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Q.
The function $ f(x) = \sec ^{-1} \frac {x} {\sqrt {x- [x]}}$ is defined for
Inverse Trigonometric Functions
Solution:
$sec^{-1}$ 1 x is defined for x $\geq\,1\,$ or $\,x=\leq\,-1$ i.e., for R-(-1. 1)
$\therefore $ f(x) is defined if $\frac{x}{\sqrt{x-[x]}}\leq\,-1\,$ or $\,\geq\,1$
Also x - [x] $\neq$ 0 $\therefore $ X is not an integer. [$\because$ [x] = n and n -[n]= 0 for all integer n].
Hence domain of $sec^{-1} \frac{x}{\sqrt{x-[x]}}$ = R - {(-1, 1) ($\cup$ Z} where Z is the set of all integers.