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

The function $$f : \mathbb{N} \to \mathbb{N}$$ is defined as:

$$f(n) = \begin{cases} 2n, & n = 2, 4, 6, 8, \ldots \text{ (even numbers)} \\ n-1, & n = 3, 7, 11, 15, \ldots \text{ (} n \equiv 3 \pmod{4}\text{)} \\ \frac{n+1}{2}, & n = 1, 5, 9, 13, \ldots \text{ (} n \equiv 1 \pmod{4}\text{)} \end{cases}$$

For even inputs $$n = 2,4,6,\ldots$$, we have $$f(n)=2n$$, which yields $$4,8,12,16,\ldots$$—all multiples of 4, i.e.\ $$\{n\in\mathbb{N}:n\equiv0\pmod4\}$$. If $$n\equiv3\pmod4$$ (so $$n=3,7,11,\ldots$$), then $$f(n)=n-1$$ gives $$2,6,10,14,\ldots$$—the set $$\{n\in\mathbb{N}:n\equiv2\pmod4\}$$. Finally, if $$n\equiv1\pmod4$$ (so $$n=1,5,9,\ldots$$), then $$f(n)=\frac{n+1}{2}$$ produces $$1,3,5,7,\ldots$$—all odd natural numbers.

Thus the range of $$f$$ is $$\{1,3,5,7,\ldots\}\cup\{2,6,10,14,\ldots\}\cup\{4,8,12,16,\ldots\}=\{\text{odd numbers}\}\cup\{\text{numbers}\equiv2\pmod4\}\cup\{\text{multiples of }4\}=\mathbb{N},$$ so $$f$$ is onto.

The three output sets—odd numbers, numbers congruent to 2 mod 4, and multiples of 4—are mutually disjoint, and in each case the rule is strictly increasing (hence injective). Therefore $$f$$ is one-one as well.

Therefore, the correct answer is Option A: One-one and onto.