Question

Let $f\left(x\right)=\frac{x^2-1}{x},g\left(x\right)=\frac{x+2}{x-3}$ then domain of $\frac{f\left(x\right)}{g\left(x\right)}$ is

• A. $R-\{0,-2\}$
• B. $R-\{-2,0,3\}$
• C. $R$
• D. $R-\{0,-3\}$

Answer 1

The correct option is B $R-\{-2,0,3\}$

Let $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$
$\therefore h\left(x\right)=\frac{\frac{x^2-1}{x}}{\frac{x+2}{x-3}}=\frac{(x^2-1)(x-3)}{x(x+2)}$
But for $x=3,g\left(x\right)$ not defined
Hence $h\left(x\right)$ is not defined for $x=0,-2,3$
Hence required domain is
$R-\{-2,0,3\}$