RI P2 MS

2012 RI H2 Mathematics Preliminary Examination Paper 2 Qn Solution 1 [5] Let Pn be the statement 1 1sin sin 22cos , 2sin 2 n r n rn        . When 1n , LHS cos  3sin sin22RHS 2sin 2 1 3 1 32cos sin2 2 2 2 2 2 2sin 2 2cos sin 2 cos LHS 2sin 2                                  Hence 1P is true. Assume Pk is true for some k  , i.e. 1 1sin sin 22cos 2sin 2 k r k r       . To prove 1Pk is true, i.e. 1 1 3sin sin 22cos 2sin 2 k r k r        .

2 1 LHS cos cos( 1) 1sin sin 22 cos( 1) 2sin 2 1sin sin 2cos( 1) sin2 2 2 2sin 2 1sin sin sin ( 1) sin ( 1)2 2 2 2 2sin 2 1sin sin sin22 k r rk k k kk k k k kk                                                      31 sin22 2sin 2 3sin sin 22 RHS 2sin 2 k k                  Hence Pk is true implies 1Pk is true. Since 1P is true, and Pk is true implies 1Pk is true, by Mathematical induction, Pn is true for all n  2(i) [6] d ,0d x kx kt   1 ddx k tx   ln , 0x kt c x    e ktxA  When 0t , 80x , thus 80A . When 3t , 20x , 320 80e k 1 1 1ln ln 43 4 3k   . Thus ln 4380e t x   2(ii) [3] 2ln 4 1When 6, 80e (80) 16tx    Just before the (n+1)th injection, the amount of drug present in the blood stream = 1 16 nu

3 1 Immediately after the ( +1)th injection, 1amount of drug present, 80 16 nn n uu  [1] In the long run, the amount present approaches the value 85.3 (3 s.f.). 3a [3] 3 1 0 22 / 7 1 4 9 4 5 3 2 cc                                           223 (1) 7 8 4 4 (2) 5 3 2 (3)cc                   From (1) and (2) we have 22 3 (4)7 4 4 8 (5)         Solve (4) and (5) by GC 13 7,33  Sub into (5)  13 75 3 2 433 c c c                . [3] Let the foot of perpendicular be F . Then     1 40 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                                                           The coordinates of F are  4, 5, 8 . 3b [2] 1cd 3bi [3] (i) Shortest distance is

4 22 22 0 3 1 31 9 1 4 84 1 5 3 43 261 4 3 40 1 5 26 20 5 8 1 4 45 26 26                                                          3bii [2] 8 0 8 1 9 1 4 1 4 8 15 4                                    r r 3bii i [2]     1 2 : 8 4 5 : 8 4 2 5 8 4 3 8 13 4 p x y z p x y z x y z              r 4(a) [2] 6 3 1 , 2, 1,0,1, 2,3 ki z z e k       4bi [2] 1kkP OP = 2 n  .