Question
State whether the function is one-one, onto or bijective. Justify your answer. $f: R \rightarrow R$ defined by $f(x) = 1+ x^2$

Answer

Let $x_1, x_2 \in R$
If $f(x_1) = f(x_2)$
$1 + x_1^2 = 1 + x_1^2$
$x_1^2 = x_1^2$
${x_1} = \pm {x_2}$
Hence not one $-$ one
$y = 1 + x^2$
$x = \pm \left( {\sqrt {1 - y} } \right)$
$f\left( {\sqrt {1 - y} } \right) = 1 + (1 - y) = 2 - y \ne y$
Therefore, $f$ is not onto.

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