Question

The smallest positive root of the equation $tanx-x=0$ lies is

• A. $(0,\pi /2)$
• B. $(\pi /2,\pi )$
• C. $(\pi ,3\pi /2)$
• D. $(3\pi /2,2\pi )$

Answer 1

The correct option is C $(\pi ,3\pi /2)$

Let $f\left(x\right)=\tan x-x$
For $0<x<\frac{\pi }{2}$
$\tan x>x$
$\therefore f\left(x\right)=\tan x-x$ has no roots in $(0,\frac{\pi }{2})$
For $\frac{\pi }{2}<x<\pi$ $\tan x$ is negative
$\therefore f\left(x\right)=\tan x-x<0$
So $f\left(x\right)=0$ has no roots in $(\frac{\pi }{2},\pi )$
For $\frac{3\pi }{2}<x<2\pi ,\tan x$ is negative
$\therefore f\left(x\right)=\tan x-x<0$
So $f\left(x\right)=0$ has no roots in $(\frac{3\pi }{2},2\pi )$
We have $f\left(\pi \right)=0-\pi <0$
and $f\left(\frac{3\pi }{2}\right)=\tan \frac{3\pi }{2}-\frac{3\pi }{2}>0$
$\therefore f\left(x\right)=0$ has atleast one root between $\pi$ and $\frac{3\pi }{2}$