2025 JPJC H2 Math Prelim P1 Solution

Jurong Pioneer Junior College H2 Mathematics JC2-2025 Preliminary Exam Paper 1 Solution Q1 (i) ( ) 12 2 1 44y x x xx − = = −− ( ) ( ) ( ) 22 22 d 2 4 4 4 2d 4 yx x x xx xx − −=− − − = − At turning point, ( ) ( ) 22d 4 4 2 0 2d y x x x xx − =− − − =  = ( ) ( ) ( ) ( ) 2 23 222 2 d 4 2 2 4 4 2d y x x x x xx −−=− − − − − − When 2x= , ( ) ( ) 2 22 2 d1 4(2) (2) 2 0d8 y x −=− − − =  Or Using first derivative test, x 1.9 2 2.1 d d y x 0.0126− 0 0.0126 tangent \ − / The turning point at 2x= is a minimum point. (ii) Area 2 21 1 d4 xxx= − ( ) ( ) 22 2 2 211 11 d or d 44 2 2 xx xxxx == −− − + − ( ) 22 2211 1 1 1 1dd 4422 xx xxx = = + −−− ( ) ( ) 2 2 1 1 2211 ln ln ln 42(2) 2 2 4 x xxx  +−= =  − −  −− 22 11 11ln ln4 4 4 4 xx xx    ==   −−    1 1 1 1ln ln4 3 4 3=− =− 11ln 3 ln 344==

2 Q2 (i) ( ) 2 1 xa xa =− − ( ) ( ) ( ) ( ) ( ) 22 33 11 or 11 11 11 x a x a x a x a x a x a x a x a x a x a = − =− − −− − = − =− − = − =− = + = − (ii) For ( ) 2 1 xa xa − − , 1 1 ,a x a x a−   +  Or 1a x a−   or 1a x a  + x y O

3 Q3 Let h m be the vertical distance between the top of the ladder and the floor Let x m be the horizontal distance between the foot of the ladder and the corner of the wall Method 1 By Pythagoras’ theorem, 2 2 2 3.12hx+= By implicit differentiation w.r.t. t, dd2 2 0dd hxhx tt+= When 1.2,h= d 0.2d x t = and 2 2 2 3.12 1.2 2.88 (Since 0) x xx =− = Hence, ( ) ( )( )d2 1.2 2 2.88 0.2 0d d 0.48d h t h t += =− Hence, the top of ladder is sliding down at a rate of 0.48 m/s. Method 2 By Pythagoras’ theorem, 2 2 2 3.12hx+= By implicit differentiation w.r.t. x, or By implicit differentiation w.r.t. h dd2 2 0dd h h xhx x x h+ =  =− When 1.2,h= d 0.2d x t = and 2 2 2 3.12 1.2 2.88 (Since 0) x xx =− = Hence, d d d d d d h h x t x t= 2.88 0.21.2 0.48 =−  =− Hence, the top of ladder is sliding down at a rate of 0.48 m/s. T F h x ladder 0.2 m/s

4 Q4 (i) Smallest value of a = 2. (For 1f− to exist, f must be a one-one function.) (ii) Let 243y x x= + − 22 22 2 2 2 2 3 4 or 3 4 0 3 ( 3) 4(1)( 4 )33 42 2 2 3 25 43 25 2 4 2 3 25 3 25 2 4 2 4 3 25 24 y x x x x y yxx yxx y x x y xy =− − − − − + =   − − − +   =− − − − − =        −=− − − = =− − + =  −  − = − 1 1 gg 3 25 24 3 25 24 3 3 25since , 2 2 4 3 25 25Thus, g : , 2 4 4 25D R , 4 xy xy x x y x x x − − − = − =  −  = − − − −  = = −  (iii)  )1 ffR D 2,− = =  1 gg 25D R , 4 − = = −  Since 11fgRD−−  , 11gf−− does not exist. (2, 2) y x O

5 Q5 (a) (i) (ii) For 1C and 2C to intersect, 3k  (since 0k  ). (iii) Common lines of symmetry for both 1C and 2C are 0x= and 0y= . (b) (i) ( ) ( ) ( ) ( )f 46 f 41 f 36 ...... f 1 2(1) 6 4= = = = =− + = (ii) (iii) For 55 x−   . the roots of ( )f0 x−= are 3 and 2− . x y O 3 5 −2 −5 6 y x O

6 Q6 (i) 22ed xxx −  2 2 21 e e d2 xxx x x−−=− +  2 2 2 21 1 1e e e d2 2 2 x x xx x x− − −=− − +  2 2 2 21 1 1e e e2 2 4 x x xx x C− − −=− − − + 221 e 2 2 14 x x x C− =− + + + (ii) Volume ( ) 221 0 1ed e xx x x − =−  or ( ) ( ) 221 0 11e d 1 3e xxx − −  1 2 2 2 20 1ed e xx x x −=− ( ) 1 12 2 3 2 0 0 1 e 2 2 14 3e x x x x  − = − + + −  ( ) ( )( ) ( ) 232 2 1 e 2 l 2 1 1 14 4 3e  −= − + + + − 22 5 4e 4 3e   =− + − 2 19 4 12e =− ( ) 23 19e12  −=− Q7 (a)(i) 2 d 25 x x x− ( ) 1 2 21 2 25 d2 x x x − =− − − 225 xC=− − + 22 2 2 2 d ed d1 2ed2 d ed d1 1ed2 x x x x vux x u xvx vux x u vx − − − − == = =− == = =−

7 (a)(ii) 4 2 d3 25 x x x = − 04 22 0 d d 3 25 25 xx xx xx − + = −− 04 22 0 25 25 3xx     − − − + − − =    04 22 0 25 25 3xx     − − − =      25 25 3 5 3− − − − = 225 4−= 2 9 = 3 = Since 0  , 3 =− (b) 4 tanx = 2d 4secd x  = When 4, tan 1 4x = =  = When 0, tan 0 0x = =  = 24 20 d16 x xx+ 2 24 20 16 tan 4sec d16 16 tan   = + 2 24 20 tan 4sec dsec   = 24 0 tan4 sec dsec   =  or 24 0 4 sin sec d    =  4 0 4 tan sec d    =    24 0 4 sin cos d     − =− −   4 04 sec  = ( ) 1 4 0 cos4 1   − =−  − 4 sec sec04 =−  ( ) 1 1 4 cos cos 04  − −=−  ( )4 2 1=− ( )4 2 1=−

8 Q8 (a) 2 1 2 3 9 (1) 4 2 27 (2) 9 3 55 (3) nu an bn c u a b c u a b c u a b c = + + = + + = −−−−− = + + = −−− = + + = −−− Using GC, 5, 3, 1a b c= = = 25 3 1nu n n= + + (b)(i) ( ) ( ) ( ) 33 1 1 1 22 22 2 5 2 5 125 4 1 52 n n n r r r rr nn n nn n = = = + = +  +=+  +=+    (b)(ii) Method 1 ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 2 2 3 4 23 33 11 2 2 2 2 22 2 2 5 2 2 2 5 ... 2 2 5 25 2 5 2 5 2 3 3 4 5 2 5 322 23 5 772 n r n r n rr rn r rr nn n nn n = + = + == + + = + + + + + + =+ = + − + ++ = + + − +  ++= + −   

9 Method 2 ( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 2 2 3 4 2 3 2 33 1 1 4 2 2 2 2 22 2 2 5 2 2 2 5 ... 2 2 5 25 25 2 3 3 42 5 1 44 23 5 772 n r n r nn r r r rn r rr nn n nn n = + = ++ = = = + + = + + + + + + =+ = − +  ++= − + −  ++= + −      (b)(iii) From (i) ( ) 22 1 1 52 n r r nnun = +=+ As ,n→ ( ) 22 1 ,52 nn n+ → → Hence, ( ) 22 1 1 52 n r r nnun = += + → , series 1 r r u  =  does not converge.

10 Q9 (i) 1 2 2 1 1 1 3d d tan 4 6249 9 9 xx x K x x −= = ++ +  (ii) 12 21 19f ( ) (4 9 ) 1 44 xxx − − = + = +  ( ) ( ) 222 2 4 24 121 9 91 1 ...4 4 2 4 1 9 811 ...4 4 16 1 9 81 4 16 64 xx x x xx      −−= + − + +           = − + +  − + (iii) From (i) 1 2 31tan 6 d , where 62 49 x x C C K x − = + =− + From (ii) 1 2 43 1 9 81tan 6 d2 4 16 64 x x x x C− = − + + 35 35 1 3 816 4 16 320 3 9 243 2 8 160 x x x D x x x D = − + + = − + + 10, tan 0 0 0xD −= =  = 1 3 53 3 9 243tan 2 2 8 160 x x x x− = − + (iv) 0.5 0.5 1 3 5 00 3 3 9 243 tan d d 0.174 (to 3 dp) (Using GC)2 2 8 160 x x x x x x− = − + = (v) From GC, 0.5 1 0 3 tan d 0.173 (to 3 dp)2 x x− = (vi) The estimate in (iv) is accurate up to 2 decimal places but not to 3 decimal places. To improve the estimate, we can include higher-order terms in the Maclaurin series expansion of 1 3tan 2 x− .