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.