Question

Function $g\left(x\right)=9x-4\tan x,x\in (0,\frac{\pi }{2})$ will have

• A. No local extremum
• B. Local maximum at $x=\frac{1}{2}$
• C. Local maximum at ${\cos }^{-1}(-\frac{2}{3})$
• D. Local maximum at ${\cos }^{-1}\left(\frac{2}{3}\right)$

Answer 1

The correct option is D Local maximum at ${\cos }^{-1}\left(\frac{2}{3}\right)$

Given, $g\left(x\right)=9x-4\tan x$
$\Rightarrow g^{\prime }\left(x\right)=9-4{\sec }^{2}x=0$
$\Rightarrow {\sec }^{2}x=\frac{9}{4}$
$i.e.x={\cos }^{-1}\left(\frac{2}{3}\right)$
$\{\because x\in (0,\frac{\pi }{2})\}$
Also, $g^{\prime \prime }\left(x\right)={\sec }^{2}x(-8\tan x)$
$\Rightarrow g^{\prime \prime }\left({\cos }^{-1}\frac{2}{3}\right)=-8\left(\frac{9}{4}\right)\left(\frac{\sqrt{5}}{2}\right)<0$
Thus, $g$ has local maximum only at $x={\cos }^{-1}\left(\frac{2}{3}\right)$