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(D) Given,$f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$.For $f(x)$ to be defined,we must have $\frac{2-|x|}{3-|x|} \geq 0$.Let $t = |x|$,where $t \geq 0$. The i

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$(-\infty, -3) \cup [-2, 2] \cup (3, \infty)$

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(D) Given,$f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$.For $f(x)$ to be defined,we must have $\frac{2-|x|}{3-|x|} \geq 0$.Let $t = |x|$,where $t \geq 0$. The inequality becomes $\frac{2-t}{3-t} \geq 0$,which is equivalent to $\frac{t-2}{t-3} \geq 0$.Using the wavy curve method for $t \geq 0$,the solution for $t$ is $t \in [0, 2] \cup (3, \infty)$.Now,substitute $t = |x|$ back:Case $I$: $0 \leq |x| \leq 2 \Rightarrow x \in [-2, 2]$.Case $II$: $|x| > 3 \Rightarrow x \in (-\infty, -3) \cup (3, \infty)$.Combining these,the domain is $x \in (-\infty, -3) \cup [-2, 2] \cup (3, \infty)$.

The domain of the real-valued function $f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$ is

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