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(C) Given the function $f: N ightarrow N$ defined by $f(n) = \begin{cases} \frac{n+1}{2} & n 	ext{ is odd} \ \frac{n}{2} & n 	ext{ is even} \end{

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onto but not one-one

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(C) Given the function $f: N ightarrow N$ defined by $f(n) = \begin{cases} \frac{n+1}{2} & n 	ext{ is odd} \ \frac{n}{2} & n 	ext{ is even} \end{cases}$.To check for one-one: Consider $n=1$ (odd) and $n=2$ (even).$f(1) = \frac{1+1}{2} = 1$.$f(2) = \frac{2}{2} = 1$.Since $f(1) = f(2)$ but $1 
eq 2$,the function $f$ is not one-one.To check for onto: For any $y \in N$,we need to find $n \in N$ such that $f(n) = y$.If $y$ is any natural number,we can choose $n = 2y$ (which is even). Then $f(2y) = \frac{2y}{2} = y$.Since for every $y \in N$,there exists an $n \in N$ such that $f(n) = y$,the function $f$ is onto.Thus,$f$ is onto but not one-one.

Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$. Then $f$ is:

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