Q58 / 170 · Manipal (MET) 2020 · Mathematics
If $f^{\prime}(x)\gt0 \forall x \in R, f^{\prime}(3)=0$ and $g(x)=f\left(\tan ^2 x-2 \tan x+4\right), 0 \lt x \lt \frac{\pi}{2}$, then $g(x)$ is increasing in
- A$\left(0, \frac{\pi}{4}\right)$
- B$\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$
- C$\left(0, \frac{\pi}{3}\right)$
- Correct answerD$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Solution
Given, $g(x)=f\left(\tan ^2 x-2 \tan x+4\right)$
On differentiating w.r.t. $x$, we get
$\begin{aligned} g^{\prime}(x)=f^{\prime}\left(\tan ^2 x-2 \tan x\right. & +4) \\ & \times\left(2 \tan x \sec ^2 x-2 \sec ^2 x\right)\end{aligned}$
$=f^{\prime}\left(\tan ^2 x-2 \tan x+4\right) 2 \sec ^2 x(\tan x-1)$
$\because \quad f^{\prime}(x)\gt0 \forall x \in\left(0, \frac{\pi}{2}\right)$
Also, $2 \sec ^2 x\gt0 \forall x \in\left(0, \frac{\pi}{2}\right)\left[\because x \in\left(0, \frac{\pi}{2}\right)\right.$, given $]$
But $\tan x-1\gt0 \forall x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
$\therefore \quad g^{\prime}(x)\gt0 \forall x \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Hence, $g(x)$ is increasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.
Hence, the correct answer is option: 'D'.