Question:
If the function $f: R \rightarrow A$ given by $f(x)=\frac{x^{2}}{x^{2}+1}$ is a surjection, then $\mathrm{A}=$
(a) $R$
(b) $[0,1]$
(c) $[0,1)$
(d) $[0,1)$
Solution:
As $f$ is surjective, r ange of $f=$ co-domain of $f$
$\Rightarrow A=$ range of $f$
$\because f(x)=\frac{x^{2}}{x^{2}+1}$
$y=\frac{x^{2}}{x^{2}+1}$
$\Rightarrow y\left(x^{2}+1\right)=x^{2}$
$\Rightarrow(y-1) x^{2}+y=0$
$\Rightarrow x^{2}=\frac{-y}{(y-1)}$
$\Rightarrow x=\sqrt{\frac{y}{(1-y)}}$
$\Rightarrow \frac{y}{(1-y)} \geq 0$
$\Rightarrow y \in[0,1)$
$\Rightarrow$ Range of $f=[0,1)$
$\Rightarrow A=[0,1)$
So, the answer is (d).