If the function f : R→A given by

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).