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Q.
The domain of $f(x) = \cot^{-1} \frac {x} {\sqrt x^2-[x^2]}$ $ x \in $ R is
Inverse Trigonometric Functions
Solution:
Domain of $\cot^{-1}$ x is R and $\frac{x}{\sqrt{x^2-[x^2]}}$ is defined if $x^2 \neq [(x^2)]$. i,e., $x^2$ is not integer $(\because x^2 \geq [x^2])$
Hence $x^2 \ne$ non-negative integer i.e., 0 or $+ve$ integer.
Hence domain = R $-\{(\sqrt {n} : n \geq \,0,n\,\in \,Z)\}$