Question

If $f^{\prime \prime }\left(x\right)>0,\forall x\in R,f^{\prime }\left(3\right)=0$ 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. $(\frac{\pi }{6},\frac{\pi }{3})$
• B. $(0,\frac{\pi }{3})$
• C. $(0,\frac{\pi }{4})$
• D. $(\frac{\pi }{4},\frac{\pi }{2})$

Answer 1

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

$g\left(x\right)=f({\tan }^{2}x-2\tan x+4),0<x<\frac{\pi }{2}$

$g^{\prime }\left(x\right)=f^{\prime }({\tan }^{2}x-2\tan x+4)\times (2\tan x{\sec }^{2}x-2{\sec }^{2}x)$

For $g\left(x\right)$ to be increasing

$g^{\prime }\left(x\right)>0$

$f^{\prime }({\tan }^{2}x-2\tan x+4)(2\tan x{\sec }^{2}x-2{\sec }^{2}x)>0$

$2\tan x{\sec }^{2}x>2{\sec }^{2}x$

$\tan x>1$

$\tan x>1$ for $xϵ(\frac{\pi }{4},\frac{\pi }{2})$ when $[0<x<\frac{\pi }{2}]$