If `f:NrarrZ` defined as `f(n)={{:((n-1)/(2),":"," if n is odd"),((-n)/(2),":", " if n is even"):}` and `g:NrarrN` defined as `g(n)=n-(-1)^(n)`, then fog is (where, N is the set of natural numbers and Z is the set of integers)

Options

A. one - one and onto
B. one - one and into
C. many - one and onto
D. many - one and into

Answer

To solve the problem, we need to find the composition of the functions \( f \) and \( g \), denoted as \( f \circ g \).

### Step-by-Step Solution:

1. **Define the Functions**:
- The function \( f: \mathbb{N} \rightarrow \mathbb{Z} \) is defined as:
\[ f(n) = \begin{cases} \frac{n-1}{2} & \text{if } n \text{ is odd} \\ -\frac{n}{2} & \text{if } n \text{ is even} \end{cases} \]
- The function \( g: \mathbb{N} \rightarrow \mathbb{N} \) is defined as:
\[ g(n) = n - (-1)^n \]

2. **Calculate \( g(n) \)** for different values of \( n \):
- For \( n = 1 \):
\[ g(1) = 1 - (-1)^1 = 1 + 1 = 2 \]
- For \( n = 2 \):
\[ g(2) = 2 - (-1)^2 = 2 - 1 = 1 \]
- For \( n = 3 \):
\[ g(3) = 3 - (-1)^3 = 3 + 1 = 4 \]
- For \( n = 4 \):
\[ g(4) = 4 - (-1)^4 = 4 - 1 = 3 \]
- For \( n = 5 \):
\[ g(5) = 5 - (-1)^5 = 5 + 1 = 6 \]

3. **Calculate \( f(g(n)) \)** for the values of \( n \):
- For \( n = 1 \):
\[ f(g(1)) = f(2) = -\frac{2}{2} = -1 \quad (\text{since } 2 \text{ is even}) \]
- For \( n = 2 \):
\[ f(g(2)) = f(1) = \frac{1-1}{2} = 0 \quad (\text{since } 1 \text{ is odd}) \]
- For \( n = 3 \):
\[ f(g(3)) = f(4) = -\frac{4}{2} = -2 \quad (\text{since } 4 \text{ is even}) \]
- For \( n = 4 \):
\[ f(g(4)) = f(3) = \frac{3-1}{2} = 1 \quad (\text{since } 3 \text{ is odd}) \]
- For \( n = 5 \):
\[ f(g(5)) = f(6) = -\frac{6}{2} = -3 \quad (\text{since } 6 \text{ is even}) \]

4. **Summarize the Results**:
- We have the following results:
- \( f(g(1)) = -1 \)
- \( f(g(2)) = 0 \)
- \( f(g(3)) = -2 \)
- \( f(g(4)) = 1 \)
- \( f(g(5)) = -3 \)

5. **Identify the Pattern**:
- For odd \( n \), \( f(g(n)) \) yields negative integers: \( -1, -2, -3, \ldots \)
- For even \( n \), \( f(g(n)) \) yields non-negative integers: \( 0, 1, \ldots \)
- Therefore, the range of \( f \circ g \) includes all integers \( \mathbb{Z} \).

6. **Conclusion**:
- The function \( f \circ g \) is onto \( \mathbb{Z} \) because it covers all integers.