Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 1 · Variant 2

1 2 7 1 Find the coefficient of in the expansion of 3x . [4] x2 + 3x2 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 For a correctly selected term in 2 1 x : (3x)4 or 34 B1 Components of coefficient added together 0/4 B1 expect 81 × 2 2 ³ 3x       or (2/3)3 B1 B1 expect 8/27 × 7C3 or 7C4 B1 B1 expect 35 → 840 or 2 840 x B1 All of the first three marks can be scored if the correct term is seen in an expansion and it is selected but then wrongly simplified. SC: A completely correct unsimplified term seen in an expansion but not correctly selected can be awarded B2. 4

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4 @ A 2 2 Showing all necessary working, find x dx. [4] x Ô1 + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Integrate → 3 2 3 2 x + 2 1 2 1 2 x (+C) B1 B1 required. 4 3 1 2 2 1 2 3 1 2 2 x x     +       → 40 14 3 3 − M1 Evidence of 4 and 1 used correctly in their integrand ie at least one power increased by 1. = 26 3 oe A1 Allow 8.67 awrt. No integrand implies use of integration function on calculator 0/4. Beware a correct answer from wrong working. 4

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3 y y = 5x Q R P y = x 9 −x2 x O The diagram shows part of the curve y x 9 and the line y 5x, intersecting at the origin O and the point R. Point P lies on the line y = 5x between−x2 O and R and= the x-coordinate of P is t. Point Q lies on the curve and PQ is parallel to the= y-axis. (i) Express the length of PQ in terms of t, simplifying your answer. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Given that t can vary, find the maximum value of the length of PQ. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) B1 B1 subsequent working. B1 for PQ allow 4 – ³ t t or ³ – 4 t t . Note: 4x – x3 from equating line and curve 0/2 even if x then replaced by t. [2] Question Answer Marks Guidance 3(ii) ( ) d d PQ t = 4 – 3t² B1FT B1FT for differentiation of their PQ, which MUST be a cubic expression, but can be ( ) d f x dx from (i) but not the equation of the curve. = 0 → t = + 2 3 √ M1 Setting their differential of PQ to 0 and attempt to solve for t or x. → Maximum PQ = 16 3 3 √ or 16 3 9 A1 Allow 3.08 awrt. If answer comes from wrong method in (i) award A0. Correct answer from correct expression by T&I scores 3/3. 3

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5 The first three terms of an arithmetic progression are 4, x and y respectively. The first three terms of a geometric progression are x, y and 18 respectively. It is given that both x and y are positive. (i) Find the value of x and the value of y. 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(ii) Find the fourth term of each progression. 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Mark scheme: 5(i) From the AP: 4 x y x B1 Or equivalent statement e.g. y = 2x – 4 or 4 2 y x = . From the GP: 18 y x y = B1 Or equivalent statement e.g. y2 = 18x or 2 18 y x = . Simultaneous equations: ² 9 36 0 y y − − = or 2 ² 17 8 0 x x − + = M1 Elimination of either x or y to give a three term quadratic (= 0) OR 4+d =x, 4+2d=y → 4 2 4 d r d + = + oe B1 ( ) 2 4 2 4 18 4 d d d +   + =   +   → 2 2 28 0 d d − − = M1 Uses ar2 = 18 to give a three term quadratic (= 0) d = 4 B1 Condone inclusion of d = 7 2 − oe Question Answer Marks Guidance 5(i) OR From the GP 18 y x y = B1 → 2 18 y x = → 4+ d = 2 18 y →d = 2 18 y – 4 B1 2 4 2 4 18 y y   + − =     → ² 9 36 0 y y − − = M1 x = 8, y = 12. A1 Needs both x and y. Condone 1 , 3 2   −     included in final answer. Fully correct answer www 4/4. 4 5(ii) AP 4th term = 16 B1 Condone inclusion of 13 2 − oe GP 4th term = 8 × 3 12 8       M1 A valid method using their x and y from (i). = 27 A1 Condone inclusion of –108 Note: Answers from fortuitous x = 8, y = 12 in (i) can only score M1. Unidentified correct answer(s) with no working seen after valid x = 8, y = 12 to be credited with appropriate marks. 3

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6 B 20 cm 1Å A C 9 cm D The diagram shows a triangle ABC in which BC 20 cm and angle ABC The perpendicular from B to AC meets AC at D and AD 9 cm. Angle= BCA = 90Å. = = 1Å. (i) By expressing the length of BD in terms of in each of the triangles ABD and DBC, show that 20 9 cos 1 [4] sin21 = 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence, showing all necessary working, calculate [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) In ∆ ABD, tanθ = 9 BD → BD = 9 tanθ or 9tan(90 – θ) or 9 cotθ or ( ) 2 2 20 9 θ   −   tan (Pythag) or ( ) 9sin 90 sin θ θ − (Sine rule) B1 In ∆ DBC, sinθ= 20 BD → BD = 20sinθ B1 20sinθ = 9 tanθ M1 Equates their expressions for BD and uses sinθ/cosθ = tanθ or cosθ/sinθ = cot θ if necessary. → 20sin²θ = 9cosθ AG A1 Correct manipulation of their expression to arrive at given answer. SC: In ∆ DBC, sinθ = 20 BD → BD = 20sinθ B1 In ∆ ABD, BA = 9 sinθ and cosθ = BD BA cosθ = 20sin 9 / sin θ θ → cosθ = 20 ² 9 sin θ M1 → 20sin²θ = 9cosθ A1 Scores 3/4 4 6(ii) Uses s² + c² = 1 → 20cos²θ + 9cosθ – 20 (= 0) M1 Uses s² + c² = 1 to form a three term quadratic in cosθ → cosθ = 0.8 A1 www → θ = 36.9º awrt A1 www. Allow 0.644c awrt. Ignore 323.1º or 2.50c. Note: correct answer without working scores 0/3. 3

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7 R P Q M 12 N C 4 k j A B 4 O i The diagram shows a solid cylinder standing on a horizontal circular base with centre O and radius 4 units. Points A, B and C lie on the circumference of the base such that AB is a diameter and angle BOC Points P, Q and R lie on the upper surface of the cylinder vertically above A, B and C respectively.= 90Å. The height of the cylinder is 12 units. The mid-point of CR is M and N lies on BQ with BN 4 units. = Unit vectors i and j are parallel to OB and OC respectively and the unit vector k is vertically upwards. Evaluate PN. PM and hence find angle MPN. 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Mark scheme: 7 PN = 8i – 8k B1 PM JJJJG = 4i + 4j – 6k B2,1,0 Loses 1 mark for each component incorrect SC: PN JJJG = – 8i + 8k and PM JJJJG = – 4i – 4j +6k scores 2/3. . PN PM JJJG JJJJG = 32 + 0 + 48 = 80 M1 Evaluates x1x2 +y1y2+z1z2 for correct vectors or one or both reversed. PN PM × = √128 × √68 (= 16 34 ) M1 Product of their moduli – may be seen in cosine rule √128 × √68 cos M ˆP N = 80 M1 All linked correctly. Angle M ˆP N = 31.0◦ awrt A1 Answer must come directly from +ve cosine ratio. Cosine rule not accepted as a complete method. Allow 0.540c awrt. Note: Correct answer from incorrect vectors scores A0 (XP) 7

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8 Y X B 8 cm 8 cm A 12 cm C The diagram shows an isosceles triangle ACB in which AB BC 8 cm and AC 12 cm. The arc XC is part of a circle with centre A and radius 12 cm, and the=arc YC= is part of a circle= with centre B and radius 8 cm. The points A, B, X and Y lie on a straight line. (i) Show that angle CBY 1.445 radians, correct to 4 significant figures. 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(ii) Find the perimeter of the shaded region. 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Mark scheme: 8(i) A ˆB C using cosine rule giving cos-1( 1 8 −) or 2sin−1(¾) or 2 1 7 cos 2 −        or B ˆA C = cos-1(¾) or B ˆA C = sin-1 7 4 or B ˆA C = 1 7 tan 3 − M1 Or for B ˆA C, expect 0.723cawrt C ˆB Y = π – A ˆB C or 2×C ˆA B M1 For attempt at C ˆB Y = π – A ˆB C or C ˆB Y = 2 × C ˆA B OR Find CY from ∆ ACY using Pythagoras or similar ∆s M1 Expect 4 7 C ˆB Y = ( ) 2 2 2 1 8 8 cos 2 8 8 theirCY −  + −     × ×   M1 Correct use of cosine rule C ˆB Y = 1.445c AG A1 Numerical values for angles in radians, if given, need to be correct to 3 decimal places. Method marks can be awarded for working in degrees. Need 82.8° awrt converted to radians for A1. Identification of angles must be consistent for A1. 3 8(ii) Arc CY = 8 × 1.445 B1 Use of s=8θ for arc CY, Expect 11.56 B ˆA C = ½(π – A ˆB C) or cos−1(¾) *M1 For a valid attempt at B ˆA C, may be from (i). Expect 0.7227c Arc XC = 12 × (their B ˆA C) DM1 Expect 8.673 Perimeter = 11.56 + 8.673 + 4 = 24.2 cm awrt www A1 Omission of ‘+4’ only penalised here. 4

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12 10 The equation of a curve is y 2x and the equation of a line is y x k, where k is a constant. x = + + = (i) Find the set of values of k for which the line does not meet the curve. 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In the case where k 15, the curve intersects the line at points A and B. = (ii) Find the coordinates of A and B. 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(iii) Find the equation of the perpendicular bisector of the line joining A and B. 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Mark scheme: 10(i) 12 2x k x x + = − or ( ) 12 2 y k y k y = − + − → 3 term quadratic. Expect 3x² − kx + 12 or 3y2 – 5ky + (2k2 + 12) (= 0) Use of b² − 4ac → k² − 144 < 0 DM1 Using the discriminant, allow ⩽ , = 0; expect 12 and −12 − 12 < k < 12 A1 Do NOT accept ⩽ . Separate statements OK. 3 10(ii) Using k = 15 in their 3 term quadratic M1 From (i) or restart. Expect 3x² − 15x + 12 or 3y2 – 75y + 462 (= 0) x = 1,4 or y = 11, 14 A1 Either pair of x or y values correct.. (1, 14) and (4, 11) A1 Both pairs of coordinates 3 10(iii) Gradient of AB = −1 → Perpendicular gradient = +1 B1FT Use of m1m2=−1 to give +1 or ft from their A and B. Finding their midpoint using their (1, 14) and (4, 11) M1 Expect (2½, 12½) Equation: y – 12½ = (x – 2½) [y = x + 10] A1 Accept correct unsimplified and isw 3

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11 y M y = 3 4x + 1 −2x A x O The diagram shows part of the curve y 3 4x 1 The curve crosses the y-axis at A and the stationary point on the curve is M. = + −2x. dy (i) Obtain expressions for and y dx. 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(ii) Find the coordinates of M. 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(iii) Find, showing all necessary working, the area of the shaded region. 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Mark scheme: 11(i) d d y x = ( ) 1 2 3 4 1 2 x −   × +     [×4] [− 2] 6 2 4 1 x   −   +   B2,1,0 d y x ∫ = ( ) 3 2 3 3 4 1 2 x   + ÷     [ ÷ 4 ] [ − 2 ² 2 x ] (+ C) ( ) 3 2 2 4 1 2 x x   +   = −       B1 B1 B1 B1 for ( ) 3 2 3 3 4 1 2 x + ÷ B1 for ‘÷4’. B1 for ‘− 2 2 2 x ’. Ignore omission of + C. If included isw any attempt at evaluating. 5 11(ii) At M, d d y x = 0 → 6 4 1 x + = 2 M1 Sets their 2 term d d y x to 0 and attempts to solve (as far as x = k) x = 2, y = 5 A1 A1 3 Question Answer Marks Guidance 11(iii) Area under the curve = ( ) 2 3 2 0 1 4 1 ² 2 x x   + −     M1 Uses their integral and their ‘2’ and 0 correctly (13.5 – 4) – 0.5 or 9.5 – 0.5 = 9 A1 No working implies use of integration function on calculator M0A0. Area under the chord = trapezium = ½ × 2 × (3 + 5) = 8 Or 2 2 0 3 8 2 x x   + =     M1 Either using the area of a trapezium with their 2, 3 and 5 or ( ) 3 their x dx ∫ + using their ‘2’ and 0 correctly. (Shaded area = 9 – 8) = 1 A1 Dependent on both method marks, OR Area between the chord and the curve is: ( ) 2 0 3 4 1 2 3 x x x dx + − − + ∫ 2 0 3 4 1 3 3 x x dx = + − − ∫ M1 Subtracts their line from given curve and uses their ‘2’ and 0 correctly. ( ) 2 2 3 2 0 1 3 4 1 6 2 x x x   = + − −     A1 All integration correct and limits 2 and 0. 27 1 3 2 2 6 6       = − − −             M1 Evidence of substituting their ‘2’ and 0 into their integral. 1 1 1 3 3 1 2 6 3    = − = =       A1 No working implies use of a calculator M0A0. [4]

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