Question

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

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

Answer 1

Answer: (D)

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