Question
Assertion $(A) :$ The function $f: R \rightarrow[0,1)$ defined by $f(x)=\frac{x^2}{x^2+1}$ is surjective.
Reason $(R) :$ For surjection, Range of $f(x)=$ codomain of $f(x)$
Reason $(R) :$ For surjection, Range of $f(x)=$ codomain of $f(x)$
✓
Answer
$(a) :$ For onto function, codomain of $f=$ range of $f$.
We have, $f(x)=\frac{x^2}{1+x^2}$. Then $y \geq 0$.
$\Rightarrow y\left(1+x^2\right)=x^2 \Rightarrow x^2(y-1)=-y$
$\Rightarrow x^2=\frac{y}{1-y} \geq 0$
$\Rightarrow \frac{y-0}{y-1} \leq 0 \Rightarrow 0 \leq y<1 (\because y \neq 1)$
$\Rightarrow$ codomain of $f=$ Range of $f$, as $f: R \rightarrow[0,1)$.
We have, $f(x)=\frac{x^2}{1+x^2}$. Then $y \geq 0$.
$\Rightarrow y\left(1+x^2\right)=x^2 \Rightarrow x^2(y-1)=-y$
$\Rightarrow x^2=\frac{y}{1-y} \geq 0$
$\Rightarrow \frac{y-0}{y-1} \leq 0 \Rightarrow 0 \leq y<1 (\because y \neq 1)$
$\Rightarrow$ codomain of $f=$ Range of $f$, as $f: R \rightarrow[0,1)$.
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