Which of the following functions form Z to itself are bijections?
(a) $f(x)=x^{3}$
(b) $f(x)=x+2$
(c) $f(x)=2 x+1$
(d) $f(x)=x^{2}+x$
(a) $f$ is not onto because for $y=3 \in$ Co-domain $(Z)$, there is no value of $\mathrm{x} \in$ Domain(Z)
$x^{3}=3$
$\Rightarrow x=\sqrt[3]{3} \notin Z$
$\Rightarrow f$ is not onto.
So, fis not a bijection.
(b) Injectivity:
Let x and y be two elements of the domain (Z), such that
$x+2=y+2$
$\Rightarrow x=y$
So, $f$ is one-one.
Surjectivity:
Let y be an element in the co-domain (Z), such that
$y=f(x)$
$\Rightarrow y=x+2$
$\Rightarrow x=y-2 \in Z$ (Domain)
$\Rightarrow f$ is onto.
So, $f$ is a bijection.
(c) $f(x)=2 x+1$ is not onto because if we take $4 \in Z$ (co domain), then $4=f(x)$
$\Rightarrow 4=2 x+1$
$\Rightarrow 2 x=3$
$\Rightarrow x=\frac{3}{2} \notin Z$
So, $f$ is not a bijection.
(d) $f(0)=0^{2}+0=0$
and $f(-1)=(-1)^{2}+(-1)=1-1=0$
$\Rightarrow 0$ and $-1$ have the same image.
$\Rightarrow f$ is not one-one.
So, $f$ is not a bijection.
So, the answer is (b).
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