ACJC 2023 JC1 H1 MYA Solutions

2023 JC1 H1 Math MYA Solutions Qn Solutions 1 23log 5 5 log 5x x −= Let log 5xu = ( ) 2 2 1 23 3 235 3 2 5 3 5 2 0 3 1 ( 2) 0 1 or 23 1log 5 or log 5 23 5 or 5 5 or 5 (reject 5 ) 1 or125 xx u u uu uu uu uu xx xx x − − −= −= − − = + − = =− = =− = == = = − = 5 x= 2 21xy+ =− ----- (1) 2 8y x mx= − + ----- (2) From (1): 12yx=− − ----- (3) Sub (3) into (2), ( ) 2 2 2 1 2 8 2 9 0 2 9 0 x x mx x x mx x m x − − = − + + − + = + − + = Since the line and the curve intersect at two distinct points, Discriminant > 0 ( ) ( ) ( )( ) ( )( ) 2 2 2 4(1)(9) 0 2 36 0 2 6 2 6 0 4 8 0 4 or 8 m m mm mm mm − −  − −  − − − +  − − −  −  Since the line is tangent to the curve, 4 or 8mm=− =

Qn Solutions 3(i) Area of garden 14= 22 22 2 2 2 2 1 1 1 1 142 2 2 2 11 2822 11 2844 28 ----- (1) y y x y y x y y xy x xy x      − − =          − − =       − − + = −= 3(ii) 112 10 22 2 10 210 ----- (2) x y y x x x y y x x           + + − =       − + = = + − 3(iii) Sub (2) into (1): 2 2 210 28 210 28 0 Using GC, 3.6465 3.65 or 12.061 1 2.1 11.325 11.3 or 14.383 14.4 x x x x xx xx yy    + − − = − − = = = = = = = = = From the diagram, 1 02 yx− , 3.65, 11.3xy = =

Qn Solutions 4(i) 2 22 2 2 2 44d 2 d dd d 44d 4 eee ee e xxx x x x x xxx x    − +− =   = − +  =− 4(ii) ( ) ( ) ( ) 12 5 2 5 2 125 2 25 dd 4 3 4 3dd 15 4 3 3 2 35 2 4 3 ee e e xx x x xxxx x x − −− − −  + − = + −   = − + − − = − − − Qn Solutions 5(i) 63 2y x=− + − 5(ii) 63 2y x=− + −  ( ) 2 d6 d 2 y x x = − When 26x=+ , 36y=− − , d 1d y x = O x y y = – 3 x = 2

Using ( )11y y m x x− = − or y mx c=+ , Equation of tangent: ( ) ( )3 6 1 2 6yx− − − = − − 3 6 2 6yx+ + = − − 5 2 6yx= − − 1 and 5 2 6mc = =− − . 5(iii) Since curve C has a vertical asymptote at x = 2, and curve D has a vertical asymptote at bx a=− , 2b a − = 2ba=− For curve D, when 0,y = 9 .4x= 90 ln 4ab  = +   9 14 9 214 4, 8 ab aa ab   +=    −=   = =− Qn Solutions 6(i) Using G.C, When t = 2, P = 118. 6(ii) ( ) ( ) 43 22 5 5 2 15 470 925, for 2 12P t t t t t= − + − − + −   6(iii) Using G.C, when P = 500, 3.378t = or 8.689t = . For 500P , 3.38 8.69t 118 – O t P 2 12 1014.046