MCQ
The range of the function $f(x)=\frac{1}{\sqrt{x-[x]}}$ is
- ✓$(1, \infty)$
- B$\left( { - \infty ,\infty } \right)$
- C$\;\left( {0,\infty } \right)$
- D$\emptyset $
Domain of $f:$
We know that, $0 \leq \mathrm{x}-[\mathrm{x}]<1$ for all $\mathrm{x} \in \mathrm{R}$
and $x-[x]=0$ for $x \in Z$
$\mathrm{So}, 0<\mathrm{x}-[\mathrm{x}]<1$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$
Hence, domain of $\mathrm{f}=\mathrm{R}-\mathrm{Z}$
Range of $f:$
We have,
$0<\mathrm{x}-[\mathrm{x}]<1 \text { for all } \mathrm{x} \in \mathrm{R}-\mathrm{Z}$
$\Rightarrow 0<\sqrt{\mathrm{x}-[\mathrm{x}]}<1$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$
$\Rightarrow 1<\frac{1}{\sqrt{x-|x|}}<\infty$ for all $x \in R-Z$
$\Rightarrow 1<\mathrm{f}(\mathrm{x})<\infty$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$
Hence, range of $\mathrm{f}=(1, \infty)$
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