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

7 B C D r 1 A O r In the diagram, AOB is a quarter circle with centre O and radius r. The point C lies on the arc AB and the point D lies on OB. The line CD is parallel to AO and angle AOC = 1 radians. (i) Express the perimeter of the shaded region in terms of r, 1 and 0. [4] (ii) For the case where r = 5 cm and 1 = 0.6, find the area of the shaded region. [3]

Mark scheme: 7 (i) CD = rcosθ, BD = r – rsinθ oe B1 B1 allow degrees but not for last B1 1 Arc CB = r ( π – θ) oe B1 2 1 → P = rcosθ + r – rsinθ + r ( π – θ) oe B1 sum – assuming trig used 2 [4] 1 1 1 (ii) Sector = .5².( π – 0.6) (12.135) M1 Uses r²θ 2 2 2 1 1 Triangle = .5cos0.6.5sin0.6 (5.825) M1 Uses bh with some use of trig. 2 2 → Area = 6.31 A1 1 (or circle − triangle – sector) [3] 4 4

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10 Relative to an origin O, the position vectors of points A, B and C are given by ` a ` a ` a 2 5 2 −−→ −−→ −−→ OA = 1 , OB = −1 and OC = 6 −2 k −3 respectively, where k is a constant. (i) Find the value of k in the case where angle AOB = 90Å. [2] (ii) Find the possible values of k for which the lengths of AB and OC are equal. [4] −−→ −−→ The point D is such that OD is in the same direction as OA and has magnitude 9 units. The point E −−→ −−→ is such that OE is in the same direction as OC and has magnitude 14 units. −−→ (iii) Find the magnitude of DE in the form n where n is an integer. [4]

Mark scheme:      10 (i) OA = 1 , OB = −1 , OC = 6              −2   k   −3  1 10 – 1 – 2k = 0 → k = 4 M1 A1 Use of scalar product = 0. 2 [2]  3    (ii) AB = −2 , B1      k + 2  OC = 7 (seen or implied) B1 Correct method. Both correct. 3² + (−2)² + (k + 2)² = 49 M1 A1 → k = 4 or −8 [4] Condone sign error in AB (iii) OA = 3  6    OD = 3 OA = 3 and OE = 2 M1 A1 Scaling from magnitudes / unit vector     – oe.  −6   4    OC = 12      −6   − 2    DE = OE − OD = 9 , M1 Correct vector subtraction.      0  → Magnitude of √85. A1 [4] π π

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