What will be the domain of $f(x)$ where,

$f(x) = \frac{\sqrt{12+x-x^2}}{x(x-2)}$

Given :

$f(x) = \frac{\sqrt{12+x-x^2}}{x(x-2)}$

As we can observe, for the function to be valid

${12+x-x^2} \gt=0 \rightarrow$ (1)

$x(x-2)\neq{0} \rightarrow$ (2)

$x(x-2)\neq{0} $

so, $ x\neq{0}$ and $x\neq{2} \rightarrow$ (3)

Using (1)

${12+x-x^2} \gt=0$

${x^2-x-12} \lt=0$

$(x-4)(x+3)\lt=0$

We have

$ x \in{[-3, 4]} \rightarrow$ (4)

$ x\in [-3,0) {\displaystyle \cup} (0,2) {\displaystyle \cup} (2,4]$ Answer