Cambridge A Level Mathematics 9709 — 2021 Feb/March Paper 1 · Variant 2

tan 1 + 2 sin 1 3 Solve the equation = 3 for 0Å < 1 < 180Å. [4] tan 1 −2 sin 1 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 tan 2sin 3tan 6sin θ θ θ θ + = − leading to [ ] 2tan 8sin 0 θ θ − = M1 OE ( ) 2sin 8sin cos 0 θ θ θ − = leading to [ ] ( ) [ ] 2 sin 1 4cos 0 θ θ − = M1 1 cos 4 θ = A1 Ignore sin 0 θ = 75.5 θ = ° only A1 4

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4 A line has equation y = 3x + k and a curve has equation y = x2 + kx + 6, where k is a constant. Find the set of values of k for which the line and curve have two distinct points of intersection. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 2 6 3 x kx x k + + = + leading to ( ) ( ) [ ] 2 3 6 0 x x k k + − + − = ( ) ( )[ ] 2 3 4 6 0 k k − − − > M1 OE. Apply 2 4 − b ac . [ ] 2 2 15 0 k k − − > A1 Form 3-term quadratic. ( )( )[ ] 3 5 0 k k + − > A1 Or 3, 5 = − k from use of formula or completing square. 3, 5 k k < − > A1 FT Or any correct alternative notation, do not allow , . FT for their outside regions. 5

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dy 6 6 A curve is such that = and A 1, −3 lies on the curve. A point is moving along the curve dx 3x −2 3 and at A the y-coordinate of the point is increasing at 3 units per second. (a) Find the rate of increase at A of the x-coordinate of the point. 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(b) Find the equation of the curve. 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Mark scheme: 6(a) At x = 1, d 6 d = y x B1 d d d 1 1 3 d d d 6 2 x x y t y t = × = × =       M1 A1 Chain rule used correctly. Allow alternative and minimal notation. 3 Question Answer Marks Guidance 6(b) [ ] ( ) ( ) [ ] 2 6 3 2 3 2 x y c − − = ÷ + −         B1 B1 3 1 c −= −+ M1 Substitute 1, 3. x y = = − c must be present. ( ) 2 3 2 2 y x − = − − − A1 OE. Allow f(x)= 4

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8 The points A 7, 1 , B 7, 9 and C 1, 9 are on the circumference of a circle. (a) Find an equation of the circle. 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(b) Find an equation of the tangent to the circle at B. 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Mark scheme: 8(a) Centre of circle is (4, 5) B1 B1 ( ) ( ) 2 2 2 7 4 1 5 r = − + − M1 OE. Either using their centre and A or C or using A and C and dividing by 2. 5 r = A1 FT FT on their (4, 5) if used. Equation is ( ) ( ) 2 2 4 5 25 x y − + − = A1 OE. Allow 52 for 25. 5 8(b) Gradient of radius = 9 5 4 7 4 3 − = − B1 FT FT for use of their centre. Equation of tangent is ( ) 3 9 7 4 − = − − y x B1 or 3 57 4 4 x y − = + 2

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9 The first term of a progression is cos 1, where 0 < 1 < 12π. 1 (a) For the case where the progression is geometric, the sum to infinity is cos 1. (i) Show that the second term is cos 1 sin21. 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(ii) Find the sum of the first 12 terms when 1 = 13π, giving your answer correct to 4 significant figures. 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(b) For the case where the progression is arithmetic, the first two terms are again cos 1 and cos 1 sin21 respectively. Find the 85th term when 1 = 13π. 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Mark scheme: 9(a)(i) cos 1 1 cos θ θ = −r B1 2 1 cos r θ − = leading to 2 1 cos r θ = − M1 Eliminate fractions 2 sin r θ = leading to 2nd term = 2 cos sin θ θ A1 AG 3 9(a)(ii) ( ) 12 2 12 12 2 cos 1 sin 0.5 1 0.75 1 0.75 1 sin π π 3 3 π 3 S − − = = − −                                     M1 Evidence of correct substitution, use of nS formula and attempt to evaluate 1.937 A1 2 9(b) [ ] 2 cos sin cos d θ θ θ = − M1 Use of 2 1 d u u = − 1 8 − A1 [85th term =] 1 1 84 2 8 + × − M1 Use of a + 84d with a calculated value of d 10 − A1 4

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10 A ka ka a E D B C The diagram shows a sector ABC which is part of a circle of radius a. The points D and E lie on AB and AC respectively and are such that AD = AE = ka, where k < 1. The line DE divides the sector into two regions which are equal in area. (a) For the case where angle BAC = 16π radians, find k correct to 4 significant figures. 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Mark scheme: 10(a) ( ) 2 1 π sin 2 6 Δ = ADE ka 2 2 1 4 k a A1 OE. Sector 2 1 π 2 6 = ABC a B1 2 2 2 1 1 π 2 4 2 6 × = k a a M1 OE. For 2 sector ADE ABC ×Δ = with at least one correct area. π 0.7236 6 k = =       A1 5 10(b) ( ) 2 2 1 1 2 sin 2 2 θ θ × = ka a M1 Condone omission of ‘2’ or ‘1/2’ on LHS for M1 only. 2 2sin k θ θ = A1 2 1 2 k > leading to 1 1 2 k < < A1 OE. Accept 1 2 k > or 0.707 k > (AWRT) or 0.707(AWRT) < k < 1 or 1 2 k > OE 3

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11 y A x O 2 The diagram shows the curve with equation y = 9 x−1 −4x−3 2 . The curve crosses the x-axis at the point A. (a) Find the x-coordinate of A. 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(b) Find the equation of the tangent to the curve at A. 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(c) Find the x-coordinate of the maximum point of the curve. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Find the area of the region bounded by the curve, the x-axis and the line x = 9. 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Mark scheme: 11(a) 1 3 2 2 9 4 0 x x − −   − =     leading to ( ) 3 2 9 4 0 x x − − = M1 OE. Set y to zero and attempt to solve. 4 x = only A1 From use of a correct method. 2 11(b) 3 5 2 2 d 1 9 6 d 2 − −   = − +     y x x x B2, 1, 0 B2; all 3 terms correct: 9, 3 2 1 2 x − − and 5 2 6x − B1; 2 of the 3 terms correct At x = 4 gradient = 1 6 9 9 16 32 8 − + =       M1 Using their x = 4 in their differentiated expression and attempt to find equation of the tangent. Equation is ( ) 9 4 8 = − y x A1 or 9 9 8 2 = − x y OE 4 11(c) 5 2 1 9 6 0 2 x x − − + =       M1 Set their d d y x to zero and an attempt to solve. 12 = x A1 Condone ( )12 ± from use of a correct method. 2 Question Answer Marks Guidance 11(d) 1 1 1 3 2 2 2 2 d 4 9 4 9 1 1 2 2 x x x x x − − − − = − −                  B2, 1, 0 B2; all 3 terms correct: 9, 1 1 2 2 4 , 1 1 2 2 x x − − − B1; 2 of the 3 terms correct ( ) 8 9 6 4 4 3     + − +         M1 Apply limits their 4 →9 to an integrated expression with no consideration of other areas. 6 A1 Use of π scores A0 4

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