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

5 2 A point is moving along the curve y = 2x + in such a way that the x-coordinate is increasing at a x constant rate of 0.02 units per second. Find the rate of change of the y-coordinate when x = 1. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 5 2 y x x = + → d 5 2 d ² y x x = − = −3 (may be implied) when x = 1. M1 A1 d d y t = d d d d y x x t × → −0.06 M1 A1 Ignore notation, but needs to multiply d d y x by 0.02. 4

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5 y B A 0, 4 x O C 8, 0 The diagram shows a kite OABC in which AC is the line of symmetry. The coordinates of A and C are 0, 4 and 8, 0 respectively and O is the origin. (i) Find the equations of AC and OB. 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(ii) Find, by calculation, the coordinates of B. 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Mark scheme: 5(i) Eqn of AC y = −1 2 x + 4 (gradient must be / y x ∆ ∆) M1A1 Gradient of OB = 2 → y = 2x (If y missing only penalise once) M1 A1 Use of m1m2 = − 1 , answers only ok. 4 Question Answer Marks Guidance 5(ii) Simultaneous equations → ((1.6, 3.2)) M1 Equate and solve for M1 and reach ⩾1 solution This is mid-point of OB. → B (3.2, 6.4) M1 A1 Uses mid-point. CAO or Let coordinates of B (h, k) OA = AB → h² = 8k − k² OC =BC → k² = 16h – h² → (3.2, 6.4) M1 for both equations, M1 for solving with 2 y x = or gradients ( 4 1 8 k k h h − × = − − ) M1 for gradient product as –1, M1 solving with 2 y x = or Pythagoras: ( ) ( ) 2 2 2 2 2 2 4 8 4 8 h k h k + − + − + = + M1 for complete equation, M1 solving with 2 y x = 3

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6 A T r cm 1 rad B O The diagram shows a circle with centre O and radius r cm. The points A and B lie on the circle and AT is a tangent to the circle. Angle AOB = 1 radians and OBT is a straight line. (i) Express the area of the shaded region in terms of r and 1. 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(ii) In the case where r = 3 and 1 = 1.2, find the perimeter of the shaded region. 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Mark scheme: 6(i) (tanθ = ) AT r → AT = r tanθ or OT = cos r θ SOI B1 → A = 1 2 r²tanθ − 1 2 r²θ B1 B1 B1 for 1 2 r²tanθ. B1 for “−1 2 r²θ” If Pythagoras used may see area of triangle as 2 2 2 1 2 r r r tan θ + or 1 2 cos r r sinθ θ       3 Question Answer Marks Guidance 6(ii) tanθ = 3 AT → AT = 7.716 M1 Correct use of trigonometry and radians in rt angle triangle Arc length = rθ = 3.6 B1 Accept 3 1.2 × OT by Pythagoras or cos1.2 = 3 OT ( = 8.279) M1 Correct method for OT Perimeter = AT + arc + OT – radius = 16.6 A1 CAO, www 4

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7 Relative to an origin O, the position vectors of the points A, B and C are given by ` a ` a ` a 1 −1 3 −−→ −−→ −−→ OA = −3 , OB = 3 and OC = 1 . 2 5 −2 −−→ (i) Find AC. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−−→ (ii) The point M is the mid-point of AC. Find the unit vector in the direction of OM. 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AC and hence find angle BAC. 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Mark scheme: 7 1 3 2 OA     = −       JJJG , 1 3 5 OB −     =       JJJG and 3 1 2 OC     =     −   JJJG 7(i) 2 4 4 AC     =     −   JJJG B1 B1 for AC JJJG . 1 Question Answer Marks Guidance 7(ii) 2 1 0 OM OA AM     = + = −       JJJJG JJJG JJJJG or 1[ 2 1 3 3 1 2 2         − +         −     ] M1 M1 for their OM OA AM = + JJJJG JJJG JJJJG oe Unit vector in direction of 1 5 OM = √ JJJJG ( ) OM JJJJG M1 A1 M1 for dividing their OM JJJJG by their modulus 3 7(iii) 2 6 3 AB −     =       JJJG , Allow ± B1 | AB JJJG |=7, | AC JJJG |=6 2 6 3 −           . 2 4 4         −   = −4 + 24 – 12 = 8 M1 M1 Product of both moduli, Scalar product of ± their AB and AC 7 × 6 cos θ = 8 → θ = 79.(0)º A1 1.38 radians ok 4

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8 (a) A geometric progression has a second term of 12 and a sum to infinity of 54. Find the possible values of the first term of the progression. 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(b) The nth term of a progression is p + qn, where p and q are constants, and Sn is the sum of the first n terms. (i) Find an expression, in terms of p, q and n, for Sn. 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(ii) Given that S4 = 40 and S6 = 72, find the values of p and q. 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Mark scheme: 8(a) ar = 12 and 1 a r − = 54 B1 B1 Eliminates a or r → 9 ² 9 2 0 r r − + = or ² 54 648 0 a a − + = M1 Elimination leading to a 3-term quadratic in a or r → r = 2 3 or 1 3 hence to a → a = 18 or 36 A1 Needs both values. 4 8(b) nth term of a progression is p + qn 8(b)(i) first term = p + q. Difference = q or last term = p + qn B1 Need first term and, last term or common difference Sn = ( ) ( ) ( ) 2 1 2 n p q n q + + − or ( ) 2 2 n p q nq + + M1A1 Use of Sn formula with their a and d. ok unsimplified for A1. 3 8(b)(ii) Hence ( ) 2 2 4 40 p q q + + = and ( ) 3 2 6 72 p q q + + = DM1 Uses their Sn formula from (i) Solution → p = 5 and q = 2 [Could use Sn with a and d → a = 7, d = 2 → p = 5, q = 2.] A1 Note: answers 7, 2 instead of 5, 2 gets M1A0 – must attempt to solve for M1 2

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10 The curve with equation y = x3 −2x2 + 5x passes through the origin. (i) Show that the curve has no stationary points. 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(ii) Denoting the gradient of the curve by m, find the stationary value of m and determine its nature. 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(iii) Showing all necessary working, find the area of the region enclosed by the curve, the x-axis and the line x = 6. 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Mark scheme: 10 y = x³ − 2x² + 5x 10(i) d d y x = 3x² − 4x + 5 B1 CAO Using ² 4 b ac − → 16 – 60 → negative → some explanation or completed square and explanation M1 A1 Uses discriminant on equation (set to 0). CAO 3 10(ii) m = 3x² − 4x + 5 d d m x = 6x – 4 (= 0) (must identify as d d m x ) B1FT FT providing differentiation is equivalent → x = 2 3 , m = 11 3 or 11 3 dy dx = Alt1: 2 2 11 3 3 3 m x   = − +     , 11 3 m = Alt2: 2 2 11 3 4 5 0, 4 0, 3 x x m b ac m − + − = − = = M1 A1 Sets to 0 and solves. A1 for correct m. Alt1: B1 for completing square, M1A1 for ans Alt2: B1 for coefficients, M1A1 for ans d²m d ²x = 6 +ve → Minimum value or refer to sketch of curve or check values of m either side of x = 2 3 , M1 A1 M1 correct method. A1 (no errors anywhere) 5 Question Answer Marks Guidance 10(iii) Integrate → 4 2 ³ 5 ² 4 3 2 x x x − + B2,1 Loses a mark for each incorrect term Uses limits 0 to 6 → 270 (may not see use of lower limit) M1 A1 Use of limits on an integral. CAO Answer only 0/4 4

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