If `f(x)` be a function such that `f(-x)= -f(x),g(x)` be a function such that `g(-x)= -g(x)` and `h(x)` be a function such that `h(-x)=h(x)`, then choose the correct statement:
I. `h(f(g(-x)))=-h(f(g(x)))`
II. `f(g(h(-x)))=f(g(h(x)))`
III. `g(f(-x))=g(f(x))`

Options

A. Only I
B. Only II
C. Only III
D. Both I and II

Answer

To solve the problem, we need to analyze each statement based on the properties of the functions given:

1. **Understanding the properties of the functions:**
- \( f(x) \) is an odd function: \( f(-x) = -f(x) \)
- \( g(x) \) is an odd function: \( g(-x) = -g(x) \)
- \( h(x) \) is an even function: \( h(-x) = h(x) \)

2. **Evaluating each statement:**

**Statement I: \( h(f(g(-x))) = -h(f(g(x))) \)**

- Start with the left-hand side:
\[ h(f(g(-x))) = h(f(-g(x))) \quad \text{(since \( g(-x) = -g(x) \))} \]
- Using the property of \( f \):
\[ h(f(-g(x))) = h(-f(g(x))) \quad \text{(since \( f(-g(x)) = -f(g(x)) \))} \]
- Now apply the property of \( h \):
\[ h(-f(g(x))) = h(f(g(x))) \quad \text{(since \( h \) is even)} \]
- Therefore, we have:
\[ h(f(g(-x))) = h(f(g(x))) \]
- This does not equal \( -h(f(g(x))) \), hence Statement I is **false**.

**Statement II: \( f(g(h(-x))) = f(g(h(x))) \)**

- Start with the left-hand side:
\[ f(g(h(-x))) = f(g(h(x))) \quad \text{(since \( h \) is even)} \]
- Therefore, we have:
\[ f(g(h(-x))) = f(g(h(x))) \]
- This means Statement II is **true**.

**Statement III: \( g(f(-x)) = g(f(x)) \)**

- Start with the left-hand side:
\[ g(f(-x)) = g(-f(x)) \quad \text{(since \( f \) is odd)} \]
- Using the property of \( g \):
\[ g(-f(x)) = -g(f(x)) \quad \text{(since \( g \) is odd)} \]
- Therefore, we have:
\[ g(f(-x)) = -g(f(x)) \]
- This does not equal \( g(f(x)) \), hence Statement III is **false**.

**Conclusion:**
- Only Statement II is correct.

### Final Answer:
The correct statement is **II**.