Question

If $f\left(x\right)=\frac{x}{1+x\tan x},x\in (0,\frac{\pi }{2})$ , then

• A. $f\left(x\right)$ has exactly one point of minima
• B. $f\left(x\right)$ has exactly one point of maxima
• C. $f\left(x\right)$ is many one in $(0,\frac{\pi }{2})$
• D. none of these

Answer 1

The correct option is B $f\left(x\right)$ has exactly one point of maxima

Given, $f\left(x\right)=\frac{x}{a+x\tan x}$

Thus $f^{\prime }\left(x\right)=\frac{(1+\tan x)-x(\tan x+x{\sec }^{2}x)}{{(1+x\tan x)}^2}$
$=\frac{1-x^2{\sec }^{2}x}{{(1+x\tan x)}^2}=\frac{(1+x\sec x)(1-x\sec x)}{{(1+\tan x)}^2}=0$ ....for extrema.
$\Rightarrow x\sec x=1$ i.e. $x=\cos x$
$\therefore f\left(x\right)$ has exactly one point of maximum viz: $x=\cos x$.