The correct option (b) [0, (π/4)]
Explanation:
f(x) = tan–1(sin x + cos x)
f'(x) = [1/{1 + (sin x + cos x)2}] (cos x – sin x)
∴ f'(x) = [(cos x – sinx)/(2 + 2 sinx cosx)] = [{cosx (1 – tanx)}/{2 + sin2x}]
Now 2 + sin2x > 0 for all x in [0, (π/4)] also cosx > 0 for all x in [0, (π/4)]
nd 1 – tanx > 0 for all x in [0, (π/4)]
∴ f'(x) > 0
∴ f(x) is strictly increasing in [0, (π/4)]