Let f: (1, 3) → R , be a function defined by f(x) = x[x] 1 + x ^ 2...

f: \left(1, 3\right) \to R, f\left(x\right) = \dfrac{x\left[x\right]}{1 + {x}^{2}}$</p><p>$f\left(x\right) = \left\{$\begin{cases}\dfrac{x}{1 + {x}^{2}}, & x \in \left(1, 2\right) \\ \dfrac{2x}{1 + {x}^{2}}, & x \in [ 2, 3 )\end{cases}$\right)

{f}^{'}\left(x\right) = \left\{\begin{cases}\dfrac{\left(1 + {x}^{2}\right)\left(1\right) - x\left(2x\right)}{{\left(1 + {x}^{2}\right)}^{2}}, & x \in \left(1, 2\right) \\ \dfrac{\left(1 + {x}^{2}\right)\left(2\right) - 2x\left(2x\right)}{{\left(1 + {x}^{2}\right)}^{2}}, & x \in [ 2, 3 )\end{cases}\right)</p><p>{f}^{'}\left(x\right) = \left\{$\begin{cases}\dfrac{1 - {x}^{2}}{1 + {x}^{2}}, & x \in \left(1, 2\right) \\ \dfrac{1 - 2{x}^{2}}{1 + {x}^{2}}, & x \in [ 2, 3 )\end{cases}$\right)

$∴ f\left(x\right)$ is decreasing function.
$∴ {R}_{f} \in \left(\dfrac{2}{5}, \dfrac{1}{2}\right) \cup \left(\dfrac{6}{10}, \dfrac{4}{5}\right]$