Let f(x) be a real function not identically zero in Z, such that for all `x,y in R` `f(x+y^(2n+1))=f(x)={f(y)^(2n+1)}, n in Z`
If `f'(0) ge 0`, then f'(6) is equal to

Options

A. 0
B. 1
C. 2
D. 6

Explanation

To solve the problem step by step, we start with the given functional equation:

1. **Given Equation**:
\[ f(x + y^{2n+1}) = f(x) = f(y)^{2n+1}, \quad \forall x, y \in \mathbb{R}, n \in \mathbb{Z} \]

2. **Substituting Values**:
Let's substitute \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0^{2n+1}) = f(0) = f(0)^{2n+1} \]
This simplifies to:
\[ f(0) = f(0)^{2n+1} \]
This implies that \( f(0) \) must be either 0 or 1. However, since \( f(x) \) is not identically zero, we conclude:
\[ f(0) = 0 \]

3. **Finding the Derivative**:
Since \( f'(0) \geq 0 \), we can use the definition of the derivative:
\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} \]
Since \( f(0) = 0 \), this means:
\[ \lim_{x \to 0} \frac{f(x)}{x} \geq 0 \]

4. **Analyzing the Function**:
From the functional equation, we can analyze the behavior of \( f(x) \). Let's substitute \( y = 1 \):
\[ f(x + 1^{2n+1}) = f(x) = f(1)^{2n+1} \]
This gives us:
\[ f(x + 1) = f(x) = f(1)^{2n+1} \]
If we set \( n = 0 \), we find:
\[ f(x + 1) = f(x) \]
This implies that \( f(x) \) is constant for all \( x \). Since \( f(0) = 0 \), we conclude:
\[ f(x) = 0 \quad \forall x \]

5. **Finding \( f'(6) \)**:
Since \( f(x) = 0 \) for all \( x \), the derivative \( f'(x) \) is:
\[ f'(x) = 0 \quad \forall x \]
Thus:
\[ f'(6) = 0 \]

6. **Conclusion**:
The value of \( f'(6) \) is:
\[ \boxed{0} \]