Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 3 · Variant 2

1 Using the substitution u ex, or otherwise, solve the equation = 1 ex = + 6e−x, giving your answer correct to 3 significant figures. [4]

Mark scheme: 1 Rearrange as e2x – ex – 6 = 0, or u2 – u – 6 = 0, or equivalent B1 Solve a 3-term quadratic for ex or for u M1 Obtain simplified solution ex = 3 or u = 3 A1 Obtain final answer x = 1.10 and no other A1 [4]

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2 The parametric equations of a curve are x y 2 cos3t. = 3(1 + sin2t), = dy Find in terms of t, simplifying your answer as far as possible. [5] dx

Mark scheme: 2 EITHER: Use chain rule M1 dx obtain = 6 sin t cos t , or equivalent A1 dt d y 2 obtain = −6 cos t sin t , or equivalent A1 d t dy dy dx Use = ÷ M1 dx dt dt d y Obtain final answer = − cos t A1 d x OR: Express y in terms of x and use chain rule M1 1 dy x ) = k ( 2 − Obtain 2 , or equivalent A1 dx 3 1 dy x ) = − ( 2 − Obtain 2 , or equivalent A1 dx 3 Express derivative in terms of t M1 d y Obtain final answer = − cos t A1 [5] d x 2 2

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3 The polynomial x4 3x3 ax 3 is denoted by It is given that is divisible by x2 1. + + + p(x). p(x) −x + (i) Find the value of a. [4] (ii) When a has this value, find the real roots of the equation 0. [2] p(x) =

Mark scheme: 3 (i) EITHER: Attempt division by x2 – x + 1 reaching a partial quotient of x2 + kx M1 Obtain quotient x2 + 4x + 3 A1 Equate remainder of form lx to zero and solve for a, or equivalent M1 Obtain answer a = 1 A1 OR: Substitute a complex zero of x2 – x + 1 in p(x) and equate to zero M1 Obtain a correct equation in a in any unsimplified form A1 Expand terms, use i2 = –1 and solve for a M1 Obtain answer a = 1 A1 [4] [SR: The first M1 is earned if inspection reaches an unknown factor x2 + Bx + C and an equation in B and/or C, or an unknown factor Ax2 + Bx + 3 and an equation in A and/or B. The second M1 is only earned if use of the equation a = B – C is seen or implied.] (ii) State answer, e.g. x = –3 B1 State answer, e.g. x = –1 and no others B1 [2]

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4 The variables x and θ are related by the differential equation dx sin 2θ cos 2θ, dθ = (x + 1) where 0 θ 12π. When θ 12π,1 x 0. Solve the differential equation, obtaining an expression for < < = = x in terms of θ, and simplifying your answer as far as possible. [7]

Mark scheme: 4 Separate variables and attempt integration of at least one side M1 Obtain term ln(x + 1) A1 Obtain term k ln sin 2θ, where k = ±1, ±2, or ± 1 M1 2 Obtain correct term 1 ln sin 2θ A1 2 Evaluate a constant, or use limits θ = 1 π, x = 0 in a solution containing terms a ln(x + 1) and 12 b ln sin 2θ M1 Obtain solution in any form, e.g. ln(x + 1) = 1 ln sin 2θ − 1 ln 1 (f.t. on k = ±1, ±2, or ± 1 ) A1√ 2 2 2 2 Rearrange and obtain x = ( 2 sin 2θ ) − 1 , or simple equivalent A1 [7] GCE AS/A LEVEL – October/November 2011 9709 32

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5 (i) By sketching a suitable pair of graphs, show that the equation secx 3 2x2, = −1 where x is in radians, has a root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 1 and 1.4. [2] (iii) Show that this root also satisfies the equation . x [1] = cos−1 6 2 −x2 (iv) Use an iterative formula based on the equation in part (iii) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 5 (i) Make recognisable sketch of a relevant graph over the given interval B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Consider the sign of sec x – (3 – 1 x2) at x = 1 and x = 1.4, or equivalent M1 2 Complete the argument with correct calculated values A1 [2] (iii) Convert the given equation to sec x = 3 – 1 x2 or work vice versa B1 [1] 2 (iv) Use a correct iterative formula correctly at least once M1 Obtain final answer 1.13 A1 Show sufficient iterations to 4 d.p. to justify 1.13 to 2 d.p., or show there is a sign change in the interval (1.125, 1.135) A1 [3] [SR: Successive evaluation of the iterative function with x = 1, 2, … scores M0.]

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α 0 and giving the exact6 (i) Express cos x 3 sin x in the form R where R + cos(x −α), > < 90◦, 0◦< value of R and the value of α correct to 2 decimal places. [3] θ 3 sin 2θ 2, for [5] (ii) Hence solve the equation cos 2θ + = < 90◦. 0◦<

Mark scheme: 6 (i) State or imply R = 10 B1 Use trig formulae to find α M1 Obtain α = 71.57° with no errors seen A1 [3] [Do not allow radians in this part. If the only trig error is a sign error in cos(x – α) give M1A0] (ii) Evaluate cos–1 ( 2 / 10 ) correctly to at least 1 d.p. (50.7684…°) (Allow 50.7° here) B1√ Carry out an appropriate method to find a value of 2θ in 0° < 2θ < 180° M1 Obtain an answer for θ in the given range, e.g. θ = 61.2° A1 Use an appropriate method to find another value of 2θ in the above range M1 Obtain second angle, e.g. θ = 10.4°, and no others in the given range A1 [5] [Ignore answers outside the given range.] [Treat answers in radians as a misread and deduct A1 from the answers for the angles.] [SR: The use of correct trig formulae to obtain a 3-term quadratic in tan θ, sin 2θ, cos 2θ,or tan 2θ earns M1; then A1 for a correct quadratic, M1 for obtaining a value of θ in the given range, and A1 + A1 for the two correct answers (candidates who square must reject the spurious roots to get the final A1).] GCE AS/A LEVEL – October/November 2011 9709 32

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7 With respect to the origin O, the position vectors of two points A and B are given by −−→OA i 2j 2k = + + and −−→OB 3i 4j. The point P lies on the line through A and B, and −−→AP λ −−→AB. = + = (i) Show that −−→OP [2] = (1 + 2λ)i + (2 + 2λ)j + (2 −2λ)k. (ii) By equating expressions for cos AOP and cos BOP in terms of λ, find the value of λ for which OP bisects the angle AOB. [5] (iii) When λ has this value, verify that AP : PB OA : OB. [1] =

Mark scheme: 7 (i) Use a correct method to express OP in terms of λ M1 Obtain the given answer A1 [2] (ii) EITHER: Use correct method to express scalar product of OA and OP , or OB and OP in terms of λ M1 Using the correct method for the moduli, divide scalar products by products of moduli and express cos AOP = cos BOP in terms of λ, or in terms of λ and OP M1* OR1: Use correct method to express OA2 + OP2 – AP2, or OB2 + OP2 – BP2 in terms of λ M1 Using the correct method for the moduli, divide each expression by twice the product of the relevant moduli and express cos AOP = cos BOP in terms of λ, or λ and OP M1* 9 + 2 λ 11 + 14 λ Obtain a correct equation in any form, e.g. = A1 3 (9 + 4 λ + 12 λ 2 ) 5 (9 + 4 λ + 12 λ 2 ) Solve for λ M1(dep*) Obtain λ = 3 A1 [5] 8 [SR: The M1* can also be earned by equating cos AOP or cos BOP to a sound attempt at cos 1 AOB and obtaining an equation in λ. The exact value of the cosine is (13 / 15) , 2 but accept non-exact working giving a value of λ which rounds to 0.375, provided the spurious negative root of the quadratic in λ is rejected.] [SR: Allow a solution reaching λ = 3 after cancelling identical incorrect expressions for 8 OP to score 4/5. The marking will run M1M1A0M1A1, or M1M1A1M1A0 in such cases.] (iii) Verify the given statement correctly B1 [1]

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9 y x e O M The diagram shows the curve y x2 ln x and its minimum point M. = (i) Find the exact values of the coordinates of M. [5] (ii) Find the exact value of the area of the shaded region bounded by the curve, the x-axis and the line x e. [5] =

Mark scheme: 9 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero1 and solve for x M1 Obtain answer x = e– 2 , or equivalent A1 Obtain answer y = – 1 e–1, or equivalent A1 [5] 2 1 (ii) Attempt integration by parts reaching kx3 ln x ± k ∫ x 3 . x dx M1* Obtain 1 x 3 ln x − 1 x 2 d x , or equivalent A1 3 3 ∫ Integrate again and obtain 1 x 3 ln x − 1 x 3 , or equivalent A1 3 9 Use limits x = 1 and x = e, having integrated twice M1(dep*) Obtain answer 1 (2e3 + 1), or exact equivalent A1 [5] 9 [SR: An attempt reaching ax2 (x ln x – x) + b ∫ 2 x ( x ln x −)x dx scores M1. Then give the first A1 for I = x2 (x ln x – x) – 2I + ∫ 2x 2 dx, or equivalent.]

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10 (a) Showing your working, find the two square roots of the complex number 1 Give your −(2√6)i. answers in the form x iy, where x and y are exact. [5] + (b) On a sketch of an Argand diagram, shade the region whose points represent the complex numbers which satisfy the inequality Find the greatest value of arg for points in this region. ß |ß −3i| ≤2. ß [5]

Mark scheme: 10 (a) EITHER: Square x + iy and equate real and imaginary parts to 1 and − 2 6 respectively M1* Obtain x2 – y2 = 1 and 2xy = − 2 6 A1 Eliminate one variable and find an equation in the other M1(dep*) Obtain x4 – x2 – 6 = 0 or y4 + y2 – 6 = 0, or 3-term equivalent A1 Obtain answers ± ( 3 − i 2 ) A1 [5] OR: Denoting 1− 2 i6 by Rcisθ, state, or imply, square roots are ± R cis ( 1 θ ) 2 and find values of R and either cos θ or sin θ or tan θ M1* Obtain ± 5 (cos 1 θ + i sin 1 θ ) , and cos θ = 1 or sin θ = − 2 6 or 2 2 5 5 tan θ = −2 6 A1 Use correct method to find an exact value of cos 1 θ or sin 1 θ M1(dep*) 2 2 Obtain cos 1 θ = ± 3 and sin 1 θ = ± 2 , or equivalent A1 2 5 2 5 Obtain answers ± ( 3 − i 2 ) , or equivalent A1 [Condone omission of ± except in the final answers.] (b) Show point representing 3i on a sketch of an Argand diagram B1 Show a circle with centre at the point representing 3i and radius 2 B1√ Shade the interior of the circle B1√ Carry out a complete method for finding the greatest value of arg z M1 Obtain answer 131.8° or 2.30 (or 2.3) radians A1 [5] [The f.t. is on solutions where the centre is at the point representing –3i.]

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