1. Understand the functions

Given

with domain $\mathbb N$ and codomain $\mathbb Z$.

Also,

with domain $\mathbb Z$ and codomain $\mathbb N$.

We must test the statements about $f$, $g$, $g\circ f$, and $f\circ g$.

2. Analyze $f: \mathbb N \to \mathbb Z$

Let us list some values:

• $f(1)=\dfrac{2}{2}=1$
• $f(2)=\dfrac{4-2}{2}=1$
• $f(3)=\dfrac{4}{2}=2$
• $f(4)=\dfrac{4-4}{2}=0$
• $f(5)=\dfrac{6}{2}=3$
• $f(6)=\dfrac{4-6}{2}=-1$
• $f(7)=4$
• $f(8)=-2$

So the outputs go as $1,1,2,0,3,-1,4,-2,5,-3,\dots$

(i) Is $f$ one-one?

No, because $f(1)=1=f(2), \quad 1\ne 2.$ So $f$ is not one-one.

(ii) Is $f$ onto $\mathbb Z$?

We show every integer occurs.

• If $z\ge 1$, choose odd $n=2z-1$. Then $f(2z-1)=\frac{(2z-1)+1}{2}=z.$
• If $z=0$, choose $n=4$. Then $f(4)=0.$
• If $z<0$, write $z=-k$ where $k\in\mathbb N$. Choose even $n=2k+4$. Then $f(2k+4)=\frac{4-(2k+4)}{2}=\frac{-2k}{2}=-k=z.$

Hence every integer is attained, so $f$ is onto.

Therefore Option D is true.

3. Analyze $g: \mathbb Z \to \mathbb N$

Let us compute a few values:

• For $n\ge 0$: $g(n)=3+2n$, so values are $3,5,7,9,\dots$
• For $n<0$: $g(n)=-2n$, so values are $2,4,6,8,\dots$

Thus the range is $\{2,3,4,5,6,7,\dots\}=\mathbb N\setminus\{1\}$ (assuming $\mathbb N=\{1,2,3,\dots\}$ as usual in JEE).

(i) Is $g$ one-one?

• On $n\ge 0$, $3+2n$ is strictly increasing.
• On $n<0$, $-2n$ is also distinct for distinct negative integers.
• The first branch gives odd numbers $\ge 3$.
• The second branch gives even positive numbers.

These two sets do not overlap. Hence different inputs give different outputs. So $g$ is one-one.

(ii) Is $g$ onto $\mathbb N$?

No, since $1\in\mathbb N$ is never attained. So $g$ is not onto.

Therefore Option C is false.

4. Analyze $g\circ f : \mathbb N \to \mathbb N$

Since $f$ is not one-one, let us test whether composition is one-one:

$f(1)=1, \quad f(2)=1.$ Therefore $(g\circ f)(1)=g(1)=5, \qquad (g\circ f)(2)=g(1)=5.$ So $g\circ f$ is not one-one.

Now determine whether it is onto $\mathbb N$.

Because $f$ is onto $\mathbb Z$, the range of $g\circ f$ is exactly the range of $g$, namely $\mathbb N\setminus\{1\}.$ So $1$ is not attained. Hence $g\circ f$ is not onto.

Therefore Option A is true.

5. Analyze $f\circ g : \mathbb Z \to \mathbb Z$

We compute separately.

Case 1: $n\ge 0$

Then $g(n)=3+2n,$ which is odd. Hence $(f\circ g)(n)=f(3+2n)=\frac{(3+2n)+1}{2}=n+2.$

Case 2: $n<0$

Then $g(n)=-2n,$ which is even. Hence $(f\circ g)(n)=f(-2n)=\frac{4-(-2n)}{2}=\frac{4+2n}{2}=n+2.$

So for all $n\in\mathbb Z$, $(f\circ g)(n)=n+2.$

This is clearly:

• one-one, because if $n_1+2=n_2+2$, then $n_1=n_2$.
• onto $\mathbb Z$, because for any $m\in\mathbb Z$, choose $n=m-2$, then $(f\circ g)(n)=m$.

Thus Option B, which says "not one-one but onto," is false.

6. Final evaluation of options

• A: True
• B: False
• C: False
• D: True

So the correct choices are: $\boxed{A,\ D}$

7. Comparison with stored correct answer

Stored correct answer: A, D

Our derived answer matches the stored answer exactly.