Let $$\mathbb{N}$$ denote the set of all natural numbers, and $$\mathbb{Z}$$ denote the set of all integers. Consider the functions $$f: \mathbb{N} \to \mathbb{Z}$$ and $$g: \mathbb{Z} \to \mathbb{N}$$ defined by

$$f(n) = \begin{cases} (n+1)/2 & \text{if } n \text{ is odd}, \\ (4-n)/2 & \text{if } n \text{ is even}, \end{cases}$$

and

$$g(n) = \begin{cases} 3 + 2n & \text{if } n \geq 0, \\ -2n & \text{if } n < 0. \end{cases}$$

Define $$(g \circ f)(n) = g(f(n))$$ for all $$n \in \mathbb{N}$$, and $$(f \circ g)(n) = f(g(n))$$ for all $$n \in \mathbb{Z}$$.

Then which of the following statements is (are) TRUE?

For convenience let $$\mathbb{N}=\{1,2,3,\dots\}$$ and $$\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}$$.

1. Properties of $$f:\mathbb{N}\to\mathbb{Z}$$
  • If $$n$$ is odd, write $$n=2k-1\;(k\ge 1)$$. Then $$f(n)=\frac{n+1}{2}=\frac{2k-1+1}{2}=k,$$ which produces all positive integers $$1,2,3,\dots$$.
  • If $$n$$ is even, write $$n=2k\;(k\ge 1)$$. Then $$f(n)=\frac{4-n}{2}=\frac{4-2k}{2}=2-k,$$ which produces the values $$1,0,-1,-2,-3,\dots$$.
Combining the two cases, the image of $$f$$ is all of $$\mathbb{Z}$$, so $$f$$ is onto.

However, $$f(1)=\frac{1+1}{2}=1,\quad f(2)=\frac{4-2}{2}=1,$$ and $$1\ne 2$$. Hence distinct inputs can give the same output, so $$f$$ is not one-one.

Conclusion for $$f$$: not one-one, but onto  ⇒  Option D is true.

2. Properties of $$g:\mathbb{Z}\to\mathbb{N}$$
  • $$n\ge 0$$: $$g(n)=3+2n=3,5,7,9,\dots$$ (all odd numbers $$\ge 3$$).
  • $$n\lt 0$$: write $$n=-k\;(k\ge 1)$$, then $$g(n)=-2(-k)=2k=2,4,6,8,\dots$$ (all even numbers $$\ge 2$$).
Thus $$\operatorname{Im}(g)=\{2,3,4,5,6,\dots\}=\mathbb{N}\setminus\{1\}$$, so $$g$$ is not onto.

To test injectivity, suppose $$g(n_1)=g(n_2)$$.
  • If both $$n_1,n_2\ge 0$$, then $$3+2n_1=3+2n_2\Rightarrow n_1=n_2$$(strictly increasing).
  • If both $$n_1,n_2\lt 0$$, write them as $$-k_1,-k_2$$. Then $$2k_1=2k_2\Rightarrow k_1=k_2\Rightarrow n_1=n_2$$.
  • If one is $$\ge 0$$ and the other $$\lt 0$$, their images have opposite parity (odd vs even), so equality is impossible.
Therefore $$g$$ is one-one.

Conclusion for $$g$$: one-one, but not onto  ⇒  Option C is false.

3. The composition $$g\circ f:\mathbb{N}\to\mathbb{N}$$
First compute the value explicitly.

$$f(n)=k\quad(\text{positive})$$
$$g(f(n))=g(k)=3+2k$$ Hence $$g\circ f(n)=3+2k$$, i.e. the numbers $$5,7,9,\dots$$.

$$f(n)=2-k$$
If $$k=1$$, $$f(n)=1,\;g(1)=5$$ (already obtained).
If $$k=2$$, $$f(n)=0,\;g(0)=3$$.
If $$k\ge 3$$, $$f(n)=2-k\le -1,\;g(2-k)=2k-4$$ (even numbers $$2,4,6,8,\dots$$).

The total image of $$g\circ f$$ is therefore $$\{2,3,4,5,6,7,8,9,\dots\}=\mathbb{N}\setminus\{1\},$$ so $$g\circ f$$ is not onto.

Also $$g\circ f(1)=5$$ and $$g\circ f(2)=5$$, yet $$1\ne 2$$, so the composition is not one-one.

Conclusion for $$g\circ f$$: neither one-one nor onto  ⇒  Option A is true.

4. The composition $$f\circ g:\mathbb{Z}\to\mathbb{Z}$$

$$g(n)=3+2n\;(\text{odd})$$
$$f(g(n))=\frac{(3+2n)+1}{2}=n+2.$$ So for $$n=0,1,2,\dots$$ we obtain $$2,3,4,\dots$$.

$$g(n)=2k\;(\text{even})$$
$$f(g(n))=\frac{4-2k}{2}=2-k.$$ For $$k=1,2,3,\dots$$ this gives $$1,0,-1,-2,\dots$$.

The union of the two cases yields every integer, so $$f\circ g$$ is onto. To check injectivity, observe:   • $$n\ge 0\; \Rightarrow f\circ g(n)=n+2$$ (strictly increasing).   • $$n\lt 0\; \Rightarrow f\circ g(n)=2-n$$ (strictly decreasing). No non-negative input can share an image with a negative input, so different arguments always give different values. Hence $$f\circ g$$ is one-one. Therefore Option B is false.

Final result: Only Option A and Option D are correct.