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1. h = r so 2r2 = 100 r2 = 50 l = 10 = 2r 2π 50 = 10 2 π5 2 = 10 = 2 = 4.44 (3s.f.) Note: Accept either answer. (M1) (M1) (A1) (A1) [4] 2. (a) (b) 1 2 1 r (152)(2) 2 2 2 = 225 (cm ) Area = 1 2 15 sin 2 = 102.3 2 Area = 225 – 102.3 = 122.7 (cm2) = 123 (3 s.f.) Area ∆OAB = (M1) (A1) (C2) (A1) (A1) (C2) [4] 3. (a) l = r (b) AÔB (obtuse) = 2 – 2 1 1 Area = r 2 = (2 – 2) (15)2 2 2 or ACB = 2 × OA = 30cm = 482 cm2 (3 s.f.) (M1) (A1) (C2) (A1) (M1)(A1) (A1) (C4) [6] 4. Note: Do not penalize missing units in this question. (a) AB2 = 122 + 122 – 2 × 12 × 12 × cos 75° (A1) = 122(2 – 2 cos 75°) (A1) 2 = 12 × 2(1 cos 75°) AB = 12 2(1 cos 75) (AG) 2 Note: The second (A1) is for transforming the initial expression to any simplified expression from which the given result can be clearly seen. 1 (b) PÔB = 37.5° BP = 12 tan 37.5° = 9.21 cm (A1) (M1) (A1) OR BP̂A = 105° BÂP = 37.5° AB BP sin105 sin 37.5 ABsin37.5 BP = = 9.21(cm) sin 105 (c) (i) (ii) (d) (e) 1 1 12 9.21 or 12 12 tan 37.5 2 2 = 55.3(cm2) (accept 55.2 cm2) Area ∆OBP = 1 (9.21)2 sin105° 2 = 41.0(cm2) (accept 40.9 cm2) Area ∆ABP = 1 π 75 12 2 75 π 12 2 or 2 180 360 2 = 94.2 (cm ) (accept 30π or 94.3 (cm2))) Area of sector = Shaded area = 2 × area ∆OPB – area sector = 16.4 (cm2) (accept 16.2 cm2, 16.3 cm2) (A1) (M1) (A1) 3 (M1) (A1) (M1) (A1) 4 (M1) (A1) 2 (M1) (A1) 2 [13] 5. sin A = 5 12 cos A = 13 13 But A is obtuse cos A = – (A1) 12 13 sin 2A = 2 sin A cos A 5 12 =2× 13 13 120 =– 169 (A1) (M1) (A1) (C4) [4] 6. (a) 3sin2 x + 4cos x = 3(1 – cos2 x) + 4cos x = 3 – 3cos2 + 4 cos x (A1) (C1) 2 (b) 3sin2 x + 4 cos x – 4 = 0 3 – 3cos2 x + 4 cos x – 4 = 0 3cos2 x – 4cos x + 1 = 0 (A1) (3cos x – 1)(cos x – 1) = 0 1 cos x = or cos x = 1 3 x = 70.5° or x = 0 (A1)(A1) (C3) Note: Award (C1) for each correct radian answer, i.e. x = 1.23 or x = 0 [4] 7. (a) (b) (i) AP = ( x 8) 2 (10 6) 2 x 2 16 x 80 (ii) OP = ( x 0) 2 (10 0) 2 x 2 100 AP 2 OP 2 OA 2 2AP OP 2 ( x 16 x 80) ( x 2 100 ) (8 2 6 2 ) (A1) cos OP̂A (M1) = (M1) = 2 x 2 16 x 80 x 2 100 2 x 2 16 x 80 2 x 2 16 x 80 x 2 100 cos OP̂A (c) (M1) (AG) x 2 8 x 40 {( x 2 16 x 80)( x 2 100 )} For x = 8, cos OP̂A = 0.780869 arccos 0.780869 = 38.7 (3 s.f.) 2 (M1) (AG) 3 (M1) (A1) OR 8 10 OP̂A = arctan (0.8) = 38.7 (3 s.f.) tan OP̂A (d) (A1) 2 OP̂A = 60° cos OP̂A = 0.5 0.5 = x 2 8 x 40 {( x 2 16 x 80)( x 2 100)} 2x2 – 16x + 80 – x = 5.63 (e) (M1) (i) {( x 2 16 x 80)( x 2 100 )} = 0 f(x) = 1 when cos OP̂A = 1 hence, when OP̂A = 0. This occurs when the points O, A, P are collinear. (M1) (M1) (G2) 4 (R1) (R1) (R1) 3 (ii) The line (OA) has equation y = 3x 4 (M1) 40 (= 13 13 ) 3 When y = 10, x = (A1) OR 40 (= 13 13 ) 3 Note: Award (G1) for 13.3. x= (G2) 5 [16] 8. (a) x is an acute angle => cosx is positive. 2 2 cos x + sin x = 1 => cosx = => cos x = = (b) 1 1– 3 8 9 (= (M1) 1 – sin x 2 (M1) 2 (A1) 2 2 ) 3 1 3 (A1) (C4) 2 cos2x = 1 – 2sin2x = 1 – 2 = (M1) 7 9 (A1) (C2) Notes: (a) Award (M1)(M0)(A1)(A0) for. 1 cos sin –1 = 0.943 3 (b) 1 Award (M1)(A0) for.cos 2 sin –1 = 0.778. 3 4