# 2012 RI H2 Math Prelim Paper 2 (Solutions).pdf - 2012 RI H2...

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2012 RI H2 Mathematics Preliminary Examination Paper 2 Qn Solution 1 [5] Let P n be the statement 1 1 sin sin 2 2 cos , 2sin 2 n r n r n . When 1 n , LHS cos 1 3 1 3 3 2cos sin sin sin 2cos sin 2 2 2 2 2 2 2 2 2 RHS cos LHS 2sin 2sin 2sin 2 2 2 Hence 1 P is true. Assume P k is true for some k , i.e. 1 1 sin sin 2 2 cos 2sin 2 k r k r . To prove 1 P k is true, i.e. 1 1 3 sin sin 2 2 cos 2sin 2 k r k r . 1 1 sin sin 2 2 LHS [cos cos( 1) ] cos( 1) 2sin 2 1 sin sin 2cos( 1) sin 2 2 2 2sin 2 1 sin sin sin ( 1) sin ( 1) 2 2 2 2 2sin 2 1 sin sin sin 2 2 k r k r k k k k k k k k 3 1 sin 2 2 2sin 2 3 sin sin 2 2 RHS 2sin 2 k k k Hence P k is true implies 1 P k is true. Since 1 P is true, and P k is true implies 1 P k is true, by Mathematical induction, P n is true for all n
2 2(i) [6] d , 0 d x kx k t   1 d d x k t x ln , 0 x kt c x   e kt x A When 0 t , 80 x , thus 80 A . When 3 t , 20 x , 3 20 80e k 1 1 1 ln ln 4 3 4 3 k   . Thus ln4 3 80e t x 2(ii) [3] 2ln4 1 When 6, 80e (80) 16 t x Just before the ( n +1)th injection, the amount of drug present in the blood stream = 1 16 n u 1 Immediately after the ( +1)th injection, 1 amount of drug present, 80 16 n n n u u [1] In the long run, the amount present approaches the value 85.3 (3 s.f.). 3(a) [3] 3 1 0 22/ 7 1 4 9 4 5 3 2 c c   22 3 (1) 7 8 4 4 (2) 5 3 2 (3) c c                From (1) and (2) we have 22 3 (4) 7 4 4 8 (5)         Solve (4) and (5) by GC 13 7 , 3 3   Sub into (5) 13 7 5 3 2 4 3 3 c c c .
3 [3] Let the foot of perpendicular be F . Then 1 4 0 3 3 1 0 1 1 4 9 4 0 5 3 4 3 3 4 15 1 16 9 36 12 0 1 AF  