Question

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

• A. $(\frac{2}{5},\frac{3}{5}]\cup (\frac{3}{4},\frac{4}{5})$
• B. $(\frac{2}{5},\frac{4}{5}]$
• C. $(\frac{3}{5},\frac{4}{5})$
• D. $(\frac{2}{5},\frac{1}{2})\cup (\frac{3}{5},\frac{4}{5}]$

Answer 1

The correct option is D $(\frac{2}{5},\frac{1}{2})\cup (\frac{3}{5},\frac{4}{5}]$

$f\left(x\right)=\frac{x\left[x\right]}{x^2+1}$

$\Rightarrow f\left(x\right)=\{\begin{array}{cc}\frac{x}{x^2+1};& 1<x<2\\ \frac{2x}{x^2+1};& 2\le x<3\end{array}$

Let $y=\frac{x}{x^2+1}$
$y^{\prime }=\frac{1-x^2}{(x^2+1{)}^2}<0$ for all $x\in (1,2)$
So, $f$ is decreasing in $(1,2)$
$f\left(1\right)=\frac{1}{2},f\left(2\right)=\frac{2}{5}$
$\therefore f\in (\frac{2}{5},\frac{1}{2})$ for all $x\in (1,2)$

We can also conclude from above differentiation that $f$ is decreasing in $[2,3)$ also.
$f\left(2\right)=\frac{4}{5},f\left(3\right)=\frac{3}{5}$
$\therefore f\in (\frac{3}{5},\frac{4}{5}]$ for all $x\in [2,3)$

$\Rightarrow$ Range of $f\left(x\right)$ is $(\frac{2}{5},\frac{1}{2})\cup (\frac{3}{5},\frac{4}{5}]$