Question

The set of all $x$ for which the none of functions is defined$f\left(x\right)=lo{g}_{\frac{x-1}{x+3}}2$ and$g\left(x\right)=\frac{1}{\sqrt{x^2-9}}$ is

• A. $(-3,2)$
• B. $[-3,2)$
• C. $(-3,2]$
• D. $[-3,2]$

Answer 1

The correct option is D

$[-3,2]$

Explanation for the correct option.

Step 1:Solve for the value of $x$ for which $f\left(x\right)={\log }_{\frac{x-1}{x+3}}2$is not defined.

$f\left(x\right)={\log }_{\frac{x-1}{x+3}}2$

$\frac{(x–1)}{(x+3)}>0$

$x\in (-\infty ,-3)\cup (1,\infty )—\left(1\right)$

$\frac{(x–1)}{(x+3)}\ne 1$

$\frac{(x–1)}{(x+3)}–1\ne 0$

$\frac{x-1-x-3}{x+3}\ne 0$

$–\frac{4}{x+3}\ne 0$

$$\begin{gathered} x\to \infty \\ x\in [-3,1] \end{gathered}$$

Step 2: Solve for the value of $x$ for which $g\left(x\right)=\frac{1}{\sqrt{x^2-9}}$is not defined.

For $g\left(x\right)$ to not be defined, its denominator must equal $0$.

$\Rightarrow$ $\sqrt{x^2-9}>0$

$\Rightarrow$ $x^2–9>0$

$\Rightarrow$$(x+3)(x–3)>0$

$\Rightarrow -3\le x\le 3$

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

Taking common of both intervals we get

$$\begin{gathered} -3\le x\le 2 \\ \therefore x\in [-3,2] \end{gathered}$$

Hence, option (D) is the correct option.