Cambridge A Level Mathematics 9709 — 2016 May/June Paper 1 · Variant 2

5 A 1 130 B C x M x In the diagram, triangle ABC is right-angled at C and M is the mid-point of BC. It is given that angle ABC 1 radians and angle BAM radians. Denoting the lengths of BM and MC by x, = 30 = 1 (i) find AM in terms of x, [3] @ A 1 1 (ii) show that . [2] 1 = 60 −tan−1 2ï3

Mark scheme: f ( ) ( 5 − 2 x ) 2 ( ) 1

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6 O ! rad r cm Q T P The diagram shows a circle with radius r cm and centre O. The line PT is the tangent to the circle at P and angle POT radians. The line OT meets the circle at Q. = ! (i) Express the perimeter of the shaded region PQT in terms of r and [3] !. (ii) In the case where 1 and r 10, find the area of the shaded region correct to 2 significant ! = 30 = figures. [3]

Mark scheme: 6 (i) PT = r tanα B1 r QT = OT – OQ = − r cosα or r ² + r ²tan²α − r B1 Perimeter = sum of the 3 parts including rα B1 [3] π (ii) Area of triangle = ½ × 10 × 10 tan M1 Correct formula used, 50 3, 86.6 3 50π Area of sector = ½ × 10² × ⅓π M1 Correct formula used, , 52.36 3 Shaded region has area 34 (2sf) A1 [3]

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10 y 8 y = + 2x x M x O 8 The diagram shows the part of the curve y 2x for x 0, and the minimum point M. = x + > dy d2y (i) Find expressions for dx, dx2 and Ó y2 dx. [5] (ii) Find the coordinates of M and determine the coordinates and nature of the stationary point on the part of the curve for which x 0. [5] < (iii) Find the volume obtained when the region bounded by the curve, the x-axis and the lines x 1 = and x 2 is rotated through about the x-axis. [2] = 360Å

Mark scheme: ff(x) = 10 − 3 (10 − 3x ) B1 Correct unsimplified expression 10 gf(2) = (= −2) B1 Correct unsimplified expression ( ( ) )

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