question_answer

Prove that the function f given by \[f(x)=\log \,\,\cos \,x\]is strictly decreasing.

Answer:

Given, \[f(x)=\log \,\,\cos \,x\] On differentiating both sides w.r.t. x, we get             \[f'(x)=\frac{1}{\cos \,x}\cdot \frac{d}{dx}(\cos \,x)\]             \[=\frac{1}{\cos \,x}\cdot (-\sin \,x)=-\tan \,x\] We know that, for \[x\in \left( 0,\,\,\frac{\pi }{2} \right),\,\,\tan \,x>0\] \[\therefore \]      \[f'(x)=-\tan \,x<0\] Hence, \[f(x)\] is strictly decreasing.  Hence proved.