Question

Range of the function $f\left(x\right)=\frac{x}{1+x^2}$ is

• A. $(-\infty ,\infty )$
• B. $[-1,1]$
• C. $[\frac{-1}{2},\frac{1}{2}]$
• D. $[-\sqrt{2},\sqrt{2}]$

Answer 1

The correct option is C

$[\frac{-1}{2},\frac{1}{2}]$

Explanation for the correct option:

Given function is

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

$f\left(x\right)$ is defined for all real numbers since the denominator will not be $0$ for any real number,

Let the given function as

$$\begin{gathered} \frac{x}{1+x^2}=y \\ \Rightarrow x=y+y{x}^2 \\ \Rightarrow x^{2}y-x+y=0 \end{gathered}$$

Now, the discriminant of the above quadratic equation must be greater than or equal to zero for the real solution, therefore

$$\begin{gathered} {(-1)}^2-4y^2\ge 0 \\ \Rightarrow 1-4y^2\ge 0 \\ \Rightarrow (1-2y)(1+2y)\ge 0 \end{gathered}$$

Now using wavy curve method, we see that

$y\in [\frac{-1}{2},\frac{1}{2}]$

Hence, option (C) is the correct answer