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 will find the composition of the functions \( f \) and \( g \), denoted as \( f \circ g \).

### Step 1: Understand the functions \( f \) and \( g \)

The function \( f \) is defined as:
- \( f(n) = \frac{n-1}{2} \) if \( n \) is odd
- \( f(n) = -\frac{n}{2} \) if \( n \) is even

The function \( g \) is defined as:
- \( g(n) = n - (-1)^n \)

### Step 2: Analyze \( g(n) \)

We need to evaluate \( g(n) \) for both odd and even \( n \):
- If \( n \) is odd, then \( (-1)^n = -1 \):
\[ g(n) = n - (-1) = n + 1 \]

- If \( n \) is even, then \( (-1)^n = 1 \):
\[ g(n) = n - 1 \]

### Step 3: Find \( f(g(n)) \)

Now we will find \( f(g(n)) \) for both cases of \( n \).

**Case 1: \( n \) is odd**
- From Step 2, \( g(n) = n + 1 \) (which is even since odd + 1 = even).
- Now we apply \( f \):
\[ f(g(n)) = f(n + 1) = -\frac{(n + 1)}{2} \]

**Case 2: \( n \) is even**
- From Step 2, \( g(n) = n - 1 \) (which is odd since even - 1 = odd).
- Now we apply \( f \):
\[ f(g(n)) = f(n - 1) = \frac{(n - 1) - 1}{2} = \frac{n - 2}{2} \]

### Step 4: Combine the results

We can summarize the results of \( f(g(n)) \) as follows:
- If \( n \) is odd:
\[ f(g(n)) = -\frac{(n + 1)}{2} \]

- If \( n \) is even:
\[ f(g(n)) = \frac{(n - 2)}{2} \]

### Final Result
Thus, the function \( f \circ g \) can be expressed as:
\[ f \circ g(n) = \begin{cases} -\frac{(n + 1)}{2} & \text{if } n \text{ is odd} \\ \frac{(n - 2)}{2} & \text{if } n \text{ is even} \end{cases} \]