Question

If $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$, where $f\left(x\right)=x^2-2x$ and $g\left(x\right)=x^3-3x^2+2x,$ then the domain of $h\left(x\right)$ is

• A. $R-\left\{2\right\}$
• B. $R-\left\{1\right\}$
• C. $R-\{0,2\}$
• D. $R-\{0,1,2\}$

Answer 1

The correct option is D $R-\{0,1,2\}$

$h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2-2x}{x^3-3x^2+2x}\Rightarrow h\left(x\right)=\frac{x(x-2)}{x(x^2-3x+2)}\Rightarrow h\left(x\right)=\frac{x(x-2)}{x(x-1)(x-2)}$
Here, $x(x-1)(x-2)\ne 0$
$\therefore$ The domain of function $h\left(x\right)$ is $R-\{0,1,2\}$