The domain of the function f(x)

(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 $(2)$

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