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Question

# Which of the following functions form Z to itself are bijections? (a) $f\left(x\right)={x}^{3}$ (b) $f\left(x\right)=x+2$ (c) $f\left(x\right)=2x+1$ (d) $f\left(x\right)={x}^{2}+x$

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Solution

## $\left(a\right)f\text{is not onto because for}\mathit{\text{y}}\text{= 3∈Co-domain(}\mathit{\text{Z}}\text{), there is no value of x∈Domain(Z)}\phantom{\rule{0ex}{0ex}}{x}^{3}=3\phantom{\rule{0ex}{0ex}}⇒x=\sqrt[3]{3}\notin Z\phantom{\rule{0ex}{0ex}}\text{⇒}f\text{is not onto.}\phantom{\rule{0ex}{0ex}}\text{So,}f\text{is not a bijection.}\phantom{\rule{0ex}{0ex}}$ (b) Injectivity: Let x and y be two elements of the domain (Z), such that $x+2=y+2\phantom{\rule{0ex}{0ex}}⇒x=y$ So, f is one-one. Surjectivity: Let y be an element in the co-domain (Z), such that $y=f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒y=x+2\phantom{\rule{0ex}{0ex}}⇒x=y-2\in Z\left(\text{Domain}\right)$ $⇒$f is onto. So, f is a bijection. $\left(c\right)f\left(x\right)=2x+1\text{is not onto because if we take 4 ∈}Z\left(\text{co domain}\right),\text{then}4=f\left(x\right)\phantom{\rule{0ex}{0ex}}⇒4=2x+1\phantom{\rule{0ex}{0ex}}⇒2x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{2}\notin Z\phantom{\rule{0ex}{0ex}}\text{So,}f\text{is not a bijection.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(d\right)f\left(0\right)={0}^{2}+0=0\phantom{\rule{0ex}{0ex}}\mathrm{and}f\left(-1\right)={\left(-1\right)}^{2}+\left(-1\right)=1-1=0\phantom{\rule{0ex}{0ex}}\text{⇒0 and -1 have the same image.}\phantom{\rule{0ex}{0ex}}\text{⇒}f\text{is not one-one.}\phantom{\rule{0ex}{0ex}}\text{So,}\mathit{\text{f}}\text{is not a bijection.}\phantom{\rule{0ex}{0ex}}$ So, the answer is (b).

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