2022 H2 Math RI Prelim Paper 1 Solns

RAFFLES INSTITUTION 2022 Year 6 H2 Mathematics Preliminary Examination Paper 1 Questions and Solutions with comments Page 1 of 18 1 (i) On the same axes, sketch the graphs of xy xa= − and 2 1,y xa= −+ where a is a positive constant. [3] (ii) Hence solve the inequality 2 1,x xaxa > −+− leaving your answer in terms of a. [3] (i) Note that the asymptotes intersect at (a, 1), and the vertex of the modulus graph is at the same point. Also note the curve passes through the origin. (ii) Let the 2 graphs intersect at the point T. To solve for T, consider 2 2 2( ) 1 2( ) () 2 2 x xaxa x xa xa axa axa = −+− = − +− −= = ± Since xa> at the point T, 2 2 axa= + . Thus the solution is 2 .2 aaxa<<+ The graphs are in (i), so it is easier to solve the equation to find the x- coordinate of the intersection point, then refer to the graphs in (i) to write down the solution to the inequality. 0

Page 2 of 18 2 A function is defined as 2 2 12f ( ) , , 0, 2.2( 2 ) xxx xx xx +−= ∈≠ −  (i) Show that f( )x can be written in the form 2 2 2( ) ( )1 xpq xp  −+  +− , where p and q are constants to be found. [2] (ii) Hence describe a sequence of transformations that will transform the graph of 2 2 2 1 xy x −= − onto the graph of f( ).yx= [2] (iii) Determine, with the help of a sketch or otherwise, the set of values of k for which the equation 2 2 2 1 x kx − =− has no real roots. [2] (i) 2 2 2 2 2 2 12 2( 2 ) 1 2 ( 2 1) 2 ( 2 1) 1 1 2 ( 1) 2 ( 1) 1 xx xx xx xx x x +− −  −−+=  − +−  −−=  −− Thus 11 and . 2pq= −= Do show enough working for a ‘show’ question. Note that a ‘show’ question is different from a ‘verify’ question, so here the given answer should not be expanded to compare with the expression for f. (ii) The 2 transformations (in either order) are 1. A translation of 1 unit in the positive x-direction 2. A scaling parallel to the y-axis by a factor of 1 .2 Remember to use standard mathematical terms and descriptions. Unacceptable ones include ‘along the x-axis’, ‘in the x-axis’, ‘on the x-axis’. (iii) The equation 2 2 2 1 x kx − =− has no real roots for values of k in the set ( ]2, 1−− . It is usually easier to use the method as suggested in the question. Here the question suggested a sketch, and the correct answer was mainly obtained by those who referred to the graph instead of the discriminant method. 1y=− 1x=− 1x= 2 2 2 1 xy x −= −

Page 3 of 18 Alternative Solution: 2 2 22 2 2 1 2 ( 1) ( 2) 0 x kx x kx k kxk − =− −= − + −+= If 1k =− , 2( 1) ( 2) 0kxk+ −+= becomes 10−= , so there is no solution. If 1k ≠− , the quadratic equation will have no real roots if [ ]0 4 ( 2)( 1) 0 ( 2)( 1) 0 21 kk kk k −−+ + < + +< − < <− So 21 k−