IJC_H2_MATH_P1_(9740)_Solutions

1 Let nP denote     0 ! 1 ! 1 n r r r r     for , 0nn .     0 When 0, LHS 0 0! 0 RHS 0 1 ! 1 0 LHS Therefore P is true. n               0 1 1 0 Assume P is true for some , 0, i.e. ! 1 ! 1 Want to prove that P is true, i.e. ! 2 ! 1 k k r k k r kk r r k r r k                                    1 0 0 LHS ! ! 1 1 ! 1 ! 1 1 1 ! 1 ! 1 1 1 1 ! 2 1 2 ! 1 RHS k r k r rr r r k k k k k kk kk k                              1 01 Thus P is true P is true. Since P is true, and P is true P is true, by mathematical induction, P is true for all , 0. kk kk n nn      2 (i)                   2 24 e cos d e cos e sin( ) d e cos e sin e cos( ) d e cos sin e cos d x n xx nn x x x n n n xx nn nx x n nx n n nx x n nx n n nx n n nx x n nx n nx n nx x                               42 2 4 1 e cos d e cos sin e cos d e cos sin 1 xx nn xx nn n nx x n nx n nx nnx x nx n nx C n          

(ii)                        2 22 2 4 4 4 4 4 e cos d e cos sin 1 e cos 2 e cos 1 e e cos 1 e e 1 if is even 1 e e 1 if is odd 1 xx nn nn nn nn nn nnx x nx n nx n n nn n n n n n n n n n n                        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 ins