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 B A A B                Hence 0.2 0.71.4e 1.2e 5ttb    (iii) [2] Using GC, when 0.2 0.7 1.4 e 1.2e 5 5.69, tt    4.07hours (3s.f.)t 4(i) (ii)   cot cot e cos d e cot cos sin 0d tan cot 2 x x               1 cote

(iii)         cot cot cot cot e sin d e cot sin cosd e cot sin cosd d e cot cos sin cot tan 1 cot tan tan tan 1 tan tan tan y y y x                             Acute angle between the tangent at P and the line OP = 22       . OR Acute angle between the tangent at P and the line OP =        . (iv) Total length of arc

2 2 0 2 cot 2 cot 2 0 cot 2 0 2 cot 2 0 2 cot 22 0 d dd e e cot d e 1 cot d tan 1 e tan cot tan 1 etan cot rr                              cotsec e 1   OR Total length of arc       22 0 222 cot 2 cot 0 cot 2 2 2 2 2 0 cot 2 0 2 cot 2 0 2 cot 22 0 dd ddd e cot cos sin e cot sin cos d e cot cos sin cos sin d e cot 1 d tan 1 e tan cot tan 1 etan cot xy                                                             cotsec e 1  

5(a) Improved Euler with step size 2h  ,     2g , cos 2x y x x xy 11 0, 0,xy    2 1 1 1 g , 0 0 0 2u y h x y            2 1 1 1 2 2 2 g , g ,2 0 0 cos 2 04 2 2 2 hy y x y x u                          2 2 2 cos82y        22 2 2 4 2 2 3 2 2 2 2 2 2 2 2 2 2 g, cos cos 2 cos8 2 2 2 2 2 8 2 cos cos cos8 2 4 2 16 2 cos 616 2 u y h x y                                                                        3 2 2 2 3 3 g , g ,2 hy y x y x u        2 2 2 2 22 2 2 2 cos82 cos 2 cos2 2 2 8 2 4 cos 2 cos 6 16 2                                               2 2 42 28cos 8 cos32 7 2         (b) (i)     2 2 22 22 0 1 0 12 00 12 0 ed ed 11e e 1 d22 11 e 1 e d22 nx n nx n x x n n n x I x x x x x x n x x n x x                         21 2 11 e 1 (shown)22 n nnI n I      

(ii) 22 5 3 d 12d2 d 2d4 yx x xyx yx xy x xx        2 22 2 2 2 2 2 2 2 2 d 5 3 35 0 1 3 5 4 1 3 3 4 13 2 2 4 1 0 1 d4 1( ) (0)e d 4 Since (0) 0, 1() 4 11 e242 31 e28 3 1 1 ee Integrating Factor: e e e e 2 e e e e 28 e x x xx x x xx xy x x x y y x x x y y I I I I I I I x I II                                         2 2224 1 5 3 1 ee2 4 8I     22 1 0 11Since d e , e 22     xIx x 2 2 2 2 2 2 2 24 24 e 5 1 1 3 1( ) e e e2 2 2 4 8 5 5 3 1e e e4 4 4 8 y                   2 245 5 3 1( ) e 4 4 4 8y         Substituting 0.1  , for part (a),        2 2 4 22 0.1f 0.1 8cos 0.1 8 cos32 2 0.00495, 0.1 0.1 7 0.1          for part (b)(ii),     2 240.15 5 3 1f (0.1) e 0.1 0.1 0.004954 4 4 8      Both approximations are comparable. However, to obtain the estimate given by the small angle approximation, it requires the solving of the differential equation, whereas the improved Euler method does not.

Section B: Probability and Statistics [50 marks] 6 Given ˆ ˆ ˆ ˆˆˆ 1.9600 , 1.9600 (0.229,0.371)A A A A AA p q p qpp nn     , 0.229 0.371ˆ 0.32 Ap  0.3 0.7 0.371 0.2291.9600 0.071 2 160.03 160 n n    Based on the 2 samples,  160 0.3 36 84 21ˆ 160 100 260 65p     A symmetric % confidence interval for p = ˆ ˆ ˆ ˆˆˆ ,pq pqp z p z nn   21 44 65 652 0.1 260 1.7240 z z    100 P( 1.7 2 4 0 1.7 2 4 0) 91.5 2 9 Z        Largest possible value of 91.5  (1 d.p.) 7 0H: The two factors, level of reading activity and literacy skills, are independent. 1H: The two factors, level of reading activity and literacy skills, are not independent. Perform a 2 test on independence. Observed Frequencies (Oij) Level of Reading Activity High Medium Low Total Good 25 12 7 44 Literacy Skills Average 35 57 27 119 Poor 9 13 20 42 Total 69 82 54 205

Under 0H , Expected Frequencies (Eij) Level of Reading Activity High Medium Low Total Good 14.81 17.6 11.59 44 Literacy Skills Average 40.05 47.6 31.35 119 Poor 14.14 16.8 11.06 42 Total 69 82 54 205 Contributions to  2 cal   2 11 mn ij ij ij ij OE E     : Level of Reading Activity High Medium Low Literacy Skills Good 7.01167 1.78182 1.81794 Average 0.63763 1.85630 0.60264 Poor