For $f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$ the domain and range of $f(x)$ in $R$ are
For $f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$ the domain and range of $f(x)$ in $R$ are
$R-\{0, \pm 1\}$ and $[-1,1]$, respectively
$R-[-1,1]$ and $[-1,1]$ respectively
$R-\{0, \pm 1\}$ and $[0,1]$, respectively
$R-[-1,1]$ and $[0,1]$, respectively
Solution
Given function $f(x)=\sin \left(\frac{1}{|x| \sqrt{x^2-1}}\right)$
For the domain, $|x| \neq 0 \Rightarrow x \neq 0$
and $x^2-1>0 \Rightarrow x \in \mathbf{R}-[-1,1]$
So domain of $f$ is $R-[-1,1]$
and we know the range of sine function is $[-1,1]$.