Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 1 · Variant 1

4 cos 1 4 (i) Show that the equation + 15 = 0 can be expressed as tan 1 4 sin21 −15 sin 1 −4 = 0. 4 cos 1 (ii) Hence solve the equation + 15 = 0 for 0Å ≤1 ≤360Å. [3] tan 1

Mark scheme: 2 sin θ 4 (i) 4 cos θ + 15 sin θ = 0 M1 Replace tan θ by and multiply by cos θ sin θ or equivalent 41( − s 2 ) + 15 s = 0 → 4 sin 2 θ − 15 sin θ − 4 = 0 M1A1 Use c 2 = 1 − s 2 and rearrange to AG [3] (www) (ii) sin θ = − 1 / 4 B1 θ = 194 5. or 345 5. B1B1 Ignore other solution [3] Ft from 1st solution, SC B1 both angles in rads (3.39 and 6.03) dy 8

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8 5 A curve has equation y = + 2x. x dy d2y (i) Find and . [3] dx dx2 (ii) Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point. [5]

Mark scheme: d y 8 5 (i) = − + 2 cao B1B1 d x x 2 d 2 y 16 = cao B1 dx 2 x 3 [3] 8 2 (ii) − + 2 = 0 → 2 x − 8 = 0 M1 Set = 0 and rearrange to quadratic form 2 x x = ± 2 A1 y = ± 8 A1 If A0A0 scored, SCA1 for just (2, 8) d 2 y  Ft for " correct" conclusion if  > 0 when x = 2 hence MINIMUM B1 2 2 dx  d y  2  2 incorrect or  d y < 0 when x = − 2 hence MAXIMUM B1  dx  dx 2 [5] any valid method inc. a good sketch  2 2

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7 C r A B E D The diagram shows a circle with centre A and radius r. Diameters CAD and BAE are perpendicular to each other. A larger circle has centre B and passes through C and D. (i) Show that the radius of the larger circle is rï2. [1] (ii) Find the area of the shaded region in terms of r. [6]

Mark scheme: 7 (i) BC 2 = r 2 + r 2 = 2 r 2 → BC = r 2 B1 AG [1] 1 1 2 (ii) Area sector BCFD = π ( r 2 2) soi M1 Expect πr 4 2 1 2 Area ∆ BCAD = ( 2 r ) r M1 Expect r (could be embedded) 2 1 2 2 Area segment CFDA = πr − r .oe A1 2 1 2 Area semi-circle CADE = πr B1 2 1 2  1 2 2  Shaded area πr −  πr − r  2  2  2  1 2  1 2 2   or πr −  πr +  πr − r   DM1 Depends on the area ∆ BCD  2  2   = r 2 A1 [6] 2

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8 The first term of a progression is 4x and the second term is x2. (i) For the case where the progression is arithmetic with a common difference of 12, find the possible values of x and the corresponding values of the third term. [4] (ii) For the case where the progression is geometric with a sum to infinity of 8, find the third term. [4]

Mark scheme: 8 (i) x 2 −x4 = 12 M1 4 x −x 2 = 12 scores M1A0 x = − 2 or 6 A1 3rd term = ( −2) 2 + 12 = 16 or 6 2 + 12 = 48 A1A1 SC1 for 16, 48 after x = ,2 − 6 [4] 2 x 2 x  (ii) r = =  soi M1 4 x  4  4 x = 8 M1 Accept use of unsimplified 1 −x x 2 4 x 4 4 or or 4 x x 2 x 4 1 x = or r = A1 3 3 16 3rd term = (or 0.593) A1 27 [4] ALT 4 x 1 4 x = 8 → r = 1 − x or = 8 → x = 21( − r ) M1 1 − r 2 1 − r 2 1  21( − r ) x = 4 x − 1 x  r = M1  2  4 4 1 x = r = A1 3 3

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