The range of the function f(x) = frac{1}{sqrt | vedclass

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(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.Domain of $f$:We know that $0 \leq x-[x] < 1$ for all $x \in \mathbb{R}$.Also,$x-[x] =

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$(1, \infty)$

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(A) The function is defined as $f(x) = \frac{1}{\sqrt{x-[x]}}$.Domain of $f$:We know that $0 \leq x-[x] Also,$x-[x] = 0$ when $x \in \mathbb{Z}$.Since the denominator cannot be zero,$x-[x] > 0$,which implies $x 
otin \mathbb{Z}$.Thus,the domain is $\mathbb{R} - \mathbb{Z}$.Range of $f$:For $x \in \mathbb{R} - \mathbb{Z}$,we have $0 Taking the square root,$0 Taking the reciprocal,$\frac{1}{\sqrt{x-[x]}} > 1$.Therefore,$f(x) \in (1, \infty)$.Hence,the range of $f$ is $(1, \infty)$.

The range of the function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

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