Answer :

(i) It is given that f(x) =









Now, if f’(x) =0


cos x = 0 or cosx = 4


But, cosx = 4 is not possible


Therefore, cosx =0


x =


Now, x = divides (0,2π) into three disjoints intervals



In the intervals and, f’(x)>0


Therefore, f(x) is increasing for 0< x < and < x < 2π.


In interval, f’(x)<0


Therefore, f(x) is decreasing for < x < .


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