Cambridge A Level Mathematics 9709 — 2018 May/June Paper 1 · Variant 3

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

Mark scheme: 2 5C3 3 2 2 −       x x SOI ‒80 Accept 80 − x B1 +80 without clear working scores SCB1 3

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3 The common ratio of a geometric progression is 0.99. Express the sum of the first 100 terms as a percentage of the sum to infinity, giving your answer correct to 2 significant figures. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 1 1 1   −    ÷   − −       n a r a r r DM1 Allow numerical a (M1M1). 3rd M1 is for division ∞ n S S (or ratio) SOI 100 1 0.99 − SOI OR ( ) ( ) 63 100 a a SOI A1 Could be shown multiplied by 100(%). Dep. on DM1 63(%) Allow 63.4 or 0.63 but not 2 infringements (e.g. 0.634, 0.63%) A1 n = 99 used scores Max M3. Condone a = 0.99 throughout ∞ = n S S ( ) without division shown scores 2 / 5 5

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5 A 5 cm 6 cm O C B The diagram shows a triangle OAB in which angle OAB = 90Å and OA = 5 cm. The arc AC is part of a circle with centre O. The arc has length 6 cm and it meets OB at C. Find the area of the shaded region. 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Mark scheme: 5 Angle AOC = 6 5 or 1.2 M1 Allow 68.8º. Allow 5 6 AB = 5 tan( 1 .2) their × OR by e.g. Sine Rule Expect 12.86 DM1 OR 5 cos 1.2 = OB their . Expect 13.80 Area 1 5 1 2.86 2 OAB their ∆ = × × Expect 32.15 DM1 OR 1 2 × 5 × their OB × sin their 1.2 Area sector 2 1 5 1 .2 2 their × × Expect 15 DM1 All DM marks are dependent on the first M1 Shaded region = 32.15 ‒ 15 = 17.2 A1 Allow degrees used appropriately throughout. 17.25 scores A0 5

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6 The coordinates of points A and B are −3k −1, k + 3 and k + 3, 3k + 5 respectively, where k is a constant (k ≠−1). (i) Find and simplify the gradient of AB, showing that it is independent of k. 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(ii) Find and simplify the equation of the perpendicular bisector of AB. 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Mark scheme: 6(i) Gradient, m, of AB = ( ) 3 5 3 2 2 OE 3 3 1 4 4 k k k k k k + − + +   =   + −− − +   = 1 2 2 6(ii) Mid-pt = [ 1 2 (‒3k ‒ 1 + k + 3), 1 2 (3k + 5 + k + 3)] = 2 2 4 8 , 2 2 − + +       k k SOI B1B1 B1 for 2 2 2 − + k , B1 for 4 8 2 + k (ISW) or better, i.e. ( ) 1, 2 4 −+ + k k Gradient of perpendicular bisector is 1 their m − SOI Expect ‒2 M1 Could appear in subsequent equation and/or could be in terms of k Equation: ( ) ( ) 2 4 2 1 − + = −−−+    y k x k OE DM1 Through their mid-point and with their 1 m − (now numerical) 2 6 + = y x A1 Use of numerical k in (ii) throughout scores SC2/5 for correct answer 5

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tan21 −1 7 (a) (i) Express in the form a sin21 + b, where a and b are constants to be found. 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(ii) Hence, or otherwise, and showing all necessary working, solve the equation tan21 −1 1 = tan21 + 1 4 for −90Å ≤1 ≤0Å. 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(b) y A y = sin x O x −0 0 B y = 2 cos x The diagram shows the graphs of y = sin x and y = 2 cos x for −0 ≤x ≤0. The graphs intersect at the points A and B. (i) Find the x-coordinate of A. 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(ii) Find the y-coordinate of B. 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Mark scheme: 7(a)(i) 2 2 2 2 2 2 sin 1 tan 1 cos tan 1 sin 1 cos − − = + + θ θ θ θ θ θ 2 2 2 2 sin cos sin cos − = + θ θ θ θ A1 multiplying by 2 cosθ Intermediate stage can be omitted by multiplying directly by 2 cosθ 2 2 sin cos = − θ θ ( ) 2 2 2 sin 1 sin 2sin 1 = − − = − θ θ θ A1 Using 2 2 sin cos 1 + = θ θ twice. Accept 2, 1 = = − a b ALT 1 2 2 sec 2 sec θ θ − M1 ALT 2 2 2 tan 1 sec θ θ − 2 2 1 sec θ − = 2 1 2cos θ − A1 2 2 (tan 1)cos θ θ − ( ) 2 2 1 2 1 sin 2sin 1 θ θ − − = − A1 ( ) 2 2 2 2 2 sin cos sin 1 sin 2sin 1 − = − − = − θ θ θ θ θ 3 7(a)(ii) 2 1 2sin 1 4 θ −= → ( ) ( ) 5 sin or 0.7906 8 θ = ± ± M1 OR ( ) ( ) 2 2 2 1 1 5 3 5 or 1.2910 4 3 1 t t t t t −= → = →= ± = ± + 52.2 θ = − A1 2 Question Answer Marks Guidance 7(b)(i) sin 2cos tan 2 = → = x x x M1 Or sinx = 4 5 or cosx = 1 5 1.11 = x with no additional solutions A1 Accept 0.352π or 0.353π. Accept in co-ord form ignoring y co-ord 2 7(b)(ii) Negative answer in range 1 0.8 −< < − y B1 0.894 or 0.895 or 0.896 − − − B1 2

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9 D 7 C B 6 k j E 2 O i 8 A The diagram shows a pyramid OABCD with a horizontal rectangular base OABC. The sides OA and AB have lengths of 8 units and 6 units respectively. The point E on OB is such that OE = 2 units. The point D of the pyramid is 7 units vertically above E. Unit vectors i, j and k are parallel to OA, OC and ED respectively. −−→ (i) Show that OE = 1.6i + 1.2j. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use a scalar product to find angle BDO. 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Mark scheme: 9(i) OE = 2 10 (8i + 6j) = 1.6i + 1.2j AG required 2 9(ii) OD = 1.6i + 1.2j + 7k B1 Allow reversal of one or both of OD, BD. BD = ‒8i ‒ 6j + 1.6i + 1.2j + 7k OE = ‒6.4i ‒4.8j + 7k M1A1 For M mark allow sign errors. Also if 2 out of 3 components correct Correct method for ±OD.±BD (using their answers) M1 Expect 1.6 × ‒6.4 + 1.2 × ‒4.8 + 49 = 33 or 825 25 825 / 25. Correct method for |OD| or |BD| (using their answers) M1 Expect 2 2 2 2 2 2 1.6 1.2 7 or 6.4 4.8 7 + + + + = √53 or √113 Cos BDO = their . × OD BD OD BD DM1 Expect 33 77.4 . Dep. on all previous M marks and either B1 or A1 64.8º Allow 1.13(rad) A1 Can’t score A1 if 1 vector only is reversed unless explained well 7

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