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    − −− − − − − − − = = − − −