Let N be the set of numbers and two functions f and g be defined as `f,g:N to N` such that `f(n)={((n+1)/(2), ,"if n is odd"),((n)/(2),,"if n is even"):}` and `g(n)=n-(-1)^(n)`. Then, fog is

Options

A. one-one but not onto
B. onto but not one-one
C. both one-one and onto
D. neither one-one nor onto

Explanation

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 \( f(n) \) 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 \( g(n) \) is defined as:
\[ g(n) = n - (-1)^n \]

2. **Evaluate \( g(n) \)**:
- For \( n \) odd:
\[ g(n) = n - (-1) = n + 1 \]
- For \( n \) even:
\[ g(n) = n - 1 \]

3. **Compute \( f(g(n)) \)**:
- **Case 1**: When \( n \) is odd:
- Here, \( g(n) = n + 1 \) (which is even).
- Thus, we use the even case of \( f \):
\[ f(g(n)) = f(n + 1) = \frac{n + 1}{2} \]

- **Case 2**: When \( n \) is even:
- Here, \( g(n) = n - 1 \) (which is odd).
- Thus, we use the odd case of \( f \):
\[ f(g(n)) = f(n - 1) = \frac{(n - 1) + 1}{2} = \frac{n}{2} \]

4. **Combine the Results**:
- Therefore, we can summarize \( f(g(n)) \) as:
\[ f(g(n)) = \begin{cases} \frac{n + 1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases} \]

5. **Check if \( f \circ g \) is One-to-One and Onto**:
- **One-to-One**:
- For \( n \) odd, \( f(g(n)) = \frac{n + 1}{2} \) gives the same output for different odd inputs, hence it is not one-to-one.
- For \( n \) even, \( f(g(n)) = \frac{n}{2} \) also gives the same output for different even inputs, hence it is not one-to-one.
- **Onto**:
- The range of \( f(g(n)) \) does not cover all natural numbers, hence it is not onto.

### Conclusion:
The function \( f \circ g \) is neither one-to-one nor onto.