Question

The domain of the function $f\left(x\right)={\log }_{10}\left(\sqrt{x-4}+\sqrt{6-x}\right)$ is

• A. $\left[4,6\right]$
• B. $\left(-\infty ,6\right)$
• C. $[2,3)$
• D. None of these

Answer 1

The correct option is A

$\left[4,6\right]$

Explanation for correct answer Option(s)

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

The given function is defined as:

$\sqrt{x-4}+\sqrt{6-x}>0$

Case 1:

$$\begin{gathered} \Rightarrow x-4=0 \\ \Rightarrow x=4 \end{gathered}$$

Then,

$$\begin{gathered} f\left(x\right)=\log \left(\sqrt{4-4}+\sqrt{6-4}\right) \\ f\left(x\right)=\log \left(0+\sqrt{2}\right) \\ f\left(x\right)=\log \left(\sqrt{2}\right) \end{gathered}$$

Case 2:

$$\begin{gathered} \Rightarrow 6-x=0 \\ \Rightarrow x=6 \end{gathered}$$

Then,

$$\begin{gathered} f\left(x\right)=\log \left(\sqrt{6-4}+\sqrt{6-6}\right) \\ f\left(x\right)=\log \left(\sqrt{2}+0\right) \\ f\left(x\right)=\log \left(\sqrt{2}\right) \end{gathered}$$

Thus, $x\in \left[4,6\right]$

Hence, the domain of the function $f\left(x\right)$ is $\left[4,6\right]$

Therefore, the correct answer is Option A.