In each of the following cases,state whether the function is one-one,onto or bijective. Justify your answer. $f : R \rightarrow R$ defined by $f(x) = 1 + x^2$.

(D) The function $f : R \rightarrow R$ is defined as $f(x) = 1 + x^2$.
To check for one-one:
Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.
$\Rightarrow 1 + x_1^2 = 1 + x_2^2$
$\Rightarrow x_1^2 = x_2^2$
$\Rightarrow x_1 = \pm x_2$.
Since $f(1) = 1 + (1)^2 = 2$ and $f(-1) = 1 + (-1)^2 = 2$,we have $f(1) = f(-1)$ but $1 \neq -1$.
Therefore,$f$ is not one-one.
To check for onto:
Consider an element $-2$ in the codomain $R$.
Since $x^2 \geq 0$ for all $x \in R$,$f(x) = 1 + x^2 \geq 1$ for all $x \in R$.
Thus,there is no $x \in R$ such that $f(x) = -2$.
Therefore,$f$ is not onto.
Conclusion:
Since the function is neither one-one nor onto,it is not bijective.