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Q.
Let $f=\left\{\left(x,\frac{x^{2}}{1+x^{2}}\right): x \in\, R\right\}$ be a function from $R$ into $R$. The range of $f$ is
Relations and Functions
Solution:
We have, $f=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right), x\in R\right\}$.
Clearly, domain of $f$ is $R$.
Let $y=\frac{x^{2}}{1+x^{2}}\cdot$ It is clear that $\frac{x^{2}}{1+x^{2}} \ge\,0$
$(\because\, x^{2} \ge\,0$ and $1+x^{2}\ge\,0)$
and $x^{2} <\,1+x^{2}$
$\Rightarrow \frac{x^{2}}{1+x^{2}}<\,1$
So this implies $0 \le\, y <\,1$.
Hence, range of $f$ is $[0,1)$.