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ધારો કે $f:(1,3) \rightarrow \mathrm{R}$ એ $f(\mathrm{x})=\frac{\mathrm{x}[\mathrm{x}]}{1+\mathrm{x}^{2}},$ મુજબ વિધેય વ્યાખ્યાતિ છે કે જ્યાં $[\mathrm{x}]$ એ મહતમ પૃણાંક વિધેય છે તો વિધેય $f$ નો વિસ્તાર મેળવો.
$\left(\frac{3}{5}, \frac{4}{5}\right)$
$\left(\frac{2}{5}, \frac{3}{5}\right] \cup\left(\frac{3}{4}, \frac{4}{5}\right)$
$\left(\frac{2}{5}, \frac{4}{5}\right]$
$\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$
Solution
$f(\mathrm{x})=\left\{\begin{array}{ll}{\frac{\mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in(1,2)} \\ {\frac{2 \mathrm{x}}{\mathrm{x}^{2}+1}} & {; \quad \mathrm{x} \in[2,3)}\end{array}\right.$
$f(\mathrm{x})$ is decreasing function
$f(x) \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{3}{5}, \frac{4}{5}\right]$
