HCI_8865_2022_Prelim_Solutions_vetted

Suggested Solutions for 2022 C2 H1 Preliminary Examinations No. Suggested Solutions 1 2 30x kx k− + + = has real roots if Discriminant 0 Discriminant ( ) ( )( ) 2 2 43 4 12 26 kk kk kk = − + = − − = + − For discriminant 0 , we have { : 2 or 6 }k k k −  2a ( ) 33 15 266 61dd 52 xx x x x x x x C x −− = − = − + 2(b) (i) ( ) ( )( ) 3 2 2 2 2 d 6ln 3 4d 6d ln 3 43d 62 34 12 34 xx xx x x x x + =+ =  + = + Method 2 ( ) ( ) 3 2 2 2 3 3 2 2 d 6ln 3 4d 1 3 4 636 34 12 34 xx xx x x x − +  + =  + = + (ii) ( ) ( ) 22 3 2 3 22 1 12 d d3 4 12 3 4 1 6ln 3 412 11 = ln 3 4 or ln 3 426 xx xxxx xc x c x c =++ = + + + + + +  3(i) 22 2AD x x x= + = ( )1 2 2 2 4 102PerimeterAEBCGFD x AD DF x= + + + = 2 2 2 4 10x x DF x + + + = -2 6

( )10 2 2 4 2 =5 2 22 x DF x   − + + = − + + (ii) Let area of window be A ( ) ( ) 2 22 2 2 2 2 2 2 2 2 11 2 4 422 1 3 5 2 2 422 1 3 20 2 4 2 82 3 5 4 2 202 A x x x x DF x x x x x x x x x x x x x x x     = + + +  = + + − + +   = + + − − − =− − − + ( ) 2 2 d 3 10 8 2 20d d 20For max , 0 d 3 10 8 2 d 3 10 8 2 0d A x x xx AAx x A x    =− − − + =  = ++ =− + +  2   is maximum whe 20 3 0 8 n 1 A x  = ++ 4(i) (ii) 33e xy −=− At x-axis, 3 10 e 3 ln 3 3 xyx −=  =  =− 3d 3ed xy x −= x y

(iii) When 1 ln 33x=− , 13 ln3 ln33d 3e 3e 9d y x −−= = = Equation of tangent, 10 9 ln 3 3 9 3ln 3 yx yx  − = − −   =+ Area ( ) ( ) ( ) ( ) ( ) 0 3 ln 3 3 03 2 ln 3 3 ln 3 2 2 2 2 113ln 3 ln 3 3 e d23 1e ln 3 323 1 1 eln 3 ln 32 3 3 1 1 3ln 3 ln 32 3 3 12ln 3 ln 3 units23 x x x x − − − − = − −  = − +   = − − − +  = − + − +  = − +  5(i) Let x be the selling price of a toy boat. Let y be the selling price of a toy car. Let z be the selling price of a toy plane. x y

2 9 5 3 335 3 4 38 2 xy x y z x y x z = + + = + = + + Rearranging the equations, we get 2 0 0 9 5 3 335 2 4 2 38 x y z x y z x y z − + = + + = + − = From GC, 22.4, 11.2, 25.8x y z= = = Hence, the selling price of a toy plane is $25.80. (ii) Substitute D = 20 and p = 0 into 0.083e pDa −=+ , 020 3 e 17 a a =+ = (iii) (iv) 0.083 17e pD −=+ 0.08 0.08 d 17( 0.08)ed 1.36e p p D p − − =− =− (v) ( ) 2 0.5 9 312pt= − + p (dollars) D (thousands) D = 3

( ) ( ) ( ) ( ) ( ) 1 2 2 1 2 2 2 d1 0.5 9 312 (0.5)(2) 9d2 1 9 0.5 9 3122 9 2 0.5 9 312 p ttt tt t t − − = − + − = − − +  −= −+ ( ) 0.08 2 d d d d d d 91.36e 2 0.5 9 312 p D D p t p t t t − = −=−  −+ When 20t= , ( ) 2 0.5 20 9 312 19.30025907 p= − + = ( ) ( ) 0.08 19.30025907 2 d 20 9 1.36ed 2 0.5 20 9 312 0.290388 0.2849703 0.0828 (3sf) D t − −=−  −+ =−  =−