Question

The domain of the function $f\left(x\right)=\sqrt{\cos x}$is

• A. $\left[\frac{3\mathrm{\pi }}{2},2\mathrm{\pi }\right]$
• B. $\left[0,\frac{\mathrm{\pi }}{2}\right]$
• C. $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
• D. $\left[0,\frac{\mathrm{\pi }}{2}\right]\cup \left[\frac{3\mathrm{\pi }}{2},2\mathrm{\pi }\right]$

Answer 1

The correct option is C

$\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Finding the domain of the function $f\left(x\right)$:

We have the given function as: $f\left(x\right)=\sqrt{\cos x}$

Define the above function as:

$\Rightarrow \cos x\ge 0$, since if the value if less than zero, it will become an imaginary function.

$\Rightarrow 0\le \cos x\le 1$, Since the value of $\cos x$ lies between $[0,1]$.

$\cos x$ lies between $0\text{to}1$. Then, $x$ lies between $1^{st}$ or $4^{th}$ quadrant

$$\begin{gathered} \Rightarrow 0\le x\le \frac{\mathrm{\pi }}{2}\text{or}\frac{3\mathrm{\pi }}{2}\le x\le 2\mathrm{\pi } \\ \Rightarrow \mathrm{x}\in \left[0,\frac{\mathrm{\pi }}{2}\right]\cup \left[\frac{3\mathrm{\pi }}{2},2\mathrm{\pi }\right] \end{gathered}$$

We can rewrite it as:

$x\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Hence, the domain of the given function is $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Therefore, the correct answer is option (C).