The domain of defined of the function $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is
The domain of defined of the function $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is
$[-1,1] \cup(-\infty,-2] \cup[2, \infty)$
$[-1,1] \cup(-\infty,-2) \cup(2, \infty)$
$(\infty, 2) \cup(2, \infty)$
$R$
Solution
$f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$
For domain
$
\frac{1-|x|}{2-|x|} \geq 0
$
and $2-|x| \neq 0$
$
\begin{aligned}
|x| & \neq 2 \\
x & = \pm 2
\end{aligned}
$
Case I When, $x \geq 0$
So, Eq. (i) becomes
$
\frac{1-x}{2-x} \geq 0 \quad\left\{\because|x|=\left[\begin{array}{cc}
x & x \geq 0 \\
-x & x < 0
\end{array}\right\}\right.
$
Now, critical points are
$
x=1,2
$
Using wavy curve method
But $x \geq 0$
$\therefore$ Solution is $x \in[0,1] \cup(2, \infty)$.
Case II When, $x < 0$
So, Eq. (i) becomes
$
\frac{1+x}{2+x} \geq 0
$
Critical points are $x=-1,-2$
Using wavy curve method
But $x < 0$
Solution is $x \in(-\infty,-2) \cup[-1,0)$
$\therefore$ Required solution is union of case I and case II.
$\therefore$ Required solution is
$x \in(-\infty,-2) \cup[-1,1] \cup(2, \infty)$
