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Q.
If $f''(x)>0, \forall x \in R, f'(3)=0$ and $g(x)=f\left(\tan ^{2} x-2 \tan x+4\right), 0< x< \frac{\pi}{2}$, then $g(x)$ is increasing in
Application of Derivatives
Solution:
$g'(x)=\left(f'\left((\tan x-1)^{2}+3\right)\right) 2(\tan x-1) \sec ^{2} x$
Since $f''(x)>0 $
$\Rightarrow f'(x)$ is increasing
So, $f'\left((\tan x-1)^{2}+3\right)>f'(3)=0$
$\forall x \in\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Also, $(\tan x-1)>0 x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
So, $g(x)$ is increasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$