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

1 Find the set of values of m for which the line with equation y = mx −3 and the curve with equation y = 2x2 + 5 do not meet. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 ( ) 2 2 2 5 3 2 8 0 + = − → − + = x mx x mx B1 Form 3-term quadratic 2 64 − m M1 Find 2 4 − b ac . 8 8 −< < m A1 Accept (-8, 8) and equality included 3

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3 Air is being pumped into a balloon in the shape of a sphere so that its volume is increasing at a constant rate of 50 cm3 s−1. Find the rate at which the radius of the balloon is increasing when the radius is 10 cm. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 (Derivative =) 2 4πr (→ 400π) B1 SOI Award this mark for d d r V 50 derivative their M1 Can be in terms of r 1 8π or 0.0398 A1 AWRT 3

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4 y 5 4 3 2 1 y = cos 1 1 0 π 2π 3π 4π −1 In the diagram, the lower curve has equation y = cos 1. The upper curve shows the result of applying a combination of transformations to y = cos 1. Find, in terms of a cosine function, the equation of the upper curve. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 ( )[ ] [ ] 1 3 2 cos 2 y   = +     θ B1 B1 B1 3

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6 The equation of a curve is y = 2 + 25 −x2. Find the coordinates of the point on the curve at which the gradient is 3.4 [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 6 ( ) [ ] 1/2 2 d 1 25 2 d 2 −   = − × −     y x x x B1 B1 ( ) 2 1/2 2 2 4 16 3 9 25 25 − = → = − − x x x x M1 Set = 4 3 and square both sides ( ) ( ) 2 2 2 16 25 9 25 400 4 − = → = → = ± x x x x A1 When x = ‒ 4, y = 5 → (‒ 4, 5) A1 5

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sin 1 sin 1 7 (a) Show that −  2 tan21. [3] 1 −sin 1 1 + sin 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ sin 1 sin 1 (b) Hence solve the equation − = 8, for 0Å < 1 < 180Å. [3] 1 −sin 1 1 + sin 1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 2 sin sin sin (1 sin ) sin (1 sin ) 1 sin 1 sin 1 sin + − −   − =   − + −   θ θ θ θ θ θ θ θ θ *M1 Put over a single common denominator 2 2 2sin cos θ θ DM1 Replace 2 2 1 sin by cos − θ θ and simplify numerator 2 2tan θ A1 AG 3 7(b) ( ) 2 2tan 8 tan 2 θ θ = → = ± B1 SOI ( ) 63.4 , 1 16.6 θ = ° ° B1 B1 FT FT on 180 – 1st solution (with justification) 3

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8 A geometric progression has first term a, common ratio r and sum to infinity S. A second geometric progression has first term a, common ratio R and sum to infinity 2S. (a) Show that r = 2R −1. 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It is now given that the 3rd term of the first progression is equal to the 2nd term of the second progression. (b) Express S in terms of a. 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Mark scheme: 8(a) , 2 1 1 a a S S r R = = − − B1 SOI at least one correct 2 1 1 = − − a a r R M1 SOI 2 2 1 2 1 − = − → = − R r r R A1 AG 3 Question Answer Marks Guidance 8(b) ( )( ) ( ) 2 2 2 1 = → − = ar aR a R R a *M1 ( ) ( )( ) ( ) 2 4 5 1 0 4 1 1 0 − + = → − − = R R R R DM1 Allow use of formula or completing square. 1 4 R = A1 Allow R = 1 in addition 2 3 = a S A1 Alternative method for question 8(b) ( ) ( )( ) 2 2 ½ 1 = → = + ar aR a r r a *M1 Eliminating 1 variable ( ) ( )( ) ( ) 2 2 1 0 2 1 1 0 − − = → + − = r r r r DM1 Allow use of formula or completing square. Must solve a quadratic. 1 2 = − r A1 Allow r = 1 in addition 2 3 = a S A1 4

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9 C A B D The diagram shows a circle with centre A passing through the point B. A second circle has centre B and passes through A. The tangent at B to the first circle intersects the second circle at C and D. The coordinates of A are −1, 4 and the coordinates of B are 3, 2 . (a) Find the equation of the tangent CBD. 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(b) Find an equation of the circle with centre B. 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(c) Find, by calculation, the x-coordinates of C and D. 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Mark scheme: 9(a) 4 2 1 1 3 2 − = = − −− AB m B1 Equation of tangent is ( ) 2 2 3 − = − y x B1 FT (3, 2) with their gradient 1 − AB m 2 9(b) 2 2 2 4 2 20 = + = AB or 2 20 or 20 or 20 r r AB = = = B1 Equation of circle centre B is ( ) ( ) 2 2 3 2 20 − + − = x y M1 A1 FT their 20 for M1 3 9(c) ( ) ( ) 2 2 3 2 6 20 x x their − + − = M1 Substitute their 2 2 6 y x − = − into their circle, centre B ( ) 2 2 5 30 25 0 or 5 3 20 x x x − + = − = A1 ( )( )( ) 5 5 1 or 3 2 x x x  − − − = ±    5, 1 x = A1 3

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10 C 1 rad F E r O A B D The diagram shows a sector CAB which is part of a circle with centre C. A circle with centre O and radius r lies within the sector and touches it at D, E and F, where COD is a straight line and angle ACD is 1 radians. (a) Find CD in terms of r and sin 1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is now given that r = 4 and 1 = 16π. (b) Find the perimeter of sector CAB in terms of π. 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(c) Find the area of the shaded region in terms of π and 3. 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Mark scheme: 10(a) sin r OC   = →     θ sin r OC = θ M1 A1 sinθ = + r CD r A1 3 10(b) Radius of arc AB = 4 4 4 8 12 π sin 6 + = + = B1 SOI (Arc AB =) 2π 12 6 their × or 1 π 12 2 6 AB their    = ×       M1 Expect 4π, must use their CD, not 4 Perimeter = 24 4π + A1 3 Question Answer Marks Guidance 10(c) Area FOC = 1 π 4 sin 2 3 their OC × × × M1 8 3 A1 Area sector FOE = 2 1 2π 16π 4 2 3 3 × × = B1 Shaded area = 16π 16 3 3 − A1 Alternative method for question 10(c) FC = ( ) 2 2 4 their OC − M1 48 or 4 3 Area FOC = 1 4 4 3 2 × × = 8 3 A1 Area of half sector FOE = 2 1 π 8π 4 2 3 3 × × = B1 Shaded area = 16π 16 3 3 − A1 4

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12 y B A 4, 0 O x 1 2 −2x y = 4x y = 3 −x C 1 The diagram shows a curve with equation y = 4x 2 −2x for x ≥0, and a straight line with equation y = 3 −x. The curve crosses the x-axis at A 4, 0 and crosses the straight line at B and C. (a) Find, by calculation, the x-coordinates of B and C. 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(b) Show that B is a stationary point on the curve. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the area of the shaded region. 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Mark scheme: 12(a) ( ) 1 1 2 2 4 2 3 4 3 0 − = − → − + = x x x x x *M1 3-term quadratic. Can be expressed as e.g. 2 4 3 − + u u (=0) ( ) ( )( )( ) 1 1 2 2 1 3 0 or 1 3 0 x x u u    − − = − − =          DM1 Or quadratic formula or completing square 1 2 1 , 3 = x A1 SOI 1, 9 x = A1 Alternative method for question 12(a) ( ) 2 1 2 2 4 3   = +       x x *M1 Isolate 1 2 x ( ) 2 2 16 9 6 10 9 0 = + + → − + = x x x x x A1 3-term quadratic ( )( ) ( ) 1 9 0 − − = x x DM1 Or formula or completing square on a quadratic obtained by a correct method 1, 9 = x A1 4 12(b) 1/2 d 2 2 d = − y x x *B1 1/2 d or 2 2 0 d y x x − = when x =1 hence B is a stationary point DB1 2 Question Answer Marks Guidance 12(c) Area of correct triangle = 1 2 (9 ‒ 3) × 6 M1 or ( )( ) 9 2 3 1 3 d 3 18 2 x x x x   − = − →−      ( ) 3 1 2 2 2 4 (4 2 ) d 3 2      − = −       x x x x x B1 B1 ( ) 64 72 81 16 3   − − −     M1 Apply limits 4 → their 9 to an integrated expression 1 3 14 − A1 OE Shaded region = 1 2 3 3 18 14 3 − = A1 OE 6

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