Cambridge A Level Mathematics 9709 — 2016 Oct/Nov Paper 3 · Variant 1

3x 21 Solve the equation + 8, giving your answer correct to 3 decimal places. [3] 3x = −2

Mark scheme: 1 Solve for3x and obtain 3x = 187 B1 Use correct method for solving an equation of the form 3x = a , where a > 0 M1 Obtain answer x = 0.860 3 d.p. only A1 [3] 3

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3 Express the equation sec 3 cos tan as a quadratic equation in sin Hence solve this equation 1 = 1 + 1 1. for [5] −90Å < 1 < 90Å.

Mark scheme: 3 EITHER: Correctly restate the equation in terms ofsinθand cosθ B1 Correct method to obtain a horizontal equation insinθ M1 Reduce the equation to a correct quadratic in any form, e.g. 3sin 2 θ− sinθ− 2 = 0 A1 Solve a three-term quadratic forsinθ M1 Obtain final answer θ= −41.8° only A1 [Ignore answers outside the given interval.] OR 1: Square both sides of the equation and use 1 + tan 2 θ = sec 2 θ B1 Correct method to obtain a horizontal equation insinθ M1 Reduce the equation to a correct quadratic in any form, e.g. 9sin 2 θ − 6sin θ − 8 = 0 A1 Solve a three-term quadratic for sinθ M1 Obtain final answer θ= −41.8° only A1 OR 2: Multiply through by (secθ + tanθ) M1 Use sec2θ – tan2θ = 1 B1 Obtain 1 = 3 + 3sinθ A1 Solve for sinθ M1 Obtain final answer θ= −41.8° only A1 [5] 2 dy

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5 (i) Prove the identity tan tan sec [4] 21 −tan 1  1 21. 160 1 3 (ii) Hence show that tan sec ln [4] 2 2. Ó 0 1 21 d1 =

Mark scheme: 5 (i) EITHER: Use tan 2A formula to express LHS in terms of tanθ M1 Express as a single fraction in any correct form A1 Use Pythagoras or cos 2A formula M1 Obtain the given result correctly A1 OR: Express LHS in terms of sin 2θ, cos 2θ, sin θand cos θ M1 Express as a single fraction in any correct form A1 Use Pythagoras or cos 2A formula or sin(A – B) formula M1 Obtain the given result correctly A1 [4] (ii) Integrate and obtain a term of the form a ln(cos2θ) or b ln(cosθ) (or secant equivalents) M1* Obtain integral − 12 ln(cos 2θ) + ln(cosθ) , or equivalent A1 Substitute limits correctly (expect to see use of both limits) DM1 Obtain the given answer following full and correct working A1 [4]

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6 (i) By sketching a suitable pair of graphs, show that the equation 1 cosec 2x1 = 13x + has one root in the interval 0 x [2] < ≤0. (ii) Show by calculation that this root lies between 1.4 and 1.6. [2] (iii) Show that, if a sequence of values in the interval 0 x given by the iterative formula < ≤0 @ A 3 2 sin−1 xn+1 = xn 3 + converges, then it converges to the root of the equation in part (i). [2] (iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

Mark scheme: 6 (i) Make recognizable sketch of a relevant graph B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Use calculations to consider the value of a relevant expression at x = 1.4 and x = 1.6, or the values of relevant expressions at x = 1.4 and x = 1.6 M1 Complete the argument correctly with correct calculated values A1 [2] −1 3  (iii) State x = 2sin   B1  x + 3  Rearrange this in the form cosec 12 x = 13 x + 1 B1 [2] x  3  If working in reverse, need sin = for first B1   2  x + 3  (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.471 A1 Show sufficient iterations to 5 d.p. to justify 1.471 to 3 d.p., or show there is a sign change in the interval (1.4705, 1.4715) A1 [3]

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7 y M x O 2 12x The diagram shows part of the curve y 2x e and its maximum point M. = −x2 (i) Find the exact x-coordinate of M. [4] (ii) Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis. [5]

Mark scheme: 7 (i) Use the correct product rule M1 1 2 x 1 2 12 x Obtain correct derivative in any form, e.g. (2 − 2 x )e + 2 (2 x − x )e A1 Equate derivative to zero and solve for x M1 Obtain x = 5 − 1 only A1 [4] (2 − 2 x )e 1 (ii) Integrate by parts and reach a (2 x − x 2 )e 1 2 x dx M1* 2 x + b ∫ 1 (2 − 2 x )e d x , or equivalent A1 Obtain 2 e 2 x (2 x − x 2 ) − 2 ∫ 12 x 2 x , or equivalent A1 Complete the integration correctly, obtaining (12 x − 2 x 2 − 24)e 1 Use limits x = 0, x = 2 correctly having integrated by parts twice DM1 Obtain answer 24 – 8e, or exact simplified equivalent A1 [5]

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8 Two planes have equations 3x y 2 and x 2z 3. + −z = −y + = (i) Show that the planes are perpendicular. [3] (ii) Find a vector equation for the line of intersection of the two planes. [6]

Mark scheme: 8 (i) State or imply a correct normal vector to either plane, e.g. 3i + j − k or i − j + 2 k B1 Use correct method to calculate their scalar product M1 Show value is zero and planes are perpendicular A1 [3] (ii) EITHER: Carry out a complete strategy for finding a point on l the line of intersection M1 Obtain such a point, e.g. (0, 7, 5) , (1, 0, 1), (5/4, –7/4, 0) A1 EITHER: State two equations for a direction vector a i + b j + ck for l, e.g. 3a + b − c = 0 and a − b + 2 c = 0 B1 Solve for one ratio, e.g. a : b M1 Obtain a : b : c = 1 : −7 : −4, or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Obtain a second point on l, e.g. (1, 0, 1) B1 Subtract vectors and obtain a direction vector for l M1 Obtain −+i 7 j + 4k , or equivalent A1 State a correct answer, e.g. r = i + k + λ( −+i 7 j + 4k ) A1 OR2: Attempt to find the vector product of the two normal vectors M1 Obtain two correct components of the product A1 Obtain i − 7 j − 4k , or equivalent A1 State a correct answer, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR1: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. y = 7 – 7x A1 Express the third variable in terms of the second M1 Obtain a correct simplified expression, e.g. z = 5 – 4x A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 7 j + 5k + λ( i − 7 j − 4k ) A1 OR2: Express one variable in terms of a second variable M1 Obtain a correct simplified expression, e.g. z = 5 − 4 x A1 Express the same variable in terms of the third M1 Obtain a correct simplified expression e.g. z = (7 + 4 y ) / 7 A1 Form a vector equation for the line M1 Obtain a correct equation, e.g. r = 54 i − 74 j + λ( − 14 i + 74 j + k ) A1 [6]

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9 Throughout this question the use of a calculator is not permitted. (a) Solve the equation 1 2i w2 4w 1 0, giving your answers in the form x iy, where + + − −2i = + x and y are real. [5] (b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequalities z and [5] −1 −i ≤2 −140 ≤argz ≤140.

Mark scheme: 9 (a) EITHER: Use quadratic formula to solve for w M1 Use 2i = −1 M1 1 5 Obtain one of the answers w = and w = − A1 2i + 1 2i + 1 Multiply numerator and denominator of an answer by –2i + 1, or equivalent M1 Obtain final answers 15 − 52 i and –1 + 2i A1 OR1: Multiply the equation by 1 – 2i M1 Use 2i = −1 M1 Obtain 5 w 2 + 4 w (1 − 2 i ) − (1 − 2 i ) 2 = 0 , or equivalent A1 Use quadratic formula or factorise to solve for w M1 Obtain final answers 15 − 52 i and –1 + 2i A1 OR2: Substitute w = x + iy and form equations for real and imaginary parts M1 Use 2i = −1 M1 Obtain ( x 2 − y 2 ) − 4 xy + 4 x − 1 = 0 and 2 ( x 2 − y 2 ) + 2 xy + 4 y + 2 = 0 o.e. A1 Form equation in x only or y only and solve M1 Obtain final answers 15 − 52 i and –1 + 2i A1 [5] (b) Show a circle with centre 1 + i B1 Show a circle with radius 2 B1 Show half-line arg z = 14π B1 Show half-line arg z = − 14π B1 Shade the correct region B1 [5]

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10 A large field of area 4 km2 is becoming infected with a soil disease. At time t years the area infected is dx x km2 and the rate of growth of the infected area is given by the differential equation kx 4 , dt = −x where k is a positive constant. It is given that when t 0, x 0.4 and that when t 2, x 2. = = = = (i) Solve the differential equation and show that k 1 ln 3. [9] = 4 (ii) Find the value of t when 90% of the area of the field is infected. [2]

Mark scheme: 10 (i) Separate variables correctly and integrate at least one side M1 Integrate and obtain term kt, or equivalent A1 1 A B Carry out a relevant method to obtain A and B such that ≡ + , or equivalent M1* x (4 − x ) x 4 − x 1 Obtain A = B = , or equivalent A1 4 1 1 Integrate and obtain terms ln x − ln(4 − x ) , or equivalent A1 4 4 EITHER: Use a pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 Obtain correct answer in any form, e.g.ln x − ln(4 − x ) = 4 kt − ln9 ,  x = or ln   4 kt − 8k A1  4 − x  Use a second pair of limits and determine k DM1 Obtain the given exact answer correctly A1 OR: Use both pairs of limits in a definite integral M1* Obtain the given exact answer correctly A1 Substitute k and either pair of limits in an expression containing plnx, qln(4 – x) and rt and evaluate a constant DM1 x Obtain ln = t ln 3 − ln 9 or equivalent A1 [9] 4 − x (ii) Substitute x = 3.6 and solve for t M1 Obtain answer t = 4 A1 [2]

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