IJC_H2_MATH_P1_(9758)_Solutions

1 2 2 22 2 4 7 1 231 4 7 1 ( 2)(3 1) 031 4 7 1 (3 6 2) 031 1 031 ( 1)( 1) 031 xx xx x x x x x x x x x x x x x xx x                    1x  or 1 13 x   4 7 1 2 31 xx x x    Replace x with √𝑥, 1 x  or 1 13 x   (rejected as 0x  ) Since 0x  , 1 1 0 13 01 xx x        2 (i)                              1 1 2 1/21 2 2 2 1/222 1 1 2 2 cos d cos d 1 1cos 2 1 d 2 11cos 12 2 cos 1 n nx x nnx nx nx x nx nx nx n x n x x nxnx nx C nx nx n x C                          + − − + 1 1

(ii)           1 12 0 1 21 2 2 0 1 cos d cos 1 1 1 1cos 1 0 12 2 4 3 2 316 2 6 2 n n n nx x nx nx n x or                 3 (i) (2 5 ) (2 5 ) 4 10 10 25 20 2 20 1 1 0 20 2 b a a ab b                                   p q p q p p p q q p q q pq Alternative:   22 (2 5 ) (2 5 ) 2 1 5 1 2 1 5 1 00 4 5 4 5 37 22 6 14 8 10 8 10 28 35 12 15 20 20 40 20 bb aa bb aa aa a ab a ab bb a ab b                                                                                           p q p q 20 2 a ab b            Given that the i- and j- components of the vector 20 2 a ab b     are equal, 0 ( 1) 0 Since 0, thus 1 a ab ab a ab ab     

(ii) 2 2 2 (2 5 ) (2 5 ) 80 20 80 2 4 21 2 9 4 2 9 16 7 2 7 14 22 a ab b a a a a a a or                 p q p q (iii) Since 25pq and 25pq are perpendicular,        22 22 2 2 2 5 2 5 0 4 25 0 25 4 25 114 25 2 52 2           p q p q pq pq p Alternative: 2 2 4 5 4 5 (2 5 ) (2 5 ) 3 7 22 16 25 21 4 4 30 aa a a                           p q p q Since 25pq and 25pq are perpendicular, 2 2 (2 5 ) (2 5 ) 0 4 30 0 15 2 a a      p q p q 22 15 25 5 22 1 5 2 2 2a      p

4 (a) Method 1 Since the coefficients are real, 2iw is another root of the equation.        22 2 2 2 i 2 i 2 i 4 4 1 45 w w w ww ww               32 30 0w pw qw      2 4 5 6 0w w w    (By inspection) Comparing coefficients of 2w , 6 4 2p   Comparing coefficients of w , 24 5 19q   Method 2 Substitute 2iw ( or 2iw ) into the given eqn, 32(2 i) (2 i) (2 i) 30 0 (3 4i)(2 i) (3 4i) (2 i) 30 0 (6 3i 8i 4) (3 4i) (2 i) 30 0 (32 3 2 ) ( 11 4 )i 0 pq pq pq p q p q                                Comparing the real parts, 3 2 32pq  --- (1) Comparing the imaginary parts, 4 11pq  ---- (2) (1)  (2)  2: 3 8 32 11 2 5 10 2 pp p p        From (2):