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#### Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 31 maths

$f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})$

Given:

$f(x)=log\: cos\: x$

To prove:

$\text { We have to prove that } f(x) \text { is strictly increasing on }(-\frac{\pi }{2},0) \text { and strictly decreasing on }(0,\frac{\pi }{2})$

Hint:

1. for f(x) to be increasing we must have f'(x)>0
2. for f(x) to be decreasing we must have f'(x)<0

Solution:

Given

$f(x)=log\: cos\: x$

On differentiating both sides w.r.t x we get

\begin{aligned} &\Rightarrow f^{\prime}(x)=\frac{d}{d x}[\log \cos x] \\ &\Rightarrow f^{\prime}(x)=\frac{1}{\cos x} \times(-\sin x) \\ &\Rightarrow f^{\prime}(x)=-\tan x \\ &\text { In interval }\left(0, \frac{\pi}{2}\right), \tan x>0 \\ &\Rightarrow-\tan x<0 \end{aligned}

By applying negative sign change comparison sign

$\therefore f^{\prime}(x)<0 \text { on }\left(0, \frac{\pi}{2}\right)$

$\text { Hence } f(x) \text { is strictly decreasing on }(0,\frac{\pi }{2})$

\begin{aligned} &\text { In interval }\left(\frac{\pi}{2}, \pi\right), \tan x<0 \\ &\Rightarrow-\tan x>0 \end{aligned}

By applying negative sign change comparison sign

$\therefore f^{\prime}(x)>0 \text { on }\left(\frac{\pi}{2}, \pi\right)$

$\text { Hence } f(x) \text { is strictly increasing on }(\frac{\pi }{2},\pi )$