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 =− − − = + − = =− =