Question

The domain of definition of \(f(x)=\sqrt{\frac{1-|x|}{2-|x|}}\) is
(A) \((-\infty,-1) \cup(2, \infty)\)
(B) \([-1,1] \cup(2, \infty) \cup(-\infty,-2)\)
(C) \((-\infty, 1) \cup(2, \infty)\)
(D) \([-1,1] \cup(2, \infty)\)
Here \((a, b) \equiv\{x: a<x<b\} \&[a, b] \equiv\{x: a \leq x \leq b\}\)

Answer 1

Ans: (B)
Hint \(: \frac{1-|x|}{2-|x|} \geq 0 \Rightarrow \frac{|x|-1}{|x|-2} \geq 0 \Rightarrow|x| \leq 1\) as \(|x|>2\)
$$
\Rightarrow x \in(-\infty,-2) \cup(2, \infty) \cup[-1,1]
$$