Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 2 · Variant 2

1 Solve the inequality [4] |x + 3| > |2x|.

Mark scheme: 1 EITHER: Obtain a non-modular inequality from (x + 3)2 > (2x)2, or corresponding equation, or pair of linear equations (x + 3) = ± 2x M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –1 and x = 3 A1 State answer –1 I x I 3 A1 OR: Obtain critical value x = 3 from a graphical method, or by inspection, or by solving a linear inequality or linear equation B1 Obtain the critical value x = –1 similarly B2 State answer –1 I x I 3 B1 [4] 2

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2 It is given that y 2 ln x. Express y in terms of x, in a form not involving logarithms. ln(y + 5) −ln = [4]

Mark scheme: 2 State or imply 2 ln x = ln(x2) B1 Use law for the logarithm of a quotient or a product M1 Remove logarithms and obtain yx2 = y + 5, or equivalent A1 5 Obtain answer y = A1 [4] x 2 − 1

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3 (i) Use the trapezium rule with two intervals to estimate the value of 13π sec x dx, ã 0 giving your answer correct to 2 decimal places. [3] (ii) Using a sketch of the graph of y sec x for 0 3π, explain whether the trapezium rule gives an under-estimate or an over-estimate= of the true≤x value≤1 of the integral in part (i). [2]

Mark scheme: 3 (i) Show or imply correct ordinates 1, 1.15470…, 2 B1 Use correct formula, or equivalent, with h = 1 π and three ordinates M1 6 Obtain answer 1.39 with no errors seen A1 [3] (ii) Make recognisable sketch of y = sec x for 0 @ x @ 1 π B1 3 Using a correct graph, explain that the rule gives an over-estimate B1 [2] dx t d y t t

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4 The parametric equations of a curve are 1 et x y = −e−t, = + e−t. dy (i) Show that e2t [3] dx = −1. (ii) Hence find the exact value of t at the point on the curve at which the gradient is 2. [2]

Mark scheme: dx d y 4 (i) State = e–t or = et – e–t B1 dt d t dy d y dx Use = ÷ M1 dx d t dt Obtain given answer correctly A1 [3] dy (ii) Substitute = 2 and use correct method for solving an equation of the form e2t = a, dx where a > 0 M1 Obtain answer t = 1 ln 3, or equivalent A1 [2] 2

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5 The polynomial ax3 bx2 2, where a and b are constants, is denoted by It is given that and are+ factors−5xof+ p(x). (x + 1) (x −2) p(x). (i) Find the values of a and b. [5] (ii) When a and b have these values, find the other linear factor of [2] p(x).

Mark scheme: 5 (i) Substitute x = –1 or x = 2 and equate to zero M1 Obtain a correct equation, e.g. –a + b + 5 + 2 = 0 A1 Obtain a second correct equation, e.g. 8a + 4b – 10 + 2 = 0 A1 Solve for a or b M1 Obtain a = 3 and b = –4 A1 [5] (ii) Substitute for a and b and attempt division by (x + 1)(x – 2) or attempt third factor by inspection M1 Obtain answer 3x – 1 A1 [2] GCE A/AS LEVEL – October/November 2009 9709 22

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6 (i) Express 3 cos x 4 sin x in the form R where R 0 and α 90◦, stating the exact value of R and giving+ the value of α correctcos(x −α),to 2 decimal places.> 0◦< < [3] (ii) Hence solve the equation 3 cos x 4 sin x 4.5, + = giving all solutions in the interval [4] x 0◦< < 360◦.

Mark scheme: 6 (i) State answer R = 5 B1 Use trig formula to find a M1 Obtain a = 53.13° A1 [3] (ii) Evaluate cos–1 (4.5/5) ≈ 25.84° M1 Obtain answer 79.0° A1 Carry out correct method for second answer M1 Obtain answer 27.3° and no others in the given range A1√ [4] [Treat the giving of answers in radians as a misread. Ignore answers outside the given range.]

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7 y M 1 x O 2p The diagram shows the curve y x2 cos x, for 0 2π, and its maximum point M. = ≤x ≤1 (i) Show by differentiation that the x-coordinate of M satisfies the equation 2 tan x x. = [4] (ii) Verify by calculation that this equation has a root (in radians) between 1 and 1.2. [2] 2 (iii) Use the iterative formula tan−1 to determine this root correct to 2 decimal places. xn+1 = xn Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and express tan x in terms of x M1 Obtain given answer A1 [4] 2 (ii) Consider sign of tan x – at x = 1 and x = 1.2, or equivalent M1 x Complete the argument with correct calcuations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.08 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.075, 1.085) A1 [3] 1

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13π 8 (a) Find the exact value of 2x dx. [5] ã 0 (sin + sec2x) 4 1 1 (b) Show that dx ln 5. [4] 2x x 1 + = ä1 +

Mark scheme: 8 (a) Integrate and obtain term k cos 2x, where k = ± 1 or ±1 M1 2 Obtain term – 1 cos 2x A1 2 Obtain term tan x B1 Substitute correct limits correctly M1 3 Obtain exact answer + 3 A1 [5] 4 (b) Integrate and obtain 1 ln x + ln(x + 1) or 1 ln 2x + ln(x + 1) B1 + B1 2 2 Substitute correct limits correctly M1 Obtain given answer following full and correct working A1 [4]

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