Cambridge A Level Mathematics 9709 — 2018 May/June Paper 1 · Variant 2

2 The equation of a curve is y = x2 −6x + k, where k is a constant. (i) Find the set of values of k for which the whole of the curve lies above the x-axis. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the value of k for which the line y + 2x = 7 is a tangent to the curve. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) A complete method as far as finding a set of values for k by: Either (x – 3)² + k – 9 >0, k – 9 >0 Either completing the square and using ‘their k – 9’ > or ⩾ 0 OR or 2x – 6 = 0 → (3, k ─ 9), k – 9 >0 M1 Differentiating and setting to 0, using ‘their x=3’ to find y and using ‘their k – 9’ > or ⩾0 OR or b² < 4ac oe → 36 < 4k Use of discriminant < or ⩽ 0. Beware use of > and incorrect algebra. → k > 9 Note: not ⩾ A1 T&I leading to (or no working) correct answer 2/2 otherwise 0/2. 2 Question Answer Marks Guidance 2(ii) EITHER 2 – 6 x x k + = 7 – 2x → 2 – 4 –7 x x k + (= 0) *M1 Equates and collects terms. Use of b² – 4ac = 0 (16 – 4(k – 7) = 0) DM1 Correct use of discriminant = 0, involving k from a 3 term quadratic. OR 2x – 6 = – 2 → x = 2 (y = 3) *M1 Equates their d d y x to ± 2, finds a value for x. (their 3) or 7– 2(their 2) = (their 2)2 – 6(their 2) + k DM1 Substitutes their value(s) into the appropriate equation. → k = 11 A1 3

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3 A company producing salt from sea water changed to a new process. The amount of salt obtained each week increased by 2% of the amount obtained in the preceding week. It is given that in the first week after the change the company obtained 8000 kg of salt. (i) Find the amount of salt obtained in the 12th week after the change. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the total amount of salt obtained in the first 12 weeks after the change. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) r = 1.02 or 102 100 used in a GP in some way. Amount in 12th week = 8000 (their r)11 or (their a from 8000 . their r ) ( their r )12 M1 Use of arn – 1 with a = 8000 & n = 12 or with a = 8000 1.02 and n = 13. = 9950 (kg) awrt A1 Note: Final answer of either 9943 or 9940 implies M1. Full marks can be awarded for a correct answer from a list of terms. 3 Question Answer Marks Guidance 3(ii) In 12 weeks, total is ( ) ( ) ( ) ( ) 12 8000 1 1 their r their r − − M1 Use of Sn with a = 8000 and n = 12 or addition of 12 terms. = 107000 (kg) awrt A1 Correct answer but no working 2/2 2

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5 C D B k j O i A The diagram shows a three-dimensional shape. The base OAB is a horizontal triangle in which angle AOB is 90Å. The side OBCD is a rectangle and the side OAD lies in a vertical plane. Unit vectors i and j are parallel to OA and OB respectively and the unit vector k is vertical. The position −−→ −−→ −−→ vectors of A, B and D are given by OA = 8i, OB = 5j and OD = 2i + 4k. −−→ −−→ (i) Express each of the vectors DA and CA in terms of i, j and k. 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(ii) Use a scalar product to find angle CAD. 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Mark scheme: 5(i) 6 –4 DA = i k 6 –5 –4 CA = i j k JJJG B1 2 5(ii) Method marks awarded only for their vectors ± CA JJJG & ± DA JJJG Full marks can be obtained using AC JJJG & AD JJJG CA JJJG . DA JJJG = 36 + 16 ( = 52) M1 Using x1x2+y1y2+z1z2 52 DA = JJJG , 77 CA = JJJG M1 Uses modulus twice 52 = √77√52cos ˆ CAD oe M1 All linked correctly Cos ˆ CAD = 0.82178..→ ˆ CAD = 34.7º or 0.606ᶜ awrt A1 Answer must come from +ve cosine ratio 4

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6 B r O 21 rad T A The diagram shows points A and B on a circle with centre O and radius r. The tangents to the circle at A and B meet at T. The shaded region is bounded by the minor arc AB and the lines AT and BT. Angle AOB is 21 radians. (i) In the case where the area of the sector AOB is the same as the area of the shaded region, show that tan 1 = 21. 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(ii) In the case where r = 8 cm and the length of the minor arc AB is 19.2 cm, find the area of the shaded region. 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Mark scheme: 6(i) AT or BT = rtanθ or OT = r cosϑ ½r²2θ, & ½×r×(rtanθ or AT) or ½×r×( r cosϑ or OT) sinθ M1 Both formulae, (½r²θ, ½bh or ½absinθ), seen with 2θ used when needed. ½r²2θ = 2×½×r×rtanθ – ½r²2θ oe → 2θ = tanθ AG A1 Fully correct working from a correct statement. Note: ½r²2θ = ½ r²tanθ is a valid statement. 3 Question Answer Marks Guidance 6(ii) θ = 1.2 or sector area = 76.8 B1 Area of kite = 165 awrt B1 164.6 – 76.8 = 87.8 awrt B1 awrt 87.8 with little or no working can be awarded 3/3. SC Final answers that round to 88 with little or no working can be awarded 2/3. 3

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10 (i) Solve the equation 2 cosx + 3 sin x = 0, for 0Å ≤x ≤360Å. 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(ii) Sketch, on the same diagram, the graphs of y = 2 cos x and y = −3 sin x for 0Å ≤x ≤360Å. [3] (iii) Use your answers to parts (i) and (ii) to find the set of values of x for 0Å ≤x ≤360Å for which 2 cosx + 3 sin x > 0. 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Mark scheme: 10(i) 2cosx = –3sinx → tanx = – ⅔ M1 Use of tan=sin/cos to get tan =, or other valid method to find sin or cos =. M0 for tanx = +/ 3 – 2 → x = 146.3º or 326.3ºawrt A1 A1FT FT for 180 added to an incorrect first answer in the given range. The second A1 is withheld if any further values in the range 0°⩽ x⩽ 360° are given. Answers in radians score A0, A0. 3 Question Answer Marks Guidance 10(ii) No labels required on either axis. Assume that the diagram is 0º to 360º unless labelled otherwise. Ignore any part of the diagram outside this range. B1 Sketch of y = 2cosx. One complete cycle; start and finish at top of curve at roughly the same positive y value and go below the x axis by roughly the same distance. (Can be a poor curve but not straight lines.) B1 Sketch of y= –3sinx One complete cycle; start and finish on the x axis, must be inverted and go below and then above the x axis by roughly the same distance. (Can be a poor curve but not straight lines.) B1 Fully correct answer including the sine curve with clearly larger amplitude than cosine curve. Must now be reasonable curves. Note: Separate diagrams can score 2/3 3 10(iii) x < 146.3º, x > 326.3º B1FT B1FT Does not need to include 0º, 360º. √ from their answers in (i) Allow combined statement as long as correct inequalities if taken separately. SC For two correct values including ft but with ⩽ and ⩾ B1 2

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11 y x 6 y = + 2 x P Q y = 4 x O x 6 The diagram shows part of the curve y = + . The line y = 4 intersects the curve at the points P 2 x and Q. (i) Show that the tangents to the curve at P and Q meet at a point on the line y = x. 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(ii) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. Give your answer in terms of 0. 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Mark scheme: 11(i) 6 2 x y x = + = 4 → x = 2 or 6 2 d 1 6 d 2 y x x = − B1 Unsimplified OK When x = 2, m = ─1 → x + y = 6 When x = 6, m = 1 3 → y = 1 3 x + 2 *M1 Correct method for either tangent Attempt to solve simultaneous equations DM1 Could solve BOTH equations separately with y = x and get x = 3 both times. (3,3) A1 Statement about y = x not required. 6 Question Answer Marks Guidance 11(ii) V = (π) ² 36 6 4 ² x x   ∫ + +     (dx) *M1 Integrate using π ²d y x ∫ (doesn’t need π or dx). Allow incorrect squaring. Not awarded for π 2 6 4 d . 2 x x x     ∫ − +         Integration indicated by increase in any power by 1. Integration → x³ 36 6 12 x x + ─ A2,1 3 things wanted —1 each error, allow + C. (Doesn’t need π) Using limits ‘their 2’ to ‘their 6’ (53 1 3 π, 160 , 1 68 3 π awrt) DM1 Evidence of their values 6 and 2 from (i) substituted into their integrand and then subtracted. 48 ─ 16 3   −     is enough. Vol for line: integration or cylinder (→ 64π) M1 Use of πr²h or integration of 42 (could be from 2 6 4 2 x x     − +         ) Subtracts → 10 2 3 π oe 32 e.g. , 33.5 awrt 3 π       A1 Question Answer Marks Guidance 11(ii) OR V = (π) 2 2 6 4 2 x x   ∫ − +     (dx) M1 *M1 Integrate using π ²d y x ∫ (doesn’t need π or dx) Integration indicated by increase in any power by 1. = (π) ² 36 16 6 4 ² x x   ∫ − + +     (dx) = (π) 3 36 16 6 12 x x x x     − + −           (dx) A2,1 Or 3 36 10 12 x x x   − +     = (π) ( 48 - 37⅓) DM1 Evidence of their values 6 and 2 from (i) substituted = 10 2 3 π oe 32 eg , 33.5 awrt 3 π       A1 6

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