Question

The domain of $f\left(x\right)=\frac{1}{\sqrt{\left(\left|\cos x\right|+\cos x\right)}}$ is

• A. $\left[\left(4n-1\right)\frac{\mathrm{\pi }}{2},\left(4n+1\right)\frac{\mathrm{\pi }}{2}\right]$
• B. $\left[\left(4n+1\right)\frac{\mathrm{\pi }}{2},\left(4n-1\right)\frac{\mathrm{\pi }}{2}\right]$
• C. $\left(\mathrm{n\pi },\left(\mathrm{n}+1\right)\mathrm{\pi }\right)$
• D. $\left[2n\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{2}\right),\left(\frac{\mathrm{\pi }}{2}\right)+2\mathrm{n\pi }\right]$

Answer 1

The correct option is D

$\left[2n\mathrm{\pi }-\left(\frac{\mathrm{\pi }}{2}\right),\left(\frac{\mathrm{\pi }}{2}\right)+2\mathrm{n\pi }\right]$

Explanation for the correct option :

Find the domain of the $f\left(x\right)$,

Consider the given Equation as,

$f\left(x\right)=\frac{1}{\sqrt{\left(\left|\cos x\right|+\cos x\right)}}$

In the above Equation under root

$$\begin{gathered} \left|\cos x\right|+\cos x>0 \\ \left|\cos x\right|+\cos x\ne 0 \end{gathered}$$

Let us say

$g\left(x\right)=\left|\cos x\right|+\cos x=\left\{\begin{array}{ll}2\cos x,& \cos x>0\\ 0& \cos x<0\end{array}\right.$

From the graph,

$x\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

$\cos x$ Period is equal to $2\mathrm{\pi }$, then the intervals are

$$\begin{gathered} 2\mathrm{\pi }-\frac{\mathrm{\pi }}{2},2\mathrm{\pi }+\frac{\mathrm{\pi }}{2} \\ 4\mathrm{\pi }-\frac{\mathrm{\pi }}{2},4\mathrm{\pi }+\frac{\mathrm{\pi }}{2} \end{gathered}$$

Since,

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

Hence, the correct answer is Option D.