Answer :

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.

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