EJC 9649 2023 Prelim P2 Solutions

2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over Q1 Solution 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         Q2 Solution (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

2 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 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 (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. (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. Q3 Solution (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  (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   . (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)

3 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over Q4 Solution (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) (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: 48 27S  (iv) 2 2 2 2 22 3d 1 3ln 3 1 33ln 3 80 1 9 ln 3 x x Ix            (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 th e Trapezium rule which makes use of a linear approximation. Hence Simpson rule uses a better approximation to the curve 3xy . (vi) Absolute percentage error 100% 0.706%IS I   

4 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 Q4 Solution 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     Q5 Solution (a) 42 1 3iv  πi4 32 2ev  πi2 π4 3e k v   , where k 1 πi2 π43e k v     ππi 12 2e k v     πi 6 1 12e k  For arguments in the principal range, choose 0, 1, 2k     11π 5π π 7πi i i i12 12 12 12e , e , e , ev                       (b) Let ie cos isinpw p p   where π 12p . By De Moivre’s Theorem, for any positive integer n,

5 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over Q5 Solution 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) (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.

6 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 Q5 Solution (d) Triangle OAP is isosceles triangle. 2π π ππ 3 12 4 OAP OPA      πππ2 42AOP      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). Q6 Solutions (a) ˆ 7 9p , 0.95 1.6449 (or 1.645)z  , 900n Since  ˆ 1~N , ppP p n      approx. by CLT, 90% confidence interval for p 7 7 7 71177 9 9 9 91.645 , 1.6459 900 9 900                  0.7550, 0.8006 Answer is 4 dp as interval width is 0.0456 to 3 sf Re(z) Im(z) O A 1 P

7 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over (b)   0.95 124 ˆ 0.0ˆppz n   771991.6449 0.02 n   Using GC, 1169.06 1169n  = 1170 (3sf) Q7 Solutions 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 there is insufficient evidence that a binomial distribution with 0.6p is not a good fit. If the experiment is repeated 1000 times, the new 2 value will be 1000 8.807 9.4882001.7614   and so there is no change to the result of the test. Q8 Solutions (a) The depths of tread on