Question

The domain of ${\sin }^{-1}\left[{\log }_2\frac{x}{12}\right]$ is ?

• A. $\left[2,12\right]$
• B. $\left[-1,1\right]$
• C. $\left[\left(\frac{1}{3}\right),24\right]$
• D. $\left[\left(\frac{2}{3}\right),24\right]$
• E. $\left[6,24\right]$

Answer 1

The correct option is E

$\left[6,24\right]$

Explanation for the correct option(s)

Find the domain.

Consider the given function as,

$f\left(x\right)={\sin }^{-1}\left[{\log }_2\frac{x}{12}\right]$

We know that,

$\frac{\mathrm{\pi }}{2}\le {\sin }^{-1}x\le \frac{\mathrm{\pi }}{2}$

Then,

$$\begin{gathered} \Rightarrow \frac{\mathrm{\pi }}{2}\le {\sin }^{-1}\left[{\log }_2\frac{x}{12}\right]\le \frac{\mathrm{\pi }}{2} \\ \Rightarrow -1\le {\log }_2\frac{x}{12}\le 1 \end{gathered}$$

We know that

$$\begin{gathered} \Rightarrow {\log }_{a}x=b \\ \Rightarrow x=a^b \end{gathered}$$

$$\begin{gathered} \because {\log }_2\frac{x}{12}\ge -1 \\ \Rightarrow \frac{x}{12}=2^{-1} \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow 2^{-1}\le {\log }_2\frac{x}{12}\le 2^1 \\ \Rightarrow 2^{-1}\times 12\le {\log }_2\frac{x}{12}\le 2^1\times 12 \\ \Rightarrow \frac{1}{2}\times 12\le {\log }_2\frac{x}{12}\le 2\times 12 \\ \Rightarrow 6\le {\log }_2\frac{x}{12}\le 24 \end{gathered}$$

Hence, the domain is $\left[6,24\right]$

Therefore, the correct answer is Option D.