Cambridge A Level Mathematics 9709 — 2013 May/June Paper 1 · Variant 3

2 R Q C O S 3 cm P 6 cm The diagram shows a circle C with centre O and radius 3 cm. The radii OP and OQ are extended to S and R respectively so that ORS is a sector of a circle with centre O. Given that PS = 6 cm and that the area of the shaded region is equal to the area of circle C, (i) show that angle POQ = 140 radians, [3] (ii) find the perimeter of the shaded region. [2]

Mark scheme: 1 dy = 2 x + 5 dx

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3 (i) Express the equation 2 cos21 = tan21 as a quadratic equation in cos21. [2] (ii) Solve the equation 2 cos21 = tan21 for 0 ≤1 ≤0, giving solutions in terms of 0. [3]

Mark scheme: 3 B1 B1 Everything without “÷2”

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6 The non-zero variables x, y and u are such that u = x2y. Given that y + 3x = 9, find the stationary value of u and determine whether this is a maximum or a minimum value. [7]

Mark scheme: 6 u = x 2 y y + 3 x = 9 M1 Expressing u in terms of 1 2 variable 2  9 − y  u = x (9 − 3 x ) or   y  3  du du DM1A1 Knowing to differentiate. =18x − 9x² or =27 − 12y + y² dx dy = 0 when x = 2 or y = 3 → u = 12 DM1 Setting differential to 0. A1 d 2 u DM1 Any valid method 2 =18−18x −ve A1 dx [7]

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7 y A (2, 14) X B (14, 6) C (7, 2) x O The diagram shows three points A 2, 14 , B 14, 6 and C 7, 2 . The point X lies on AB, and CX is perpendicular to AB. Find, by calculation, (i) the coordinates of X, [6] (ii) the ratio AX : XB. [2]

Mark scheme: 7 A (2, 14), B (14, 6) and C (7, 2). (i) m of AB = −⅔ B1 m of perpendicular = 32 M1 For use of m1m2 = −1 eqn of AB y − 14 = − 23 ( x − 2) M1 Allow M1 for unsimplified eqn eqn of CX y − 2 = 32 ( x − 7 ) M1 Allow M1 for unsimplified eqn Sim Eqns → X (11, 8) M1 A1 [6] For solution of sim eqns. (ii) AX : XB = 14−8 : 8− 6 = 3 : 1 M1 A1 Vector steps or Pythagoras. Or √(9²+6²) : √(3²+2²) = 3: 1 [2]

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8 A B O C The diagram shows a parallelogram OABC in which ` a ` a 3 5 −−→ −−→ OA = 3 and OB = 0 . −4 2 (i) Use a scalar product to find angle BOC. [6] −−→ (ii) Find a vector which has magnitude 35 and is parallel to the vector OC. [2]

Mark scheme: 8  3   5      OA =  3  and OB =  0  .     − 4 2      2    (i) OC = AB = b − a =  − 3  M1 Knowing how to find OC   6   Uses OC and OB B1 Using OC.OB or CO.BO OC.OB = 22 = 7 × √29 cos BOC M1 M1 M1 Use of x1x2 + ... M1 for modulus → Angle BOC = 54.3⁰ (or 0.948 rad) M1 A1 M1 everything linked. (nb uses BO.OC loses B1 A1) (nb uses other vectors – max [6] M1M1) GCE AS/A LEVEL – May/June 2013 9709 13 (ii) Modulus of OC = 7 M1 Knows to scale by factor of 35 ÷ Vector = 35 ÷ 7 × OC Mod  2  A1√ For their OC.   → ± 5  − 3    6   [2]

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9 (a) In an arithmetic progression, the sum, Sn, of the first n terms is given by Sn = 2n2 + 8n. Find the first term and the common difference of the progression. [3] (b) The first 2 terms of a geometric progression are 64 and 48 respectively. The first 3 terms of the geometric progression are also the 1st term, the 9th term and the nth term respectively of an arithmetic progression. Find the value of n. [5]

Mark scheme: 9 (a) Sn = 2 n 2 + 8n S1 = 10 = a B1 S2 = 24 = a + (a + d) d = 4 M1 A1 [3] correct use of Sn formula. (b) GP a = 64 ar = 48 → r = ¾ B1 →3rd term is ar² = 36 M1 ar² numerical – for their r AP a = 64, a + 8d = 48 → d = −2 B1 36 = 64 + (n − 1)(−2) M1 correct use of a+(n−1)d → n = 15. A1 [5]

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