Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 3 · Variant 1

1 Solve the inequality 2 −5x > 2 x −3 . [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 Make a recognisable sketch graph of 2 3 = − y x and the line y = 2 – 5x x = 3 stated, domain at least (– 2, 5). Find x-coordinate of intersection with y = 2 – 5x M1 Find point of intersection with y = 2|x – 3| or solve 2 – 5x with 2(x – 3) or –2(x – 3) Obtain 4 3 = − x A1 State final answer 4 3 < − x A1 Do not accept x < –1.33 [Do not condone ⩽ for < in the final answer.] Alternative method for question 1 State or imply non-modular inequality/equality (2 – 5x)2 >, ⩾, =, 22(x – 3)2, or corresponding quadratic equation, or pair of linear equations (2 – 5x) >, ⩾ , =, ± 2(x – 3) B1 Two correct linear equations only Make reasonable attempt at solving a 3-term quadratic, or solve one linear equation, or linear inequality for x M1 21x2 + 4x – 32 = (3x + 4)(7x – 8) = 0 2 – 5x or –(2 – 5x) with 2(x – 3) or –2(x – 3) Obtain critical value 4 3 = − x A1 State final answer 4 3 <− x A1 Do not accept x < –1.33 [Do not condone ⩽ for < in the final answer.] 4

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2 On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z ≥2 and z −1 + i ≤1. [4]

Mark scheme: 2 Show a circle with centre the origin and radius 2 B1 Show the point representing 1 – i B1 Show a circle with centre 1 – i and radius 1 B1 FT The FT is on the position of 1 – i. Shade the appropriate region B1 FT The FT is on the position of 1 – i. Shaded region outside circle with centre the origin and radius 2 and inside circle with centre ±1 ± i and radius 1 4

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3 The parametric equations of a curve are x = 3 −cos 21, y = 21 + sin 21, for 0 < 1 < 12π. dy Show that = cot 1. [5] dx ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 State or imply d dθ x = 2sin 2θ or d dθ y = 2 2cos2θ + B1 Use d d y x = d dθ y ÷ d dθ x M1 Obtain correct answer d d y x = 2 2cos2 2sin2 θ θ + A1 OE Use correct double angle formulae M1 Obtain the given answer correctly d cot d θ = y x A1 AG. Must have simplified numerator in terms of cosθ. Alternative method for question 3 Start by using both correct double angle formulae e.g. x = 3 – (2cos2θ − 1), y = 2θ + 2sinθcosθ M1 d dθ x or d dθ y B1 d d y x = ( ) ( ) 2 2 n 2 2 cos 4cos sin si + − θ θ θ θ M1 A1 Simplify to given answer correctly d cot d θ = y x A1 AG

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2x5 (a) By sketching a suitable pair of graphs, show that the equation cosec x = 1 + e−1 has exactly two roots in the interval 0 < x < π. [2] ........................................................................................................................................................ ........................................................................................................................................................ (b) The sequence of values given by the iterative formula P Q 1 xn+1 = π −sin−1 , e−12xn + 1 with initial value x1 = 2, converges to one of these roots. Use the formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = cosec x B1 cosec x, U shaped, roughly symmetrical about π π , 1 2 2 x y  = =     and domain at least π 5π , 6 6       . Sketch a second relevant graph, e.g. 1 2 1 e − = + x y , and justify the given statement B1 Exponential graph needs y(0) = 2, negative gradient, always increasing, and y(π) > 1 Needs to mark intersections with dots, crosses, or say roots at points of intersection, or equivalent 2 5(b) Use the iterative formula correctly at least twice M1 2, 2.3217, 2.2760, 2.2824… Need to see 2 iterations and following value inserted correctly Obtain final answer 2.28 A1 Must be supported by iterations Show sufficient iterations to at least 4 d.p. to justify 2.28 to 2 d.p., or show there is a sign change in the interval (2.275, 2.285) A1 3

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7 (a) Verify that −1 + 5i is a root of the equation 2x3 + x2 + 6x −18 = 0. 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(b) Find the other roots of this equation. 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Mark scheme: 7(a) Substitute 1 5 −+ i in the equation and attempt expansions of 2x and x3 M1 All working must be seen. Allow M1 if small errors in 1 2 5i 5 − − or 1 5i 5i 5 − − − and 4 2 5i 10 − + or 4 4 5i 2 5i 10 − + + Use i2 = –1 correctly at least once M1 1 – 5 or 4 + 10 seen Complete the verification correctly A1 ( ) ( ) ( ) 2 14 2 5i 4 2 5i 6 1 5i 18 0 − + −− + −+ − = 3 7(b) State second root 1 5i −− B1 Carry out a complete method for finding a quadratic factor with zeros 1 5i −+ and 1 5i −− M1 Obtain x2 + 2x + 6 A1 Obtain root 3 2 = x A1 OE Alternative method for question 7(b) State second root 1 5i −− B1 (x + 1 – 5i ) (x + 1 + 5i ) (2x + a) = 2x3 + x2 + 6x – 18 M1 (1 – 5 i) (1 + 5 i) a = –18 A1 6a = –18 a = –3 leading to 3 2 = x A1 OE Question Answer Marks Guidance 7(b) Alternative method for question 7(b) State second root 1 5i −− B1 POR = 6 SOR = – 2 M1 Obtain x2 + 2x + 6 A1 Obtain root 3 2 = x A1 OE Alternative method for question 7(b) State second root 1 5i −− B1 POR ( )( ) 1 5i 1 5i 9 −− −+ = a M1 A1 Obtain root 3 2 = x A1 OE Alternative method for question 7(b) State second root 1 5i −− B1 SOR ( ) ( ) 1 1 5i 1 5i 2 −− + −+ + = − a M1 A1 Obtain root 3 2 = x A1 OE 4

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10 y x O M The diagram shows the curve y = 2 −x e−12x, and its minimum point M. (a) Find the exact coordinates of M. 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(b) Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e. 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Mark scheme: 10(a) Use correct product or quotient rule *M1 ( ) 1 1 2 2 d 1 2 e e d 2 − −   = − − −     x x y x x M1 requires at least one of derivatives correct Obtain correct derivative in any form A1 Equate derivative to zero and solve for x DM1 Obtain x = 4 A1 ISW Obtain y = –2e–2, or exact equivalent A1 5 Question Answer Marks Guidance 10(b) Commence integration and reach ( ) 1 1 2 2 2 e e d − − − +  x x a x b x *M1 Condone omission of dx ( ) 1 1 2 2 2 2 e 4e − − − − + x x x or 1 2 2 e −x x Obtain ( ) 1 1 2 2 2 2 e 2 e d − − − − − x x x x A1 OE Complete integration and obtain 1 2 2 e −x x A1 OE Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW Alternative method for question 10(b) 1 2 1 1 2 2 d 2 e 2e e d − − −        = − x x x x x x *M1 A1 1 2 2 e − ∴ x x A1 Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW 5

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11 Two lines have equations r = i + 2j + k + , ai + 2j −k and r = 2i + j −k + - 2i −j + k , where a is a constant. (a) Given that the two lines intersect, find the value of a and the position vector of the point of intersection. 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(b) Given instead that the acute angle between the directions of the two lines is cos−1 1 , find the 6 two possible values of a. 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Mark scheme: 11(a) Express general point of at least one line correctly in component form, i.e. (1 + aλ, 2 + 2λ, 1 – λ) or (2 + 2µ, 1 – µ, – 1 + µ) B1 Equate at least two pairs of corresponding components and solve for λ or for µ M1 May be implied 1 + aλ = 2 + 2µ 2 + 2λ = 1 – µ 1 – λ = – 1 + µ Obtain λ = – 3 or µ = 5 A1 Obtain a = –11 3 A1 Allow a = – 3.667 State that the point of intersection has position vector 12i – 4j + 4k A1 Allow coordinate form (12, – 4, 4) 5 Question Answer Marks Guidance 11(b) Use correct process for finding the scalar product of direction vectors for the two lines M1 (a, 2, – 1) . (2, – 1, 1) = 2a – 2 – 1 or 2a – 3 Using the correct process for the moduli, divide the scalar product by the product of the moduli and equate the result to 1 6 ± *M1 State a correct equation in a in any form, e.g. ( ) 2 2 2 1 1 6 6 5 − − = ± + a a A1 Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 Question Answer Marks Guidance 11(b) Alternative method for question 11(b) cos(θ) = [|a2 + 22 + (–1)2|2 + | 22 + (–1)2 + 12|2 – |(a – 2)2 + 32 + (–2)2|2] / [2|a2 + 22 + (–1)2|.|22 + (–1)2 +12|] M1 Use of cosine rule. Must be correct vectors. Equate the result to 1 6 ± *M1 A1 Allow M1* here for any two vectors Solve for a DM1 Solve 3-term quadratic for a having expanded (2a – 3)2 to produce 3 terms e.g. 36(2a – 3)2 = 6(a2 + 5) 138a2 – 432a + 294 = 0 23a2 – 72a + 49 = 0 (23a – 49) (a – 1) = 0 Obtain a = 1 A1 Obtain a = 49 23 A1 Allow a = 2.13 6

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