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If $[x]$ denotes the greatest integer $\leq x$, then the range of the real valued function $f(x)=\frac{1}{\sqrt{x-[x]}}$ is
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The correct answer is:
$(1, \infty)$
We are given that $f(x)=\frac{1}{\sqrt{x-[x]}}$ but $\mathrm{x}-[\mathrm{x}] \in(0,1)$ $\Rightarrow \mathrm{f}(\mathrm{x}) \in(1, \infty)$
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