Question

$f:R\to R$ defined by $f\left(x\right)=1+x^2$. Is it one one , onto or bijective?

Answer 1

Here, $f:R\to R$ is defined as $f\left(x\right)=1+x^2$
Let $x_1,x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 1+x_1^2=1+x_2^2\Rightarrow x_1^2\Rightarrow x_1=\pm x_2$
For instance, f(1)=f(-2)=2
Therefore, $f\left(x_1\right)=f\left(x_2\right)$ does not imply that $x_1=x_2$
Therefore, f is not one-one.
Consider an element -2 in co-domain R.
It is seen that $f\left(x\right)=1+x^2$ is positive for all x in R.
Thus, there does not exist any x in domain R such that f(x)=-2.
Therefore, f is not onto. Hence, f is neither one-one nor onto.