Question

Prove that function f given by $f\left(x\right)=log\left(cosx\right)$ is strictly decreasing on $(0,\frac{\pi }{2})$ and strictly increasing on $\left(\frac{\pi }{2}\pi \right)$.

Answer 1

Given, $f\left(x\right)=log\left(cosx\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{cosx}.(-sinx)=-tanx$ (Differentiate w.r.t x)
In interval $(0,\frac{\pi }{2}),tanx>0$ ($\because$ tan x is in 1st quadrant)
$\Rightarrow -tanx<0$ ($\because tanx$ is in 1st quadrant)
$\therefore f^{\prime }\left(x\right)<0$ in $(0,\frac{\pi }{2})$
Hence, f is strictly decreasing in $(0,\frac{\pi }{2})$
Also, in interval $\left(\frac{\pi }{2}\pi \right),tanx<0\Rightarrow -tanx>0$ ($\because tanx$ is in IInd quadrant)
$\therefore f^{\prime }\left(x\right)>0in\left(\frac{\pi }{2}\pi \right)$
Hence, f is strictly increasing in $\left(\frac{\pi }{2}\pi \right)$