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. Both one-one and onto
B. Neither one-one nor onto
C. onto but not one-one
D. One-one but not onto

Explanation

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

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

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

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

The term \( (-1)^n \) evaluates to:
- \( -1 \) when \( n \) is odd
- \( +1 \) when \( n \) is even

Thus, we can rewrite \( g(n) \):
- If \( n \) is odd:
\[ g(n) = n - (-1) = n + 1 \]
- If \( n \) is even:
\[ g(n) = n - 1 \]

So, we have:
\[ g(n) = \begin{cases} n + 1 & \text{if } n \text{ is odd} \\ n - 1 & \text{if } n \text{ is even} \end{cases} \]

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

Now, we need to find \( f(g(n)) \):
- If \( n \) is odd:
\[ g(n) = n + 1 \quad (\text{which is even}) \]
Therefore,
\[ f(g(n)) = f(n + 1) = \frac{n + 1}{2} \]

- If \( n \) is even:
\[ g(n) = n - 1 \quad (\text{which is odd}) \]
Therefore,
\[ f(g(n)) = f(n - 1) = \frac{(n - 1) + 1}{2} = \frac{n}{2} \]

### Step 4: Combine results

Now we can summarize the composition \( f \circ g \):
\[ f(g(n)) = \begin{cases} \frac{n + 1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases} \]

### Step 5: Conclusion

The final result for \( f \circ g \) is:
\[ f \circ g(n) = \begin{cases} \frac{n + 1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases} \]