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(D) The function is defined as $f(x) = \frac{x[x]}{1+x^2}$ for $x \in (1, 3)$.We split the domain based on the greatest integer function $[x]$:Case $1

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$\left(\frac{2}{5}, \frac{1}{2}\right) \cup \left(\frac{3}{5}, \frac{4}{5}\right]$

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(D) The function is defined as $f(x) = \frac{x[x]}{1+x^2}$ for $x \in (1, 3)$.We split the domain based on the greatest integer function $[x]$:Case $1$: $x \in (1, 2)$,then $[x] = 1$. So,$f(x) = \frac{x}{1+x^2}$.For $x \in (1, 2)$,$f'(x) = \frac{(1+x^2)(1) - x(2x)}{(1+x^2)^2} = \frac{1-x^2}{(1+x^2)^2} As $x 	o 1^+$,$f(x) 	o \frac{1}{1+1^2} = \frac{1}{2}$. As $x 	o 2^-$,$f(x) 	o \frac{2}{1+2^2} = \frac{2}{5}$. So,$f(x) \in (\frac{2}{5}, \frac{1}{2})$.Case $2$: $x \in [2, 3)$,then $[x] = 2$. So,$f(x) = \frac{2x}{1+x^2}$.For $x \in [2, 3)$,$f'(x) = \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2} = \frac{2-2x^2}{(1+x^2)^2} At $x = 2$,$f(2) = \frac{2(2)}{1+2^2} = \frac{4}{5}$. As $x 	o 3^-$,$f(x) 	o \frac{2(3)}{1+3^2} = \frac{6}{10} = \frac{3}{5}$. So,$f(x) \in (\frac{3}{5}, \frac{4}{5}]$.Combining both cases,the range is $(\frac{2}{5}, \frac{1}{2}) \cup (\frac{3}{5}, \frac{4}{5}]$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

Let $f:(1,3) \rightarrow R$ be a function defined by $f(x)=\frac{x[x]}{1+x^{2}},$ where $[x]$ denotes the greatest integer $\leq x.$ Then the range of $f$ is

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