The domain of the real-valued function f(x) = | vedclass

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(C) The function is defined as $f(x) = \sqrt{\frac{|x|-x}{x-[x]}}$.For the numerator $\sqrt{|x|-x}$ to be defined,we need $|x|-x \geq 0$,which implies

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$R-Z$

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(C) The function is defined as $f(x) = \sqrt{\frac{|x|-x}{x-[x]}}$.For the numerator $\sqrt{|x|-x}$ to be defined,we need $|x|-x \geq 0$,which implies $|x| \geq x$. This is true for all $x \in R$.For the denominator $\sqrt{x-[x]}$ to be defined and non-zero,we need $x-[x] > 0$.We know that $x-[x] = \{x\}$,where $\{x\}$ is the fractional part of $x$.The condition $\{x\} > 0$ holds for all $x 
otin Z$ (all real numbers except integers).If $x \in Z$,then $\{x\} = 0$,which makes the denominator zero and the function undefined.Therefore,the domain of the function is $R-Z$.

The domain of the real-valued function $f(x) = \frac{\sqrt{|x|-x}}{\sqrt{x-[x]}}$ is

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