TMJC_DHS_HCI_RI_9649_2024_Prelim_P1_Solutions

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 1 of 19 Qn Solution 1 Mathematical Induction (a) 11 0 0.5 = A 2 2 1 1.5 1 2 0.5 0 0.25 0 0.5 −   ==       A 2 3 3 1 1.75 1 2 0.5 0 0.125 0 0.5  −==   A Conjecture: ( ) ( ) 1 1 2 0.5 0 0.5 n n n − −= A (b) Let Pn be the proposition that ( ) ( ) 1 1 2 0.5 0 0.5 n n n − −= A for all positive integers n. When 1n= LHS 11 0 0.5 ==  A RHS ( ) 0 1 111 2 0.5 0 0.50 0.5  − =   = Since LHS = RHS, P1 is true. Assume Pk is true for some positive integer k i.e. ( ) ( ) 1 1 2 0.5 0 0.5 k k k − −= A When 1nk=+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 2 0.511 0 0.5 0 0.5 1 2 0.5 0.5 0 0.5 1 2 0.5 0 0.5 kk k k kk k k k + − − + +  − =    −+=  −= A = AA 1 true true.kkPP + Since P1 is true and Pk is true  Pk+1 is true, by Mathematical Induction, Pn is true for all positive integers n. Qn Solution

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 2 of 19 2 Complex Numbers (a) ( ) ( ) 5 5 4 3 2 2 3 4 5 cos5 isin 5 cos isin by De Moivre's Theore m cos i5cos sin 10cos sin i10cos sin 5cos sin isin sin 5tan 5 cos5                 + = + = + − − + + = By comparing real and imaginary parts, ( ) 4 2 3 5 5 3 2 4 35 5 24 5cos sin 10cos sin sintan 5 cos 10cos sin 5cos sin 5 10 dividing throughout by cos1 10 5 x x x xx            −+= −+ −+= −+ (b) πtan 5 0 5   =  ( ) 35 2 5 10 0 10 100 4 50 or 2 5 2 5 x x x xx − + = −== = Since πtan 5 0 5 k  =  for 2, 1,0,1,2k =− − , 2 πtan 5   is the smaller value. 2 πtan 5 2 55  =−

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 3 of 19 Qn Solution 3 Recurrence Relations (a) 2 3 1 0xx− + = ( )( )3 9 4 1 1 35 22x − == 3 5 3 5 and22 −+== (b) Note that  and  are the limits of the sequence if the sequence converges. Consider nx : ( ) ( ) ( ) ( )( ) 2 2 2 1 31n n n n nn x x x x xx  + − = − − =− − − Since nx , ( ) ( )0 and 0.nnxx −  −  Thus, ( ) ( ) 22 1 0nnxx+ − Also, ( ) ( ) ( )( ) 22 1 1 1n n n n n nx x x x x x+ + +− = − + Since nx and 1nx + are positive, then ( )1 0nnxx+ + Thus, ( )11 0n n n nx x x x++−    Since nx , we have ( ) 2 2 1 3 1 3 1nnxx + = −  − = . Thus, 11 since 0nnxx ++ Therefore if nx , then 1nnxx +   for all .n + Thus, the sequence increases and converges to .

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 4 of 19 Qn Solution 4 Recurrence Relation (a) ( ) 1 2 0 1 11 1 r r xx xx x −  = =−− = + + + = (b) ( ) 0 f r r r x u x  = = ( ) ( ) 1 1 01 22 2 02 f f rr rr rr rr rr rr x x u x u x x x u x u x  + − ==  + − == == ==   ( ) ( ) ( )   2 12 0 1 2 0 1 1 0 1 0 1 2 2 f f f r r r