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)