If f and g be two functions, then prove the following:

Options

A. f odd, g odd =(fog) is odd
B. f even, g even =(fog) is even
C. f even, g odd (fog) is even
D. f odd, g even =(fog) is even

Answer

To prove the properties of the composition of functions \( f \) and \( g \) based on whether they are odd or even, we will analyze each case step by step.

### Step-by-Step Solution:

1. **Definitions**:
- A function \( h(x) \) is **even** if \( h(-x) = h(x) \).
- A function \( h(x) \) is **odd** if \( h(-x) = -h(x) \).

2. **First Option**: Prove that if \( f \) is odd and \( g \) is odd, then \( f(g(x)) \) is odd.

- Since \( f \) is odd, we have:
\[ f(-x) = -f(x) \]
- Since \( g \) is odd, we have:
\[ g(-x) = -g(x) \]
- Now, consider \( f(g(-x)) \):
\[ f(g(-x)) = f(-g(x)) \quad \text{(since \( g \) is odd)} \]
- Using the property of \( f \):
\[ f(-g(x)) = -f(g(x)) \quad \text{(since \( f \) is odd)} \]
- Thus, we have:
\[ f(g(-x)) = -f(g(x)) \]
- This shows that \( f(g(x)) \) is odd.

3. **Second Option**: Prove that if \( f \) is even and \( g \) is even, then \( f(g(x)) \) is even.

- Since \( f \) is even, we have:
\[ f(-x) = f(x) \]
- Since \( g \) is even, we have:
\[ g(-x) = g(x) \]
- Now, consider \( f(g(-x)) \):
\[ f(g(-x)) = f(g(x)) \quad \text{(since \( g \) is even)} \]
- This shows that \( f(g(x)) \) is even.

4. **Third Option**: Prove that if \( f \) is even and \( g \) is odd, then \( f(g(x)) \) is even.

- Since \( f \) is even:
\[ f(-x) = f(x) \]
- Since \( g \) is odd:
\[ g(-x) = -g(x) \]
- Now, consider \( f(g(-x)) \):
\[ f(g(-x)) = f(-g(x)) \quad \text{(since \( g \) is odd)} \]
- Using the property of \( f \):
\[ f(-g(x)) = f(g(x)) \quad \text{(since \( f \) is even)} \]
- This shows that \( f(g(x)) \) is even.

5. **Fourth Option**: Prove that if \( f \) is odd and \( g \) is even, then \( f(g(x)) \) is odd.

- Since \( f \) is odd:
\[ f(-x) = -f(x) \]
- Since \( g \) is even:
\[ g(-x) = g(x) \]
- Now, consider \( f(g(-x)) \):
\[ f(g(-x)) = f(g(x)) \quad \text{(since \( g \) is even)} \]
- Using the property of \( f \):
\[ f(g(-x)) = -f(g(x)) \quad \text{(since \( f \) is odd)} \]
- This shows that \( f(g(x)) \) is odd.

### Summary of Results:
- If \( f \) and \( g \) are both odd, then \( f(g(x)) \) is odd.
- If \( f \) and \( g \) are both even, then \( f(g(x)) \) is even.
- If \( f \) is even and \( g \) is odd, then \( f(g(x)) \) is even.
- If \( f \) is odd and \( g \) is even, then \( f(g(x)) \) is odd.