If the domain of the function $f(x) = \frac{x}{1+\lfloor x \rfloor^2}$, where $\lfloor x \rfloor$ is greatest integer $\leq x$, is $[2,6)$, then its range is (1) $\left\{\frac{5}{26}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$ (2) $\left[\frac{5}{26}, \frac{2}{5}\right]$ (3) $\left\{\frac{5}{37}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$ (4) $\left[\frac{5}{37}, \frac{2}{5}\right]$
If the domain of the function $f(x) = \frac{x}{1+\lfloor x \rfloor^2}$, where $\lfloor x \rfloor$ is greatest integer $\leq x$, is $[2,6)$, then its range is\\
(1) $\left\{\frac{5}{26}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$\\
(2) $\left[\frac{5}{26}, \frac{2}{5}\right]$\\
(3) $\left\{\frac{5}{37}, \frac{2}{5}\right\} \cup \left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$\\
(4) $\left[\frac{5}{37}, \frac{2}{5}\right]$