(a and c)

The function f is not differentiable at x = 0,

x = π /2 as f ′ (0 – ) = –10, f ′ (0 + ) = 1;

f ′ ( π /2 – ) = 1, f ′ ( π /2 + ) = –1. The function f ′ (x) is given by

f ′ (x) =

The critical points of f are given by f ′ (x) = 0 or x = 0, π /2.

Thus critical points are

x = π /2, x = 0. Since f ′ (x) > 0, for 0 < x < π /2 and f ′ (x) < 0, for π /2 < x < π so f has local maxima at x = π /2. Also f ′ (x) < 0 for –1 ≤ x < 0 and f ′ (x) > 0 for 0 < x < π /2 so f has local minima at x = 0. Since f(–1) = –9, f( π /2) = 1, f(0) = 0 and f( π ) = 0. Thus f has absolute minimum at x = –1 and absolute maximum at x = 0