Question

The domain of $f\left(x\right)={\sin }^{-1}\left[{\log }_2\left(\frac{x}{2}\right)\right]$ is

• A. $0\le x\le 1$
• B. $0\le x\le 4$
• C. $1\le x\le 4$
• D. $\left[-2,-1\right]\cup \left[1,2\right]$

Answer 1

The correct option is D

$\left[-2,-1\right]\cup \left[1,2\right]$

Explanation for the correct option:

Find the domain of ,

Consider the given function as,

$f\left(x\right)={\sin }^{-1}\left[{\log }_2\left(\frac{x}{2}\right)\right]$

Defined the above function as,

$$\begin{gathered} \Rightarrow -1\le {\log }_2\left(\frac{x^2}{2}\right)\le 1 \\ \Rightarrow 2^{-1}\le \frac{x^2}{2}\le 2^1 \\ \Rightarrow \frac{1}{2}\le \frac{x^2}{2}\le 2 \\ \Rightarrow 1\le x^2\le 4 \end{gathered}$$

Then, $\left|x\right|$ lies between $1\text{and}2$ this can be expressed as,

$1\le \left|x\right|\le 2$

Since, the domain of the function $f\left(x\right)$ is $\left[-2,-1\right]\cup \left[1,2\right]$

Hence, the correct answer is Option D.