Cambridge A Level Mathematics 9709 — 2020 May/June Paper 1 · Variant 3

5 O 5 A B 13 P The diagram shows a cord going around a pulley and a pin. The pulley is modelled as a circle with centre O and radius 5 cm. The thickness of the cord and the size of the pin P can be neglected. The pin is situated 13 cm vertically below O. Points A and B are on the circumference of the circle such that AP and BP are tangents to the circle. The cord passes over the major arc AB of the circle and under the pin such that the cord is taut. Calculate the length of the cord. 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Mark scheme: 5 ( ) 5 cos 1.17 6 13 POA POA = → = Allow 67.4° or sin = 12 13 or tan = 12 5 M1 A1 Reflex AOB = ( ) 2 2 1.17 6 π−×their OE in degrees or minor arc AB = 5×2×their1.17(6) M1 Major arc = 5 × their 3.93(1) or 2π 5 - 11.7(6) their × M1 AP (or BP) = 2 2 13 5 1 2 − = B1 Cord length = 43.7 A1 6

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6 A point P is moving along a curve in such a way that the x-coordinate of P is increasing at a constant 1 rate of 2 units per minute. The equation of the curve is y = 5x −1 2. (a) Find the rate at which the y-coordinate is increasing when x = 1. 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(b) Find the value of x when the y-coordinate is increasing at 5 units per minute. 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Mark scheme: 6(a) ( ) [ ] 1/2 d 1 5 1 5 d 2 −   = − ×     y x x B1 B1 Use d d 2 when 1 d d   = × =     y y their x t x M1 5 2 A1 4 Question Answer Marks 6(b) ( ) 1/2 5 5 2 5 1 2 8 − × − = their x oe M1 ( ) 1/2 5 1 8 − = x A1 13 = x A1 3

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tan 1 tan 1 2 7 (a) Show that +  [4] 1 + cos 1 1 −cos 1 sin 1 cos 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ tan 1 tan 1 6 (b) Hence solve the equation + = for 0Å < 1 < 180Å. [4] 1 + cos 1 1 −cos 1 tan 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 2 tan tan tan (1 cos ) tan (1 cos ) 1 cos 1 cos 1 cos θ θ θ θ θ θ θ θ θ − + + + = + − − M1 2 2tan sin θ θ = M1 2 2sin cos sin θ θ θ = M1 2 sin cos θ θ = AG A1 4 Question Answer Marks 7(b) 2 6cos sin cos sin θ θ θ θ = M1 ( ) 2 1 cos cos 0.5774 3 θ θ = → = ± A1 54.7º, 125.3º (FT for 180º ‒ 1st solution) A1 A1FT 4

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8 The first term of a progression is sin2 1, where 0 < 1 < 12π. The second term of the progression is sin21 cos21. (a) Given that the progression is geometric, find the sum to infinity. 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It is now given instead that the progression is arithmetic. (b) (i) Find the common difference of the progression in terms of sin 1. 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(ii) Find the sum of the first 16 terms when 1 = 13π. 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Mark scheme: 8(a) 2 cos θ = r SOI M1 2 2 sin 1 cos θ θ ∞= − S M1 1 A1 3 8(b)(i) 2 2 2 sin cos sin θ θ θ = − d M1 ( ) 2 2 sin cos 1 θ θ − M1 4 sin θ − A1 3 Question Answer Marks 8(b)(ii) Use of [ ] 16 16 2 15 2 = + S a d M1 With both 3 9 and 4 16 = = − a d A1 16 1 55 2 = − S A1 3

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11 y A y = x3 −2bx2 + b2x x O a b The diagram shows part of the curve with equation y = x3 −2bx2 + b2x and the line OA, where A is the maximum point on the curve. The x-coordinate of A is a and the curve has a minimum point at b, 0 , where a and b are positive constants. (a) Show that b = 3a. 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(b) Show that the area of the shaded region between the line and the curve is ka4, where k is a fraction to be found. 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Mark scheme: 11(a) 2 2 d 3 4 d y x bx b x = − + B1 ( )( ) ( ) 2 2 3 4 0 3 0 − + = → − − = x bx b x b x b M1 or 3 b x b = A1 3 3 b a b a = → = AG A1 Alternative method for question 11(a) 2 2 d 3 4 d y x bx b x = − + B1 Sub b = 3a & obtain d 0 d y x = when x = a and when x = 3a M1 2 2 d 6 12 d y x a x = − A1 < 0 Max at x = a and > 0 Min at x = 3a. Hence ܾ= 3ܽ AG A1 4 Question Answer Marks 11(b) Area under curve = ( ) 3 2 2 6 9 d  − + x ax a x x M1 4 2 2 3 9 2 4 2 − + x a x ax B2,1,0 4 4 4 4 9 11 2 4 2 4   − + =     a a a a (M1 for applying limits 0 → a) M1 When x = a, 3 3 3 3 6 9 4 = − + = y a a a a B1 Area under line = 3 1 4 2 × a their a M1 Shaded area = 4 4 4 11 3 2 4 4 − = a a a A1 7

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