2017 A Level H2 FM 9649 P1 (Suggested Solutions)

1 2017 GCE A Level H2 FM 9649 Paper 1 Solutions [Solution] Let 9 1w 9 10w     33 10w       23 3 31 1 0w w w       3 6 31 1 0w w w     3 6 31 0 or 1 0w w w       6 roots of 63 10zz   are among the nine 9th roots of unity. Since 9 10w  has the 9 roots as shown on the Argand diagram and 3 10w  has roots  2 /3 ie  ,  4 /3 ie  and 1. Hence, 6 roots of 63 10zz   are  2 /9 ie  ,  4 /9 ie  ,  8 /9 ie  ,  10 /9 ie  ,  14 /9 ie  &  16 /9 ie  . [Solution] s a c b d    -------- (1) AI ab cd       is an eigenvalue of A  det A I 0      0a d bc      2 0a d ad bc          2 0a d ad s d s a        from (1)     22 0a d ad s a d s ad            22 0a d s a d s            0s s a d      [by long division] ors a d s      s is an eigenvalue of A and the second eigenvalue a d s   ab 1

2 [Solution]   2fe xxx (i)      1 2 2 2f 2 e e 2 1 e x x xx x x          2 2 2 2 2 2 2f 2 e 2e 2e 2 2 2 ex x x xx x x             3 3 2 2 2 2 2 2 2 3 2 2f 2 e 2 e 2 e 2 e 2 3 2 ex x x x xx x x            4 4 2 3 2 3 2 3 2 3 2 4 3 2f 2 e 2 e 2 e 2 e 2 e 2 4 2 ex x x x x xx x x       Conjecture:      12f 2 2 e n n n xx x n  , for all n  (ii) Let Let Pn be the statement:      12f 2 2 e n n n xx x n  , for all n  When n = 1, LHS =      1 2f 2 1 e xxx ; RHS     1 0 2 2= 2 2 e 2 1 e xxxx   = LHS Hence 1P is true. Assume Pk is true for some k  , i.e.,      12f 2 2 e k k k xx x k  Need to show that 1Pk is true, i.e.,      1 12f 2 1 2 e k k k xx x k     LHS =       1 dff d kk xx x     12d 2 2 ed k k xxkx     1 2 22 2 2 e 2 ek k x k xxk      1 2 22 2 e 2 ek k x k xxk     122 1 2 ek k xxk   = RHS Thus 1P is true P is also true.kk  Since 11 is true and is true is true,kkP P P  by mathematical induction, nP is true for all .n 

3 [Solution] T: 2 cos3 θ, 0 θ 2r     (i) 0r No solution. Max r =3, when cos3θ 1 3θ 0, 2 , 4 24θ 0, , 33  Min r =1, when cos3θ 1 3θ , 3 , 5   5θ , ,33  Lines of symmetry: x-axis ( θ 0,  ), The line 4θ , 33 