HCI 2024 Term 2 Common Test Practice 3 (solution)

1 Solution to Term 2 CT Practice 3 1 3 1yx+= ----------- (1) 22 2 2 25x xy y+ + = ---------- (2) From (1), 1 3yx=− ---------- (3) Substitute (3) into (2). ( ) ( )22 2 1 3 2 1 3 25x x x x+ − + − = ( )2 2 2 2 6 2 1 6 9 25x x x x x+ − + − + = 222 5 2 12 18 25x x x x− + − + = 213 10 2 25xx − + = 213 10 23 0xx − − = ( )( )13 23 1 0xx− + = 13 23 0x −= or 1 0x += 13 23x = 1x =− 10 1 13x = Substitute 10 1 13x = into (3). 10 1 3 1 13y =−  4 1 5 13=− 4 4 13=− Substitute 1x =− into (3). ( ) 1 3 1y = − − 1 3=+ 4= The coordinates of A 10 4are 1 , 413 13  − and B ( ) 1, 4=− . 13x −23 −x x +1 +x 13x2 − −x

2 2 (i) 2 6y x x= − − + ( )2 6xx= − + − 22 2 11 6 22xx     = − + + − −        2 11 6 24x = − + − 2 11 6 24x= − + + 2 11 6 42 x= − +  (ii) 2 6y x x= − − + The coefficient of x2 is −1 and the graph has a maximum point at 11 , 624 − . When the curve cuts the x-axis, 0y = . 2 11 6 024x− + + = 2 11 6 24x += 11 6 24x + =  11 6 24x += or 11 6 24x + = − 2x = 3x =− The curve cuts the x-axis at (2, 0) and (−3, 0). When the curve cuts the y-axis, 0x = . 2 0 0 6y = − + 6= The curve cuts the y-axis at (0, 6).

3 3 ( ) 213 28 3 9xx+ − =− ( ) ( )( ) 2 2 3 3 28 3 9 0 Let 3 . 3 28 9 0 3 1 9 1 or 93 13 3 93 1 or 2 xx x xx u uu uu uu xx  − + = = − + = − − = == == =− = 2 4 6 8 10 12-2-4 2 4 6 2 4 6 8 10 12-2-4 2 4 6 Y=6-X-X2 y x O -3

4 4 (a) 3 2 5x+ = + 3 4 4 5 5 (squaring both sides) 6 4 5 x x + = + +  = + Oops.. the question may be more challenging if it is changed to 32 3 4 4 (squaring both sides) 3 4 4 1 4 1 4 1 16 xx x x x x x x x x x + = + + = + + + − − = −= −= = (b) 20 27 80 12xx + = − ( ) ( ) 2 5 3 3 4 5 2 3 2 5 3 4 5 3 3 4 5 3 3 2 5 3 xx x x + = − + = − −= + ( ) ( )( ) ( ) ( ) ( ) 4 5 3 3 5 3 532 5 3 4 5 3 3 5 3 2 5 3 4 5 4 15 3 15 3 3 4 29 7 15 4 −−= −+ −− = − − − += −=

5 5 (a) ( )2 12 1 6 0 4x − −  ( ) 2 2 12 1 2 0 2x − −   112 1 2 2 1 2 022xx  − + − −      112 1 2 3 022xx  + −      The range of values of x for which ( )2 12 1 6 0 4x − −  is 3 4x − or 3 1 4x  . (b) 2 7 9 8 x x x c+ −  + 2 9 0x x c− − −  For 2 9 0x x c− − −  , the equation 2 9 0x x c− − − = has no real roots. 2 4 0b ac− ( ) ( )( )2 1 4 1 9 0 c− − − −  1 36 4 0 c+ +  37 4 0c+ 4 37c − 1 9 4c − y x O 0 x 1 2 3 4 5 6 7 8 9-1-2 1 2 3 4 5 6 7 8 9-1-2

6 6 (i) ( ) 2 3 2 ky k x kx= − − + Given that the curve has a minimum point, 3 0k − 3 ............(1)k  Given that the curve cuts the x-axis at two distinct points, 2 4 0b ac− ( ) ( )2 4 3 0 2 kkk − − −   ( )( )2 2 3 0k k k− −  22 2 6 0k k k− +  2 6 0kk− +  2 6 0kk − ( ) 6 0kk − The range of values of k is 0 6k . ……………….(2) Combining (1) and (2), the range of values of k is 3 6k . y k O 6

7 (ii) ( ) 2 3 2 ky k x kx= − − + When 4k = , ( ) 2 4 4 3 4 2y x x= − − + 2 4 2y x x= − + ---------- (1) y x c=+ ---------- (2) Substitute (1) into (2). 2 4 2 x x x c− + = + 2 5 2 0x x c− + − = Since the line meets the curve, the above equation has real roots. 2 4 0b ac− ( ) ( )( )2 5 4 1 2 0 c− − −  25 8 4 0 c− +  17 4 0c+ 4 17c − 1 4 4c − 7 The sum of their areas 2 84 cm= 2 84x xy+= ---------- (1) The sum of their perimeter 52 cm= ( )4 2 52x x y+ + = 4 2 2 52x x y+ + = 6 2 52xy+= 3 26xy+= 26 3yx=− ---------- (2) Substitute (2) into (1). ( )2 26 3 84x x x+ − = 22 3 26 84x x x− + = 22 26 84 0xx− + − = 2 13 42 0xx − − = ( )( ) 6 7 0xx− − = 6 0x −= or 7 0x −= 6x = 7x = x −6 −x x −7 −x x2 + −x

8 Substitute 6x = into (2). ( ) 26 3 6y =− 26 18=− 8= Substitute 7x = into (2). ( ) 26 3 7y =− 26 21=− 5= (rejected yx ) The dimensions of the rectangular card is 6 cm by 8 cm.