Cambridge A Level Mathematics 9709 — 2016 May/June Paper 3 · Variant 2

120 3 Find the exact value of x2 sin 2x dx. [5] Ó 0

Mark scheme: 3 Integrate by parts and reach ax 2 cos2 x + b ∫ x cos2 x dx M1* Obtain − 12 x 2 cos2 x +∫ x cos2 x , or equivalent A1 Complete the integration and obtain − 12 x 2 cos2 x + 12 x sin 2 x + 14 cos2 x , or equivalent A1 Use limits correctly having integrated twice DM1* Obtain answer 18 (π 2 − 4) , or exact equivalent, with no errors seen A1 [5] 2 2ln x

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5 (i) Prove the identity cos 8 [4] cos41 −4 21  sin41 −3. (ii) Hence solve the equation cos 4 cos 3, 41 = 21 + for [4] 0Å ≤1 ≤360Å.

Mark scheme: 5 (i) EITHER: Express cos 4θ in terms of cos 2θ and/or sin 2θ B1 Use correct double angle formulae to express LHS in terms of sin θ and/or cos θ M1 Obtain a correct expression in terms of sin θ alone A1 Reduce correctly to the given form A1 OR: Use correct double angle formula to express RHS in terms of cos 2θ M1 Express cos 2 2θ in terms of cos 4θ B1 Obtain a correct expression in terms of cos 4θ and cos 2θ A1 Reduce correctly to the given form A1 [4] (ii) Use the identity and carry out a method for finding a root M1 Obtain answer 68.5° A1 Obtain a second answer, e.g. 291.5° A1 Obtain the remaining answers, e.g. 111.5° and 248.5°, and no others in the given interval A1 [4] [Ignore answers outside the given interval. Treat answers in radians as a misread.]

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6 The variables x and satisfy the differential equation 1 dx 3 cos x sin + 21 = 21, d1 and it is given that x 3 when 1 = 1 = 40. (i) Solve the differential equation and obtain an expression for x in terms of [7] 1. (ii) State the least value taken by x. [1]

Mark scheme: 6 (i) Separate variables correctly and attempt integration of at least one side B1 Obtain term ln x B1 Obtain term of the form k ln(3 + cos2θ ) , or equivalent M1 Obtain term − 12 ln(3 + cos2θ ) , or equivalent A1 Use x = 3, θ = 14 π to evaluate a constant or as limits in a solution with terms a ln x and b ln(3 + cos2θ ) ,where ab ≠ 0 M1 State correct solution in any form, e.g. ln x = − 12 ln(3 + cos2θ ) + 32 ln3 A1  27  Rearrange in a correct form, e.g. x =   A1 [7]  3 + cos2θ  (ii) State answer x = 3 3 / 2 , or exact equivalent (accept decimal answer in [2.59, 2.60]) B1 [1] B C

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8 y x O a 0 The diagram shows the curve y cosecx for 0 x and part of the curve y When x a, the = < = e−x. = < 0 tangents to the curves are parallel. 1 dy (i) By differentiating show that if y cosecx then cotx. [3] sin x, = dx = −cosecx (ii) By equating the gradients of the curves at x a, show that = @ A ea . [2] a = tan−1 sin a (iii) Verify by calculation that a lies between 1 and 1.5. [2] (iv) Use an iterative formula based on the equation in part (ii) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

Mark scheme: 8 (i) Use correct quotient or chain rule M1 Obtain correct derivative in any form A1 Obtain the given answer correctly A1 [3] (ii) State a correct equation, e.g. − e − a = − cosec a cot a B1 Rearrange it correctly in the given form B1 [2] (iii) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 M1 Complete the argument correctly with correct calculated values A1 [2] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.317 A1 Show sufficient iterations to 5 d.p. to justify 1.317 to 3 d.p., or show there is a sign change in the interval (1.3165, 1,3175) A1 [3]

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9 The points A, B and C have position vectors, relative to the origin O, given by −−→OA i 2j 3k, = + + −−→OB 4j k and −−→OC 2i 5j A fourth point D is such that the quadrilateral ABCD is a = + = + −k. parallelogram. (i) Find the position vector of D and verify that the parallelogram is a rhombus. [5] (ii) The plane p is parallel to OA and the line BC lies in p. Find the equation of p, giving your answer in the form ax by cz d. [5] + + =

Mark scheme: 9 (i) Either state or imply AB or BC in component form, or state position vector of midpoint of AC B1 Use a correct method for finding the position vector of D M1 Obtain answer 3i + 3 j + k , or equivalent A1 EITHER: Using the correct process for the moduli, compare lengths of a pair of adjacent sides, e.g. AB and BC M1 Show that ABCD has a pair of adjacent sides that are equal A1 OR: Calculate scalar product AC . BD or equivalent M1 Show that ABCD has perpendicular diagonals A1 [5] (ii) EITHER: State a + 2b + 3c = 0 or 2 a + b − 2 c = 0 B1 Obtain two relevant equations and solve for one ratio, e.g. a : b M1 Obtain a : b : c = −7 : 8 : −3, or equivalent A1 Substitute coordinates of a relevant point in −7x + 8y −3z = d, and evaluate M1 Obtain answer −7x + 8y −3z = 29, or equivalent A1 OR1:Attempt to calculate vector product of relevant vectors, e.g. ( i + 2 j + 3k ) × (2i + j − 2k ) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −7 i + 8 j − 3k A1 Substitute coordinates of a relevant point in −7 x + 8 y − 3 z = d and evaluate d M1 Obtain answer −7 x + 8 y − 3 z = 29 or equivalent A1 OR2:Attempt to form a 2-parameter equation with relevant vectors M1 State a correct equation, e.g. r = 2 i + 5 j − k + λ ( i + 2 j + 3k ) + µ (2i + j − 2k ) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate λ and µ M1 Obtain answer −7 x + 8 y − 3 z = 29 , or equivalent A1 OR3:Using a relevant point and relevant direction vectors, form a determinant equation for the plane M1 x − 2 y − 5 z + 1 State a correct equation, e.g. 1 2 3 = 0 A1 2 1 −2 Attempt to expand the determinant M1 Obtain correct values of two cofactors A1 Obtain answer −7 x + 8 y − 3 z = 29 , or equivalent A1 [5]

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10 (a) Showing all necessary working, solve the equation iz2 2z 0, giving your answers in the + −3i = form x iy, where x and y are real and exact. [5] + (b) (i) On a sketch of an Argand diagram, show the locus representing complex numbers satisfying the equation z z . [2] = −4 −3i (ii) Find the complex number represented by the point on the locus where z is least. Find the modulus and argument of this complex number, giving the argument correct to 2 decimal places. [3]

Mark scheme: 10 (a) EITHER: Use quadratic formula to solve for z M1 Use 2i = −1 M1 Obtain a correct answer in any form, simplified as far as ( −±2 i 8) / 2i A1 Multiply numerator and denominator by i, or equivalent M1 Obtain final answers 2 + i and − 2 + i A1 OR: Substitute x + iy and equate real and imaginary parts to zero M1 Use 2i = −1 M1 Obtain −2 xy + 2 x = 0 and x 2 − y 2 + 2 y − 3 = 0 , or equivalent A1 Solve for x and y M1 Obtain final answers 2 + i and − 2 + i A1 [5] (b) (i) EITHER: Show the point representing 4 + 3i in relatively correct position B1 Show the perpendicular bisector of the line segment joining this point to the origin B1 [2] OR: Obtain correct Cartesian equation of the locus in any form, e.g. 8 x + 6 y = 25 B1 Show this line B1 [This f.t. is dependent on using a correct method to determine the equation.] (ii) State or imply the relevant point is represented by 2 + 1.5i or is at (2, 1.5) B1 Obtain modulus 2.5 B1 Obtain argument 0.64 (or 36.9°) (allow decimals in [0.64, 0.65] or [36.8, 36.9]) B1 [3]

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