Cambridge A Level Mathematics 9709 — 2014 May/June Paper 3 · Variant 2

1 Find the set of values of x satisfying the inequality x + 2a > 3 x −a , where a is a positive constant. [4]

Mark scheme: 1 EITHER: State or imply non-modular inequality ( x + 2 a ) 2 > (3( x − a )) 2 , or corresponding quadratic equation, or pair of linear equations ( x + 2 a ) = ±3( x − a ) B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for x M1 Obtain critical values x = 14 a and x = 52 a A1 State answer 14 a < x < 52 a A1 OR: Obtain critical value x = 52 a from a graphical method, or by inspection, or by solving a linear equation or inequality B1 Obtain critical value x = 14 a similarly B2 State answer 14 a < x < 52 a B1 4 [Do not condone Y for <.] 2 1

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4 The parametric equations of a curve are x = t −tan t, y = ln cost , for −120 < t < 120. dy (i) Show that = cott. [5] dx (ii) Hence find the x-coordinate of the point on the curve at which the gradient is equal to 2. Give your answer correct to 3 significant figures. [2]

Mark scheme: dx 24 (i) State = 1 − sec t , or equivalent B1 dt Use chain rule M1 dy sin t Obtain = − , or equivalent A1 dt cos t dy dy dx Use = ÷ M1 dx dt dt Obtain the given answer correctly. A1 5 (ii) State or imply t = tan −1 ( 12 ) B1 Obtain answer x = −0.0364 B1 2 GCE A LEVEL – May/June 2014 9709 32 2

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6 A B C x r O In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is equal to x radians. The shaded region is bounded by AB, AC and the circular arc with centre A joining B and C. The perimeter of the shaded region is equal to half the circumference of the circle. 0 1 0 (i) Show that x = cos−1 . [3] 4 + 4x (ii) Verify by calculation that x lies between 1 and 1.5. [2] (iii) Use the iterative formula @ A 0 xn+1 = cos−1 4 + 4xn to determine the value of x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 6 (i) Use correct arc formula and form an equation in r and x M1 Obtain a correct equation in any form A1 Rearrange in the given form A1 3 (ii) Consider sign of a relevant expression at x = 1 and x = 1.5, or compare values of relevant expressions at x = 1 and x = 5.1 M1 Complete the argument correctly with correct calculated values A1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.21 A1 Show sufficient iterations to 4 d.p. to justify 1.21 to 2 d.p., or show there is a sign change in the interval (1.205,1.215) A1 3 3

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7 (a) It is given that −1 + ï5 i is a root of the equation Ï3 + 2Ï + a = 0, where a is real. Showing your working, find the value of a, and write down the other complex root of this equation. [4] w −1 (b) The complex number w has modulus 1 and argument 21 radians. Show that = i tan 1. [4] w + 1

Mark scheme: 7 (a) EITHER: Substitute and expand ( − 1 + 5 3)i completely M1 Use i 2 = −1 correctly at least once M1 Obtain a = −12 A1 State that the other complex root is − 1− 5 i B1 OR1: State that the other complex root is − 1− 5 i B1 State the quadratic factor z 2 + 2 z + 6 B1 Divide the cubic by a 3-term quadratic, equate remainder to zero and solve for a or, using a 3-term quadratic, factorise the cubic and determine a M1 Obtain a = −12 A1 OR2: State that the other complex root is − 1− i5 B1 State or show the third root is 2 B1 Use a valid method to determine a M1 Obtain a = –12 A1 o OR3: Substitute and use De Moivre to cube 6 cis (114.1 ) , or equivalent M1 Find the real and imaginary parts of the expression M1 Obtain a = –12 A1 State that the other complex root is − 1− i5 B1 4 GCE A LEVEL – May/June 2014 9709 32 (b) EITHER: Substitute w = cos 2θ + i sin 2θ in the given expression B1 Use double angle formulae throughout M1 Express numerator and denominator in terms of cosθ and sinθ only A1 Obtain given answer correctly A1 OR: Substitute w = e 2 θi in the given expression B1 Divide numerator and denominator by ie θ , or equivalent M1 Express numerator and denominator in terms of cosθ and sinθ only A1 Obtain the given answer correctly A1 4

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8 y x 0 O The diagram shows the curve y = x cos 12x for 0 ≤x ≤0. dy 1 (i) Find and show that 4d2y + y + 4 sin = 0. [5] 2x dx dx2 (ii) Find the exact value of the area of the region enclosed by this part of the curve and the x-axis. [5]

Mark scheme: 8 (i) Use product rule M1 Obtain derivative in any correct form A1 Differentiate first derivative using the product rule M1 Obtain second derivative in any correct form, e.g. − 12 sin 12 x − 14 x cos 12 x − 12 sin 12 x A1 Verify the given statement A1 5 2 x d x M1* (ii) Integrate and reach kx sin 12 x + l ∫ sin 1 2 x d x , or equivalent A1 Obtain 2 x sin 12 x − 2 ∫ sin 1 Obtain indefinite integral 2 x sin 12 x + 4 cos 12 x A1 Use correct limits x = 0, x = π correctly M1(dep*) Obtain answer 2π − 4 , or exact equivalent A1 5 dN

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10 Referred to the origin O, the points A, B and C have position vectors given by −−→ −−→ −−→ OA = i + 2j + 3k, OB = 2i + 4j + k and OC = 3i + 5j −3k. (i) Find the exact value of the cosine of angle BAC. [4] (ii) Hence find the exact value of the area of triangle ABC. [3] (iii) Find the equation of the plane which is parallel to the y-axis and contains the line through B and C. Give your answer in the form ax + by + cÏ = d. [5]

Mark scheme: 10 (i) EITHER: State or imply AB and AC correctly in component form B1 Using the correct processes evaluate the scalar product AB. AC , or equivalent M1 Using the correct process for the moduli divide the scalar product by the product of the moduli M1 20 Obtain answer A1 21 OR: Use correct method to find lengths of all sides of triangle ABC M1 Apply cosine rule correctly to find the cosine of angle BAC M1 20 Obtain answer A1 4 21 (ii) State an exact value for the sine of angle BAC, e.g. 41/21 B1 Use correct area formula to find the area of triangle ABC M1 Obtain answer 12 41 , or exact equivalent A1 3 [SR: Allow use of a vector product, e.g. AB × AC = −6 i + 2 j − k B1 . Using correct process for the modulus, divide the modulus by 2 M1. Obtain answer 12 41 A1.] (iii) EITHER: State or obtain b = 0 B1 Equate scalar product of normal vector and BC (or CB ) to zero M1 Obtain a + b − 4c = 0 (or a − 4c = 0) A1 Substitute a relevant point in 4x + z = d and evaluate d M1 Obtain answer 4x + z = 9, or equivalent A1 OR1: Attempt to calculate vector product of relevant vectors, e.g. (j)×(i + j – 4k) M1 Obtain two correct components of the product A1 Obtain correct product, e.g. −4i − k A1 Substitute a relevant point in 4x + z = d and evaluate d M1 Obtain 4x + z = 9, or equivalent A1 OR2: Attempt to form 2-parameter equation for the plane with relevant vectors M1 State a correct equation, e.g. r = 2i +4j + k + λ(j) + µ(i + j – 4k) A1 State 3 equations in x, y, z, λ and µ A1 Eliminate µ M1 Obtain answer 4x + z = 9, or equivalent A1 OR3: State or obtain b = 0 B1 Substitute for B and C in the plane equation and obtain 2a + c = d and 3a – 3c = d (or 2a + 4b + c = d and 3a + 5b –3c = d) B1 Solve for one ratio, e.g. a : d M1 Obtain a : c : d, or equivalent M1 Obtain answer 4x + z = 9, or equivalent A1 OR4: Attempt to form a determinant equation for the plane with relevant vectors M1 x − 2 y − 4 z − 1 State a correct equation, e.g. 0 1 0 = 0 A1 1 1 − 4 Attempt to use a correct method to expand the determinant M1 Obtain two correct terms of a 3-term expansion, or equivalent A1 Obtain answer 4 x + z = 9 , or equivalent A1 5

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