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(D) The domain $N$ is partitioned into three disjoint sets based on the form of $n$:$1$. $n = 2k$ (even numbers): $f(n) = 2(2k) = 4k$$2$. $n = 4k-1$ (

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one-one and onto

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(D) The domain $N$ is partitioned into three disjoint sets based on the form of $n$:$1$. $n = 2k$ (even numbers): $f(n) = 2(2k) = 4k$$2$. $n = 4k-1$ (numbers of the form $4k-1$): $f(n) = (4k-1)-1 = 4k-2$$3$. $n = 4k-3$ (numbers of the form $4k-3$): $f(n) = \frac{(4k-3)+1}{2} = 2k-1$For any $y \in N$,we check if there exists a unique $n$ such that $f(n) = y$:- If $y$ is of the form $4k$,then $n = 2k$ is the unique preimage.- If $y$ is of the form $4k-2$,then $n = 4k-1$ is the unique preimage.- If $y$ is of the form $2k-1$ (odd numbers),then $n = 4k-3$ is the unique preimage.Since every $y \in N$ has a unique preimage $n \in N$,the function $f$ is both one-one and onto.

Let a function $f: N \rightarrow N$ be defined by
$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \dots \\ n-1, & n = 3, 7, 11, 15, \dots \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \dots \end{cases}$
Then,$f$ is

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Let a function $f: N \rightarrow N$ be defined by$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \dots \\ n-1, & n = 3, 7, 11, 15, \dots \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \dots \end{cases}$Then,$f$ is