If the domain of the function f(x) = frac{[x] | vedclass

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(D) Given $f(x) = \frac{[x]}{1+x^2}$ for $x \in (2, 6)$.Since $[x]$ is the greatest integer function,we analyze the intervals:For $x \in [2, 3)$,$[x]

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$\left(\frac{5}{37}, \frac{2}{5}\right]$

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(D) Given $f(x) = \frac{[x]}{1+x^2}$ for $x \in (2, 6)$.Since $[x]$ is the greatest integer function,we analyze the intervals:For $x \in [2, 3)$,$[x] = 2$,so $f(x) = \frac{2}{1+x^2}$. As $x$ increases from $2$ to $3$,$f(x)$ decreases from $\frac{2}{1+2^2} = \frac{2}{5}$ to $\frac{2}{1+3^2} = \frac{2}{10} = \frac{1}{5}$. Range: $(\frac{1}{5}, \frac{2}{5}]$.For $x \in [3, 4)$,$[x] = 3$,so $f(x) = \frac{3}{1+x^2}$. Range: $(\frac{3}{17}, \frac{3}{10}]$.For $x \in [4, 5)$,$[x] = 4$,so $f(x) = \frac{4}{1+x^2}$. Range: $(\frac{4}{26}, \frac{4}{17}] = (\frac{2}{13}, \frac{4}{17}]$.For $x \in [5, 6)$,$[x] = 5$,so $f(x) = \frac{5}{1+x^2}$. Range: $(\frac{5}{37}, \frac{5}{26}]$.The union of these ranges is $(\frac{5}{37}, \frac{2}{5}]$. Note that the function is strictly decreasing in each interval $[n, n+1)$,and the values at the right endpoints are excluded. The set of values is the union of these intervals,which simplifies to $(\frac{5}{37}, \frac{2}{5}]$.

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

If the domain of the function $f(x) = \frac{[x]}{1+x^2}$,where $[x]$ is the greatest integer $\leq x$,is $(2, 6)$,then its range is

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