Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 1 · Variant 3

1 1 Find the coefficient of in the expansion of x −2 7 . [3] x3 x ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 7C5 2 5 ( 2 / ) x x − soi 21 32 × − soi B1 Identified. Allow (21x2) × (‒32 x‒5). Implied by correct answer ‒672 B1 Allow 3 672 x − . If 0/3 scored, 672 scores SCB1 3

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3 A 5 cm C 4 cm D B The diagram shows an arc BC of a circle with centre A and radius 5 cm. The length of the arc BC is 4 cm. The point D is such that the line BD is perpendicular to BA and DC is parallel to BA. (i) Find angle BAC in radians. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the area of the shaded region BDC. 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Mark scheme: 3(i) 0.8 oe B1 1 3(ii) BD = 5sin 0.8 their M1 Expect 3.58(7). Methods using degrees are acceptable DC = 5 ‒ 5cos their 0.8 M1 Expect 1.51(6) Sector = ½ × 52 × their0.8 OR Seg = ½ × 52 × [their 0.8 ‒ sintheir 0.8] M1 Expect 10 for sector. Expect 1.03(3) for segment Trap = ½(5 + theirDC) × theirBD oe OR ∆BDC = ½theirBD × theirCD M1 OR (for last 2 marks) if X is on AB and XC is parallel to BD: Shaded area = 11.69 ‒ 10 OR 2.71(9) ‒ 1.03(3) = 1.69 cao A1 BDCX ‒(sector ‒ AXC ∆ ) = 5.43(8) ‒ [10 ‒ 6.24(9)] = 1.69 cao M1A1 5

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4 Two points A and B have coordinates −1, 1 and 3, 4 respectively. The line BC is perpendicular to AB and intersects the x-axis at C. (i) Find the equation of BC and the x-coordinate of C. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the distance AC, giving your answer correct to 3 decimal places. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Gradient, m, of AB = 3/4 B1 Equation of BC is ( ) 4 4 3 3 y x − − = − M1A1 Line through (3, 4) with gradient 1 m − (M1). (Expect 4 8 3 y x − = + ) 6 x = A1 Ignore any y coordinate given. 4 Question Answer Marks Guidance 4(ii) ( ) 2 2 2 7 1 7.071 AC AC = + → = M1A1 M mark for ( ) 2 6 / 1 1 their + − + . 2

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5 In an arithmetic progression the first term is a and the common difference is 3. The nth term is 94 and the sum of the first n terms is 1420. Find n and a. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 1 3 94 a n + − = ( ) [ ] 2 1 3 1420 OR 94 1420 2 2 n n a n a  + − = + =   B1 Attempt elimination of a or n M1 ( ) ( ) 2 2 3 191 2840 0 OR 3 598 0 n n a a − + = − − = A1 3-term quadratic (not necessarily all on the same side) n = 40 (only) A1 a = ‒23 (only) A1 Award 5/6 if a 2nd pair of solutions (71/3, 26) is given in addition or if given as the only answer. 6

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6 F G 2 E D 4 7 B C 6 k j A 8 O i The diagram shows a solid figure OABCDEFG with a horizontal rectangular base OABC in which OA = 8 units and AB = 6 units. The rectangle DEFG lies in a horizontal plane and is such that D is 7 units vertically above O and DE is parallel to OA. The sides DE and DG have lengths 4 units and 2 units respectively. Unit vectors i, j and k are parallel to OA, OC and OD respectively. Use a scalar 0a 1 product to find angle OBF, giving your answer in the form cos−1 , where a and b are integers. b [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ 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Mark scheme: 6 8 6 = − − BO i j ( ) 6 8 7 4 2 4 4 7 = − − + + + = − − + BF j i k i j i j k B1 OR (FB) = 4i + 4j ‒ 7k ( ) ( )( ) ( )( ) . 4 8 4 6 = − − + − − BF BO M1 OR (FB.OB) Expect 56. Accept one reversed but award final A0 2 2 2 2 2 4 4 7 8 6 × = + + × + BF BO M1 Expect 90. At least one magnitude methodically correct Angle OBF 1 1 1 56 56 28 cos cos or cos 90 90 45 their their − − −       = =             DM1A1 Or equivalent ‘integer’ fractions. All M marks dependent on use of (±)BO and (±)BF. 3rd M mark dep on both preceding M marks 6

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9 A curve has equation y = 2x2 −3x + 1 and a line has equation y = kx + k2, where k is a constant. (i) Show that, for all values of k, the curve and the line meet. 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(ii) State the value of k for which the line is a tangent to the curve and find the coordinates of the point where the line touches the curve. 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Mark scheme: 9(i) For their 3-term quad a recognisable application of 2 4 b ac − M1 Expect 2 2 2 3 1 0 x x k k − + + − = oe for the 3-term quad. ( ) ( ) ( )( ) 2 2 2 4 3 4 2 1 b ac k k − = + − − oe A1 Must be correct. Ignore any RHS 2 9 6 1 k k + + A1 Ignore any RHS ( ) 2 3 1 0 k + . Do not allow > 0. Hence curve and line meet. AG A1 Allow 2 1 (9) 0 3 k   +     . . Conclusion required. ALT Attempt solution of 3-term quadratic M1 Solutions ( ) 1, ½ 1 x k k = + − A1A1 Which exist for all values of k. Hence curve and line meet. AG A1 4 Question Answer Marks Guidance 9(ii) 1/ 3 k = − B1 ALT d / d 4 3 4 3 y x x x k = − ⇒ − = Sub (one of) their 13 k = − into either line 1 → ( ) 2 8 8 2 0 3 9 x x − + = Or into the derivative of line 1 → 4x ( )( ) 3 0 k − + = M1 Sub 4 3 k x = − into line 1 → ( ) ( ) ( ) 2 2 2 4 1 4 3 0 x x x x − + − − = x = 2/3 Do not allow unsubstantiated 2 1 , 3 9   −     following 1 3 k = − A1 x = 2/3, y = ‒1/9 (both required) [from 2 18 24 8 x x − + − (=0) oe] y = ‒1/9 Do not allow unsubstantiated 2 1 , 3 9   −     following 1 3 k = − A1 1/ 3 k = − 4

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10 y x = 32 A 3 x = 3 y = 2 3x −1 −1 x O 1 2 3 −1 The diagram shows part of the curve y = 2 3x −1 3 and the lines x = 2 and x = 3. The curve and the 3 line x = 2 intersect at the point A. 3 (i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. 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(ii) Find the equation of the normal to the curve at A, giving your answer in the form y = mx + c. 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Mark scheme: 10(i) ( ) ( ) ( ) ( ) [ ] 1/3 2/3 3 1 4 3 1 d 4 3 1/ 3 x V x x π π −   − = ∫ − =  ÷     for [ ] [ ] ( )[ ] 4 2 1 π − DM1 Expect ( )( ) 13 4 3 1 x π − 4π or 12.6 A1 Apply limits ⅔ → 3. Some working must be shown. 5 Question Answer Marks Guidance 10(ii) ( ) 4/3 d / d ( 2 / 3) 3 1 3 y x x − = − − × B1 Expect ( ) 4/3 2 3 1 x − − − When 2 / 3, 2 x y = = soi d / d 2 y x = − B1B1 2nd B1 dep. on correct expression for dy//dx Equation of normal is ( ) 23 2 ½ y x − = − M1 Line through (⅔, their 2) and with grad ‒1/m. Dep on m from diffn 1 5 2 3 y x = + A1 5

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