TMJC 9758 2025 Prelim P1 Solutions

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 1 of 22 2025 H2 MATH (9758/01) JC 2 PRELIMINARY EXAMINATION Qn Solution 1 Graphing Techniques, Equations and Inequalities (a) (b) Points of intersection: 1xa xa−= − Since xa , 0xa− ( ) 2 1 1 xa xa −= = For 1xa xa− − , from the graph, 1 or 1x a x a −  + y x O a y x a=− 1y xa= − a 0y=

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 2 of 22 Qn Solution 2 Differentiation (a) ( ) ( ) 2 2 22 2 2d2 d ) e de (1) Differentiate with respect to , 1 d 2 e 1 2 e (shown) 2 2ed (d x x x x x y x x x y y y x y yy yy x = + −− =+ −− + = + − (b) ( ) ( ) ( ) ( ) 2 2 Subt 0 into 1 0 0 or 1 Subt 0,0 o into h 2 t e equati n of the tangent is d 1d 10 yx x yy yy yy y y x y = = −= == − −= = =− ( ) ( ) ( ) the equation of the tangent is 1 3 0 31 Subt 0,1 into 2 d 3d y yx x x y =  − = − =+

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 3 of 22 Qn Solution 3 Techniques of Integration (a) ( ) 2 2 2 2 2 2 21 1 2 d d41 21 d d 41 11 d d 2 1 ln 4 2 24 4 31 1 41 1 2 C x xxx xxx xx xx x x xx xx xx x x x − + − =+ −+ −=+ −+ =+ + =+ − − − − + + −       (b) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 22 2 2 ln d d 1ln d22 1ln2 2 2 ln ln 2 22 4 C C x kk kx x x x kx x x xx kx xx x x x k xx k           =  =− =− = + − + −    (c) cosx = d sind x  =− ( ) ( ) ( ) ( ) 2 2 2 2 sin d2 sin d 2 1 4cos 1 1 d41 1 d 1 1 1 1 2 ln2 1 1 c sin o d n cos 2 2 os 1 2 1 1 2l42 cs co1 2 s 1 2 1 C C x x x x x x            = = − = − = −  +=  − + + += + − − + −       

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 4 of 22 Qn Solution 4 A&GS, Sequences & Series (a) ( ) ( ) ( ) 2 5 9 61 92 10 3 a d br a d br a d br + = −− + = −− + = −− ( ) ( ) ( ) ( ) 23 54 (2) (1) : 3 1 4 (3) (2) : 15 d br r d br r − = − −− − = − −− Substitute (5) into (4), ( ) ( ) ( ) ( ) ( ) ( ) 5 4 2 3 3 4 3 73 3 1 1 3 1 1 since , 0 3 4 1 0 shown br r br r r r r b r rr − = − − = −  − + = By GC, since the series is convergent, 1r  , 0.6639r = (4 d.p.) (b) New Series: 35, , ,...br br br ( ) ( ) ( ) ( ) 2 2 2 2 1 1.78 0.6639 1 0.6639 1.781 0.6639 1 0.6639 0.6639 1 1 0.6639 1.78 01 0.6639 1 6 11 0. 7 6 1 1. 39 8 n n n br br S b r b b r r b b r b b − − −− − −−− − −  −− −− Using GC, ( ) ( ) 2 2 0.6639 1When 14, 1 0.6639 1.7785 1.781 0.6639 1 0.6639 0.6639 1When 15, 1 0.6639 1.7818 1.781 0.6639 1 0.6639 n n n n = − − = −− = − − = −− Therefore, least 15.n=

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 5 of 22 Qn Solution 5 Recurrence Relations (a) Let nxL→ as n→ . Then 1nxL+ → . ( )( ) ( ) 2 2 2 20 1 2 0 1 or 2 shown LL LL LL LL =− − − = + − = =− = (b) Method 1: From the limits, Since 1L=− , the sequence remains constant at 1− . Method 2: Using GC, the sequence remains constant at 1− . (c) 2 1 2 2 2 n n n n nn x x x x xx + − = − − = − − If 1 2, nx−   2 20nnxx − −  from the graph. Hence 11 0.n n n nx x x x++−    x 2 2y x x= − − y

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 6 of 22 Qn Solution 6 Application of Differentiation (a) 2 m r m 6 m h m By similar triangles, 2 6 1 3 r h rh = = 2 2 2 31 1 1 1 1 1 3 3 3 3 9 27V r h h h h h h       = = = =        (b) 3 d d d d d d 0.0 d d outin in VV t V t V t t = −= − dSince 0.1,d h t = ( ) 2 2 d d d d d d 1 0.19 1 3 0.19 0.1 V V h t h t h   = = = = 40.03 .d d d 0 1 0.03 0.34d inV t V t  = = = + + Hence, water is pouring in at a rate of 0.344 m/s

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 7 of 22 Qn Solution 7 Functions Graphing Techniques Equations and Inequalities (a) For 1f − to exist, f must be one-one function. Least value of 3k = (b) Let ( ) 21 36 32yx= − − ( ) ( ) ( ) 2 22 2 2 2 36 3 4 36 3 3 36 4 yx yx xy = − − = − − − = − 23 36 4xy− = − 23 36 4xy= − − or 23 36 4xy= + − (Rejected, 3x ) ( ) 12 2 f 3 36 4 3 2 9 xx x − = + − = + − (c) (d) Range of f = (0,3] Domain of g = (0, ) Since fgRD , gf exists (e) fg f f f g 81D [3,9) R (0,3] R ,11 8 = ⎯⎯ → = ⎯⎯ → = −  fg 81R ,11 8 =− y x O (0,9) (3,3) (9,0) ( )9,9 ( ) 1ffyx −=

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 8 of 22 Qn Solution 8 Definite Integrals (a) 2 23 d 2d 54 d 8d d8 4d2 xt x t yt y tt yt tx =+ = =− =− −= =− At ( )2,4 , 12 3 2 2 3 2x t t t= +  = +  =− d1 42d2 y x =− − = Equation of normal, N: ( )142 2 1 52 yx yx − =− − =− + (b) 323 2 xx t t −= +  = Sub t into y: ( ) 2 2 2 54 354 2 53 yt x x =− −=−  = − − Alternatively, 2 554 4 yy t t −= −  = Sub t into x: 52 3 3 5 2 yxy  −=  + =  −

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 9 of 22 (c) ( ) 2 53 35 yx xy = − − =  − Since 3,x 35xy= − − ( ) 22 35 9 6 5 5 14 6 5 xy yy yy = − − = − − + − = − − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 4 2 2 4 4 4 4 3 2 4 2 4 43 2 2 4 4 4 24 V d olume of solid of revolution about Volum e of co 4 d 35 -axis 3 6 5 5 d 1 ne 1 2 1 13 y 3 414 6 5 dy 3 65 44 323 12 114 4 52 y yy y y y yyy y xy y yy         − − − − − − =+ =+ = − − + − + = − − − +   −= − − +  − = − + − + −  −     ( ) 4 3 456 8 4 56 8 108 3 28 3    = − + + + − + = x ( )2,4

TMJC/2025 JC2 Preliminary Examination Markers Comments/H2 Math (9758/01)/Math Dept Page 10 of 22 Qn Solution 9 Complex Numbers (a) Method 1: Since the equation has real coefficients, 2 2i−+ is a root 2 2i− − is also a root. Consider ( ) ( ) ( ) ( ) ( )   22 2 2 2 2i 2 2i 2 2i 2 2i 2 2i 4 4 4 48 z z z z z zz zz − − + − − − = + − + +           = + − = + + + = + + ( )( ) ( )( ) 3 2 2 3 3 2 2 2 16 2 4 8 coeff of : 1 constant: 2 2 16 2 4 8 2 2 coeff of : 2 2 4 coeff of : 8 2 8 z az bz z z pz q zp q z az bz z z z za zb + + − = + + + = =− + + − = + + − =− + =− + 4 2 2, 8 8 2 andab = − = − the roots are 2 2i, 2 2i and 2 2.− + − − Method 2: ( ) ( ) ( ) ( ) ( ) 32 6 Since 2 2i is a root, 2 2i 2 2i 2 2i 16 2 0 16 16i 8i 2 2i 16 20 i i+ i 0 Comparing real and imaginary s 16 2 16 2 part , 16 2 16 2 1 8ab z a b ab b b =− + − + + − + + − + − = + + − + − + − = −− − + − − = ( ) and 8 8 2 8 16 2 Subst 8 8 2 into 8 16 2 , 8 16 2 8 8 2 4 2 2 8 8 2 2 16 8 2 00 a b a b b a b a a b b = − = + = − = + = + − −+  = −  = − = = Since the equation has real coefficients, 2 2i−+ is a root 2 2i− − is also a root. Consider ( ) ( ) ( ) ( ) ( )   22 2 2 2 2i 2 2i 2 2i 2 2i 2 2i 4 4 4 48 z z z z z zz zz − − + − − − = + − + +           = + − = + + + = + +