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

1 Given that 13x use logarithms to show that y kx and find the value of k correct to 3 significant = (2.8)y, = figures. [3]

Mark scheme: 1 State or imply y log 2.8 = x log 13 B1 log 13 Rearrange into form y = x or equivalent B1 log 8.2 Obtain answer k = 2.49 B1 [3]

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2 y R x O 1 2 3 The diagram shows part of the curve y The shaded region R is bounded by the curve and by = xe−x. the lines x 2, x 3 and y 0. = = = (i) Use the trapezium rule with two intervals to estimate the area of R, giving your answer correct to 2 decimal places. [3] (ii) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of R. [1]

Mark scheme: 2 (i) State or imply correct ordinates 0.27067..., 0.20521..., 0.14936... B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.21 with no errors seen A1 [3] (ii) Justify statement that the trapezium rule gives an over-estimate B1 [1] 2 2

More questions on Integration

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

Mark scheme: 3 EITHER State or imply non-modular inequality (2x –1)2 < (x + 4)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 –1 and 5 A1 State correct answer –1 < x < 5 A1 [4] OR Obtain one critical value, e.g. x = 5, 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 –1 < x < 5 B1  1 

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4π 4 (a) Show that cos 2x dx 12. [2] ã 0 = (b) By using an appropriate trigonometrical identity, find the exact value of 13π 3 tan2x dx. ã 1 [4] 6π

Mark scheme:  1  4 (a) Obtain integral a sin 2x with a = ±  ,1 2 or  M1  2  1 Use limits and obtain (AG) A1 [2] 2 (b) Use tan2 x = sec2 x – 1 and attempt to integrate both terms M1 Obtain 3tan x – 3x A1 Attempt to substitute limits, using exact values M1 π Obtain answer 2 3 − A1 [4] 2

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5 The equation of a curve is y = x3e−x. (i) Show that the curve has a stationary point where x 3. [3] = (ii) Find the equation of the tangent to the curve at the point where x 1. [4] =

Mark scheme: 5 (i) Use product rule M1 Obtain correct derivative in any form A1 Show that derivative is equal to zero when x = 3 A1 [3] (ii) Substitute x = 1 into gradient function, obtaining 2e–l or equivalent M1 State or imply required y-coordinate is e–1 B1 Form equation of line through (l, e–1) with gradient found (NOT the normal) M1 Obtain equation in any correct form A1 [4] 2

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6 (i) By sketching a suitable pair of graphs, show that the equation ln x 2 = −x2 has only one root. [2] (ii) Verify by calculation that this root lies between x 1.3 and x 1.4. [2] = = (iii) Show that, if a sequence of values given by the iterative formula xn+1 = √(2 −ln xn) converges, then it converges to the root of the equation in part (i). [1] (iv) Use the iterative formula xn+1 = √(2 −ln xn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = ln x or y = 2 – x2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of In x – (2 – x2) at x = 1.3 and x = 1.4, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] GCE AS/A LEVEL – May/June 2010 9709 23 (iii) Show that given equation is equivalent to x = ( 2 − ln x ) or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.31 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.305, 1.315) B1 [3]

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7 The polynomial 2x3 ax2 bx 6, where a and b are constants, is denoted by It is given that + + + p(x). when is divided by the remainder is 30, and that when is divided by the p(x) (x −3) p(x) (x + 1) remainder is 18. (i) Find the values of a and b. [5] (ii) When a and b have these values, verify that is a factor of and hence factorise (x −2) p(x) p(x) completely. [4]

Mark scheme: 7 (i) Substitute x = 3 and equate to 30 M1 Substitute x = –1 and equate to 18 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = 1 and b = –13 A1 [5] (ii) Either show that f(2) = 0 or divide by (x – 2), obtaining a remainder of zero B1 Obtain quadratic factor 2x2 + 5x – 3 B1 Obtain linear factor 2x – 1 B1 Obtain linear factor x + 3 B1 [Condone omission of repetition that x – 2 is a factor.] [If linear factors 2x – 1, x + 3 obtained by remainder theorem or inspection, award B2 + B1.] [4]

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8 (i) Prove the identity sin x. sin(x −30◦) + cos(x −60◦) ≡(√3) [3] (ii) Hence solve the equation 1 2 sec x, sin(x −30◦) + cos(x −60◦) = for [6] x 0◦< < 360◦.

Mark scheme: 8 (i) Use correct sin(A – B) and cos(A – B) formulae M1 Substitute exact values for sin 30° etc. M1 Obtain given answer correctly A1 [3] 1 (ii) State 3 sin x = sec x B1 2 Rearrange to sin 2x = k, where k is a non-zero constant M1 1 −1  1  Carry out evaluation of sin   M1 2  3  Obtain answer 17.6° A1 Carry out correct method for second answer M1 Obtain remaining 3 answers from 17.6°, 72.4°, 197.6°, 252.4° and no others in the range A1 [6] [Ignore answers outside the given range]

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