It is given that f (x) = x¹⁰⁰ + sin x –1

Then, f’(x) = 100x⁹⁹ + cosx

In interval (0,1), cos x >0 and 100x⁹⁹ > 0

⇒ f’(x)>0

Therefore, function f is strictly increasing in interval (0,1).

In interval, cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 in

Therefore, function f is strictly increasing in interval .

In interval , cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 on

Therefore, function f is strictly increasing in interval .

Hence, function f is strictly decreasing on none of the intervals.