RI_DHS_HCI_TMJC_9649_2023_Prelim_P2_Solutions

1(i) Using de Moivre’s theorem,  cos isin cos isin n nn         6 6 4 2 2 4 6 cos6 Re cos isin cos 15cos sin 15cos sin sin                  6 5 3 3 5 sin 6 Im cos isin 6cos sin 20cos sin 6cos sin              5 3 3 5 6 4 2 2 4 6 6 6cos sin 20cos sin 6cos sintan 6 cos 15cos sin 15cos sin sin Dividing both numerator annd denominator by cos ,                  35 2 4 6 6tan 20tan 6tantan 6 1 15tan 15tan tan           (Shown) (ii) Substituting 5   , 35 2 4 6 6 tan 20 tan 6 tan6 5 5 5tan 5 1 15 tan 15 tan tan5 5 5                                             As 6tan tan55           , 35 2 4 6 6 tan 20 tan 6 tan5 5 5tan 5 1 15 tan 15 tan tan5 5 5                                             Since tan 05   , 24 2 4 6 6 20 tan 6 tan551 1 15 tan 15 tan tan5 5 5                                   Letting 2tan 5v   and rearranging, we have

               32 32 32 2 2 9 5 5 0 Let f 9 5 5 f 1 1 9 1 5 1 5 0 1 is a factor 1 10 5 0 1 rej. since tan 0 5 v v v v v v v v v v v v                             or 10 80 2 10 16 5 2 5 2 5 v    Since 22tan tan 154           , 2tan 5 2 55 .  (Shown) 2(a) [2]         cos cos sin sin coscos cos sin cos sin sinsinA rrr rrr                          Hence 1T gives an anti-clockwise rotation of angle  about the origin. (b) [1] It is a reflection in the x-axis. (c) [2] M is obtained when we let 6   in matrix A. Rotating a point about O for 2 radians brings it back to the original position. Since 2 12 , 6   the smallest positive integer 12.k  (d) [3] 1 1 1 1 1 133 102 2 2 2 ,1 1 0 1 1 1332 2 2 2                           C MBM

hence C has eigenvalues 1 and 1. Hence we have 11 1Cx x and 22 1Cx x for eigenvectors 1x and 2,x 1 1 32where 1 2     x and 2 1 2 1 32    x . So the two invariant lines are 1 3 yx and 3yx , which pass through the origin. 3(i) [5] Differentiate d d b gbt  with respect to t,   2 2 d d d d d d d1 d dd1 dd dd dd b g b t t t bg t bb btt bb btt                          Hence   2 2 dd dd bb btt        Solving auxiliary equation,      2 0 0 or                       Hence complementary function is ee ttb A B  Let the particular solution be ,bc then 1cc    

Solution is 1ee ttb A B      (ii) [2] Given that 0.2 and 0.7, and 7.6b , d 1.12d b t  when 0.t  17.6 2.6 0.2 1.12 0.2 0.7 0.7 0.5 1.4, 1.2 A B A B A