If the function f : R→A given by


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) $[0,1)$


As $f$ is surjective, r ange of $f=$ co-domain of $f$

$\Rightarrow A=$ range of $f$

$\because f(x)=\frac{x^{2}}{x^{2}+1}$


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

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

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

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

$\Rightarrow \frac{y}{(1-y)} \geq 0$

$\Rightarrow y \in[0,1)$

$\Rightarrow$ Range of $f=[0,1)$

$\Rightarrow A=[0,1)$

So, the answer is (d).


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