Question

A mapping $f:N\to N$, where $N$ is the set of natural numbers is defined as
$f(n)=n^2$ , for $n$ odd
$f(n)=2n+1$ , for $n$ even
for $n\in N$.
Then $f$ is

• A. surjective but not injective
• B. injective but not suijective
• C. bijective
• D. neither injective nor suijective

Answer 1

Answer: (B)

Since, $f(n)=<br/><br/>\{\begin{array}{ll}{n}^2& \text{if n is odd}\\ 2n+1& \text{if n is even}<br/>\end{array}$

For even functions

$2n_1+1=2n_2+1$

$\Rightarrow 2n_1=2n_2$

$\Rightarrow n_1=n_2$

and for two odd functions

$n_1^2=n_2^2$

$\Rightarrow n_1^2-n_2^2=0$

$\Rightarrow n_1-n_2=0$ or $n_1+n_2=0$
$\Rightarrow n_1=n_2\left(\because n_1+n_2\ne 0\right)$

$\therefore f\left(n\right)$ is one-one.

But $f\left(n\right)$ is not onto.

$\therefore f\left(n\right)$ is injective but not surjective.