Let f(1,3)
Question:

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:

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

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

3. (3) $\left(\frac{2}{5}, \frac{4}{5}\right)$

4. (4) $\left(\frac{3}{5}, \frac{4}{5}\right)$

Correct Option: , 2

Solution:

$f(x) \begin{cases}\frac{x}{x^{2}+1} ; & x \in(1,2) \\ \frac{2 x}{x^{2}+1} ; & x \in[2,3)\end{cases}$

$f^{\prime}(x) \begin{cases}\frac{1-x^{2}}{1+x^{2}} ; & x \in(1,2) \\ \frac{1-2 x^{2}}{1+x^{2}} ; & x \in[2,3)\end{cases}$

$\therefore f(x)$ is a decreasing function

$\therefore \quad y \in\left(\frac{2}{5}, \frac{1}{2}\right) \cup\left(\frac{6}{10}, \frac{4}{5}\right]$

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