Trigonometrical Equations 3 Question 7
8. The smallest positive root of the equation $\tan x-x=0$ lies in
(1987, 2M)
(a) $0, \frac{\pi}{2}$
(b) $\frac{\pi}{2}, \pi$
(c) $\pi, \frac{3 \pi}{2}$
(d) $\frac{3 \pi}{2}, 2 \pi$
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Answer:
Correct Answer: 8. (c)
Solution:
- Let $ f(x)=\tan x-x $
We know, for $0<x<\frac{\pi}{2}$
$ \Rightarrow \quad \tan x>x $
$\therefore \quad f(x)=\tan x-x$ has no root in $(0, \pi / 2)$
For $\pi / 2<x<\pi, \tan x$ is negative.
$ \begin{aligned} & \therefore \quad f(x)=\tan x-x<0 \\ & \text { So, } \quad f(x)=0 \text { has no root in } \frac{\pi}{2}, \pi \text {. } \\ & \text { For } \quad \frac{3 \pi}{2}<x<2 \pi, \tan x \text { is negative. } \\ & \therefore \quad f(x)=\tan x-x<0 \end{aligned} $
So, $\quad f(x)=0$ has no root in $\frac{3 \pi}{2}, 2 \pi$.
We have, $f(\pi)=0-\pi<0$
and $\quad f \frac{3 \pi}{2}=\tan \frac{3 \pi}{2}-\frac{3 \pi}{2}>0$
$\therefore f(x)=0$ has at least one root between $\pi$ and $\frac{3 \pi}{2}$.
