For the function f(x) = left[ frac{1}{[x]} ri | vedclass

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(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} ight]$.For the function to be defined,the denominator $[x]$ must not be $0$.Thus,$[x]

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The domain is $(-\infty, 0) \cup [1, \infty)$

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(C) The function is defined as $f(x) = \left[ \frac{1}{[x]} ight]$.For the function to be defined,the denominator $[x]$ must not be $0$.Thus,$[x] 
eq 0$,which implies $x Therefore,the domain is $(-\infty, 0) \cup [1, \infty)$.For $x \in [1, 2)$,$[x] = 1$,so $f(x) = [1/1] = 1$.For $x \in [2, \infty)$,$[x] \geq 2$,so $0 For $x \in [-1, 0)$,$[x] = -1$,so $f(x) = [1/(-1)] = -1$.For $x \in [-2, -1)$,$[x] = -2$,so $f(x) = [1/(-2)] = [-0.5] = -1$.Thus,the range is $\{-1, 0, 1\}$.

$x$	$-2$	$-1.5$	$-1$	$-0.5$	$0.25$	$0.5$	$1$	$1.5$	$2$
$y = \frac{1}{x}$	....	....	....	....	....	....	....	....	....

For the function $f(x) = \left[ \frac{1}{[x]} \right]$,where $[x]$ denotes the greatest integer less than or equal to $x$,which of the following statements is true?

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