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 1