Question

A mapping $f:n⇢N$, where $N$ is the set of natural numbers is defined as

$f\left(n\right)=\left\{\begin{array}{l}{n}^2,fornodd\\ 2n+1,forneven\end{array}\right.$

for $n$ belongs to $N$. Then $f$is

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

Answer 1

The correct option is D

neither injective nor surjective

Finding the value of :

Given the function:

$f\left(n\right)=\left\{\begin{array}{l}{n}^2,fornodd\\ 2n+1,forneven\end{array}\right.$

$n,n^{2}and2n+1isodd\forall n\in N$

$\Rightarrow$ $f\left(n\right)$ is not even.

$\Rightarrow$ It is not onto or not surjective.

Since, $f\left(3\right)=f\left(4\right)$,$f\left(n\right)$ is neither one-one nor injective.

So, f (n) is neither injective nor not surjective.

Hence, option (D) is the correct option.