2020 A Level H2 FM 9649 P2 (Qns & Solutions) (2 Oct 2024)

Possible Solutions to 2020 FM Paper 2 [Solution] 1 The system of equation can be represented by 1 2 3 4 5 1 1 2 1 5 0 1 3 2 1 1 0 1 0 1 1 0 0 0 1 1 1 2 0 x x x x x −−        −− =            −−     It can be reduced to 1 2 3 4 5 1 0 1 0 3 0 0 1 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 x x x x x         −− =    −             1 3 5 2 3 5 45 30 0 0 x x x x x x xx + + = − − = −= The solution is 1 3 5 2 3 5 3 3 3 5 45 55 3 13 11 10 01 01 x x x x x x x x x x xx xx −− −−               +       = = +                              The two vectors 1 1 1 0 0 −      and 3 1 0 1 1 −      are linearly independent. A basis is 13 11 ,10 01 01  − −                            (ii) dimension = 2

[Solution] (i) Considering z7 = 1 z7 = 1 = 2eik  where k = 0, 1, 2 and 3  2 7e ki z  = 7 6 5 4 1... 1 1 zz z z z z −+ + + + = − , ( ) 2 2 4 4 6 6i i i i i i7 7 7 7 7 7 71 1 e e e e e ez z z z z z z z      − − −      − = − − − − − − −              ( ) 2 2 4 4 6 6i i i i i i222 7 7 7 7 7 71 e e +1 e e +1 e e +1z z z z z z z      − − −        = − − + − + − +                    ( ) 222 2 4 61 2 cos +1 2 cos +1 2 cos +17 7 7z z z z z z z      = − − − −       7 6 5 4 1... 1 1 zz z z z z −+ + + + = − = 222 2 4 6( 2 cos 1)( 2 cos 1)( 2 cos 1)7 7 7z z z z z z   − + − + − + (i) [Method 1] Comparing the coefficient of the z term: 1 = 2 4 62cos 2cos 2cos7 7 7   − − − 2 4 6 1cos cos cos7 7 7 2   + + =−  2 3 1cos cos cos7 7 7 2   − − =− using cos( − ) = −cos .  2 3 1cos cos cos7 7 7 2   − + = [Method 2] The sum of roots of the equation 6 5 4 ... 1 0z z z z+ + + + = is −1 = 2 2 4 4 6 6 7 7 7 7 7 7e e e e e e i i i i i i     − − − + + + + +

 2 4 61 2cos 2cos 2cos7 7 7   − = + +  2 3 1cos cos cos7 7 7 2   − + = (ii) [Method 1]: Let z = i in 7 6 5 4 2 2 2 1 2 4 6... 1 ( 2 cos 1)( 2 cos 1)( 2 cos 1)1 7 7 7 zz z z z z z z z z z z   −+ + + + = = − + − + − +− 7 2 2 2 2 4 61 ( 1)( 2 cos 1)( 2 cos 1)( 2 cos 1)7 7 7i i i i i i i i   − = − − + − + − + −i – 1 = (i – 1) 2 4 6( 2 cos )( 2 cos )( 2 cos )7 7 7iii   −−− −i – 1 = 8i (i – 1) 2 4 6cos cos cos7 7 7     2 4 6 1cos cos cos7 7 7 8    =  2 3 1cos cos cos7 7 7 8    = using cos( − ) = − cos, [Method 2]: Comparing the coefficient of z3 term in