HCI 2024 Term 2 Common Test Practice 4 (solution)

1 Solution to Term 2 CT Practice 4 1 ( ) ( ) 1 0.5 0.5 15 30 525 55 25 30 5 5 25 6 5 Let 5 . 56 5 5 1 5 1 0 xx x x xx x x y yy y y x − −+= +=  += = += = = = = 2 2 2 2 ( 2) 8( 2) 15 0 4 4 8 16 15 0 4 3 0 ( 3)( 1) 0 xx x x x xx xx + − + +  + + − − +  − +  − −  The range of values of x is 13 x . 3(i) 2 22 2 2 23 21 18 3( 7 6) 7736 22 7 253 24 7 753 24 x x x x x x x − + − =− − +    =− − − +       =− − − =− − + Maximum value is 75 4 . x

2 (ii) G1 – indicating all the intercepts G1 – indicating the turning point G1 – correct shape, smooth curve, labelling axis 4 3 2 2 2 3 2 3 3 2 24 3 2 2 3 3 2 2 2 3 2 3 3 2 2 3 3 2 2 6 3 2 2 3 3 2 1 1 3 6 2 3 3 2 2 3 3 2 6 6 6 3 6 2 6 6 6 18 66 132 2 622 −− − + − − −=  − +− − − −=  −  +− − − + −=− − =− + + 5 22 22 2 1---- (1) 9 3 4 0----(2) Sub (1) into (2): 9 ( 1) 3 4 0 10 3 0 13 or 25 11When , 22 38When , 55 yx x y x x x x xx x xy xy =+ + − − = + + − − = − − = =− =− = =− = The 2 coordinates are 1 1 3 8, and , .2 2 5 5    −       Deduct 1 mark for not writing in coordinates form 1 x 6 y -18 (3.5, 18.75) x

3 6 2 22 2 2 2 2 2 12 2 ( 3) 12 2 3 ( 3) 12 2 0 40 144 4( 3)(2 ) 0 8 24 144 0 3 18 0 3 18 0 ( 6)( 3) 0 6 3 x m m x x m mx x x m x m b ac mm mm mm mm mm m − = + − = + + − + = − − +  − − +  − − +  + −  + −  −   Line meets curve means 2 40b ac− Change the sign when you divide by negative number! *7 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 (1 ) 2 (2 1) Since (1 ) 0, the graph is an inverted U sh ape. 4 (2 ) 4(1 )(2 1) 4 4(2 1 2 ) 4 8 4 4 4( 2 1 ) Since 0 and 0 , 2 1 1 . 4 y p x pqx q p b ac pq p q p q q p q p p q q p p q q p p q p q q p =− + + − + − +  − = − + + = − + + + =− − − − =− + + +   + + +  − 2 2 2 2 2 ( 2 1 ) 4 0, Hence 4 0. p q q p b ac + + + −  − 2 2 2(1 ) 2 (2 1)y p x pqx q =− + + − + is always negative for all real values of x. Must mention the 2 conditions for the graph to be always negative!

4 8 ( ) ( ) ( ) ( ) ( ) 1 2 2 4 142 2 4 14 14 24 14 2 4 2 4 2 4 14 2 4 2 16 24 42 AM AM AM   + =  + = = + −= +− − = − =− − =− cm 9 2 2 2 2 2 2 1 4 12 2 1 4 2 3 0 ( 1)( 3) 0 1 or 3 ......... (1) 4 12 16 0 ( 4)( 4) 0 4 4 ......... (2) xx xx xx xx xx x x xx x −  −  −  − − + +  + −   −  − − − +  −   Combining the 2 inequalities, The range of values of x is The range of values of is 4 1or 3 4.x x x−  −   3 is a crowd… Separate them into 2 inequalities…

5 10 ( ) ( ) ( ) 2 2 2 2 2 3 2 2 2 3 2 4 4 2 2 (squaring both sides) 2 4 2 3 2 2 4 2 (isolate the surd) 2 4 4 2 2 2 16 2 4 4 4 16 32 4 16 16 16 32 0 4 32 48 0 8 xx x x x xx x x x xx xx x x x x x x xx xx − = + − − = + − + − = + + − − − − = − − = − − = − − + = − − + − + = − + = −+ ( )( ) 12 0 6 2 0 6 or 2 xx xx = − − =  = = Check: When 2, LHS 3(2) 2 2 RHS 2 2 2 2 x= = − = = + − = When 6, LHS 3(6) 2 4 RHS 2 6 2 4 x= = − = = + − = 6 or 2xx = =