Cambridge A Level Mathematics 9709 — 2016 May/June Paper 1 · Variant 3

@1 A5 1 Find the coefficient of x in the expansion of + 3x2 . [3] x

Mark scheme:  1  2 2 3x1 5C2   ( ) B1 Can be seen in expansion  x  10 ( × 1) × 3 2 B1 Identified as leading to answer 90 (x) B1 [3] ( ) ∫ ( 3 ) 2

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dy k 3 A curve is such that = 6x2 + and passes through the point P 1, 9 . The gradient of the curve dx x3 at P is 2. (i) Find the value of the constant k. [1] (ii) Find the equation of the curve. [4]

Mark scheme: 3 (i) 6 + k = 2 → k = −4 B1 [1] 6 x 3 4 2 k 2 (ii) ( y ) = x− (+c) B1B1 ft on their k. Accept + x− 3 −−2 − 2 9 = 2 + 2 + c c must be present M1 Sub (1,9) with numerical k. Dep on attempt ∫ ( y ) = 2 x 3 + 2 x−2 + 5 A1 Equation needs to be seen [4] Sub (2, 3) →c = –13½ scores M1A0 3 + 2 d 3 + 12 d 2 3 + 12 d

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4 The 1st, 3rd and 13th terms of an arithmetic progression are also the 1st, 2nd and 3rd terms respectively of a geometric progression. The first term of each progression is 3. Find the common difference of the arithmetic progression and the common ratio of the geometric progression. [5]

Mark scheme: 3 + 2 d 3 + 12 d 2 3 + 12 d4 r = or or r = B1 1 correct equation in r and d only is 3 3 + 2 d 3 sufficient ( 3 + 2 d ) 2 = 3 ( 3 + 12 d ) oe M1 Eliminate r or d using valid method OR sub 2d = 3r – 3 ( 4 ) d ( d − 6 ) = 0 DM1 Attempt to simplify and solve quadratic OR 3r 2 = 18 r − 15 → ( r − 1)( r − 5 ) d = 6 A1 Ignore d = 0 or r = 1 Do not allow −5 or ±5 r = 5 A1 [5] dy −2 ( )

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6 3 E 2 A B 3 2 F G 1 1 C The diagram shows triangle ABC where AB = 5 cm, AC = 4 cm and BC = 3 cm. Three circles with centres at A, B and C have radii 3 cm, 2 cm and 1 cm respectively. The circles touch each other at points E, F and G, lying on AB, AC and BC respectively. Find the area of the shaded region EFG. [7]

Mark scheme: 6 BAC = sin −1 (3 / 5) or cos −1 (4 / 5) or tan −1 (3 / 4) B1 Accept 36.8(7)º ABC = sin −1 (4 / 5) or cos −1 (3 / 5) or tan −1 (4 / 3) B1 Accept 53.1(3)º B1 ACB = π / 2 (Allow 90º) Shaded area = ∆ABC – sectors (AEF + BEG + M1 CFG) 1 B1 ∆ABC = × 4 × 3 oe 2 1 2 3 0.6435 Sum sectors =  ) +  2 2 2 M1 2 0.9273 + 1 1.5708] π 2 2 2 OR  3 36.8 ( 7 ) + 2 53.1 ( 3 ) + 1 90    360 A1 6 – 5.536 = 0.464 [7] dy 1/2

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8 (i) Show that 3 sin x tan x −cos x + 1 = 0 can be written as a quadratic equation in cos x and hence solve the equation 3 sin x tanx −cosx + 1 = 0 for 0 ≤x ≤0. [5] (ii) Find the solutions to the equation 3 sin 2x tan2x −cos 2x + 1 = 0 for 0 ≤x ≤0. [3]

Mark scheme: 8 (i) 3sin 2 x − cos 2 x + cos x = 0 M1 Multiply by cos x Use s 2 = 1 − c 2 and simplify to 3-term quad M1 Expect 4c 2 −−c 3 = 0 cos x = −3 / 4 and 1 A1 x = 2.42 (allow 0.77π ) or 0 (extra in range A1A1 SC1 for 0.723 (or 0.23π), π max 1) [5] following 4c 2 + c − 3 = 0 (ii) 2x = 2π −their 2.42 or 360 – 138.6 B1 Expect 2x = 3.86 x = 1.21 (0.385π), 1.93 (0.614/5π), 0, π (3.14) B1B1 Any 2 correct B1. Remaining 2 (extra max 1) [3] correct B1. SCB1for all 69.3, 110.7, 0, 180 (degrees) SCB1 for .361, π/2, 2.78 after 4c 2 + c − 3 = 0  −1   

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9 The position vectors of A, B and C relative to an origin O are given by ` a ` a ` a 2 1 5 −−→ −−→ −−→ OA = 3 , OB = 5 and OC = 0 , −4 p 2 where p is a constant. (i) Find the value of p for which the lengths of AB and CB are equal. [4] (ii) For the case where p = 1, use a scalar product to find angle ABC. [4] [Questions 10 and 11 are printed on the next page.]

Mark scheme:   9 (i) AB = OB – OA = 2 B1 Ignore labels. Allow BA or BC      p + 4   −4    CB = OB – OC = 5 B1      p − 2  2 2 1 + 4 + ( p + 4 ) = 16 + 25 + ( p − 2 ) M1 p = 2 A1 [4] (ii) AB.CB = 4+10−5 = 9 M1 Use of x1 x2 + y1 y2 + z1 z 2 │AB│= 1 + 4 + 25 = √30, │CB│ = 16 + 25 + 1 =√42 M1 Product of moduli 9 9 cos ABC = or M1 Allow one of AB, CB reversed - but 30 42 6 35 award A0 ABC = 75.3˚ or 1.31rads (ignore reflex angle 285˚) A1 [4] ( 2 ) 2

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11 Triangle ABC has vertices at A −2, −1 , B 4, 6 and C 6, −3 . (i) Show that triangle ABC is isosceles and find the exact area of this triangle. [6] (ii) The point D is the point on AB such that CD is perpendicular to AB. Calculate the x-coordinate of D. [6]

Mark scheme: 11 (i) AB 2 = 6 2 + 7 2 = 85, BC 2 = 2 2 + 9 2 = 85 (→ isosceles) B1B1 Or AB = BC = 85 etc AC 2 = 82 + 2 2 = 68 B1 2 2 2 M = (2, −2) or BM = ( 85) − (½ 68) B1 Where M is mid-point of AC BM = 2 2 + 8 2 = 68 or 85 − 17 = 68 B1 1 Area ∆ABC = 68 68 = 34 B1 2 [6] (ii) Gradient of AB = 7 / 6 B1 7 7 Equation of AB is y + 1 = ( x + 2 ) M1 Or y − 6 = ( x − 4 ) 6 6 Gradient of CD = −6 / 7 M1 − 6 Equation of CD is y + 3 = ( x − 6 ) M1 7 − 6 36 7 14 Sim Eqns 2 = x + − x − M1 7 7 6 6 34 2 x = = oe A1 85 5 [6]

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