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
$$
f(n) =
\begin{cases}
\frac{n + 1}{2}, & \text{if } n \text{ is odd} \\
\frac{4 - n}{2}, & \text{if } n \text{ is even}
\end{cases}
\quad \text{and} \quad
g(n) =
\begin{cases}
3 + 2n, & \text{if } n \ge 0 \\
-2n, & \text{if } n<0
\end{cases}
$$
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?
Show Hint
- \( g \circ f \) 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
The Correct Option is A, D
Solution and Explanation
Step 1: Understand the given functions
- \( \mathbb{N} \) is the set of natural numbers. We'll assume \( \mathbb{N} = \{1, 2, 3, \dots\} \).
- \( \mathbb{Z} \) is the set of all integers: \( \{\dots, -2, -1, 0, 1, 2, \dots\} \).
- Function \( f: \mathbb{N} \to \mathbb{Z} \) is defined as: \[ f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd}, \\ \frac{4 - n}{2} & \text{if } n \text{ is even} \end{cases} \]
- Function \( g: \mathbb{Z} \to \mathbb{N} \) is defined as: \[ g(n) = \begin{cases} 3 + 2n & \text{if } n \geq 0, \\ -2n & \text{if } n < 0 \end{cases} \]
- We are to evaluate \( g \circ f \) and \( f \circ g \) for injectivity and surjectivity.
Step 2: Analyze \( f(n) \)
Odd inputs: \( f(1) = 1, f(3) = 2, f(5) = 3 \Rightarrow f(2k-1) = k \Rightarrow \{1, 2, 3, \dots\} \)
Even inputs: \( f(2) = 1, f(4) = 0, f(6) = -1, f(8) = -2 \Rightarrow f(2k) = 2 - k \Rightarrow \{1, 0, -1, -2, \dots\} \)
Conclusion: \( f \) is onto \( \mathbb{Z} \), but not one-to-one (e.g., \( f(1) = f(2) = 1 \)).
Step 3: Analyze \( g(n) \)
\( n \geq 0 \Rightarrow g(n) = 3 + 2n \Rightarrow \{3, 5, 7, \dots\} \)
\( n < 0 \Rightarrow g(n) = -2n \Rightarrow \{2, 4, 6, \dots\} \)
Conclusion: \( g \) is one-to-one, but not onto \( \mathbb{N} \) (missing 1).
Step 4: Analyze \( g \circ f: \mathbb{N} \to \mathbb{N} \)
For odd \( n = 2k-1 \): \( f(n) = k \Rightarrow g(k) = 3 + 2k \) (odd \( \geq 5 \))
For even \( n = 2k \): \( f(n) = 2 - k \). Cases:
- if \( k = 1 \): \( f(2) = 1 \Rightarrow g(1) = 5 \)
- if \( k = 2 \): \( f(4) = 0 \Rightarrow g(0) = 3 \)
- if \( k = 3 \): \( f(6) = -1 \Rightarrow g(-1) = 2 \)
Conclusion: Range = \( \{2, 3, 4, 5, \dots\} = \mathbb{N} \setminus \{1\} \); Not onto and not one-to-one (e.g., \( (g \circ f)(1) = (g \circ f)(2) = 5 \)).
Step 5: Analyze \( f \circ g: \mathbb{Z} \to \mathbb{Z} \)
For \( n \geq 0 \): \( g(n) = 3 + 2n \) → odd → \( f(g(n)) = \frac{g(n)+1}{2} = 2 + n \)
For \( n < 0 \): \( g(n) = -2n \) → even → \( f(g(n)) = \frac{4 - (-2n)}{2} = 2 - n \)
Conclusion: Range = \( \mathbb{Z} \), one-to-one and onto.
Step 6: Evaluate Options
- (A) \( g \circ f \): not one-to-one and not onto → True
- (B) \( f \circ g \): is one-to-one and onto → False
- (C) \( g \): one-to-one but not onto → False
- (D) \( f \): onto but not one-to-one → True
Final Answer: \( \boxed{\text{(A), (D)}} \)
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