Question

The domain of the function $f\left(x\right)={\cos }^{-1}\left[{\log }_3\left(\frac{x}{4}\right)\right]$ is

• A. $\left[4,12\right]$
• B. $\left[0,3\right]$
• C. $\left[\frac{4}{3},4\right]$
• D. $\left[\frac{4}{3},12\right]$

Answer 1

The correct option is D

$\left[\frac{4}{3},12\right]$

Explanation for the correct options:

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

We know that ${\cos }^{-1}\left(x\right)$ exists if $-1\le x\le 1$

therefore, we can define the above function as:

$\Rightarrow -1\le {\log }_3\left(\frac{x}{4}\right)\le 1$

We know that

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

Since,

$$\begin{gathered} \Rightarrow {\log }_3\left(\frac{x}{4}\right)\ge -1 \\ \therefore \left(\frac{x}{4}\right)\ge 3^{-1} \end{gathered}$$

Then, we can rewrite the inequality as:

$$\begin{gathered} \Rightarrow \frac{x}{4}\le 3^1 \\ \Rightarrow x\le 3^1\times 4 \\ \Rightarrow x\le 12 \end{gathered}$$

And

$$\begin{gathered} \Rightarrow \left(\frac{x}{4}\right)\ge 3^{-1} \\ \Rightarrow \left(\frac{x}{4}\right)\ge \frac{1}{3} \\ \Rightarrow x\ge \frac{4}{3} \end{gathered}$$

Hence, the domain of the function $f\left(x\right)$ is $\left[\frac{4}{3},12\right]$.

Therefore the correct answer is option D.