Question

The domain of the function $f\left(x\right)=\{\begin{array}{c}(x^2-9)/(x-3),ifx\ne 3\\ 6,ifx=3\end{array}$ is

• A. $(0,3)$
• B. $(-\infty ,3)$
• C. $(-\infty ,\infty )$
• D. $(3,\infty )$
• E. $(-3,3)$

Answer 1

The correct option is C $(-\infty ,\infty )$

Given $f\left(x\right)=\{\begin{array}{c}\frac{x^2-9}{x-3},x\ne 3\\ 6,x=3\end{array}$ at $x=3$
$LHL={lim}_{x\to 3^{-}}\frac{x^2-9}{x-3}={lim}_{x\to 3^{-}}(x+3)={lim}_{h\to 0}(3-h+3)=6$
$RHL={lim}_{x\to 3^{+}}\frac{x^2-9}{x-3}={lim}_{x\to 3^{+-}}(x+3)={lim}_{h\to 0}(3+h+3)=6$
$\Rightarrow f\left(3\right)=6$
$\therefore$ LHL=RHL
Hence, $f\left(x\right)$ is continuous at $x=3$
$\therefore$ Domain of $f\left(x\right)=(-\infty ,\infty )$