Show that the function $f: R \rightarrow R: f(x)=1+x^{2}$ is many-one into.
To prove: function is many-one into
Given: $f: R \rightarrow R: f(x)=1+x^{2}$
We have,
$f(x)=1+x^{2}$
For, $f\left(x_{1}\right)=f\left(x_{2}\right)$
$\Rightarrow 1+x_{1}^{2}=1+x_{2}^{2}$
$\Rightarrow x_{1}^{2}=x_{2}^{2}$
$\Rightarrow x_{1}^{2}-x_{2}^{2}=0$
$\Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)=0$
$\Rightarrow x_{1}=x_{2}$ or, $x_{1}=-x_{2}$
Clearly $x_{1}$ has more than one image
$\therefore f(x)$ is many-one
$f(x)=1+x^{2}$
Let $f(x)=y$ such that $y \in R$
$\Rightarrow y=1+x^{2}$
$\Rightarrow x^{2}=y-1$
$\Rightarrow x=\sqrt{y-1}$
If $y=3$, as $y \in R$
Then $x$ will be undefined as we can't place the negative value under the square root
Hence $f(x)$ is into
Hence Proved
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