This question was also reported to be attempted by majority of the candidates and their performance was described as fair.  Candidates were reported to have performed better in part (a) than in any other part of the question. 

In part (a), since the function was quadratic, the minimum/maximum value would either be at the turning points or at the end points of the domain i.e. -3 ≤ x ≤ 4.  Therefore candidates were expected to substitute the end points of the domain -3≤ x ≤ 4 to obtain f(-3) = 29, f(4) = 50.  The minimum value of the function was obtained by differentiating the function to obtain f1(x) = 6   At the turning
point  = 0. f(0) = 2.  Therefore, 2 is the minimum value of the function and since x = 0 is part of the given interval, the range of f = 2≤ f() ≤ 50.

In part (b), candidates were expected to show that if g-1 is the inverse of g, then g(g-1(x)) =  i.e.  = .  Solving this equation gave g-1 = ,  x  1.  g-1 of (-2) =   =   = .