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 r r r r r r r r r r r x x x x x u x u x u x u x u x u x u u u x x    −− = = =  −− = − − = − − = + − + − − =     Thus, ( )( ) ( ) 2 2 f1 f 1 x x x x xx xx − − = −= +− ( ) 2f 1 x A Bx x x x x  −= = ++ − − − where 1 5 1 5 and .22 − − − +== Solving: 1 5 1 5 1 5 1 5 2 2 2 25 1 5 1 5 1 5 1 5 2 2 2 25 A B − − − − − + − −=−  − =  − + − + − − − +=−  − =− 

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 5 of 19 ( ) ( ) 00 0 f 11 1 1 using result from part 11 rr rr rr r r ABx xx AB x x A x B x AB x              ==  = =+ −−   =− −     − −    =− −     = − −       a Comparing with ( ) 0 f r r r x u x  = = , we get 11 1 2 1 2 5 1 5 5 1 5 rr r nn n ABu u     =− −      =− +    − − − +   

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 6 of 19 Qn Solution 5 Numerical Methods (a) Note that f is continuous for all .x ( ) ( ) 1f 0 0 3 f 1 0.825 0 =−  = ( )f 2 cos 0 for all x x x = −   Since f is continuous and is a strictly increasing function, and ( ) ( )f 0 f 1 0,  C cuts the x-axis exactly once in the interval 0 1.x (b) Using ( ) ( ) 10 f with 0 ,f n nn n xx x x x + = − =  1 2 3 4 0.333333 0.327515 0.327509 0.327509 x x x x = = = = To verify: ( ) ( ) 5 5 f 0.32745 6.29 10 0 f 0.32755 4.24 10 0 − − =−   =   Thus, x-intercept of C ( )0.3275 4 d.p.= (c) Possible Explanation 1: For ( )f,yx= there will be a kink at 0.3275x= for the first quadratic segment from 0x= to 0.5x= when using Simpson’s Rule with four strips. Thus, the estimation might not be good. Possible Explanation 2: For ( )f,yx= the curve is below the x-axis from 0x= to 0.3275x= and above the x-axis from 0.3275x= to 1.x= The use of Simpson’s Rule with four strips will split the region into 2 parts: f rom 0x= to 0.5x= and from 0.5x= to 1x= for estimation. Thus, this estimation might not be good as the regions under and above the x-axis are not properly accounted for using Simpson’s Rule with four strips. (d) Method 1: Using ( )fyx= ( ) ( ) ( ) ( ) 1 0 f d 1 1 0 1 1 1 2 5f 0 4 f 2 f 4 f 2 f 4 f f 13 6 6 3 2 3 6 0.31774 5 d.p. A xx= −            + + + + + +                       = 

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 7 of 19 Method 2: Using ( )fyx= ( ) ( ) ( ) ( ) ( ) 1 13 10 3 f d f d 1 1 0 1 1 f 0 4f f3 6 6 3 1 1 0 1 1 2 5f 4f 2f 4f f 13 6 3 2 3 6 0.31705 5 d.p. A x x x x− + −       − + +           −           + + + + +                   = 

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 8 of 19 Qn Solution 6 Matrices and Linear Spaces (a)(i) ( ) ( ) ( ) ( ) 1 1 1 2 1 2 2 2 1 2 1 2 11 22 1 2 1 2 0 0 0 0 0 0 ' 0,0 ', ', ', k k ak bk ak bkab k k ck dk ck dkcd O P ak ck Q bk dk R ak bk ck dk +    =    +      ++ (ii) Note that ' ' ' 'O P R Q is a parallelogram Area of parallelogram ( ) 12 12 1 2 1 2 1 2 1 2 12 ' ' ' ' 0 0 00 det area of rectangle O P R Q ak bk ck dk adk k bck k adk k bck k ad bc k k OPRQ            =  =           −      =− =− = A (b)(i) Let A and B be orthogonal matrices. ( )( ) T TT T = = = AB AB ABB A AIA I ( ) ( ) T TT T = = = AB AB B A AB B IB I Therefore, AB is an orthogonal matrix. (ii) ( ) ( ) ( ) 2 det det 1 det det 1 det 1 det 1 or det 1 T T == = = = =− MM I MM M MM (iii) 11 10 11det 110 but 1 1 1 1 2 1 1 0 1 0 1 1    =−      =        

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 9 of 19 Hence, 11 10   is not orthogonal (iv) 22 22 cos 0 sin cos 0 sin 0 1 0 0 1 0 sin 0 cos sin 0 cos cos sin 0 cos sin cos sin 0 1 0 cos sin cos sin 0 sin cos 1 0 0 0 1 0 0 0 1                     −           −    + − + = − + +  =  Alternative 22 22 cos 0 sin cos 0 sin 0 1 0 0 1 0 sin 0 cos sin 0 cos cos sin 0 cos sin cos sin 0 1 0 cos sin cos sin 0 sin cos 1 0 0 0 1 0 0 0 1                     −            −    +− =  −+  =  1 cos 0 sin 0 1 0 sin 0 cos T   − − ==   BB Since T =BB I and 1− =BB I imply that 1 .T− =BB

TMJC/2024 JC2 Prelim Exam Marking Scheme/H2 Further Math P1 (9649/01) Page 10 of 19 Qn Solution 7 Matrices and Linear Spaces (a) Since 12 3 1  − is an eigenvector of M, 3 1 3 12 12 1 1 3 3 1 3 1 1 k k           − − − = −             for some  . 3 1 3 12 12 1 1 3 3 1 3 1 1 36 12 93 31 k k k k            − − − = −                    − + = − 