HCI 2024 Term 2 Common Test Practice 2 Solution (updated)

1 Solution to Common Test Practice 2 1 The equation of a curve is y = mx2 + (c – m)x – (c + 3). Given that m > 0, show that the curve cuts the x-axis at two distinct points. [3] Let mx2 + (c – m)x – (c + 3) = 0 Discriminant = (c - m)2 - 4m(-c - 3) = c2 - 2cm+ m2 + 4mc +12m = c2 + 2cm+ m2 +12m = (c + m)2 +12m Since m > 0 and (c + m)2 ³ 0, the discriminant > 0. Hence, the curve cuts the x-axis at two distinct points. 2 (a) Find the exact value of x if 218 32x+− = . [3] (b) Simplify 11 5 3 5 3 + +− . [2] (c) Find the area of a right angled triangle ABC, given that angle ABC is a right angle, ( )5 3 2 cmAB=+ and 54 cm 2 BC =− . Leave your answer in the form 2ab+ . [3] (a) 218 32x−+ = 5 3 2 1 2(2 ) 2 x−+ = 563 2x− + = 1 12x= (b) ( )( ) 1 1 5 3 5 3 5 3 5 3 5 3 5 3 − + ++= +− +− 10 25 3 10 22 5 11 = − = =

2 (c) 15Area of (5 3 2) 4 2 2 ABC  = + −  1 2520 12 2 152 2 = − + − 2 1 25 220 12 2 1522 12 5 22 52 cm24 = − + − =−  =− 3 (a) Find the range of values of x for which (2 1) 2522 xx x+  + . [3] (b) Given that the curve whose equation is 2()y p x q= − − crosses the x-axis at the points ( )1,0 and ( )3,0 , (i) find the value of p and of q, [4] (ii) the maximum value of y. [1] (a) (2 1) 2522 xx x+  + ( )( ) 2 2 2522 25 5 5 0 5 or 5 xxx x xx xx +  +  + −   −  (b) (i) At ( )1,0 , 20 (1 )pq= − − 2(1 )pq=− At ( )3,0 , 20 (3 )pq= − − 2(3 )pq=− 22 22 2 2 (1 ) (3 ) 1 2 9 6 4 8 2 When 2, (1 ) (1 2) 1 qq q q q q qq q p q − = − − + = − + =  = = = − =− = 1, 2pq = = (b) (ii) Since 21 ( 2) ,yx= − − the maximum value of is 1.y

3 4 Find the values of q for which the line 8y qx=− is a tangent to the curve 2 4xy= . Leave your answer in exact values. [4] 8y qx=− ………. (1) and 2 4xy= ………. (2) Substitute (1) into (2), 2 4( 8)x qx=− 2 4 32 0x qx− + = Since 8y qx=− is a tangent to curve, discriminant 0= 2( 4 ) 4(32) 0q− − = 216 128 0q −= 2 80q −= 8 2 2 q = = 5 Solve the simultaneous equations: 27 x ÷ 3 y = 9, 2 2x × 4 1–y = 64. [4] 3 2 1– 2 2– 2 2 6 27 3 9  3    3      3 2 3 2  ........... (1)                          2 4 64 2 2 ......... (2)   Substitute (1) into (2). Sub 3 2 2 2 2 6 2 2 4 2 (3 2) 2 2 2 2 0 xy xy xy xy x xy xy xy xx x y yx x = − = + − = −= −= − =  −= = = = −= −+ =  ( ) 2 stitute into (0 30 ). 2 1x y = =− =− .0, 2 xy = =−

4 6 Sketch the graph of 23 8 9,y x x= − + showing clearly the coordinates of the turning point and the y-intercept. [5] ( ) 2 2 22 2 2 2 3 8 9 833 3 8 4 433 3 3 3 4 1633 39 4 113 33 4 11Turning point is , .33 -intercept is 0,9 . y x x xx xx x x y = − + = − +     = − + − − − +        = − − + = − +   7 Using a suitable substitution, solve the equation 12 2 3xx+−+= . [4] 1 1 2 2 2 3 12 2 3 2 Let 2 123 2 3 1 0 (2 1)( 1) 0 1 or 12 12 or 2 12 1 or 0 xx x x x xx u u u uu uu uu xx +−+=  + = = += − + = − − = == == =− =

5 8 Solve the equation : 3 4 0xx− − = . [4] ( ) ( ) 22 2 2 3 4 0 4 3 isolate surd 43 8 16 9 17 16 0 ( 16)( 1) 0 16 or 1 (n.a.) Check: When 1, LHS 1 3 1 4 6 RHS When 16, LHS 16 3 16 4 0 RHS 16 xx xx xx x x x xx xx xx x x x − − = −= −= −+= − + = − − = == = = − − =−  = = − − = = = If there are 2 answers, must check the validity of both asnwers!