2023 ACJC JC1 H1 CA2 solutions

H1 CA2 Qn Solutions 1 ( ) 2 2 d 12ln 5 6d 56 xxx x − = − − Qn Solutions 2 From G.C, the numerical value of gradient of D at point x = 1 is 2. Equation of tangent at D is: ( )0 2 1yx− = − 22yx=− Qn Solutions 3(a) ( ) 22 42 3 42 31 d 9 6 1 d 19 6 d 9 3 ln4 x xx xx xx x x x x x x x c − −+= = − + = − + +    3(b) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 d 21 1 1d2 11 12 2 1 x x xx x c x c       − − =− −=+ − −=+ −  

Qn Solutions 4 0 12 1 2 2 11 22 0 111 (2) 1 (0) 2222 e 3 d e3 1 2 2 e 3(2) e 3(0) 11 22 22 4 2e x x R x x x − − −− =+  =+ −        = + − +       −−    =+  4, epq== Since 11 2e3 x mx − =− has no real roots, 1.2494 1.25 (to 3 s.f.) m m   Qn Solutions 5(i) Curve surface area = 2 4ππr r  = 24π r Area of rectangle = 2 4π2r r  = 8π r Total surface area of the trash bin = 2 28π 4π 1 π2 rrr++ (ii) A = 2 28π 4π 1 π2 rrr++ 2 22 d8 π 4π πd A rr rr =− − + To minimize surface area of the trash bin, d 0d A r = 2 22 8π 4π π0r rr − − + = 238π 4π = π r+ 3 4π8r =+ [Hence] Ratio of diameter of semi-circular surface to height of the trash bin = 2 4π2:r r = 3 :2 πr = 4π 8 : 2π+ = 2π 4 : π+

Qn Solution 6(i) ( )( ) ( )( ) 0.0135 500 0.95 5 30e xCx= + − 0.0117500 4.75 30e xCx = + − 6(ii) ( ) 0.01 0.01 d 4.75 30 0.01 ed 4.75 0.3e x x C x =− =− For maximum or minimum, d 0d C x = . 0.014.75 0.3e 0 x−= 276.212x= When 276.212x= , C = $18337.01. Using 1st derivative test, OR Using 2nd derivative test, 2 0.01 0.01 2 d 0.003e 0 (since e 0)d xxC x =−   Hence maximum C. x 276.10 276.212 276.23 d d C x 0.0053045 0 - 0.004194 Slope / - \