JJC_H2_MATH_P2_Soln

1 Jurong Junior College 2015 JC2 H2 Mathematics Preliminary Examination Paper 2 Solutions Qn Solution Mark Remarks 1(a)(i) (ii)     arg i arg i arg i i 2 ab ba ab         1(b) i2 0 ( 1 )  zw *3 3 * ( 2 )     zw z w  Sub (2) into (1) i3 * 2 0 3i i * 2 0 Let i 3 i i ( i) 2 ( i) 0 (2 ) i ( 3 2 ) 0          ww ww wx y xy xy yx xy Comparing real and imaginary parts, 20 ( 3 ) 32 0 ( 4 ) Sub (3) into (4) 34 0 1 2 12 i 312 i 4 2 i                yx xy xx x y w z 2 Given   d d  t      25  ,   d d  t =  25k 1 dd25 kt      ln 25    = kt c 25    =  ekt c     25    = e e  ck t  =   e2 5 ktA , where A = e c When t = 0,  = 32 32 =   0e2 5A   A = 7 When t = 1,  = 30 30 = 7e 25  k  5ln 7k  =  ln 5 77e 25  t  = 572 57 t   (Shown) When  = 37, t =   12 7 5 7 ln 1.60ln   (3 sf)

3(i) 3(ii) The time o f r = 32r  ff r r = 323rr  = 323rr  = 323r r = 323rr    2 1  n r r   = = = = =   Let nP be for all n LHS of 1P RHS of 1P 1P is Assume P i.e.,  1 4 k r r   of death is 1 3 23rr 1r 2 2r r   2 2r r   2 32r rr  23 2rr =  1 1 f6   n r r 1 f f6 n   31 236 n n  21 26 nn   1 16 nn  the stateme  . 1P = 43  1P =  11  true. kP is true for  32 3r rr 10.24pm 313r r 32 33r r r 3 266rr  23rr =  f1r r f 0   2n n 31n  21n ent “  3 1 4 n r r   18   3 18  r some k  =  1kk  21 r   213r r 223r r  = 26r     3 23rr   ,  3 . 1    2 21r  63rr  3 1nn  ”  1r   1   ”

3 We want to prove that 1kP is true, i.e.,  32 1 43 k r rr r   =  3 11 1kk    =   3 12kk 1LHS of kP =    3232 1 43 41 31 1 k r rr r k k k     =         332 14 13 1 1kk k k k      22 11 4 1 3 1 1kk k k k      =    2212 1 4 2 1 3 3 1kk k k k k k        =   32 212 4 8 4 3 3 1kk k k k k k       =   3216 1 2 8kk k k   =   21( 2 ) 4 4kk k k   =   3 12kk = 1RHS of kP Since 1P is true, and kP is true  1kP is true,  by mathematical induction, nP is true for all n  .   3(iii)  32 1 43 n r rr r   =  3 1nn  32 11 1 43 nn n rr r rr r     =  3 1nn  3 1 4 n r r   =  3 2 11 13 nn rr nn r r     =      3 113 1 2 1 162 nnn nn n n      =    2 1111 2 1 22nn n n    =   212 1 1nn n n n    =    21nn n n  =    11nn n n =  2 1nn    3 1 n r r   =  21 14 nn    or  2 1 12 nn   

4(a) (i) (ii)   OR  = 23 5 pq d = OR OP OP   

 = 23 5 pq p p

= 23 5 pp qp p = 2 23 5 pp q p