Question

The set of all $x$for which $f\left(x\right)=lo{g}_{\frac{x-3}{x+4}}5$ and $g\left(x\right)=\frac{1}{\sqrt{x^2-16}}$ are both not defined is

• A. $[-4,3]$
• B. $[-4,4]$
• C. $[0,3]$
• D. $[0,4]$

Answer 1

The correct option is A

$[-4,3]$

The explanation for the correct answer.

Step 1: Solve for the value of $x$for which $f\left(x\right)=lo{g}_{\frac{x-3}{x+4}}5$ is not defined.

$$\begin{gathered} \frac{x-3}{x+4}>0 \\ \Rightarrow x-3>0 \\ \Rightarrow x>3 \end{gathered}$$

But we require $x\le 3$

$\Rightarrow x\in (-\infty ,3]$

Step 2: Solve for the value of $x$ where $g\left(x\right)=\frac{1}{\sqrt{x^2-16}}$ is not defined.

$g\left(x\right)=\frac{1}{\sqrt{x^2-16}}$

$x\in \left[-4,4\right]$

So, the required answer is $x\in (-\infty ,3]\cap \left[-4,4\right]$

$\Rightarrow x\in [-4,3]$

Hence, option $A$ is the correct answer.