ACJC 9649 2023 Prelim P2 Solutions

1 Let P( )N be the statement 1 3 1 1 2( 1)( 2)2 ( 2)2 N nN n n n n N     for integers 1N  . When 1N  : LHS  1 3 1 (1 1)(1 2)(2) 3   RHS  1 1 1 1 12 (1 2)(2) 6 3    LHS  RHS, hence P(1) is true. Assume P( )k is true for some 1k  , i.e. 1 3 1 1 2( 1)( 2)2 ( 2)2 k nk n n n n k     . Claim P( 1)k is true, i.e. 1 1 1 3 1 1 2( 1)( 2)2 ( 3)2 k nk n n n n k        . Proof: LHS 1 1 3 ( 1)( 2)2 k n n n nn     1 1 34 ( 1)( 2)2 ( 2)( 3)2 k nk n nk n n k k        1 1 1 4 2 ( 2)2 ( 2)( 3)2kk k k k k        1 ( 3)(2) ( 4)1 2 ( 2)( 3)2 k kk kk      1 1 2 6 4 2 ( 2)( 3)2 k kk kk      1 12 2 ( 2)( 3)2 k k kk    1 11 2 ( 3)2 kk   RHS Hence P( )k is true  P( 1)k is true. Since P(1) is true, and if P( )k is true then P( 1)k is also true, then by mathematical induction, P( )N is true for all positive integers 1N  . 1 3 1 1 1 lim 22( 1)( 2)2 ( 2)2nN Nn n n n N         2(i) The distributive axiom ()c c c  u v u v is violated. 1 1 1 1 2 2 2 2 () u v cu vcc u v cu v              uv

2 11 11 2 22 22 cu cv c u vcc cu cv c u v                uv So ()c c c  u v u v in general Alternative The axiom ()c d c d  u u u is violated. 1 2 ()() () c d ucd c d u    u 2 11 1 2 22 2 cu du cducd cu du cdu                uu So ()c d c d  u u u in general 2(ii)(a) Let 1A and 2A be matrices such that 11 A B BA and 22 A B BA  1 2 1 2 1 2 1 2 ()      A A B A B A B BA BA B A A        1 1 1 1k k k k   A B A B BA B A The set is closed under addition and scalar multiplication. Also, the set is non-empty since 0B B0 0 . Hence it is a subspace. 2(ii)(b) T I I I but TT(2 ) (2 ) 4( ) 4  I I I I I I The set is not closed under scalar multiplication. Hence it is not a subspace. 3(i) 2 2 3dxIx   Let f ( ) 3 xx  and 2 ( 2) 14h  n nt f ( )nnyt  0 2 1 9 1 1 1 3 2 0 1 3 1 3 4 2 9 Let T denotes the approximation to 2 2 3dxIx   , found using trapezium rule with 5 ordinates.  0 1 2 3 42 2 22 hT y y y y y     --- (1) 88 9T  3(ii) f ( ) 3 xx    f ( ) ln 3 3 xx 

    2 f ( ) ln 3 3 0 xx  for 22 x   f ( ) 3 xx  is concave upwards over the interval  2, 2 Trapezium rule produces an overestimate T to 2 2 3dxIx   . 3(iii) Let S denotes the approximation to 2 2 3dxIx   , found using Simpson rule with 5 ordinates.  0 1 2 3 4 1 4 2 43S h y y y y y     --- (2) 48 27S  3(iv) 2 2 2 2 22 3d 1 3ln 3 1 33ln 3 80 1 9 ln 3 x x Ix            3(v) Numerical integration using the Simpson rule produces a more accurate approximation compared to the Trapezium rule, with the same number of ordinates. The Simpson rule makes use of a quadratic approximation as opposes to the Trapezium rule which makes use of a linear approximation. Hence Simpson rule uses a better approximation to the curve 3xy . 3(vi) Absolute percentage error 100% 0.706%IS I    4(a) Differentiate (1) with respect to x: 2 2 dd d4 2 0ddd yy z xxx    From (2), d 5 16d z y z xx       2 2 dd 4 2 5 16 0dd yy y z xxx       2 2 dd 4 2 10 32 0dd yy y z xxx      From (1), d2 4 8d yzy x    2 2 d d d 4 2 5 4 8 32 0ddd y y y y y xxxx       

 2 2 d d d 4 2 5 20 40 32 0ddd y y y y y xxxx         2 2 dd 9 18 32 40dd yy yxxx     (shown) 4(b) Auxiliary equation: 2 9 18 0mm    ( 3)( 6) 0mm    6m or 3 Complementary function: 63ee xxy A B  for arbitrary constants A, B For particular integral, let y cx d  d d y cx  , 2 2 d 0d y x  Substitute into DE: 0 9 18( ) 32 40c cx d x     Comparing coefficients: x: 18 32c  16 9c 0x : 9 18 40cd  40 16 4 18 3d   General solution for y is 63 16 4ee 93 xxy A B x    63d 166 e 3 ed9 xxy ABx    Substitute into (1): 6 3 6 3 16 16 46 e 3 e 4 e e 2 8 9 9 3 x x x xA B A B x z               63 64 82 e e 2 99 xxA B x z      63 32 4ee 2 9 9 xx Bz A x     Sub. 0x , 0y : 4 03AB    4 3AB  ............. (3) Sub. 0x , 0z : 4 029 BA     82 9AB   ......... (4) Using GC to solve (3) and (4), 20 27A , 16 27B  Solutions for y and z are: 6320 16 16 4ee27 27 9 3 xxyx     6320 8 32 4ee27 27 9 9 xxzx     5(a) 42 1 3iv 

πi4 32 2ev  πi2 π4 3e k v   , where k 1 πi2 π43e k v     ππi 12 2e k v   For arguments in the principal range, choose 0, 1, 2k     11π 5π π 7πi i i i12 12 12 12e , e , e , ez                       5(b) Let ie cos isinpw p p   where π 12p . By De Moivre’s Theorem, for any positive integer n, 1 cos isin cos( ) isin( ) cos isin cos isin 2cos (shown) n n n nw w ww np np np np np np np np np               12cos pw w    4 4 12cos pw w   416cos p 42 42 11 46ww ww                2cos 4 4 2cos 2 6pp   ππ2cos 8cos 636   132 8 622      7 4 3  4 7 4 3cos 16p  (shown) 5(c) 1zw Locus is a circle centred at the origin O with radius 1 unit. πarg( ) 3wz    πarg ( ) 3zw    πarg( 1) arg( ) 3zw     πarg( ) π3zw    2πarg( ) 3zw 

Locus is a half-line starting from (and excluding the point A representing w), at an argument of 2π 3 rad. 5(d) Triangle OAP is isosceles triangle. 2πππ 3 12 π 4 OAP OPA      ππ2 4 π 2 AOP      The argument of the complex number represented by P is π π 5π 12 2 12  Since P lies on the circle centred at origin with radius 1, the complex number represented by P is 5πi 12e   , which is one of the roots of the equation in (a). 6(a) 7 9 sp  , 0.95 1.645z  , 900n Confidence interval     2211 11 , s s s s ss P P P PP z P z nn      7 7 7 71177 9 9 9 91.645 , 1.6459 900 9 900                 Re(z) Im(z) O A 1 P 1 12 π

 0.7550, 0.8006 Answer is 4 dp as interval width is 0.0456 to 3 sf 6(b)   21 1 0.02ssPPz n    771991.645 0.02 771991.645 0.02 771991.645 0.02 n n n       1169.27 1169n  7 H0: the number of heads obtained follows a Binomial distribution with 0.6p . H1: the number of heads obtained does not follow a Binomial distribution with 0.6p . Level of significance: 5% No of heads 0 1 2 3 4 Frequency 5 35 64 66 30 Expected frequency 5.12 30.72 69.12 69.12 25.92 Degree of freedom is 4.     2 22 4 0 ~4ii ii OE E   By GC, the p-value is 0.780 > 0.05. Hence, we do not reject H 0 and conclude at 5% significance level that a Binomial distribution with 0.6p is a good fit. If the experiment is repeated 1000 times, the new 2 value will be 5 8.807 9.4881.7614   and so there is no change to the result of the test. 8(a) We may not be able to assume that the difference in the depths of tread on the front and rear tyres is normally distributed. 8(b) A B C D E F G H 2.4 1.5 2.3 2.4 2.6 2.5 2.1 2.6 2.3 1.9 2.1 1.8 1.8 2.8 1.4 2.1

Diff –0.1 0.4 –0.2 –0.6 –0.8 0.3 –0.7 –0.5 Rank –1 4 –2 –6 –8 3 –7 –5 H0: 0dm  H1: 0dm  7P , 29Q , so 7TP , the 10% two -tail critical region for 8n is 5T  . Therefore, we do not reject H 0 at 10% significance level and conclude there is insufficient evidence that there is a difference between the average wear for the front and rear tyres. 8(c) After correcting the mistakes, we have only one positive difference. For the change i