Question

If $f\text{'}\text{'}\left(x\right)>0,\forall x\in R,f\text{'}\left(3\right)=0$ and $g\left(x\right)=f({\tan }^{2}x-2\tan x+4),0<x<\frac{\pi }{2}$, then $g\left(x\right)$ is increasing in

• A. $(0,\frac{\pi }{4})$
• B. $(\frac{\pi }{6},\frac{\pi }{3})$
• C. $(0,\frac{\pi }{3})$
• D. $(\frac{\pi }{4},\frac{\pi }{2})$

Answer 1

The correct option is D $(\frac{\pi }{4},\frac{\pi }{2})$

$g^{\prime }\left(x\right)=\left(f^{\prime }\left(\right(\tan x-1{)}^2+3)\right)2(\tan x-1){\sec }^{2}x$
Since $f^{\prime \prime }\left(x\right)>0,f^{\prime }\left(x\right)$ is increasing
So,
$f^{\prime }\left(\right(\tan x-1{)}^2+3)>f^{\prime }\left(3\right)=0\forall x\in (0,\frac{\pi }{4})\cup (\frac{\pi }{4},\frac{\pi }{2})$
Also, $(\tan x-1)>0x\in (\frac{\pi }{4},\frac{\pi }{2})$
So, $g\left(x\right)$ is increasing in $(\frac{\pi }{4},\frac{\pi }{2})$.