Show that the function

Solution:

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