Question

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

• A. $(-1,0)$
• B. $(-1,1)$
• C. $[0,1)$
• D. $\left(1,1\right)$

Answer 1

The correct option is C

$[0,1)$

Explanation for the correct option:

Given function is $f\left(x\right)=\frac{x^2}{x^2+1}$ Since, In Denominator of the function the term $x^2\ge 0$ makes the whole denominator $x^2+1\ge 1$

Now, considering the given function and simplify it it as

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

Solving above quadratic equation for real roots, its discriminant must be greater than or equal to zero, therefore

$$\begin{gathered} 0-4(1-y)(-y)\ge 0 \\ 4y(1-y)\ge 0 \end{gathered}$$

Here we see that range is $y\in [0,1)$ since, $y\ne 1$ because it was so then the equation $x^2(1-y)-y=0$ will no more a quadratic equation.

Hence, option (C) is the correct answer