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

3 B 8 cm a rad O A 8 cm C In the diagram, OAB is a sector of a circle with centre O and radius 8 cm. Angle BOA is ! radians. OAC is a semicircle with diameter OA. The area of the semicircle OAC is twice the area of the sector OAB. (i) Find ! in terms of 0. [3] (ii) Find the perimeter of the complete figure in terms of 0. [2]

Mark scheme: 1 2 1 23 (i) (OAB ) = × 8 α , (OAC ) = × π × 4 B1B1 Accept 25.1 (for OAC) 2 2 π α = B1 8 [3] 1 (ii) 8 + 8 × their α + × 8 × π B1 23.7 gets B1B0 2 8 + 5 π B1 SC B1 for e.g. 5 π (omitted OB) [2] 2 5

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4 The third term of a geometric progression is −108 and the sixth term is 32. Find (i) the common ratio, [3] (ii) the first term, [1] (iii) the sum to infinity. [2]

Mark scheme: 4 (i) ar2 = –108, ar5 = 32 B1 32 8  r3 = =   M1 Eliminating a − 108 −27  2 3 2  A1 − from little or no working → www r =   or –0.666 or –0.667 3 3 −3  [3] 32  or  ft on their r − 1082 5 (ii) a = –243 B1  r r  [1] − 243 729 = − or –145.8 (iii) S ∞ = M1A1 Accept –146. For M1 r must be < 1 2 5 1 + [2] 3 sin θ (sin θ − cos θ ) + cos θ (sin θ + cos θ )

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