The part (a) of this question attracted majority of the candidates. However, many of them were unable to get the L.C.M. A few others cross multiplied both sides of the minus sign. They were expected to use (x - 2)(x - 1) as L.C.M and obtained
(x+2(x-1) – (x+3)(x-2) = 4
(x - 2) (x – 1) (x - 2)(x - 1)
In part (b), candidates’ performance was fair. However, majority of the candidates appeared not to be familiar with coordinate of points on a graph. To find point C, candidates were expected to use x = 0 and y = 0 to get c = 0. Similarly, substituting 1 for x and 4 for y as well as 2 for x and 10 for y gives two equations in A and B i.e. A + B = 4; 4A + 2B = 10. Solving these equations simultaneously gave A = 1, B = 3, which gave the equation as y = x2 + 3x.
To get the 2nd point of intersection with the x – axis, put y = 0, in the equation to get x2 + 3x = 0. Solving this equation gave x = -3, Thus the required point = (-3, 0)