Question

What is the domain of the function $f\left(x\right)=\frac{1}{\sqrt{\left|\tan x\right|-\tan x}}$is

• A. $\left(2n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{4}\right),\mathrm{n\pi }+\left(\frac{\mathrm{\pi }}{4}\right)\right),n\in I$
• B. $\left(n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{2}\right),\mathrm{n\pi }+\left(\mathrm{\pi }\right)\right),n\in I$
• C. $\left(n\mathrm{\pi },\pi +\left(\frac{\mathrm{\pi }}{2}\right)\right),n\in I$
• D. None of these

Answer 1

The correct option is B

$\left(n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{2}\right),\mathrm{n\pi }+\left(\mathrm{\pi }\right)\right),n\in I$

Explanation of correct answer :

Finding domain of the function :

Given, $f\left(x\right)=\frac{1}{\sqrt{\left|\tan x\right|-\tan x}}$

As we know, the term under square root must be non- negative while it is in denominator , So it must be positive and not equal to zero.

$\Rightarrow \left|\tan x\right|-\tan x>0$

$2$ cases are possible : $1)\tan x>02)\tan x<0$

First case :

$$\begin{gathered} \Rightarrow \tan x-\tan x>0 \\ \Rightarrow 0>0 \end{gathered}$$

This condition is not possible.

Second case:

So, for $\tan x<0$ , $f\left(x\right)$ will be defined as

$f\left(x\right)=\frac{1}{\sqrt{-2\tan x}}$

$$\begin{gathered} \Rightarrow -2\tan x>0 \\ \Rightarrow \tan x<0 \end{gathered}$$

$\tan x$should be negative,

so, $n\mathrm{\pi }+\frac{\mathrm{\pi }}{2}\le \mathrm{x}\le \mathrm{n\pi }+\mathrm{\pi },\mathrm{n}\in \mathrm{I}$

Hence, the domain of the function is $\left(n\mathrm{\pi }+\left(\frac{\mathrm{\pi }}{2}\right),\mathrm{n\pi }+\left(\mathrm{\pi }\right)\right),n\in I$.

Thus, the correct answer is option(B).