Question

A mapping from $N$ to $N$ is defined as follows
$f : N \to N$
$f (n) = (n+5)^{2}$, $n \in N$
($N$ is the set of natural numbers). Then,

• A. $f$ is not one to one
• B. $f$ is onto
• C. $f$ is both one to one and onto
• D. $f$ is one to one but not onto

Answer 1

Answer: (D)

Given, $f : N \to N, f \left(n\right)=\left(n+5\right)^{2}$
For one to one
$f\left(n_{1}\right)=f \left(n_{2}\right)$
$\Rightarrow \left(n_{1}+5\right)^{2}=\left(n_{2}+5\right)^{2}$
$\Rightarrow \left(n_{1}-n_{2}\right)\left(n_{1}+n_{2}+10\right)=0$
$\Rightarrow n_{1}=n_{2}$
$\therefore $ $f$ is one to one
When we put $n = 1, 2, 3, 4, \ldots\infty$, we will get
$f\left(1\right)=36, f \left(2\right), 49, f \left(3\right)=64, f \left(4\right)=81, \ldots$
Here, we see that we do not get any pre-images of $1, 2, 3$ etc
Hence, $f$ is not onto