DHS_H2_MATHS_P2_Solution

1 2014 Year 6 Prelim Examination Paper 2 Suggested Solution Qn Suggested Solution 1(i) Let P(n) be the proposition:     22 1 11 1 2 112 1 n r rr n n rr n    , n  . LHS of P(1) =   33 12 2  RHS of P(1) =   2 11 1 2 121 1   3 2 LHS of P(1) Hence P(1) is true. Assume P(k) is true for some k  , i.e.     22 1 11 1 2 112 1 k r rr k k rr k    To show P(k + 1) is true, i.e.      221 1 11 1 2 2 112 2 k r rr k k rr k       LHS of P(k + 1) =   2 12 121 kk k   +     2 12 1 12 kk kk    =        22 12 2 2 2 12 2 121 2 kk k k k k kk        =     22 12 2 2 121 2 kk k kk     =    2 12 2 122 kk k    = RHS of P(k + 1) Hence P(k) is true P(k+1) is true. Since P(1) is true and P(k) is true P(k+1) is true, by mathematical induction,     22 1 11 1 2 112 1 n r rr n n rr n    for n  .

2 1(ii)    2 11 1 1 1 11 1 rr rrrr rr r r    By cover-up rule, 1, 1AB    2 11 11 1 1i.e. 11 nn rr rr rrr r r        2 11 1 1 1 11 1 1 11 11 2 11 23 11 1 11 1 nn n rr r n r n r rr rrr r r r nn r n                                  2 11 1 2 11 1 1 11 12 1 1121 1 11 1 1121 1 1 2 nn n rr r rrr rr r r nn nn nn nn nn                      

3 Qn Suggested Solution 2(i) 2 32 1 sin (3 )23 33 34 kx h x x xh x   3 22 13 3 3 34 3 4 khk x xxx     22 22 22 2 22 2 22 12s i n2 ( 3 ) 2 ( ) 2 ( 3 )23 3 682 33 6823 4 38 62 323 33 86 ( s h o w n )23 A xx h x h x xxh x kxxx x x kxx x x kxx x               2 d8 12 3 3d3 A kxxx x   For stationary values, 2 d8 12 3 3 0d3 Ak xxx x     2 3 3 1 3 812 3 3 0 3 9438 8 94 3 8 94 3 kxx x xk kx kx          2 23 d1 6 12 3 3d3 Ak x x  Since 3 0x  , 0k  , 12 3 3 0

4 Alternative   2 23 d1 6 12 3 3d3 94 31612 3 3 38 12 3 3 6 4 3 36 9 3 0 Ak xx k k             area A is a minimum. 2(ii) Using 360 and 300,kA 22 33 3 3 3 8(360)300 6 23 1800 36 9 3 5760 88.185 282.19 0 From GC, since 0, 3.8442 or 6.8587 xx x xx x xx x xx         2 2 When 3.8442, 360 3 (3.8442) 6.46.3(3.8442) 4 When 6.8587, 360 3 (6.8587) 0.419 (rej. 0)3(6.8587) 4 x h x hh          3.84, 6.46.xh 

5 Qn Suggested Solution 3(i) For 42 22 0ww  , 2 2 ( 2) ( 2) 4(1)(2) 1i2w      πi2 4 πi2 π4 1 i = 2e = 2e k w    π1 i π84 7 ππ11 ii 8844 7 π1 i 84 2 e , 0, 1 2e , 2e Since the coefficients of the equation are real, the conjugates of the above roots of a