Trigonometrical Equations Ques 53

  1. 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: 53.(c)

Solution:

Formula:

Domain and Range of Trigonometric Functions:

  1. 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 $\frac{3 \pi}{2}$.



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