Cambridge A Level Mathematics 9709 — 2011 May/June Paper 2 · Variant 1

1 Solve the equation [3] |3x + 4| = |2x + 5|.

Mark scheme: 1 EITHER Attempt to square both sides obtaining three terms on each side M1 Attempt solution of three-term quadratic equation M1 9 Obtain 5 x + 4 x − 9 = 0 and hence − and 1 A1 5 OR Obtain value 1 from graphical method, inspection or linear equation B1 9 Obtain value − similarly B2 [3] 5 dx dy

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2 A curve has parametric equations x 3t sin 2t, y 4 2 cos 2t. = + = + Find the exact gradient of the curve at the point for which t 16π. [4] =

Mark scheme: dx dy 2 State = 3 + 2 cos 2t or = −4 sin 2t (or both) B1 dt dt d y d y d x Use = ÷ M1 d x d t d t − 4 sin 2t Obtain or imply A1 3 + 2 cos 2t 1 1 Substitute π to obtain − 3 or exact equivalent A1 [4] 6 2

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3 ln y (6, 10.2) (0, 2.0) ln x O The variables x and y satisfy the equation y Kxm, where K and m are constants. The graph of = ln y against ln x is a straight line passing through the points and as shown in the (0, 2.0) (6, 10.2), diagram. Find the values of K and m, correct to 2 decimal places. [5]

Mark scheme: 3 State or imply that ln y = ln K + m ln x B1 Equate intercept on axis for ln y to ln K M1 Obtain 7.39 for K A1 Attempt calculation of gradient of line M1 Obtain 1.37 for m A1 [5]

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4 The polynomial is defined by f(x) 3x3 ax2 ax a, f(x) = + + + where a is a constant. (i) Given that is a factor of find the value of a. [2] (x + 2) f(x), (ii) When a has the value found in part (i), find the quotient when is divided by [3] f(x) (x + 2).

Mark scheme: 4 (i) Substitute –2 and equate to zero or divide by x + 2 and equate remainder to zero M1 Obtain a = 8 A1 [2] (ii) Attempt to find quotient by division or inspection or use of identity M1 Obtain at least 3 x 2 + 2 x A1 Obtain 3 x 2 + 2 x + 4 with no errors seen A1 [3] 1

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dy 5 Find the value of when x 4 in each of the following cases: dx = (i) y x [4] = ln(x −3), x (ii) y −1 [3] = x 1. +

Mark scheme: 1 5 (i) Differentiate ln(x – 3) to obtain B1 x − 3 Attempt to use product rule M1 x Obtain ln ( x − 3 ) + or equivalent A1 x − 3 Substitute 4 to obtain 4 A1 [4] (ii) Use correct quotient or product rule M1 ( x + 1) − ( x − 1) Obtain correct derivative in any form, e.g. A1 ( x + 1)2 2 Substitute 4 to obtain A1 [3] 25 GCE AS/A LEVEL – May/June 2011 9709 21 3

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6 (a) Find dx. [4] ã 4ex(3 + e2x) 14π (b) Show that 2 dθ 1 [4] ã (3 + tan2θ) = 2(8 + π). −14π

Mark scheme: 6 (a) Rewrite integrand as 12ex + 4e3x B1 Integrate to obtain 12ex … B1 4 3 x Integrate to obtain … + e B1 3 Include … + c B1 [4] (b) Use identity tan2θ = sec2θ – 1 B1 Integrate to obtain 2tanθ + θ or equivalent B1 Use limits correctly for integral of form αtanθ + bθ M1 1 Confirm given answer (8 + π ) A1 [4] 2 2

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7 (i) By sketching a suitable pair of graphs, show that the equation e2x 14 = −x2 has exactly two real roots. [3] (ii) Show by calculation that the positive root lies between 1.2 and 1.3. [2] (iii) Show that this root also satisfies the equation x 1 = 2 ln(14 −x2). [1] (iv) Use an iteration process based on the equation in part (iii), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places. [3]

Mark scheme: 7 (i) Draw correct sketch of y = e2x B1 Draw correct sketch of y = 14 – x2 B1 Indicate two real roots only from correct sketches B1 [3] (ii) Consider sign of e2x + x2 – 14 for 1.2 and 1.3 or equivalent M1 Justify conclusion with correct calculations ( f(1.2) = –1.54, f(1.3) = 1.15 ) A1 [2] 1 2 (iii) Confirm given answer x = ln (14 − x ) B1 [1] 2 (iv) Use the iteration process correctly at least once M1 Obtain final answer 1.26 A1 Show sufficient iterations to 4 decimal places to justify answer or show a sign change in the interval (1.255, 1.256) A1 [3] [1.2 → 1.2653 → 1.2588 → 1.2595 ; 1.25 → 1.2604 → 1.2593 → 1.2594 ; 1.3 → 1.2522 → 1.2598 → 1.2594 ]

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α8 (i) Express 4 sin θ 0 and Give the exact cos θ in the form R where R 0◦< −6 sin(θ −α), > < 90◦. value of R and the value of α correct to 2 decimal places. [3] (ii) Solve the equation 4 sin θ cos θ 3 for [4] −6 = 0◦≤θ ≤360◦. (iii) Find the greatest and least possible values of sin θ cos 8 as θ varies. [2] (4 −6 θ)2 +

Mark scheme: 8 (i) State or imply R = 52 or 2 13 B1 Use appropriate formula to find α M1 Obtain 56.31° A1 [3] (ii) Attempt to find at least one value of θ – α M1 Obtain one correct value 80.9° of θ A1 Carry out correct method to find second answer M1 Obtain 211.7° and no others in range A1 [4] (iii) Obtain 60, following their value of R B1 √ Obtain 8. Allow quoted solution B1 [2]

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