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The smallest positive root of the equation $tanx\, -\, x = 0$ lies on
$\left( {0,\frac{\pi }{2}} \right)$
$\left( {\frac{\pi }{2},\pi } \right)$
$\left( {\pi,\frac{3\pi }{2}} \right)$
$\left( {\frac{3\pi }{2},2\pi } \right)$
Solution
Let $f(x)=\tan x-x$
For $0<x<\frac{\pi}{2}$
$\tan x>x$
$\therefore f(x)=\tan x-x$ has no roots in $\left(0, \frac{\pi}{2}\right)$
For $\frac{\pi}{2}<x<\pi \tan x$ is negative $\therefore f(x)=\tan x-x<0$
So $f(x)=0$ has no roots in $\left(\frac{\pi}{2}, \pi\right)$ For $\frac{3 \pi}{2}<x<2 \pi, \tan x$ is negative
$\therefore f(x)=\tan x-x<0$
So $f(x)=0$ has no roots in $\left(\frac{3 \pi}{2}, 2 \pi\right)$ We have $f(\pi)=0-\pi<0$
and $f\left(\frac{3 \pi}{2}\right)=\tan \frac{3 \pi}{2}-\frac{3 \pi}{2}>0$
$\therefore f(x)=0$ has atleast one root between $\pi$ and $\frac{3 \pi}{2}$
