Cambridge A Level Mathematics 9709 — 2011 May/June Paper 3 · Variant 2

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

Mark scheme: 1 EITHER: State or imply non-modular inequality x 2 < (5 + 2 x )2 , or corresponding equation, or pair of linear equations x = ± (5 + 2 x ) M1 5 Obtain critical values –5 and − only A1 3 5 Obtain final answer x < –5, x > − A1 3 OR: State one critical value e.g. –5, by solving a linear equation or inequality, or from a graphical method, or by inspection B1 5 State the other critical value, e.g. − , and no other B1 3 5 Obtain final answer x < –5, x > − B1 [3] 3 [Do not condone ≤ or ≥.]

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2 (i) Show that the equation log2(x + 5) = 5 −log2 x can be written as a quadratic equation in x. [3] (ii) Hence solve the equation log2(x + 5) = 5 −log2 x. [2]

Mark scheme: 2 (i) Use law for the logarithm of a product or quotient M1 Use log232 = 5 or 25 = 32 M1 Obtain x2 + 5x – 32 = 0, or horizontal equivalent A1 [3] (ii) Solve a 3-term quadratic equation M1 153 − 5 Obtain answer x = 3.68 only, or exact equivalent, e.g. A1 [2] 2

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3 Solve the equation cos θ + 4 cos 2θ = 3, giving all solutions in the interval 0◦≤θ ≤180◦. [5]

Mark scheme: 3 Use correct trig formula (or formulae) and obtain an equation in cosθ M1 Obtain 8cos2θ + cosθ – 7 = 0, or equivalent A1 Solve a 3-term quadratic in cosθ and reach θ = cos–1(a) M1 Obtain answer 29.0° A1 Obtain answer 180° and no others A1 [5] [Ignore answers outside the given interval. Treat answers in radians (0.505 and 3.14 or π) as a misread.] [SR: The answer 180° found by inspection can earn B1.]

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4 C r x A O B T The diagram shows a semicircle ACB with centre O and radius r. The tangent at C meets AB produced at T. The angle BOC is x radians. The area of the shaded region is equal to the area of the semicircle. (i) Show that x satisfies the equation tan x = x + π. [3] (ii) Use the iterative formula xn+1 = tan−1(xn + π) to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 4 (i) State or imply CT = r tan x or OT = r sec x , or equivalent B1 Using correct area formulae, form an equation in r and x M1 Obtain the given answer correctly A1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain the final answer 1.35 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p. , or show there is a sign change in the interval (1.345, 1.355) A1 [3] GCE AS/A LEVEL – May/June 2011 9709 32 dx 2

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5 The parametric equations of a curve are x = ln(tan t), y = sin2t, where 0 < t < 12π. dy (i) Express in terms of t. [4] dx (ii) Find the equation of the tangent to the curve at the point where x = 0. [3]

Mark scheme: dx 25 (i) EITHER: State = sec t / tan t , or equivalent B1 dt dy State = 2 sin t cos t , or equivalent B1 d t dy dy dx Use = ÷ M1 dx dt dt Obtain correct answer in any form, e.g. 2 sin 2 t cos 2 t A1 OR: Obtain y = e2x / (1 + e2x), or equivalent B1 Use correct quotient or product rule M1 Obtain correct derivative in any form, e.g. 2e2x / (1 + e2x)2 A1 Obtain correct derivative in terms of t in any form, e.g. (2tan2t) / (1 + tan2t)2 A1 [4] 1 (ii) State or imply t = π when x = 0 B1 4 Form the equation of the tangent at x = 0 M1 1 1 Obtain correct answer in any horizontal form, e.g. y = x + A1 [3] 2 2 1 [SR: If the OR method is used in part (i), give B1 for stating or implying y = or 2 d y 1 = when x = 0.] d x 2 dy

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6 A certain curve is such that its gradient at a point (x, y) is proportional to xy. At the point (1, 2) the gradient is 4. (i) By setting up and solving a differential equation, show that the equation of the curve is y = 2ex2−1. [7] (ii) State the gradient of the curve at the point (−1, 2) and sketch the curve. [2]

Mark scheme: dy 6 (i) Show that the differential equation is = 2 xy B1 dx Separate variables correctly and attempt integration of both sides M1 Obtain term ln y, or equivalent A1 Obtain term x2, or equivalent A1 Evaluate a constant, or use limits x = 1, y = 2, in a solution containing terms aln y and bx2 M1 Obtain correct solution in any form A1 Obtain the given answer correctly A1 [7] (ii) State that the gradient at (–1, 2) is –4 B1 Show the sketch of curve with correct concavity, positive y-intercept and axis of symmetry x = 0 B1 [2] [SR: A solution with k≠ 2, or not evaluated, can earn B0M1A1A1M1A1A0 in part (i).] dy [SR: If given answer is assumed valid, give B1 if is shown correctly to be equal to dx 2xy, is stated to be proportional to xy, and shown to be equal to 4 at (1, 2).] GCE AS/A LEVEL – May/June 2011 9709 32

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5 7 (a) The complex number u is defined by u = where the constant a is real. a + 2i, (i) Express u in the form x + iy, where x and y are real. [2] (ii) Find the value of a for which arg(u*) = 34π, where u* denotes the complex conjugate of u. [3] (b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers ß which satisfy both the inequalities |ß| < 2 and |ß| < |ß −2 −2i|. [4]

Mark scheme: 7 (a) (i) EITHER: Multiply numerator and denominator by a – 2i, or equivalent M1 5 a 10i Obtain final answer − , or equivalent A1 a 2 + 4 a 2 + 4 OR: Obtain two equations in x and y, solve for x or for y M1 5 a 10 Obtain final answer x = and y = , or equivalent A1 [2] a 2 + 4 a 2 + 4 3 (ii) Either state arg(u) = − π , or express u* in terms of a (f.t. on u) B1√ 4 Use correct method to form an equation in a, e.g. 5a = –10 M1 Obtain a = –2 correctly A1 [3] (b) Show a point representing 2 + 2i in relatively correct position in an Argand diagram B1 Show the circle with centre at the origin and radius 2 B1 Show the perpendicular bisector of the line segment from the origin to the point representing 2 + 2i B1√ Shade the correct region B1 [4] [SR: Give the first B1 and the B1√ for obtaining y = 2 – x, or equivalent, and sketching the attempt.] A Bx + C

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5x −x2 8 (i) Express in partial fractions. [5] (1 + x)(2 + x2) 5x −x2 (ii) Hence obtain the expansion of in ascending powers of x, up to and including the (1 + x)(2 + x2) term in x3. [5]

Mark scheme: A Bx + C 8 (i) State or imply partial fractions are of the form + B1 1 + x 2 + x 2 Use a relevant method to determine a constant M1 Obtain one of the values A = –2, B = 1, C = 4 A1 Obtain a second value A1 Obtain the third value A1 [5] (ii) Use correct method to obtain the first two terms of the expansion of (1 + x ) −1 , −1 x M1 + 1 1 x 2  or (2 + x 2 )−1 in ascending powers of  2  Obtain correct unsimplified expansion up to the term in x3 of each partial fraction A1√ + A1√ Multiply out fully by Bx + C, where BC ≠ 0 M1 5 2 7 3 Obtain final answer x − 3 x + x , or equivalent A1 [5] 2 4 − 1  [Symbolic binomial coefficients, e.g.   , are not sufficient for the first M1. The f.t. is  1  on A, B, C.] [If B or C omitted from the form of fractions, give B0M1A0A0A0 in (i); M1A1√A1√ in (ii), max 4/10.] [In the case of an attempt to expand (5x – x2)(1 + x)–1(2 + x2)–1, give M1A1A1 for the expansions, M1 for the multiplying out fully, and A1 for the final answer.] [Allow use of Maclaurin, giving M1A1√A1√ for differentiating and obtaining f(0) = 0 5 21 and f '(0) = , A1√ for f ''(0) = –6, and A1 for f '''(0) = and the final answer (the f.t. 2 2 is on A, B, C if used).] [For the identity 5 x − x 2 ≡ (2 + 2 x + x 2 + x 3 )( a + bx + cx 2 + dx 3 ) give M1A1; then M1A1 5 7 for using a relevant method to obtain two of a = 0, b = , c = –3 and d = ; then A1 for 2 4 the final answer in series form.] GCE AS/A LEVEL – May/June 2011 9709 32

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9 Two planes have equations x + 2y −2ß = 7 and 2x + y + 3ß = 5. (i) Calculate the acute angle between the planes. [4] (ii) Find a vector equation for the line of intersection of the planes. [6]

Mark scheme: 9 (i) State or imply a correct normal vector to either plane, e.g. i + 2j –2k or 2i + j + 3k B1 Carry out correct process for evaluating the scalar product of the two normals M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result M1 Obtain the final answer 79.7° (or 1.39 radians) A1 [4] (ii) EITHER: Carry out a method for finding a point on the line M1 Obtain such a point, e.g. (1, 3, 0) A1 EITHER: State two correct equations for the direction vector (a, b, c) of the line, e.g. a + 2b – 2c = 0 and 2a + b + 3c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 8 : –7 : –3, or equivalent A1 State a correct final answer, e.g. r = i + 3j + λ(8i – 7j – 3k) A1√  31 3  OR1: Obtain a second point on the line, e.g.  ,0 ,  A1  8 8  Subtract position vectors to find a direction vector M1 7 3 Obtain i − j − k, or equivalent A1 8 8 7 3 State a correct final answer, e.g. r = i + 3j + λ(i − j − k) A1√ 8 8 OR2: Attempt to calculate the vector product of two normals M1 Obtain two correct components A1 Obtain 8i – 7j – 3k, or equivalent A1 State a correct final answer, e.g. r = i + 3j + λ(8i – 7j – 3k) A1√ OR3: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. x = (31 – 8y) / 7 A1 Express the first variable in terms of a third M1 Obtain a correct simplified expression, e.g. x = (3 – 8z) / 3 A1 Form a vector equation of the line M1 31 3 State a correct final answer, e.g. r = j + k + λ(8i – 7j – 3k) A1√ 8 8 OR4: Express one variable in terms of a second M1 Obtain a correct simplified expression, e.g. y = (31 – 7x) / 7 A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. z = (3 – 3x) / 8 A1 Form a vector equation of the line M1 31 3 State a correct final answer, e.g. r = j + k + λ(– 8i + 7j + 3k) A1√ [6] 8 8 [The f.t. is dependent on all M marks having been earned.] GCE AS/A LEVEL – May/June 2011 9709 32 2 ∫

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10 y M P x O 3 The diagram shows the curve y = x2e−x. (i) Show that the area of the shaded region bounded by the curve, the x-axis and the line x = 3 is equal to 2 −17 . [5] e3 (ii) Find the x-coordinate of the maximum point M on the curve. [4] (iii) Find the x-coordinate of the point P at which the tangent to the curve passes through the origin. [2]

Mark scheme: 10 (i) Attempt integration by parts and reach ± x 2 e −±x ∫ 2 xe − x dx M1* Obtain − x 2 e −+x ∫ 2 xe − x d x , or equivalent A1 Integrate and obtain –x2e–x – 2xe–x – 2e–x, or equivalent A1 Use limits x = 0 and x = 3, having integrated by parts twice M1(dep*) Obtain the given answer correctly A1 [5] (ii) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for non-zero x M1 Obtain x = 2 with no errors send A1 [4] (iii) Carry out a complete method for finding the x-coordinate of P M1 Obtain answer x =1 A1 [2]

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