Cambridge A Level Mathematics 9709 — 2020 Feb/March Paper 1 · Variant 2

3 y 5 y = x2 + 1 1 x O The diagram shows part of the curve with equation y = x2 + 1. The shaded region enclosed by the curve, the y-axis and the line y = 5 is rotated through 360Å about the y-axis. Find the volume obtained. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 ( ) ( ) π 1 d y y  − *M1 SOI Attempt to integrate x2 or ( ) 1 y − ( ) 2 π 2 y y   −     A1 ( ) 25 1 π 5 1 2 2       − − −             DM1 Apply limits 1 → 5 to an integrated expression 8π or AWRT 25.1 A1 4

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4 A curve has equation y = x2 −2x −3. A point is moving along the curve in such a way that at P the y-coordinate is increasing at 4 units per second and the x-coordinate is increasing at 6 units per second. Find the x-coordinate of P. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 d 2 2 d y x x = − B1 d 4 d 6 y x = B1 OE, SOI ( ) 4 2 2 6 their x their − = M1 LHS and RHS must be their d d y x expression and value 4 3 x = oe A1 4

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5 Solve the equation tan 1 + 3 sin 1 + 2 = 2 tan 1 −3 sin 1 + 1 for 0Å ≤1 ≤90Å. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 ( ) 2tan 6sin 2 tan 3sin 2 tan 9sin 0 θ θ θ θ θ θ − + = + + → − = M1 Multiply by denominator and simplify ( ) sin 9sin cos 0 θ θ θ − = M1 Multiply by cosθ ( ) 1 sin (1 9cos ) 0 sin 0, cos 9 θ θ θ θ − = → = = M1 Factorise and attempt to solve at least one of the factors = 0 0 or 83.6 θ = ° (only answers in the given range) A1A1 5

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1 a 5 6 The coefficient of in the expansion of 2x + is 720. x x2 (a) Find the possible values of the constant a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 (b) Hence find the coefficient of in the expansion. [2] x7 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 5C2 ( ) ( ) 2 3 2 2 a x x            B1 SOI Can include correct x's 3 2 4 1 10 8 720 x a x x     × × =         B1 SOI Can include correct x's 3 a = ± B1 3 6(b) 5C4 ( ) ( ) 4 2 2 their a x x            B1 SOI Their a can be just one of their values (e.g. just 3). Can gain mark from within an expansion but must use their value of a 810 identified B1 Allow with 7 x− 2

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7 A D 0.8 rad O B C 6 cm The diagram shows a sector AOB which is part of a circle with centre O and radius 6 cm and with angle AOB = 0.8 radians. The point C on OB is such that AC is perpendicular to OB. The arc CD is part of a circle with centre O, where D lies on OA. Find the area of the shaded region. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 7 M1A1 SOI Area sector OCD = ( ) 2 1 4.18 0.8 2 their × *M1 OE ΔOCA = 1 6 4.18 sin0.8 2 their × × × M1 OE Required area = their ΔOCA ‒ their sectorOCD DM1 SOI. If not seen their areas of sector and triangle must be seen 2.01 A1 CWO. Allow or better e.g. 2.0064 6

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8 A woman’s basic salary for her first year with a particular company is $30 000 and at the end of the year she also gets a bonus of $600. (a) For her first year, express her bonus as a percentage of her basic salary. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ At the end of each complete year, the woman’s basic salary will increase by 3% and her bonus will increase by $100. (b) Express the bonus she will be paid at the end of her 24th year as a percentage of the basic salary paid during that year. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) 2% B1 1 8(b) Bonus = 600 + 23× 100 = 2900 B1 Salary = 23 30000 1.03 × M1 Allow 24 30000 1.03 × (60984) = 59207.60 A1 Allow answers of 3significant figure accuracy or better 2900 59200 their their M1 SOI 4.9(0)% A1 5

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dy 1 10 The gradient of a curve at the point x, y is given by = 2 x + 3 2 −x. The curve has a stationary dx point at a, 14 , where a is a positive constant. (a) Find the value of a. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Determine the nature of the stationary point. 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(c) Find the equation of the curve. 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Mark scheme: 10(a) ( ) 1 2 2 3 0 a a + − = M1 SOI. Set d d y x = 0 when x = a. Can be implied by an answer in terms of a ( ) 2 4 3 a a + = 2 4 12 0 a a → − − = M1 Take a to RHS and square. Form 3-term quadratic ( )( ) 6 2 6 a a a − + → = A1 Must show factors, or formula or completing square. Ignore a = ‒2 SC If a is never used maximum of M1A1 for 6 x = ,with visible solution 3 10(b) ( ) 1 2 2 2 d 3 1 d y x x − = + − B1 Sub their a → 2 2 d 1 2 1 ( 0) 3 3 d y or x = −= − < →MAX M1A1 A mark only if completely correct If the second differential is not 2 3 − correct conclusion must be drawn to award the M1 3 10(c) ( ) ( ) ( ) 3 2 2 3 2 2 3 1 2 x y x c + = − + B1B1 Sub x = their a and y = 14 ( ) 3 2 4 1 4 9 18 3 c → = − + M1 Substitute into an integrated expression. c must be present. Expect c = ‒4 ( ) 3 2 2 4 1 3 4 3 2 y x x = + − − A1 Allow ( ) . f x =… 4

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11 (a) Solve the equation 3 tan2x −5 tan x −2 = 0 for 0Å ≤x ≤180Å. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the set of values of k for which the equation 3 tan2x −5 tan x + k = 0 has no solutions. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) For the equation 3 tan2x −5 tan x + k = 0, state the value of k for which there are three solutions in the interval 0Å ≤x ≤180Å, and find these solutions. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) ( ) (tan 2)(3tan 1) 0 . x x − + = or formula or completing square M1 Allow reversal of signs in the factors. Must see a method 1 tan 2 or -3 x = A1 ( ) 63.4 only value in range or1 61.6 (only value in range x = ° ° ) B1FT B1FT 4 11(b) Apply 2 4 0 b ac − < M1 SOI. Expect ( )( ) 25 4 3 k − < 0, tan x must not be in coefficients 25 12 k > A1 Allow 2 4 0 b ac − = leading to correct 25 12 k > for M1A1 2 11(c) k = 0 M1 SOI 5 tan 0 or 3 x = A1 0 or 1 80 or 59.0 x = ° ° ° A1 All three required 3

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12 A diameter of a circle C1 has end-points at −3, −5 and 7, 3 . (a) Find an equation of the circle C1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ y C2 R C1 x O S @ A 8 The circle C1 is translated by to give circle C2, as shown in the diagram. 4 (b) Find an equation of the circle C2. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The two circles intersect at points R and S. (c) Show that the equation of the line RS is y = −2x + 13. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Hence show that the x-coordinates of R and S satisfy the equation 5x2 −60x + 159 = 0. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 12(a) Centre = (2, ‒1) B1 ( ) ( ) [ ] [ ] 2 2 2 2 2 2 3 1 5 or 2 7 1 3 r = −− + −−−  − + −−     OE M1 OR ( ) ( ) 2 2 1 3 7 5 3 2   −− + −−   OE ( ) ( ) 2 2 2 1 41 x y − + + = A1 Must not involve surd form SCB3 ( )( ) ( )( ) 3 7 5 3 0 x x y y + − + + − = 3 12(b) Centre = their (2, ‒1) + 8 4    = (10, 3) B1FT SOI FT on their (2, ‒1) ( ) ( ) 2 2 10 3 41 x y their − + − = B1FT FT on their 41 even if in surd form SCB2 ( )( ) ( )( ) 5 15 1 7 0 x x y y − − + + − = 2 Question Answer Marks Guidance 12(c) Gradient m of line joining centres = 4 8 OE B1 Attempt to find mid-point of line. M1 Expect (6, 1) Equation of RS is ( ) 1 2 6 y x −= − − M1 Through their (6, 1) with gradient 1 m − 2 13 y x = − + A1 AG Alternative method for question 12(c) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 41 10 3 41 x y x y − + + − = − + − − OE M1 2 2 2 2 4 4 2 1 20 100 6 9 x x y y x x y y − + + + + = − + + − + OE A1 Condone 1 error or errors caused by 1 error in the first line 16 8 104 x y + = A1 2 13 y x = − + A1 AG 4 12(d) ( ) ( ) 2 2 10 2 13 3 41 x x − + − + − = M1 Or eliminate y between C1 and C2 2 2 2 20 100 4 40 100 41 5 60 159 0 x x x x x x − + + − + = → − + = A1 AG 2

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