HCI 2024 Term 2 Common Test Practice 1 Solution (updated)

1 Solution to Common Test Practice 1 1 The line x + y = 3 meets the curve x2 – 2x + 2y2 = 3 at two distinct points. Find the coordinates of these two points. [4] 22 22 22 22 2 3 3 .........(1) – 2 2 3 ......(2) Substitute (1) into (2) 2 2(3 ) 3 2 2(9 6 ) 3 2 18 12 2 3 0 3 14 15 0 (3 5)( 3) 0 5 or 33 53 or 3 33 4 3 xy yx x x y x x x x x x x x x x x xx xx xx yy += =− += − + − = − + − + = − + − + − = − + = − − = == = − = − = ( ) 0 54Points are , and 3,0 .33 =   2 Find the possible values of k for which the line 1−= kxy does not intersect the curve 2 3.yx=+ [3] ( ) 2 2 2 2 2 1 ........... (1) 3 ........ (2) 31 40 Discriminant 0 4(1)(4) 0 16 0 ( 4)( 4) 0 4 4 y kx yx x kx x kx k k kk k =− =+ + = − − + =  − −  − − +   −  

2 3 Using a suitable substitution, solve the equation 14 16 66xx− += . [4] ( ) 1 2 2 2 23 4 16 66 Let 4 664 4 264 4 264 0 (4 33)( 8) 0 338 or n.a. 4 4 8 22 3 2 xx x x x u u u uu uu uu uu x − += = += += + − = + − = −== = = = 4 Sketch the graph of 24 24 20y x x= − + , indicating clearly the coordinates of the turning point, x and y intercepts. [4] ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 4 24 20 intercept is 0, 20 . 4 24 20 0 4( 6 5) 0 4( 5)( 1) 0 5 or 1 intercepts are 5,0 and 1,0 . 4 24 20 4( 6 5) 4[( 3) 9 5] 4[( 3) 4] Minimum point is 3, 16 . y x x y xx xx xx xx x y x x xx x x = − + − − + = − + = − − =  = = − = − + = − + = − − + = − − − Not labelling the axes and graph results in loss of 1 mark! Not writing the values of the 3 critical points results in loss of another mark! Curve not drawn symmetrical about the turning point results in another 1 mark gone!

3 5 The area of a rectangle ABCD is given by m2, and the length of side AB is m. Find, without using a calculator, the length of side BC in the form of m, where a and b are real numbers. [3] 1 3 2Length of 28 BC += + ( )( ) 1 3 2 2 8 2 8 2 8 1 3 2 2 2 2 48 2 2 2 6 2 12 4 10 4 2 4 5 2 2 2 5 2 m2 +−= +− +− = − −+−= − −+= − −= =− 6 Given that 3 1 4 48 2 3 3 3 − +− + can be expressed in the form of 3ab+ , find the value of a and of b . [4] 3 1 4 3 1 2 3 3 4 348 4 3 32 3 3 3 2 3 3 2 3 3 − − −+ − =  + − + + − 6 5 3 3 4 3 4312 9 3 9 5 3 4 3 4333 5 3 4 33 4 333 33 −+= + −− −= + − = − + − =+ 3, 1.ab = = ( )1 3 2+ ( )28+ ( 2)ab+

4 7 Show that 24 12 10xx−+ is always positive for all real values of x . [3] Method 1 2 2 2 2 4 12 10 10 4 3 4 3 9 10 4 2 4 4 3 4 1 1 2 y x x xx x x = − + = − + = − − + = − +  Method 2 (sketch graph) 2   is always positive for all real values of .4 12 10 x xx − + 8 Show that the roots of 2 ( 2) 2x p x p+ − = are real. [3] ( ) 2 2 2 ( 2) 2 ( 2) 2 0 Discriminant = 2 4(1)( 2 ) x p x p x p x p pp + − = + − − = − − − 2 2 2 4 4 8 44 ( 2) 0 p p p pp p = − + + = + + = +  2the roots of   are . ( 2) 2 realx p x p + − = 9 Solve the equation 25 50 2 12xx− − − = . [3] ( ) 25 50 2 12 25 2 2 12 5 2 2 12 4 2 12 2 3 2 9 11 xx xx xx x x x x − − − = − − − = − − − = −= −= −= =

5 10 Find the solution set of each of the following inequalities: (i) 1832 + xx [2] (ii) 9)2( 2 +x [2] Hence, state the set of values of x which satisfy both of these inequalities. [1] (i) 1832 + xx 2 3 18 0 ( 6)( 3) 0 3 or 6 xx xx xx − −  − +  −  (ii) 9)2( 2 +x 2 2 4 4 9 0 4 5 0 ( 5)( 1) 0 51 xx xx xx x + + −  + −  + −  −   To satisfy (1) and (2), 53 x−  − can draw the 2 inequalities on the number number for the final solution!

6 11 The given diagram shows a quadratic curve with a y-intercept of 48 which crosses the x-axis at 4x=− and 6x= . (i) State the equation of the line of symmetry for the curve. [1] (ii) Find the equation of the parabola in the form 2 where 0y ax bx c a= + +  . [3] (i) 46 2x −+= 1= (ii) Let ( 4)( 6)y a x x= + − At (0,48), 2 2 48 ( 24) 2 2( 4)( 6) 2( 2 24) 2 4 48 a a y x x xx xx =−  =− =− + − =− − − =− + + x y