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

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B

injective but not surjective

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C

bijective

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D

neither injective nor surjective

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Solution

## The correct option is D neither injective nor surjectiveFinding 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$$⇒$ $f\left(n\right)$ is not even.$⇒$ 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.

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