Question
$f: R → R$ defined by $f(x) = 1 + x^2$

Answer

$f: R → R$ is defined as
$f(x) = 1 + x^2$
Let $\text{x}_1,\text{x}_2\in\text{R}$ such that $f(x_1) = f(x_2)$
$\Rightarrow1+\text{x}_{1}^{2}=1+\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}^{2}=\text{x}_{2}^{2}$
$\Rightarrow\text{x}_{1}=\pm\text{x}_{2}$
$\therefore f(x_1) = f(x_2)$ does not imply that $x_1 = x_2.$
For instance,
$f(1) = f(-1) = 2$
$\therefore$ f is not one-one.
Consider an element $-2$ in co-domain $R.$
It is seen that $f(x) = 1 + x^2$ is positive for all $\text{x}\in\text{R}.$
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.

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