Question:
The domain of the function $f(x)=\sqrt{\frac{(x+1)(x-3)}{x-2}}$ is
(a) [−1, 2) ∪ [3, ∞)
(b) (−1, 2) ∪ [3, ∞)
(c) [−1, 2] ∪ [3, ∞)
(d) None of these
Solution:
(a) [−1, 2) ∪ [3, ∞)
$f(x)=\sqrt{\frac{(x+1)(x-3)}{x-2}}$
For $f(x)$ to be defined,
$(x-2) \neq 0$
$\Rightarrow \mathrm{x} \neq 2$ ....(1)
Also,
$\frac{(x+1)(x-3)}{(x-2)} \geq 0$
$\Rightarrow \frac{(x+1)(x-3)(x-2)}{(x-2)^{2}} \geq 0$
$\Rightarrow(\mathrm{x}+1)(\mathrm{x}-3)(\mathrm{x}-2) \geq 0$
$\Rightarrow \mathrm{x} \in[-1,2) \cup[3, \infty) \quad \ldots . .(2)$
From $(1)$ and $(2)$
$x \in[-1,2) \cup[3, \infty)$