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?
$g \circ f$ is NOT one-one and $g \circ f$ is NOT onto
$f \circ g$ is NOT one-one but $f \circ g$ is onto
$g$ is one-one and $g$ is onto
$f$ is NOT one-one but $f$ is onto
Step-by-Step Solution
Key Concept: Checking injectivity and surjectivity for composed piecewise functions by tracing ranges carefully
Analyze $f$: for odd $n$: $f(1)=1, f(3)=2, f(5)=3, \ldots$ and for even $n$: $f(2)=1, f(4)=0, f(6)=-1, \ldots$
$f(1)=1$ and $f(2)=1$, so $f$ is NOT one-one.
$f$ maps $\mathbb{N}$ onto: odd $n$ gives all positive integers, even $n$ gives $1, 0, -1, -2, \ldots$ (all non-positive integers $\cup \{1\}$). Together range is all of $\mathbb{Z}$. So $f$ IS onto. (D is TRUE)
Analyze $g$: for $n \geq 0$: $g(0)=3, g(1)=5, g(2)=7, \ldots$ (odd numbers $\geq 3$); for $n<0$: $g(-1)=2, g(-2)=4, \ldots$ (even numbers $\geq 2$). Range of $g = \{2,3,4,5,\ldots\} = \mathbb{N} \setminus \{1\}$. So $g$ is NOT onto $\mathbb{N}$. (C is FALSE). $g$ is one-one (strictly different outputs for each input).
$g \circ f: \mathbb{N} \to \mathbb{N}$. Since $f(1)=f(2)=1$, we get $g(f(1))=g(f(2))=g(1)=5$, so $g\circ f$ is NOT one-one.
Range of $g\circ f \subseteq$ range of $g = \mathbb{N}\setminus\{1\}$, so $1 \in \mathbb{N}$ is never achieved. $g\circ f$ is NOT onto. (A is TRUE)
$f\circ g$: since $g$ is one-one and $f$ is not, $f\circ g$ need not be one-one. $f\circ g(0)=f(3)=2$, $f\circ g(-1)=f(2)=1$. Range of $f\circ g$ = $f$(range of $g$) = $f(\mathbb{N}\setminus\{1\})$. $f$ is onto $\mathbb{Z}$, and $\mathbb{N}\setminus\{1\}$ still covers enough values, so $f\circ g$ is onto $\mathbb{Z}$? Check: every $k\in\mathbb{Z}$ can be reached. For $k\geq 2$: odd $n=2k-1\geq3\in\mathbb{N}\setminus\{1\}$, $f(2k-1)=k$; for $k\leq0$: even $n=4-2k\geq4\in\mathbb{N}\setminus\{1\}$, $f(4-2k)=k$; $k=1$: need odd $n=1$ but $1\notin\mathbb{N}\setminus\{1\}$; alternatively even $n=2$, $f(2)=1$ ✓ since $2\in\mathbb{N}\setminus\{1\}$. So $f\circ g$ is onto. (B claims not one-one but onto — partially matches, but we need to verify not one-one: $f\circ g(0)=f(3)=2$ and $f\circ g(-1)=f(2)=1$; try $f\circ g(1)=f(5)=3$, $f\circ g(2)=f(7)=4$; and $f\circ g(-2)=f(4)=0$, $f\circ g(-3)=f(6)=-1$. All distinct so far. Actually $g$ is injective and maps to distinct values; $f$ restricted to range of $g$ might be injective. Check: range of $g=\{2,3,4,5,\ldots\}$; on even values $f(n)=(4-n)/2$ injective; on odd values $f(n)=(n+1)/2$ injective; but $f(2)=1$ and $f(1)=1$ — but $1\notin$ range of $g$, so no collision. Thus $f\circ g$ IS one-one. (B is FALSE)
Correct Answer: A, D