Question

State whether the function $f:R\to R$ defined by $f\left(x\right)=1+x^2$ is one-one onto or bijective.

Answer 1

Given, function $f:R\to R$ such that $f\left(x\right)=1+x^2$,
Let A and B be two sets of real numbers.
Let $x_1,x_2\in A$ such that $f\left(x_1\right)=f\left(x_2\right)$.
$\Rightarrow 1+x_1^2=1+x_2^2\Rightarrow x_1^2-x_2^2=0\Rightarrow (x_1-x_2)(x_1+x_2)=0$
$\Rightarrow x_1=\pm x_2.$ Thus $f\left(x_1\right)=f\left(x_2\right)$ does not imply that $x_1=x_2$.
For instance, $f\left(1\right)=f(-1)=2$, i.e. , two elements (1, -1) of A have the same image in B. So, $f$ is many-one function.
Now, $y=1+x^2\Rightarrow x=\sqrt{y-1}\Rightarrow$elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, $f$ is not onto.
Hence, $f$ is neither one-one onto. So, it is not bijective.