Question

If the function $f:R\to A$ given by $f\left(x\right)=\frac{x^2}{x^2+1}$ is a surjection, then $A$ is equal to

• A. $R$
• B. $[0,1]$
• C. $(0,1)$
• D. $[0,1)$

Answer 1

The correct option is B $[0,1]$

The domain of $f\left(x\right)$ is the complete set of real number. Since, $f:R\to A$ is a surjection, therefore $A$ is the range of $f\left(x\right)$

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

$\Rightarrow \frac{x^2+1}{x^2}=\frac{1}{y}$

$\Rightarrow \frac{1}{x^2}=\frac{1-y}{y}$

$\Rightarrow x=\sqrt{\frac{y}{1-y}}$

Now ,

$x\in R$ if $\sqrt{\frac{y}{1-y}}$ is real

$\Rightarrow \frac{y}{1-y}\ge 0$

$\Rightarrow 0\le y\le 1$

Therefore, the range of $f\left(x\right)$ is $[0,1]$

Hence, $A=[0,1]$