This was one of the questions where candidates’ performance was commended especially in part (a).
In part (a), candidates were reported to show that if (x + 2) and (x – 1) are factors of f(x), then f(-2)  = 0. i.e 6(-2)4 + m(-2)3 – 13(-2)2 + n(-2) + 14 = 0 or 96 – 8m – 52 – 2n + 14 = 0 i.e 4m + n = 29 .... (i)  Similarly 6(1)4 + m(1)3 – 13(1)2 + n(1) + 14 = 0
i.e m+ n = -7...(ii).  Solving these two equations simultaneously gave m = 12 and n = -19.

In part (b), a few candidates were unable to obtain the points where x = 1.  To obtain these points, candidates were expected to substitute 1 for x in the equation of the circle and solve the resulting equation to obtain y i.e when x  = 1, (1)2 + y2 – 2(1) + 2y + 1 = 0 i.e y2 + 2y = 0 Þ y = 0 nd -2.  Hence the points are (1,0) and (1, -2).  Differentiating
 x2 + y2 – 2x + 2y + 1 = 0 with respect to x gave  =  .  At (1,0),  = 0.  Similarly at (1, -2),  = 0.  This implied that the gradient at these points is zero.  Thus, any tangent drawn at these points is parallel to the x – axis.