Which of the following functions form Z to itself are bijections?

Solution:

(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).