1. Understand the definition of the function

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

$$f(n)= \begin{cases} 2n, & n=2,4,6,8,\dots \\ n-1, & n=3,7,11,15,\dots \\ \dfrac{n+1}{2}, & n=1,5,9,13,\dots \end{cases}$$

Let us rewrite the three cases more systematically:

• If $n$ is even, then $f(n)=2n$.
• If $n \equiv 3 \pmod 4$, then $f(n)=n-1$.
• If $n \equiv 1 \pmod 4$, then $f(n)=\dfrac{n+1}{2}$.

So every natural number belongs to exactly one of these three classes.

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2. Check injectivity (one-one)

A function is one-one if different inputs give different outputs.

We compute a few values:

• $f(1)=\dfrac{1+1}{2}=1$
• $f(2)=2\cdot 2=4$
• $f(3)=3-1=2$
• $f(4)=2\cdot 4=8$
• $f(5)=\dfrac{5+1}{2}=3$
• $f(6)=2\cdot 6=12$
• $f(7)=7-1=6$
• $f(8)=2\cdot 8=16$
• $f(9)=\dfrac{9+1}{2}=5$

This gives the pattern:

• Numbers of the form $4k+1$ map to odd numbers: $f(4k+1)=\frac{4k+2}{2}=2k+1$
• Numbers of the form $4k+3$ map to numbers $4k+2$: $f(4k+3)=4k+2$
• Even numbers $2m$ map to multiples of $4$: $f(2m)=4m$

Thus the image sets are:

• From $n\equiv 1\pmod 4$: all odd natural numbers
• From $n\equiv 3\pmod 4$: numbers congruent to $2 \pmod 4$
• From even $n$: multiples of $4$

These three image sets are disjoint:

• odd numbers,
• numbers $\equiv 2 \pmod 4$,
• numbers $\equiv 0 \pmod 4$.

Since each class maps into a different disjoint set, two different inputs from different classes cannot have the same output.

Now within each class:

• $f(4k+1)=2k+1$ is strictly increasing in $k$,
• $f(4k+3)=4k+2$ is strictly increasing in $k$,
• $f(2m)=4m$ is strictly increasing in $m$.

Hence no two distinct inputs in the same class have the same output.

Therefore, $f$ is one-one.

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3. Check surjectivity (onto)

A function $f: \mathbb{N}\to\mathbb{N}$ is onto if every natural number has a preimage.

Take any $y\in\mathbb{N}$. We consider three cases.

Case 1: $y$ is odd

Let $y=2k+1$. Choose

$$n=4k+1.$$

Then

$$f(n)=f(4k+1)=\frac{4k+2}{2}=2k+1=y.$$

Case 2: $y\equiv 2 \pmod 4$

Let $y=4k+2$. Choose

$$n=4k+3.$$

Then

$$f(n)=f(4k+3)=4k+2=y.$$

Case 3: $y$ is a multiple of $4$

Let $y=4k$. Choose

$$n=2k.$$

Then

$$f(n)=f(2k)=4k=y.$$

Thus every natural number is hit by the function. So $f$ is onto.

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4. Conclusion

The function is both one-one and onto.

Hence, $f$ is bijective.

So the correct option is:

$$\boxed{\text{D: one-one and onto}}$$

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5. Comparison with stored correct answer

Stored correct answer: $\boxed{D}$

Our derived answer: $\boxed{D}$

So the answer agrees with the stored correct answer.