Answer :

Given:- Function f(x) = sin x


Theorem:- Let f be a differentiable real function defined on an open interval (a,b).


(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)


(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)


Algorithm:-


(i) Obtain the function and put it equal to f(x)


(ii) Find f’(x)


(iii) Put f’(x) > 0 and solve this inequation.


For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.


Here we have,


f(x) = sin x



f’(x) = cosx


Taking different region from 0 to 2π


a) let


cos(x) > 0


f’(x) > 0


Thus f(x) is increasing in


b) let


cos(x) < 0


f’(x) < 0


Thus f(x) is decreasing in


Therefore, from above condition we find that


f(x) is increasing in and decreasing in


Hence, condition for f(x) neither increasing nor decreasing in (0,π)


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