Cambridge A Level Mathematics 9709 — 2021 May/June Paper 1 · Variant 3

3 A line with equation y = mx −6 is a tangent to the curve with equation y = x2 −4x + 3. Find the possible values of the constant m, and the corresponding coordinates of the points at which the line touches the curve. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 ( ) 2 2 4 3 6 leading to 4 9 x x mx x x m − + = − − + + May be implied on the next line. ( ) 2 2 4 leading to 4 4 9 b ac m − + − × DM1 SOI. Use of the discriminant with their a, b and c ( )( ) 4 6 or 2 10 0 leading to 2 or 10 m m m m + = ± − + = = − A1 Must come from 2 4 0 − = b ac SOI Substitute both their m values into their equation in line 1 DM1 m = 2 leading to x =3 ; m = ‒10 leading to x = ‒3 A1 (3, 0), (‒3, 24) A1 Accept 'when x = 3, y = 0; when x = ‒3, y = 24' If final A0A0 scored, SC B1 for one point correct WWW Alternative method for Question 3 2 4 2 4 = − → − = dy x x m dx *M1 ( ) 2 4 3 2 4 6 − + = − − x x x x DM1 2 2 2 4 3 2 4 6 9 3 − + = − −→= → = ± x x x x x x A1 0, = y 24 or (3, 0), (‒3, 24) A1 Substitute both their x values into their equation in line 1 DM1 Or substitute both their ( ) , x y into 6 = − y mx When x = 3, m = 2; when x = ‒3, m = ‒10 A1 If A0, DM1, A0 scored, SC B1 for one point correct WWW 6

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4 (a) Show that the equation tan x + sin x = k, tan x −sin x where k is a constant, may be expressed as 1 + cos x = k. [2] 1 −cos x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence express cos x in terms of k. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ tan x + sin x (c) Hence solve the equation = 4 for −π < x < π. [2] tan x −sin x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) [ ] [ ] tan sin sin sin cos leading to tan sin sin sin cos x x x x x k k x x x x x + + = = − − or 1 1 cos 1 1 cos x x + − [=k] or tanx tanxcosx tanx tanxcosx + − [=k] or divide numerator and denominator by tan x or sin x ( ) ( ) ( ) ( ) [ ] 1 1 sin 1 cos tanx 1 cosx cos 1 cos cos or . or leading to 1 sin 1 cos cos tanx 1 cosx 1 cos 1 cos x x x x x k x x x x x + + + + = − − − − A1 AG, WWW 2 4(b) cos 1 cos leading to 1 cos cos k k x x k k x x − = + −= + M1 Gather like terms on LHS and RHS ( ) 1 1 1 cos leading to cos 1 k k k x x k − −= + = + A1 WWW, OE 2 4(c) Obtaining cos x from their (b) or (a) M1 Expect 3 cos 5 = x ±0.927 (only solutions in the given range) A1 AWRT. Accept ±0.295π 2

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5 A D 4 cm B C The diagram shows a triangle ABC, in which angle ABC = 90Å and AB = 4 cm. The sector ABD is part of a circle with centre A. The area of the sector is 10 cm2. (a) Find angle BAD in radians. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the perimeter of the shaded region. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 2 ½ 4 angle BAD 10 × × = Angle BAD 1.25 = A1 OE. Accept 0.398π, 71.6o for SC B1 only 2 5(b) Arc 4 1.25 = × BD their M1 Use of arc length formula. Expect 5. ( ) 4tan 1.25 = BC their M1 Expect 12.0(4). May use ACB=0.321 or 18.4o ( ) 4 4 cos 1.25 = − CD their or ( ) 2 2 4 4 + − their BC M1 Expect 12.69 ‒ 4 = 8.69. May use ACB. Perimeter = 5 + 12.0(4) + 8.69 = 25.7 (cm) A1 AWRT 4

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10 Points A −2, 3 , B 3, 0 and C 6, 5 lie on the circumference of a circle with centre D. (a) Show that angle ABC = 90Å. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence state the coordinates of D. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find an equation of the circle. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The point E lies on the circumference of the circle such that BE is a diameter. (d) Find an equation of the tangent to the circle at E. 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Mark scheme: 10(a) Gradient of AB = 3 5 − , gradient of BC = 5 3 or lengths of all 3 sides or vectors M1 Attempting to find required gradients, sides or vectors 1 = − ab bc m m or Pythagoras or .  AB BC =0 or cos 0 = ABC from cosine rule A1 WWW 2 10(b) Centre = mid-point of AC = (2,4) B1 1 Question Answer Marks Guidance 10(c) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 c c x c c x their y their y r or their x x their y y r     − + − = − + − =     M1 Use of circle equation with their centre ( ) ( ) 2 2 2 4 17 − + − = x y A1 Accept 2 2 4 8 3 0 − + − + = x x y y OE 2 10(d) ( ) 3 0 , 2, 4 2 2 + +  =     x y or BE = 2BD = 2 1 4 −       Or Equation of BE is ( ) ( ) 4 3 or 4 4 2 leading to 4 12 y x y x y x = − − − = − − = − + Substitute equation of BE into circle and form a 3-term quadratic. M1 Use of mid-point formula, vectors, steps on a diagram May be seen to find x coordinate at E ( ) ( ) , 1,8 = x y or OE = 3 2 0 8 −    +       = 1 8    A1 E = (1, 8) Accept without working for both marks SC B2 Gradient of BD, m, = ‒4 or gradient AC = 1 4 = gradient of tangent B1 Or gradient of BE = -4 Equation of tangent is ( ) 8 ¼ 1 − = − y x OE M1 A1 For M1, equation through their E or (1, 8) (not, A, B or C) and with gradient 1 4 − − their 5

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11 y 1 2 y = x 2 + k2x−1 x O 4k29 4k2 1 The diagram shows part of the curve with equation y = x 2 + k2x−12, where k is a positive constant. (a) Find the coordinates of the minimum point of the curve, giving your answer in terms of k. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The tangent at the point on the curve where x = 4k2 intersects the y-axis at P. (b) Find the y-coordinate of P in terms of k. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The shaded region is bounded by the curve, the x-axis and the lines x = 94k2 and x = 4k2. (c) Find the area of the shaded region in terms of k. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) 1/2 2 3/2 d 1 1 d 2 2 − − = − y x k x x B1 B1 Allow any correct unsimplified form 1/2 2 3/2 1/2 2 3/2 1 1 1 1 0 leading to 2 2 2 2 x k x x k x − − − − − = = M1 OE. Set to zero and one correct algebraic step towards the solutions. d d y x must only have 2 terms. ( ) 2 , 2 k k A1 4 11(b) When x = 4k2, d 1 1 3 d 4 16 16   = − =     y x k k k B1 OE 2 1 5 2 2 2   = + × =     k y k k k B1 OE. Accept 2 2 + k k Equation of tangent is ( ) 2 5 3 4 2 16 − = − k y x k k or ( ) 2 5 3 4 2 16 = + → = + k y mx c k c k M1 Use of line equation with their gradient and ( 2 4 , ) k their y , When 5 3 7 0, 2 4 4 k k k x y   = = − =     or from 7 , 4 k y mx c c = + = A1 OE 4 Question Answer Marks Guidance 11(c) 3 1 1 1 2 2 2 2 2 2 2 d 2 3 −    + = +       x x k x x k x B1 Any unsimplified form 3 3 3 3 16 9 4 3 3 4     + − +         k k k k M1 Apply limits 2 2 9 4 4 → k k to an integration of y. M0 if volume attempted. 3 49 12 k A1 OE. Accept 4.08 3 k 3

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