TMJC H2 Mathematics Prelims Paper 2 (A)

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 1 of 16 2024 H2 MATH (9758/02) JC 2 PRELIMINARY EXAMINATION – SUGGESTED SOLUTIONS Qn Solution 1 Maclaurin’s Series (i) ( ) ( ) 2 2 2 ln 1 3 2 d 3 4 d 1 3 2 d1 3 2 3 4 d y x x yx x x x yx x x x = + + += ++ + + = + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 2 2 2 3 2 2 32 2 32 diff wrt , dd1 3 2 3 4 4 showndd diff wrt , d d d d1 3 2 3 4 4 3 4 0d d d d d d d1 3 2 6 8 4 0 d d d x yyx x x xx x y y y yx x x x x x x x y y yx x x x x x + + + + = + + + + + + + = + + + + + = 23 23 23 23 d d dwhen 0, 0, 3, 5, 18,d d d 5 183 ... 2! 3! 5 3 3 ... 2 y y yxy x x x y x x x x x x = = = =− = = − + + = − + + (ii) ( ) ( ) ( ) ( ) ( ) 2 2322 2 2 2 3 4 3 23 ln 1 3 2 3 2 3 2 3 2 ... 23 11 3 2 9 12 4 27 ...23 5 3 3 ... (verified)2 y x x x x x x xx x x x x x x x x x = + + ++ = + − + + = + − + + + + = − + + Alternative: ( ) ( )( ) 2ln 1 3 2 ln 1 1 2y x x x x= + + = + +  ( ) ( )ln 1 ln 1 2y x x= + + + Using the standard series in MF26, ( ) ( ) 2323 23 22... 22 3 2 3 5 3 3 ... (verified)2 xxxxy x x x x x  = − + + + − +  = − + + Useful technique: Multiply both sides by the denominator before further differentiation. AVOID using quotient rule for further differentiation as much as possible. Use implicit differentiation Use implicit differentiation Find the values of f(0), f’(0), f”(0) and f ’” (0). Substitute into the general form of the Maclaurin Series that can be found in MF26 Use the standard series in MF26 for ( )ln 1 x+ up to and including the term in 3x and show clearly that it is the same as the Maclaurin Series found in (i).

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 2 of 16 (iii) ( ) 2 2 3 5ln 1 3 2 3 3 ... 2x x x x x+ + = − + + Sub 1 ,2x= 2 2 3 1 1 1 5 1 1ln 1 3 2 3 3 ... 2 2 2 2 2 2           + + = − + +                     5ln 3 4 Substitute 1 2x= into both sides of the Maclurin Series to get the approximate value for ln 3

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 3 of 16 Qn Solution 2 Vectors (i) 3 2 By ratio theorem, 41 55 OC OD =− =+ a ab Area of triangle 1 2 1 3 4 1 2 2 5 5 3 4 1 4 5 5 3 4 1 4 5 5 31 45 3 (shown)20 OCD OC OD= = −  +  =  + =  +  = +  = a a b a a b a a a b 0 a b ab (ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 22 2 22 , 2 0 AO AB ABAP AB AB  = − − −= −−  −+=− −−  −+= − = = −+ = − = + a b a b a b a b a a b a a bab a b a a b a b a a a a b b b b a b a a b a a b ab 2 22 = + a ab and OA OC are in opposite directions, thus the negative sign is important This is a SHOW question, there is a need to apply the distributive law and show the expansion clearly Must explain that =a a 0 For the projection vector formula, there is NO modulus for AO AB AB as the direction is important ( ) ( ) 2 2 =  − = − − a a a b a b a b a

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 4 of 16 Qn Solution 3 Complex Numbers (a)(i) NOTE: Do not use graphing calculator when answering this question. Answers obtained using GC will not be awarded marks. Since coefficients of polynomial are all real, 2 3i is a root 2 3i is also a root.zz= +  = − ( )( ) ( )( ) ( )( ) 22 2 Quadratic factor 2 3i 2 3i ( 2) 3i ( 2) 3i ( 2) ( 3i) 47 zz zz z zz = − + − − = − − − + = − − = − + Let be the third root.zb= (Note that the last root must be real.) ( )( ) 3 2 28 23 4 7z z z k z z z b− + + = − + − Comparing coefficient of 2z : 4 8 4bb− − =−  = Comparing coefficient of 0z : ( )7 4 28 k = − =− Thus, the roots are 2 3i, 2 3i and 4.+− Alternative Method (Not recommended) 32 8 23 0z z z k− + + == Substitute 2 3i+ into equation: ( ) ( ) ( ) ( ) ( )( ) ( ) 32 23 2 2 3i 8 2 3i 23 2 3i 0 8 12 3i+6 3i 3i 8 4 4 3i 3 46 23 3i 0 8 12 3i 18 3 3i 32 32 3i 24 46 23 3i 0 28 0 28 k k k k k + − + + + + = + + − + − + + + = + − − − − + + + + = += =− Since coefficients of polynomial are all real, 2 3i is a root 2 3i is also a root.zz= +  = − ( )( ) ( )( ) ( )( ) 22 2 Quadratic factor 2 3i 2 3i ( 2) 3i ( 2) 3i ( 2) ( 3i) 47 zz zz z zz = − + − − = − − − + = − − = − + Let be the third root.zb= (Note that the last root must be real.) State Conjugate Root Theorem properly Answer the question GC is not allowed so working needs to be shown clearly No mark if working is not shown clearly State Conjugate Root Theorem properly

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 5 of 16 ( )( ) 3 2 28 23 28 4 7z z z z z z b− + − = − + − Comparing constant 4 8 4bb− − =−  = Thus, the roots are 2 3i, 2 3i and 4.+− (a)(ii) Replace z with iz i 2 3i or i 2 3i or i 4 3 2i or 3 2i or 4i z z z z z z = + = − = = − = − =− (b) icos isin ew = + = ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) i 2 i2 i ii i i2 i2 2 2 *e 1 e1 e e e e e 2cos 1 sec e2 *1 sec21 *arg 2 1 w w w w w w            − − − − − = + + = + = = = +  =− + Answer the question “Hence” so use previous part’s answer and do a replacement Useful result: This is of the form where is the modulus and is the argument Tip: Since this involves division of 2 complex numbers, use exponential form

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 6 of 16 Qn Solution 4 Integration (a)(i) ( ) cos cosd e sin ed xx xx =− (ii) Note: ( ) cos cos cos cosd e sin e sin e d ed x x x x x x x Cx =−  =− +  ( ) ( ) ( )( ) ( ) ( ) cos cos cos cos cos cos cos cos cos cos s n 2 e d e in 2 sin co 2 d 2 cos sin e d 2 cos e sin e d 2 cos e sin e d co 1 s s e e 2e cos show x x x xx xx xx x x x x x x x x x x x x x x x C xC x= = = − − − − = − − = − + + = − +      (iii) π cos 0 sin 2 e d xxx ( ) ( ) ( ) ( ) ( ) ( ) π πcos cos2 π0 2 π πcos cos 2 π0 2 0 1 1 0 1 sin 2 e d sin 2 e d 2e 1 cos 2e 1 cos 2e 1 0 2e 1 1 2e 1 1 2e 1 0 4 4e xx xx x x x x xx − − =−    = − − −    = − − − − + − −  =−  ( ) cos cos cos sin e d sin d ed x x u x v x u x v xx == =− =−  Double angle formula (MF26): sin 2𝐴 = 2 sin 𝐴 cos 𝐴 K I − D I For integration questions, if question asks you to differentiate something first, it is to guide you to see the integration Keyword: exact ⇒Cannot just use G.C. Need to remove the modulus before you can integrate O π π 2 𝑦 = sin 2𝑥 ecos 𝑥 From the graph of 𝑦 = sin 2𝑥 ecos 𝑥, observe that when y ≤ 0, π 2 ≤ 𝑥 ≤ π cos cos cos πsin 2 e ,0 2sin 2 e πsin 2 e , π2 x x x xx x xx  =  −   Use the show result from (ii) For integration by parts, Highly recommended to work out the u and v at the side, then apply KI – DI to fill in each part accordingly

TMJC/2024 JC2 Preliminary Examination Suggested Solutions/H2 Math (9758/02)/Math Dept Page 7 of 16 (b) dsec sec tand xx    =  = ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 d 1 1 sec tan d sec 1 1 sec tan d tan 1 πsec tan d since 0tan 2 sec d ln sec tan πln sec tan since 0 sec 0, tan 0 2 ln 1 x x C C x x C                    − = − = =   = = + + = + +      = + − +         You should remove the modulus when possible Recall: tan2 𝜃 + 1 = sec2 𝜃 MF26 Steps for integration by substitution: 1) Differentiate the given substitution ➢ Hidden working: 2) Substitute everything to θ in the integration in 1 step: ➢ Expression (including the dx) ➢ Upper & Lower limits (if any) 3) Simplify and integrate accordingly 4) For indefinite integrals, substitute back the x accordingly Everything involving x Everything involving θ 𝑥 = sec 𝜃 ⇒ cos 𝜃 = 1 𝑥