Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 1 · Variant 3

1 (a) Express x2 6x 5 in the form x a 2 b, where a and b are constants. [2] + + + + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The curve with equation y x2 is transformed to the curve with equation y x2 6x 5. = = + + Describe fully the transformation(s) involved. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) ( ) [ ] 2 3 4   + −   x B1 B1 2 1(b) [Translation or shift] 3 4 −     −   B1 B1 FT Accept [translation/shift] −       their a theirb OR translation ‒3 units in x-direction and (translation) ‒4 units in y-direction. 2

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2 3 Solve the equation 3 1 for [5] tan21 + = 0Å < 1 < 180Å. tan21 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 ( ) 4 2 3tan tan 2 0 θ θ + − = M1 SOI 3-term quartic, condone sign errors for this mark only ( )( ) ( ) 2 2 3tan 2 tan 1 0 θ θ − + = M1 Attempt to factorise or solve 3-term quadratic in 2 tan θ . ( ) ( ) ( ) 2 tan or 0.816 or 0.817 3 = ± ± ± θ A1 SOI Implied by final answer = 39.2° after 1st M1 scored 39.2°, 140.8° A1 A1 FT FT for 2nd solution =180° ‒ 1st solution 5

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4 A curve has equation y 3x2 4 and a straight line has equation y mx m where m is a constant. = −4x + = + −1, Find the set of values of m for which the curve and the line have two distinct points of intersection. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 ( ) ( ) ( ) 2 2 3 4 4 1 3 4 5 0 x x mx m x m x m − + = + − → − + + − = M1 3-term quadratic ( ) ( ) 2 2 4 4 4 3 5 − = + −× × − b ac m m M1 Find 2 4 − b ac for their quadratic 2 20 44 + − m m A1 (m + 22)(m ‒ 2) A1 Or use of formula or completing square. This step must be seen 2 , 22 > < − m m A1 Allow 2 , 22 > < − x x 5

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1 7 The first and second terms of an arithmetic progression are and respectively, where −tan2 0 1 cos21 cos21, < 1 < 2π. 1 (a) Show that the common difference is [4] − cos41. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the exact value of the 13th term when 1 [3] 1 = 6π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) ( ) 2 2 2 tan 1 cos cos θ θ θ = − − d B1 Allow sign error(s). Award only at form (d =)... stage 2 4 2 sin 1 cos cos − − θ θ θ or 2 2 sec cos − θ θ M1 Allow sign error(s). Can imply B1 2 2 4 sin cos cos − − θ θ θ or 2 2 1 cos cos − θ θ M1 4 1 cos − θ A1 AG, WWW 4 7(b) 4 3 a = , 16 9 d = − B1 SOI, both required. Allow a = 1 3 4 , d = 6 1 9 1 − 13 2 4 1 12 4 16 12 3 9 cos cos θ θ −   = − = +     u M1 Use of correct formula with their a and their d. The first 2 steps could be reversed ‒20 A1 WWW 3

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1 8 The equation of a curve is y 2x 1 for x 2. 2x 1 = + + > −1 + dy d2y (a) Find and . [3] dx dx2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the coordinates of the stationary point and determine the nature of the stationary point. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) [ ] ( ) 2 d 2 [ 2 2 1 d − = − + y x x ] B1 B1 ( ) 2 3 2 d 8 2 1 d y x x − = + B1 3 8(b) Set their d d y x = 0 and attempt solution M1 (2x + 1)2 = 1 → 2x + 1 = ( ) ± 1 or 4x2 + 4x = 0 → (4)x(x + 1) = 0 M1 Solving as far as x = .... x = 0 A1 WWW. Ignore other solution. (0, 2) A1 One solution only. Accept x = 0, y = 2 only. 2 2 d d y x > 0 from a solution 1 2 > − x hence minimum B1 Ignore other solution. Condone arithmetic slip in value of 2 2 d d y x . Their 2 2 d d y x must be of the form ( ) 3 2 1 − + k x 5

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9 B 12 cm 8 cm A 8 cm O C In the diagram, arc AB is part of a circle with centre O and radius 8 cm. Arc BC is part of a circle with centre A and radius 12 cm, where AOC is a straight line. (a) Find angle BAO in radians. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the area of the shaded region. 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(c) Find the perimeter of the shaded region. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(a) 6 cos 8 = BAO or 2 2 2 8 12 8 2 8 12 + − × × M1 Or other correct method BAO = 0.723 A1 2 Question Answer Marks Guidance 9(b) Sector ABC = 2 ½ 12 0.7227 × ×their *M1 Accept 52.1 ( ) Triangle ½ 8 12sin 0.7227 AOB their = × × or ½×12×√28 *M1 or ( ) ½ 8 8sin 2 0.7227 π × × −×their . Expect 31.7 or 31.8 Shaded area = their 52.0 31.7 their − = 20.3 DM1 A1 M1 dependent on both previous M marks 4 9(c) Arc BC = 12 0.7227 ×their *M1 Expect 8.67 Perimeter = 8 + 4 + their 8.67 = 20.7 DM1 A1 3

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1 1 1 2 where x 0 and k is a positive constant.10 A curve has equation y x 2 x−1 k = + + k2 > (a) It is given that when x 14, the gradient of the curve is 3. = Find the value of k. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ k2 @1 1 A 1 13 2 2(b) It is given instead that Ô 1 k x + x−1 + k2 dx = 12. 4k2 Find the value of k. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a) [ ] ( ) 1/2 3/2 d 0 d 2 2 − −     = − +         y x x x k B2, 1, 0 [ ] ( ) 0 implies that more than 2 terms counts as an error Sub d 3 d = y x when 1 1 Expect 3 4 4 = = − x k M1 k = 1 7 (or 0.143) A1 4 Question Answer Marks Guidance 10(b) 3/2 1/2 1/2 1/2 2 2 1 1 2 2 3 −        + + = + +           x x x x x k k k k B2, 1, 0 OE 2 2 2 1 2 1 3 12 4     + + − + +         k k k k M1 Apply limits 2 2 4 → k k to an integrated expression. Expect 2 7 3 12 4 + + k k 2 7 3 13 12 4 12 k k + + = M1 Equate to 13 12 and simplify to quadratic. OE, ( ) 2 expect 7 12 4 0 k k + − = k = 2 7 only (or 0.286) A1 Dependent on ( )( ) ( ) 7 2 2 0 − + = k k or formula or completing square. 5

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11 A circle with centre C has equation x 2 y 2 100. −8 + −4 = (a) Show that the point T 6 is outside the circle. [3] −6, ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Two tangents from T to the circle are drawn. (b) Show that the angle between one of the tangents and CT is exactly [2] 45Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The two tangents touch the circle at A and B. (c) Find the equation of the line AB, giving your answer in the form y mx c. [4] = + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Find the x-coordinates of A and B. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) ( ) ( ) 2 2 6 8 6 4 −− + − M1 OE 200 = A1 200 > 10, hence outside circle A1 AG (‘Shown’ not sufficient). Accept equivalents of 200 > 10 Alternative method for question 11(a) Radius = 10 and C = (8, 4) B1 Min(x) on circle = 8 ‒ 10 = ‒2 M1 Hence outside circle A1 AG 3 11(b) angle = 1 1 0 sin 1 0 2 −      their their M1 Allow decimals for 10√2 at this stage. If cosine used, angle ACT or BCT must be identified, or implied by use of 90°‒ 45°. angle = 1 1 2 sin ( or 2 2 − or 10 10 or 10 2 200 ) = 45º A1 AG Do not allow decimals Alternative method for question 11(b) 2 2 2 (10 2) 10 = + TA M1 TA = 10 → 45º A1 AG 2 Question Answer Marks Guidance 11(c) Gradient, m, of CT = 1 7 − B1 OE Attempt to find mid-point (M) of CT *M1 Expect (1, 5) Equation of AB is ( ) 5 7 1 − = − y x DM1 Through their (1, 5) with gradient 1 m − 7 2 = − y x A1 4 11(d) ( ) ( ) 2 2 8 7 2 4 100 − + − − = x x or equivalent in terms of y M1 Substitute their equation of AB into equation of circle. ( ) 2 50 100 0 − = x x A1 x = 0 and 2 A1 WWW Alternative method for question 11(d) MC = 7 1     −   M1 1 1 0 5 7 2 −      + =      − −      , 1 1 2 5 7 12     + =         A1 x = 0 and 2 A1 3

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