Question

The domain of the function $f\left(x\right)=\sqrt{\frac{(x+1)(x-3)}{x-2}}$ is

• A. $[-1,2)\cup [3,\infty )$
• B. $(-1,2)\cup [3,\infty )$
• C. $[-1,2]\cup [3,\infty )$
• D. None of these

Answer 1

The correct option is A

$[-1,2)\cup [3,\infty )$

$f\left(x\right)=\sqrt{\frac{(x+1)(x-3)}{x-2}}$
For f(x) to be defined,
$(x-2)\ne 0$
$\Rightarrow x\ne 2\dots \dots \left(1\right)$
Also,
$\frac{(x+1)(x-3)}{(x-2)}\ge 0\Rightarrow \frac{(x+1)(x-3)(x-2)}{(x-2{)}^2}\ge 0\Rightarrow xϵ[-1,2)\cup [3,\infty )\dots \dots \left(2\right)$
From (1) and (2),
$xϵ[-1,2)\cup [3,\infty )$