If `f(x)` is an even function of x and g(x) is an odd function then which of the following is true?

Options

A. A. `f[g(x)]` is odd, `g[f(x)]` is odd
B. B. `f[g(x)` is even, `g[f(x)]` is even
C. C. `f[g(x)]` is even, `g[f(x)]` is odd
D. D. none of these

Answer

To solve the problem, we need to analyze the properties of the functions \( f(x) \) and \( g(x) \) based on their definitions as even and odd functions.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- An **even function** \( f(x) \) satisfies the property:
\[ f(-x) = f(x) \]
- An **odd function** \( g(x) \) satisfies the property:
\[ g(-x) = -g(x) \]

2. **Determine \( f(g(x)) \)**:
- We need to find \( f(g(-x)) \):
\[ g(-x) = -g(x) \quad \text{(since \( g(x) \) is odd)} \]
- Now substitute this into \( f \):
\[ f(g(-x)) = f(-g(x)) \]
- Since \( f(x) \) is even, we have:
\[ f(-g(x)) = f(g(x)) \]
- Thus, we conclude:
\[ f(g(-x)) = f(g(x)) \]
- This shows that \( f(g(x)) \) is an even function.

3. **Determine \( g(f(x)) \)**:
- Now we need to find \( g(f(-x)) \):
\[ f(-x) = f(x) \quad \text{(since \( f(x) \) is even)} \]
- Substitute this into \( g \):
\[ g(f(-x)) = g(f(x)) \]
- Since \( g(x) \) is odd, we have:
\[ g(-f(x)) = -g(f(x)) \]
- Therefore:
\[ g(f(-x)) = g(f(x)) \]
- This shows that \( g(f(x)) \) is also an odd function.

4. **Conclusion**:
- From the analysis, we find that:
- \( f(g(x)) \) is an even function.
- \( g(f(x)) \) is an odd function.

### Final Answer:
- The correct conclusion is that \( f(g(x)) \) is even and \( g(f(x)) \) is odd.