It was reported that majority of the candidates who attempted this question performed well in it.  Candidates were expected to expand (1 + mx)n in ascending powers of x up to the first three terms as: (1 + mx)n = 1     They would then compare this expression to 1+14x  + 84x2 to obtain mn = 14 --- (i) and  =
84 ---(ii). Solving these two equations simultaneously gave n = 7 and m = 2.  Hence,
 (1 + mx)n = (1+2x)7.

In part (b), candidates were expected to compare (1+2x)7 with (1.06)n to obtain              2x = 0.06.  This implied that x = 0.03. This value of x should then be substituted into the expression 1 + 14x + 84x2 to obtain 1 + 14(0.03) + 84(0.03)2 = 1 + 0.42 + 0.0756 = 1.50, correct to 3 significant figures.