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

1 The sum of the first nine terms of an arithmetic progression is 117. The sum of the next four terms is 91. Find the first term and the common difference of the progression. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 117 = ( ) 9 2 8 2 + a d B1 Either 91 = S4 with ‘a’ as a + 4d or 117 + 91 = S13 (M1 for overall approach. M1 for Sn) M1M1 Simultaneous Equations → a = 7, d = 1.5 A1 4

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1 1 5 82 The coefficient of in the expansion of kx + + 1 −2 is 74. x x x Find the value of the positive constant k. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 5 1   +     kx x + 8 2 1   −     x Coefficient in 5 1   +     kx x = 10 × k² (B1 for 10. B1 for k²) B1B1 Coefficient in 8 2 1   −     x = 8 × −2 B2,1,0 10k² − 16 = 74 → k = 3 B1 5

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3 Each year the selling price of a diamond necklace increases by 5% of the price the year before. The selling price of the necklace in the year 2000 was $36 000. (a) Write down an expression for the selling price of the necklace n years later and hence find the selling price in 2008. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The company that makes the necklace only sells one each year. Find the total amount of money obtained in the ten-year period starting in the year 2000. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) (B1 for r = 1.05. M1 method for rth term) B1M1 $53 200 after 8 years. A1 3 3(b) S10 = ( ) ( ) 10 1.05 1 36000 1.05 1 − − M1 $453 000 A1 2

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5 The equation of a line is y = mx + c, where m and c are constants, and the equation of a curve is xy = 16. (a) Given that the line is a tangent to the curve, express m in terms of c. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given instead that m = −4, find the set of values of c for which the line intersects the curve at two distinct points. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) ( ) + x mx c = 16 → 2 16 0 + − = mx cx B1 Use of b² − 4ac = c² + 64m M1 Sets to 0 → m = ² 64 −c A1 3 5(b) ( ) 4 − + x x c = 16 Use of b² − 4ac → c² − 256 M1 c > 16 and c < −16 A1 A1 3

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1 + sin 1 cos 1 2 7 (a) Prove the identity +  [3] cos 1 1 + sin 1 cos 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 + sin 1 cos 1 3 (b) Hence solve the equation + = for 0 ≤1 ≤2π. [3] cos 1 1 + sin 1 sin 1, ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) ( ) ( ) 2 2 1 sin cos cos 1 sin θ θ θ θ + + + M1 Use of 2 2 sin cos 1 θ θ + = → ( ) 2 2sin cos 1 sin θ θ θ + + → 2 cosθ . M1A1 3 7(b) 2 cosθ = 3 sinθ → tanθ = 1.5 M1 θ = 0.983 or 4.12 (FT on second value for 1st value + π) A1 A1FT 3

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8 15 cm B A C 6 cm O X In the diagram, ABC is a semicircle with diameter AC, centre O and radius 6 cm. The length of the arc AB is 15 cm. The point X lies on AC and BX is perpendicular to AX. Find the perimeter of the shaded region BXC. 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Mark scheme: 8 Angle AOB = 15 ÷ 6 = 2.5 radians B1 Angle BOC = π – 2.5 (FT on angle AOB) B1FT BC = 6(π – 2.5) (BC = 3.850) M1 sin(π – 2.5) = BX ÷6 (BX = 3.59) M1 Either OX = 6cos(π – 2.5) or Pythagoras (OX = 4.807) M1 XC = 6 – OX (XC = 1.193) → P = 8.63 A1 6

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9 The equation of a curve is y = 3 −2x 3 + 24x. dy d2y (a) Find expressions for and . [4] dx dx2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the coordinates of each of the stationary points on the curve. 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(c) Determine the nature of each stationary point. 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Mark scheme: 9(a) d d y x = 3(3−2x)² × −2 + 24 = ( ) 2 6 3 2 24 − − + x (B1 without ×−2. B1 for ×−2) B1B1 d² d ² y x = ( ) 12 3 2 2 − − ×− x = 24(3 – 2x) (B1FT from ୢ௬ ୢ௫ without – 2) B1FT B1 4 9(b) d 0 d = y x when ( ) 2 6 3 2 24 − = x → 3 2 2 x − = ± M1 x = ½, y = 20 or x = 2½, y = 52 (A1 for both x values or a correct pair) A1A1 3 9(c) If x = ½, d² d ² y x = 48 Minimum B1FT If x = 2½, d² d ² y x = −48 Maximum B1FT 2

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10 The coordinates of the points A and B are −1, −2 and 7, 4 respectively. (a) Find the equation of the circle, C, for which AB is a diameter. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the tangent, T, to circle C at the point B. 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(c) Find the equation of the circle which is the reflection of circle C in the line T. 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Mark scheme: 10(a) Centre is (3, 1) B1 Radius = 5 (Pythagoras) B1 Equation of C is ( ) ( ) 2 2 3 1 25 − + − = x y (FT on their centre) M1 A1FT 4 10(b) Gradient from (3, 1) to (7, 4) = ¾ (this is the normal) B1 Gradient of tangent = −4 3 M1 Equation is ( ) 4 4 7 or 3 4 40 3 y x y x − = − − + = M1A1 4 10(c) B is centre of line joining centres → (11, 7) B1 Radius = 5 New equation is ( ) ( ) 2 2 11 7 25 − + − = x y (FT on coordinates of B) M1 A1FT 3

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11 y A 2y + x = 8 C B 8 y = x + 2 x O 8 The diagram shows part of the curve y = and the line 2y + x = 8, intersecting at points A and B. x + 2 The point C lies on the curve and the tangent to the curve at C is parallel to AB. (a) Find, by calculation, the coordinates of A, B and C. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the volume generated when the shaded region, bounded by the curve and the line, is rotated through 360Å about the x-axis. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) Simultaneous equations 8 2 + x = 4 − ½x M1 x = 0 or x = 6 → A (0, 4) and B (6, 1) B1A1 At C ( ) 8 1 2 ² 2 − = − + x → C (2, 2) (B1 for the differentiation. M1 for equating and solving) B1 M1A1 6 11(b) Volume under line = π ( ) 1 2 4 ²d x x − + = π ³ 2 ² 16 12   − +     x x x = (42π) (M1 for volume formula. A2,1 for integration) M1 A2,1 Volume under curve = 2 8 π d 2 x x     +    = π 64 2 −     +   x = (24π) A1 Subtracts and uses 0 to 6 → 18π M1A1 6

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