EJC_8865_2022_Prelim_Solutions_vetted

Error! Reference source not found. EJC Q1 For ( ) 2 1 2 1kx k x k+ − + + always negative, Discriminant < 0 and 0k  . ( ) 2 2 22 2 40 1 4 (2 1) 0 2 1 8 4 0 7 6 1 0 (7 1)( 1) 0 b ac k k k k k k k kk kk − − − +  − + − −  − − +  − +  11 or 7kk−  and 0k  Combining solutions, 1k − Set of values of k is { : 1 }kk − Q2 (i) Let the mass (in kg) of arabica, robusta and liberica coffee beans brought in by Starluck Coffee on International Coffee Day be x, y and z respectively. 550 ............(1)x y z+ + = 4 1.5 3.5 1685 ............(2)x y z+ + = ( ) ( ) ( )4 4 44.00 1.50 70 3.50 13155 9 5 23.2 2.8 245 13153 26 1560 ............(3)3 x y x x y x xy + + − =  + + − = += Using the GC, 240, 180, 130x y z= = = (ii) Profit ( ) ( ) ( )1315 240 3.70 180 1.10 130 3.20 187= − − − =− Starluck Coffee made a loss of $187 on International Coffee Day -1 1/7 -1 1/7 0

Q3 (a) 52 3 421 33 2 d2 3d 225 3 2 −  −     = − +        xxx xx xx (b) ( ) ( ) 1 0 11 2 0 11 2 0 11 22 1 d 32 3 2 d 32 ( 2)(0.5) (3 2) (3 0) [1 3] 31 x x xx x − − =−  −=− =− − − −  =− − =−   Q4 (a) 20 px−= 2x p= (b) At A, y=0 3 3 0 3 ln(2 ) ln(2 ) 3 2e 2e px px px x p = − − −= −= −= x-coordinate of A is 32e p − . (c) d d 2 2 y p p x px px −=− =−− At point A,

Error! Reference source not found. 3 3 3 d d 22 22 yp x ep p p e p e =  −−  = −+ = Equation of tangent at A: 3 3 20 peyx ep  −− = −   3 33 33 2 21 peyx ee pyx ee −=− = + − Q5 (i) (ii) y = 1 x = 0 (-9, 0) O x y = 1 x = 0 (-9, 0) O

From the GC, the point of intersection is ( )1.5,7 Area   3 2 30 2 3 2 92 4 d 1 d 8.25 9ln 338.25 9ln 9ln 22 36.75 9ln 9ln 2 26.75 9ln 3 a a x x x x xx aa aa aa = + + + = + + = + + − − = + + − = + +  Alternative (using area of trapezium) ( )   3 2 3 2 1 3 9 4 7 1 d22 8.25 9ln 338.25 9ln 9ln 22 36.75 9ln 9ln 2 26.75 9ln 3 a a xx xx aa aa aa = + + + = + + = + + − − = + + − = + +  Q6 (i) 0.5(0) 1 When 0, 112e 0.4(0) ee − = = − − = t C 1 e million dollars (ii)

Error! Reference source not found. 0.5 1 0.5 1 d 2e (0.5) 0.4 0d e 0.4 0.5 1 ln 0.4 0.16742 0.167 0.36515 0.365 − − = − = = −= =  = t t C t t t t C Method 1: Using 1st derivative test t 0.167- 0.167 0.167+ d d C t E.g. use 0.166=t 0.5(0.166) 1 4 d 2e (0.5) 0.4d 2.84 10 0 − − =− =−   C t 0 E.g. use 0.168=t 0.5(0.168) 1 4 d 2e (0.5) 0.4d 1.16 10 0 − − =− =   C t Slope of tangent \ - / Value is a minimum Method 1: Using 2nd derivative test 2 0.5 1 2 2 2 d e (0.5)d dwhen 0.16742, 0.20000 0d Value is a minimum −= = =   tC t Ct t (iii) (iv) (0.167, 0.733) (0, 1/e) C (millions) t (years) O

3 0.5 1 1 12e 0.4 e d 1.8330 1.83− −− = t t t The total cost incurred be