Question

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

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

Answer 1

The correct option is C $[0,1)$

The domain of $f\left(x\right)$ is all real numbers.
$\because f:R\to A$ is surjective, therefore $A$ must be the range of $f\left(x\right)$
$\Rightarrow f\left(x\right)=1-\frac{1}{1+x^2}$
We know that,
$x^2+1\ge 1\Rightarrow 0<\frac{1}{x^2+1}\le 1\Rightarrow -1\le -\frac{1}{x^2+1}<0\Rightarrow 0\le f\left(x\right)<1,\forall x\in R$