Cambridge A Level Mathematics 9709 — 2014 May/June Paper 1 · Variant 3

1 Find the coefficient of x in the expansion of x2 −2 5 . [3] x

Mark scheme:   2 2 1  x −x    3 3  2   2  Term in x is 10 × (x²)² × −  B1 B1 B1 10 or 5C2 or 5C3, B1 −   x  x     Coefficient = −80(x) B1 co Must be identified [3]

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2 The first term in a progression is 36 and the second term is 32. (i) Given that the progression is geometric, find the sum to infinity. [2] (ii) Given instead that the progression is arithmetic, find the number of terms in the progression if the sum of all the terms is 0. [3]

Mark scheme: 2 36, 32, ... (i) r = 8 S∞ = (their a) ÷ (1 – their r) M1 Method for r and S∞ ok. (│r│< 1) 9 S∞ = 36 ÷ 1 = 324 A1 co 9 [2] (ii) d = −4 B1 co n  0 = (72 + (n – 1)(–4)) M1 Sn formula ok and a value for d  8  2 ≠9  → n = 19 A1 Condone n = 0 but no other soln [3]

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3 O 6 cm 2.2 rad A B The diagram shows part of a circle with centre O and radius 6 cm. The chord AB is such that angle AOB = 2.2 radians. Calculate (i) the perimeter of the shaded region, [3] (ii) the ratio of the area of the shaded region to the area of the triangle AOB, giving your answer in the form k : 1. [3]

Mark scheme: 3 (i) s = r θ M1 Used with major or minor arc Angle of major arc = 2π – 2.2 = (4.083) B1 Could be gained in (ii). Perimeter = 12 + 24.5 = 36.5 or 12π − 1.2 A1 co (or full circle − minor arc B1) [3] 1 (ii) Area of major sector = r ² θ = (73.49) M1 Used with major / minor sector. 2 1 Area of triangle = . 6 ² sin 2.2 = (14.55) M1 Correct formula or method. 2 (2π – 2.2) / sin 2.2 gets M1M1 Ratio = 5.05 : 1 (Allow 5.03 → 5.06) A1 co [3] tan x + 1

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tan x + 1 4 (i) Prove the identity  sin x + cos x. [3] sin x tan x + cosx tan x + 1 (ii) Hence solve the equation = 3 sin x −2 cosx for 0 ≤x ≤20. [3] sin x tan x + cos x

Mark scheme: tan x + 14 ≡ sin x + cos x sin x tan x + cos x  s    + 1  c  s + c M1 Use of t = s / c twice (i) LHS =  s 2 + c 2 M1 Correct algebra and use of s² + c² = 1  s 2 + c    c  = RHS A1 AG all ok [3] (ii) s + c = 3s − 2c → tanx = 3 Allow cos2 = 4 , sin2 = 9 M1 Uses (i) and t = s t = 2 or 0 is M0 2 13 13 c 3 → x = 0.983 and 4.12 or 4.13 A1 A1 co. 1st + π, providing no excess solns in range. Allow 0.313π, 1.31π [3] GCE AS/A LEVEL – May/June 2014 9709 13 15

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dy 12 6 A curve is such that = , where a is a constant. The point P 2, 14 lies on the curve and dx 4x + a the normal to the curve at P is 3y + x = 5. (i) Show that a = 8. [3] (ii) Find the equation of the curve. [4]

Mark scheme: dy 12 6 = P (2, 14) Normal 3y + x = 44 dx 4 x + a (i) m of normal = 1 B1 co −3 dy 12 = 3 = → a = 8 M1 A1 Use of m1m₂ = −1. AG. dx 4 x + a [3] 1 1 (ii) ∫ y = 12(4x + a) 2 ÷ ÷ 4 (+c) B1 B1 Correct without “÷4”. for “÷4”. 2 Uses (2, 14) M1 Uses in an integral only. Dep ‘c’. c = −10 A1 co All 4 marks can be given in (i) [4] GCE AS/A LEVEL – May/June 2014 9709 13

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7 The position vectors of points A, B and C relative to an origin O are given by ` a ` a ` a 2 6 2 −−→ −−→ −−→ OA = 1 , OB = −1 and OC = 4 . 3 7 7 (i) Show that angle BAC = cos−1 1 . [5] 3 (ii) Use the result in part (i) to find the exact value of the area of triangle ABC. [3]

Mark scheme: 7 (i) Angle BAC needs sides AB,AC or BA,CA AB.AC = (b − a).(c − a) Ignore their labels:   4   0      B1 One of AB, BA, AC, CA correct  − 2  3 =  .  = 10         M1 Use of x₁x₂ + y₁y₂, etc.  4   4      = √36 × √25 cos BAC M1M1 M1 prod of moduli. M1 all linked → BAC = cos−1 1 AG A1 If e.g. BA.OC max B1M1M1. If both 3 vectors wrong 0 / 5. If e.g. BA.AC  1   final mark A0  used → cos–1 −3 [5] 1 (ii) sinBAC = 1− B1 Use of s² + c² = 1 − not decimals 9 1 8 Area = × 6 × 5 × = 5√8 oe M1 A1 Correct formula for area. Decimals seen A0 2 9 [3] 2 2

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9 The base of a cuboid has sides of length x cm and 3x cm. The volume of the cuboid is 288 cm3. (i) Show that the total surface area of the cuboid, A cm2, is given by 768 A = 6x2 + . x (ii) Given that x can vary, find the stationary value of A and determine its nature. [5]

Mark scheme: 9 (i) 3x²y = 288 y is the height B1 co A = 2(3x² + xy + 3xy) M1 Considers at least 5 faces (y ≠ x) Sub for y → A = 6x2 + 768 A1 co answer given x [3] d 768 A = 12 x − (ii) B1 co d x x 2 = 0 when x = 4 → A = 288. Allow (4 , 288) M1 A1 Sets differential to 0 + solution. co 2 d 1536 A = 12 + M1 Any valid method 2 d x x 3 (= 36) > 0 Minimum A1 co www dep on correct f″ and x = 4 [5] GCE AS/A LEVEL – May/June 2014 9709 13

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10 y y = 2x + 1 y = −x2 + 12x −20 x O The diagram shows the curve y = −x2 + 12x −20 and the line y = 2x + 1. Find, showing all necessary working, the area of the shaded region. [8] [Question 11 is printed on the next page.]

Mark scheme: 10 pts of intersection 2 x + 1= −x² + 12x − 20 M1A1 Attempt at soln of sim eqns. co → x = 3, 7 1 Area of trapezium = (4)(7 + 15) = 44 M1A1 Either method ok. co 2 (or ∫ (2x+1) dx from 3 to 7 = 44) 1 B2,1 −1 each term incorrect Area under curve = −3 x³ + 6x² − 20x 2 Uses 3 to 7 → (54 ) DM1 Correct use of limits (Dep 1st M1) 3 2 Shaded area = 10 A1 co 3 [8] OR 7  2 x 3 2  Functions subtracted before integration  ∫ 3  − x + 10 x − 21) = − 3 + 5 x − 21x  M1 subtraction, A1A1A1 for integrated terms, Subtraction reversed allow A3A0. DM1 correct use of limits, A1 Limits reversed allow DM1A0

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