TMJC DHS HCI RI 9649 2024 Prelim P2 Solutions (updated)

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 1 of 15 Qn Solution 1 Differential Equations (a) 2 d 1d 24 PP rP htk r k rkPhk = − − =− − + − Since there are two distinct and positive equilibrium population values, 0 4 rkh . (b) 2 d 0d 2 4 4122 P r k rk Phtk k k hP kr =− − + − = =  − Let 441 , 12 2 2 2 k k h k k h kr kr= − − = + −

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 2 of 15 Qn Solution 2 Recurrence Relations 2(a) ( ) ( ) ( ) 11 11 2 1 10 . .: 1 0 k k k k k k R pR p R pR R p R A E pm m p +− +− = + − − + − = − + − = ( ) ( ) 2 1 1 4 1 2 1 1 2 2 ppm p p p  − −= −= When 1 ,12pm== ( ) 0 . .: 00 111 k N k G S R Ak B RB R AN A N kR N =+ =  = =  =  = = When 11, or 12 ppm p −= G.S.: 1 k k pR C D p  −=+  0 0 0 (1) 11 1 (2) N N R C D pR C D p =  + = −−−  −=  + = −−− Solving (1) and (2): 11 , 11 11 NNCD pp pp = =−    −− −−       1 1 1 1 1 1 1 1 1 1 1 k k k N N N p ppR pp p p p p p      −    −  −     = − =          − − −    − − −                 (b) In a fair game i.e. 1 ,2p= 0k kR N=→ since k is finite but N is infinite. Adam is likely to lose.

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 3 of 15 If 1 2p , 1 1 110 since 1 1 1 k N k N p p ppR ppp p  − − −−= →   →  − − Adam is likely to lose. If 1 1 1,0 1 02 N ppp pp −−    →  1 1 11 1 1 k k k N p p pR pp p  − −  − = → −  − − which increases as k increases. Adam has a chance to win if k is large enough.

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 4 of 15 Qn Solution 3 Complex Numbers (a) ( ) ( )( ) 22 22 22 2 2 22 2 1 2 3 i 1 2 i 3 1 4 3 3 22 35 3 0 11 1633 33 11 4 33 ww x y x y x y x y x x y xy xy − = − + − = + − − + = − + − + + =  − + =    − + =       1 2 3ww− = − describes a circle centred at 11,03   with radius 4 3 units. (b)(i) ( ) ( ) ( ) 2 3i 2 5iPoint on line: 4i 2 2 3i 2 5i 4 2i Direction of line: i 4 2i 2 4i + + − + = + − − + = − − = + To find the closest point of 1L to 2L , ( ) ( )5 i 4 2i 4i 2 4i , ,   + + − = + +  Comparing real and imaginary components, ( ) ( ) 5 4 2 1 1 2 4 4 2   + = −−−−− − = + −−−−− Solving, 13 1,10 10=− =− . 1 1 18 i55z =− + (b)(ii ) Distance between point and centre of circle ( )( ) ( ) 2 2 5 2 1 5 65 = − − + − = 11 43tan sin7 65  −− =−     ( )0.138 rad or 0.900 rad 3 s.f. =− −

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 5 of 15 Qn Solution 4 Definite Integral (a)(i) ( )( ) ( ) ( ) ( ) 6 66 6 1 0 12 0 0 1 22 0 1 2 1 2 cos cos d sin cos sin 1 sin cos d 13 1 1 cos cos d22 13 122 1 3 1 22 n n nn n n n nn n nn I x x x x x x n x x x n x x x nI n I nII nn    − −− − − − − − − = = − − − = + − − = + −  −=+     (ii) 34cos ,for 0 33r  =    2 2 6 6 6 6 2 0 6 0 6 0 5 4 0 3 2 0 0 1Required area d 2 8 cos d 3 24 cos d substituting 3 1 3 524 cos d12 2 6 9 3 1 3 320 cos d16 8 2 4 9 3 15 3 1 3 1 15 1 d16 16 4 2 2 r x x x xx xx x           = =  ==   =+    = + +    = + + +        27 3 5 84    =+

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 6 of 15 (b) Consider 2 2 2x y a+= d2 2 0 d d d yxy x y x xy += =− Let the thickness of each slice of cake be d.   ( ) 1 1 1 1 1 1 1 1 2 22 1 1d d d independent of - shown xd x x xd x xd x xd x xS y x y y x x ax ax a d x      + + + + = + −  =+ = = =   

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 7 of 15 Qn Solution 5 Differential Equations (a) ( ) 3 43 4 d d 818 dd 1 y y x x yx x y xxx x −+ + =  = + ( ) 3 1 4 8 0.1 1 n n n nn n x x yyy x +  −=+  + Given that 00 11, 3xy== and 0.1,h= When 1 1.1, x = ( ) ( ) 1 4 11 8(1)1 30.1 0.253 11 y  −  = + =  +  When 2 1.2, x = ( ) ( ) ( ) ( ) ( ) 3 2 4 1.1 8(1.1) 0.250.25 0.1 1 1.1 0.18661 to 5 s.f 0.187 to 3 s.f y  −=+  + = = (b) n nx d d nxx y x = 0 1 0.83333− 1 1.1 0.63390− 2 1.2 0.44889− Since the estimated value of d d y x is increasing over [1, 1.2], we expect that the value is an under-estimate of the actual value of y when 1.2x= . (c) ( ) 43 d18 d yx x y xx+ + = 3 44 d 8 d 11 y xx yx xx += ++ Integrating factor: ( ) ( ) 33 444 84 d 2 d 22ln 1 411e e e 1 xx xx xxxux +++= = = = +

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 8 of 15 ( ) ( ) ( ) ( )( ) ( ) ( ) 3 44 24 3 4 4 245 245 2624 d 8 d 11 d1 8 1 1 d d 1d 1d 1 26 y xx yx xx yx x x y x xx y x x xx y x x x x xxy x C += ++ + + + = + + = + + = + + = + +  When 11, 3xy== ( ) 21 1 1113 2 6 2 3 C C  + = + + = Particular solution: ( ) ( ) 2624 26 24 21 2 6 3 34 61 xxyx xxy x + = + + ++= + When 1.2x= , ( ) ( ) ( )( ) 26 24 3 1.2 1.2 4 0.19946 6 1 1.2 y ++== + (to 5 s.f) Percentage error = ( )0.19946 0.18661 100% 6.44% 3 s.f0.19946 − = Since the percentage error is only 6.44%, the estimate is close to the actual value.

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 9 of 15 Qn Solution 6 Confidence Interval (a) Let X be the height of a randomly chosen student (in cm). Since 80n= is sufficiently large, by Central Limit Theorem, X follows an approximate normal distribution. 95% Confidence Interval: 5.2064 5.2064166 1.9600 ,166 1.960080 80  −+ (165.5,166.5) (1 d.p.) Therefore, a 95% confidence for the population mean height of students in the college is (165.5,166.5). In the long run, if the process of obtaining samples is repeated an infinite number of times, the confidence intervals constructed will contain the actual population mean height of the students for 95% of the samples taken. (b) 0.598 0.801ˆ 0.69952p +== ( ) ( ) ( ) 0.6995 1 0.69950.801 0.6995 80 1.9801 P 1.9801 1.9801 0.952 95.2 3 s.f. z z Z k −=+ = −   = = Alternative method ( )ˆˆ 1ˆN,s ppPp n −   approximately ( ) ( ) P 0.598 0.801 0.95231 95.2 3 s.f. sP k   = =

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P2 (9649/02) Page 10 of 15 Qn Solution 7 Geometric / Exponential Distribution + Continuous RV (a)(i) Let X be the number of customer who checkouts in one hour. ~ Po(10)X 15 10 10P( 15) e 0.0347 15!X −= = = (ii) A customer spends 0.1 hour or 6 minutes at checkout counter. Let T be the time, in hours, customer spends at checkout counter. ~ Exp(10)T Customer spends 9 minutes or 0.15 hour at checkout counter. P( 0.15)T  = 0.15 10 0 10e d t t−  0.77687= 0.777= (to 3 sig fig) (iii) P( 0.15 | 0.1)TT by memoryless property P( 0.05)T= 0.05 10 0 1 10 e d t t−=−  0.60653= 0.607= (to 3 sig fig) (b) P( )Yn= P( 1 )Xn= + = P( 1 )n X n= −   1 1 1 ed tn n t  − − = 1 1 e n t n − − =−  ( )11 1 ee nn− − − =− + 11 1 1 1 e (1 e ) e (1 e ) nn     −−− − − − = − = −   This is the pdf of a geometric random variable with parameter 1 1 e . − − (shown)