DHS_H2_MATHS_P1_Solution

1 2014 Year 6 Prelim Examination Paper 1 Suggested Solution Qn Suggested Solution 1(i) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 800 7 21 39 20300 5 23 49 23900 ms t mst mst    Using GC, 200, 250, 350.mst Therefore the fisherman has 250 kg of salmon. (ii) m : weight of mackerel in kg s : weight of salmon in kg t : weight of tuna in kg 600 7 21 39 20300 5 23 49 23900 ms t mst mst    Using GC, 460, 990, 70.ms t   Since the weight of all fishes must be non-negative, the fisherman’s claim is not possible. Or Since the weight of salmon and tuna is more than 600kg, the fisherman’s claim is not possible.

2 Qn Suggested Solution 2(a) 2(b) 2 2(1 ) (2 ) 14 x y   Making y the subject of formula: 2(1 )21 4 xy   Let the volume of solid generated when the curve 2(1 )21 4 xy   is rotated about x-axis from 2 to 3xx be V1. V1 = 3 2 2 π dyx 2 23 2 (1 )π 21 d 4 x x     = 21.593 Let the volume of solid generated when the curve 2(1 )21 4 xy   is rotated about x-axis from 2 to 3xx be V2. V2 = 3 2 2 π dyx 2 23 2 (1 )π 21 d 4 x x     = 6.1573 Volume of required solid = V 1 – V2 = 15.4 (to 3sf) 15, 2   15, 2  O y x x = 2

3 Qn Suggested Solution 3(i) 2 2 2 2 2 2 2 e1 32 Differentiate with respect to , de3 4d Differentiate again with respect to , ddee 4dd dd 4e (shown)dd y y yy y xx x y xx x yy xx yy xx          3(ii) 2 2 2 dd 4edd yyy xx   Differentiating again with respect to x, 23 23 dd d d24 edd d d yy yy y x xx x         2 2 3 300 0 3 3 3 dd when 0, 0, 3, 5, dd de3 3 e3 5 e 0 d d 18d yyxy xx y x y x          23 2 351 8 503 3 3 2! 3! 2yx x xx x x        3(iii) 21 3 2 1.0302 1.51 (reject) or 0.01 xx xx           22 3 22 3 5ln 1 3 2 3 3 2 Using 0.01, 5ln 1 3 0.01 2 0.01 3 0.01 0.01 3 0.01 2 ln 1.0302 0.0298 (4 d.p.) yx x x x x x           Qn Suggested Solution 4(i) A C B 2 1

4 Qn Suggested Solution 1 12 0 23021 133 3 3 p p OC            ab  Area of triangle OAC = 1 2 OA OC     = 211 1123 33 pp           = 0012 63 3 366 2 1 pp pp             10 3 4(ii) Triangle OAD, triangle ADE and triangle OAC have the same height and base and thus they have the same area. Area of trapezium OAED = 1031 03   4(iii) 2 2 22 2 1 10 30cos135 10 1 1 2 10 10 2 10 10 (reject 10 since 0) p p p p pp p pp               ab

D C O 1 1 1 A E

5 Qn Suggested Solution 5(i)  πi 6 πi i6 πi 6 3i2 e 23 i 22 e e 4e 4 πarg 66 2 wz r r wr w                   