Question

Let $R$ be the set of real numbers and $f:R\to R$ be defined by $f\left(x\right)=\frac{\left\{x\right\}}{1+[x{]}^2},$ where $\left[x\right]$ is the greatest integer less than or equal to $x,$ and $\left\{x\right\}=x-\left[x\right].$ Which of the following statements are true?

I. The range of $f$ is a closed interval.
II. $f$ is continuous on $R.$
III. $f$ is one-one on $R$

• A. I only
• B. II only
• C. III only
• D. None of I, II and III

Answer 1

The correct option is D None of I, II and III

$f\left(x\right)=\frac{\left\{x\right\}}{1+[x{]}^2}=\{\begin{array}{c}\frac{x+1}{2};-1\le x<0\\ x;0\le x<1\\ \frac{x-1}{2};1\le x<2\\ \frac{x-2}{5};2\le x<3\\ .\\ .\\ .\\ \text{so on}\end{array}$

For Maximum of $f\left(x\right)=\left\{x\right\}\text{should be maximum and}\left[x\right]\text{should be minimum}$
maximum $\left\{x\right\}\to 1$
So range of $f$ is $[0,1)$ which is open interval.

Checking continuity at $x=0$
L.H.L.
$=\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1}{2}$
R.H.L.
$=\underset{x\to 0^{+}}{lim}f\left(x\right)=0$
$\therefore$ L.H.L $\ne$ R.H.L
$\therefore f$ is not continuous at $x=0$ and hence, not continuous on $R.$

For any integer $x\in I,f\left(x\right)=0.$
So, it is not one-one function.