply correct ordinates 0.125, 0.08743…, 0.21511… B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.11 with no errors seen A1 [3] (b) Attempt to expand brackets and divide by e2x M1 Integrate a term of form ke−x or ke−2x correctly A1 Obtain 2 correct terms A1 Fully correct integral x + 4e−x - 2e−2x + c A1 [4]

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7 The polynomial 2x3 ax b, where a and b are constants, is denoted by It is given that when is divided−4x2by+ + the remainder is 4, and that when is dividedp(x).by the remainder p(x)is 12. (x + 1) p(x) (x −3) (i) Find the values of a and b. [5] (ii) When a and b have these values, find the quotient and remainder when is divided by p(x) (x2 −2).[3]

Mark scheme: 7 (i) Substitute x = −1, equate to zero and obtain a correct equation in any form B1 Substitute x = 3 and equate to 12 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = −4 and b = 6 A1 [5] (ii) Attempt division by x2 − 2 and reach a partial quotient of 2 x − k M1 Obtain quotient 2 x − 4 A1 Obtain remainder −2 A1 [3]

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1 dy 8 (i) By differentiating , show that if y sec θ then tan θ sec θ. [3] cos θ dθ = = (ii) Hence show that d2y a sec3θ bsec θ, dθ2 = + giving the values of a and b. [4] (iii) Find the exact value of 14π tan2θ sec θ tan dθ. ã 0 (1 + −3 θ) [5]

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