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

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

Mark scheme: 1 EITHER: State or imply non-modular inequality (x – 3)2 > (2(x + 1))2 , or corresponding quadratic equation, or pair of linear equations (x – 3) = ± 2(x + 1) B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values −5 and 13 A1 State answer − 5 < x < 13 A1 OR: Obtain the critical value x = −5 from a graphical method, or by inspection, or by solving a linear equation or inequality B1 Obtain the critical value x = 13 similarly B2 State answer − 5 < x < 13 B1 [4] [Do not condone ≤ for <; accept 0.33 for 13 .]

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2 The variables x and y satisfy the equation y3 Ae2x, where A is a constant. The graph of ln y against = x is a straight line. (i) Find the gradient of this line. [2] (ii) Given that the line intersects the axis of ln y at the point where ln y 0.5, find the value of A = correct to 2 decimal places. [2]

Mark scheme: 2 (i) State or imply 3 ln y = ln A + 2 x at any stage B1 2 State gradient is , or equivalent B1 [2] 3 (ii) Substitute x = 0, ln y = 0.5 and solve for A M1 Obtain A = 4.48 A1 [2]

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3 Solve the equation 2 tan x, tan(45◦−x) = giving all solutions in the interval [5] x 0◦< < 180◦.

Mark scheme: 3 Attempt to use tan(A ± B) formula and obtain an equation in tan x M1 Obtain 3-term quadratic 2 tan2 x + 3 tan x – 1 = 0, or equivalent A1 Solve a 3-term quadratic and find a numerical value of x M1 Obtain answer 15.7° A1 Obtain answer 119.3° and no others in the given interval A1 [5] [Ignore answers outside the given interval. Treat answers in radians, 0.274 and 2.08, as a misread.]

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4 Given that x 1 when t 0, solve the differential equation = = dx dt = 1x −x4, obtaining an expression for x2 in terms of t. [7]

Mark scheme: 4 Separate variables correctly B1 Obtain term k ln(4 – x2), or terms k1 ln(2 – x) + k2 ln(2 + x) B1 Obtain term –2 ln(4 – x2), or –2 ln(2 – x) –2 ln(2 + x), or equivalent B1 Obtain term t, or equivalent B1 Evaluate a constant or use limits x = 1, t = 0 in a solution containing terms a ln(4 – x2) and bt or terms c ln(2 – x), d ln(2 + x) and bt M1 Obtain correct solution in any form, e.g. –2 ln(4 – x2) = t – 2 ln3 A1 Rearrange and obtain x 2 = 4 − 3exp ( − 12 t ) , or equivalent (allow use of 2 ln 3 = 2.20) A1 [7] x 2 x

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5 y M x O p The diagram shows the curve y and its maximum point M. The x-coordinate of M is = e−x −e−2x denoted by p. (i) Find the exact value of p. [4] (ii) Show that the area of the shaded region bounded by the curve, the x-axis and the line x p is 1 = equal to 8. [4]

Mark scheme: 5 (i) State derivative − e − x − ( −2e) −2 x , or equivalent B1 + B1 Equate derivative to zero and solve for x M1 Obtain p = ln 2, or exact equivalent A1 [4] (ii) State indefinite integral − e − x − ( − 12 e) −2 x , or equivalent B1 + B1 Substitute limits x = 0 and x = p correctly M1 Obtain given answer following full and correct working A1 [4] GCE AS/A LEVEL – May/June 2010 9709 33

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ln x 6 The curve y has one stationary point. = x 1 + (i) Show that the x-coordinate of this point satisfies the equation x 1 x + , = ln x and that this x-coordinate lies between 3 and 4. [5] (ii) Use the iterative formula xn 1 + xn+1 = ln xn to determine the x-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Use correct quotient or product rule M1 1 ln x Obtain correct derivative in any form, e.g. − A1 x ( x + )1 ( x + 2)1 Equate derivative to zero and obtain the given equation correctly A1 ( x + )1 Consider the sign of x − at x = 3 and x = 4, or equivalent M1 ln x Complete the argument with correct calculated values A1 [5] (ii) Use the iterative formula correctly at least once, using or reaching a value in the interval (3, 4) M1 Obtain final answer 3.59 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (3.585, 3.595) A1 [3]

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7 (i) Prove the identity cos 3θ cos3θ cos θ. [4] ≡4 −3 (ii) Using this result, find the exact value of 12π cos3θ dθ. ã 1 [4] 3π

Mark scheme: 7 (i) Use correct cos(A + B) formula to express cos 3θ in terms of trig functions of 2θ and θ M1 Use correct trig formulae and Pythagoras to express cos 3θ in terms of cosθ M1 Obtain a correct expression in terms of cosθ in any form A1 Obtain the given identity correctly A1 [4] [SR: Give M1 for using correct formulae to express RHS in terms of cos θ and cos 2θ , then M1A1 for expressing in terms of either only cos 3θ and cos θ , or only cos 2θ , sin 2θ , cos θ , and sin θ , and A1 for obtaining the given identity correctly.] 1 1 1 (ii) Use identity and integrate, obtaining terms ( sin 3θ ) and (3 sin θ ) , or equivalent B1 + B1 4 3 4 Use limits correctly in an integral of the form ksin 3θ + lsin θ M1 2 3 Obtain answer − 3 , or any exact equivalent A1 [4] 3 8 3 2

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8 (a) The equation 2x3 2x 12 0 has one real root and two complex roots. Showing your −x2 + + = working, verify that 1 i √3 is one of the complex roots. State the other complex root. [4] + (b) On a sketch of an Argand diagram, show the point representing the complex number 1 i √3. + On the same diagram, shade the region whose points represent the complex numbers which ß satisfy both the inequalities and arg 3π. [5] |ß −1 −i √3| ≤1 ß ≤1

Mark scheme: 8 (a) EITHER: Substitute 1+ i 3 , attempt complete expansions of the x3 and x2 terms M1 Use i2 = –1 correctly at least once B1 Complete the verification correctly A1 State that the other root is 1− i 3 B1 OR1: State that the other root is 1− i 3 B1 State quadratic factor x 2 −x2 + 4 B1 Divide cubic by 3-term quadratic reaching partial quotient 2x + k M1 Complete the division obtaining zero remainder A1 OR2: State factorisation ( 2 x + 3)( x 2 − 2 x + 4) , or equivalent B1 Make reasonable solution attempt at a 3-term quadratic and use i2 = –1 M1 Obtain the root 1+ i 3 A1 State that the other root is 1− i 3 B1 [4] (b) Show point representing 1+ i 3 in relatively correct position on an Argand diagram B1 Show circle with centre at 1+ i 3 and radius 1 B1√ Show line for arg z = 13 π making 13 π with the real axis B1 Show line from origin passing through centre of circle, or the diameter which would contain the origin if produced B1 Shade the relevant region B1√ [5] GCE AS/A LEVEL – May/June 2010 9709 33 A B C

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4 5x9 (i) Express + −x2 in partial fractions. [5] (1 −2x)(2 + x)2 4 5x (ii) Hence obtain the expansion of + −x2 in ascending powers of x, up to and including (1 −2x)(2 + x)2 the term in x2. [5]

Mark scheme: A B C 9 (i) State or imply partial fractions of the form + + B1 1 − 2 x 2 + x ( 2 + x 2) Use any relevant method to determine a constant M1 Obtain one of the values A = 1, B = 1, C = −2 A1 Obtain a second value A1 Obtain the third value A1 [5] A Dx + E [The form + , where A = 1, D = 1, E = 0, is acceptable 1 − 2 x ( 2 + x 2) scoring B1M1A1A1A1 as above.] (ii) Use correct method to obtain the first two terms of the expansion of 1( −x2 ) −1 , ( 2 + x ) −1 , ( 2 + x ) −2 , 1( + 12 x ) −1 , or 1( + 12 x ) −2 M1 Obtain correct unsimplified expansions up to the term in x2 of each partial fraction A1√ + A1√ + A1√ 9 15 2 Obtain answer 1 + x + x , or equivalent A1 [5] 4 4 − 1 [Symbolic binomial coefficients, e.g. , are not sufficient for the M1. The f.t. is on A, B, C.]  1  [For the A, D, E form of partial fractions, give M1A1√A1√ for the expansions then, if D ≠ 0, M1 for multiplying out fully and A1 for the final answer.] [In the case of an attempt to expand ( 4 + 5 x − x 2 )(1 − 2 x ) −1 ( 2 + x ) −2 , give M1A1A1 for the expansions, M1 for multiplying out fully, and A1 for the final answer.] [SR: If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii).] [SR: If D or E omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii).]

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10 The straight line l has equation r 2i 2j The plane p has equation 3x 9. = −j −4k + λ(i + + 2k). −y + 2ß = The line l intersects the plane p at the point A. (i) Find the position vector of A. [3] (ii) Find the acute angle between l and p. [4] (iii) Find an equation for the plane which contains l and is perpendicular to p, giving your answer in the form ax by d. [5] + + cß =

Mark scheme: 10 (i) Express general point of the line in component form, e.g. (2 + λ , −1 + 2λ , −4 + 2λ) B1 Substitute in plane equation and solve for λ M1 Obtain position vector 4i + 3j, or equivalent A1 [3] (ii) State or imply a correct vector normal to the plane, e.g. 3i – j + 2k B1 Using the correct process, evaluate the scalar product of a direction vector for l and a normal for p M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine or inverse sine of the result M1 Obtain answer 26.5° (or 0.462 radians) A1 [4] (iii) EITHER: State a + 2b + 2c = 0 or 3a – b + 2c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = 6 : 4 : −7, or equivalent A1 Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (i + 2j + 2k) × (3i – j + 2k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. 6i + 4j – 7k A1 Substitute coordinates of a relevant point in 6x + 4y – 7z = d and evaluate d M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 OR2: Attempt to form 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = 2i – j – 4k + λ(i + 2j + 2k) +µ(3i − j + 2k) A1 State three equations in x, y, z, λ, µ A1 Eliminate λ and µ M1 Obtain answer 6x + 4y – 7z = 36, or equivalent A1 [5]

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