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Q.
The domain of $f(x) = \frac{sin^{-1}x}{x}$ is
Inverse Trigonometric Functions
Solution:
$f(x) = \frac{sin^{-1}x}{x} = \frac{g(x)}{t(x)}, t(x) \ne 0$
$t(x) \ne 0 $
$\Rightarrow x\ne 0$, Again, Domain of $g(x)$ is $[-1, 1]$
Now, using concept
Domain of $f(x) = (\frac{\text{Domain of } g(x)}{t(x)})$
$=$ Domain of $g(x) - $ set of those points for which $t(x) = 0$
If $g(x), t(x)$ is defined $\forall x \in R$
$\therefore $ Domain of $f ( x ) = [-1, 1] - \{0\}$