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

$\begin{array}{1 1} \text{neither one -one nor onto} \\ \text{bijective} \\ \text{only one-one} \\ \text{only onto}\end{array}$

Can you answer this question?

Toolbox:
• A function is one-one
• if $f(x)=f(y) => x=y$
• function is onto. if then exist x such that for $f(x)=y$ foe every y
• Bijective if both one-one and onto
$f: R \to R \qquad f(x)=1+x^2$

$x_1\,x_2 \in R$

$f(x_1)=f(x_2)$

$=>1+x_1^2=1+x_2^2$

$x_1^2=x_2^2$

$x_1=\pm x_2$

This is not imply $x_1=x_2$

f is not one one

Also $f(1)=f(-1)=2$

but $1=-1$

so f is not onto

Hence f is neither one-one nor onto

answered Mar 6, 2013 by