Let f(x) be a real valued function not identically zero, which satisfied the following conditions
I. `f(x + y^(2n + 1)) = f(x) + (f(y))^(2n+1), n in N, x, y` are any real numbers.
II. `f'(0) ge 0`
The value of f(x), is

Options

A. 2x
B. `x^(2) + x + 1`
C. x
D. None of these

Answer

To solve the problem, we need to analyze the given functional equation and the conditions provided.

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the functional equation given:
\[ f(x + y^{2n + 1}) = f(x) + (f(y))^{2n + 1} \]
for all \( n \in \mathbb{N} \) and real numbers \( x, y \).

2. **Setting Initial Values**:
Let's set \( x = 0 \) and \( y = 0 \) in the functional equation:
\[ f(0 + 0^{2n + 1}) = f(0) + (f(0))^{2n + 1} \]
This simplifies to:
\[ f(0) = f(0) + (f(0))^{2n + 1} \]
Rearranging gives:
\[ (f(0))^{2n + 1} = 0 \]
Thus, we conclude that:
\[ f(0) = 0 \]

3. **Using the Derivative Condition**:
We are given that \( f'(0) \geq 0 \). This implies:
\[ \lim_{h \to 0} \frac{f(h) - f(0)}{h} \geq 0 \]
Since \( f(0) = 0 \), this simplifies to:
\[ \lim_{h \to 0} \frac{f(h)}{h} \geq 0 \]
Therefore, if \( h > 0 \), we have \( f(h) \geq 0 \).

4. **Exploring Values of \( f(1) \)**:
Now, let's set \( x = 0 \) and \( y = 1 \) in the original functional equation:
\[ f(0 + 1^{2n + 1}) = f(0) + (f(1))^{2n + 1} \]
This simplifies to:
\[ f(1) = 0 + (f(1))^{2n + 1} \]
This gives us:
\[ f(1) = (f(1))^{2n + 1} \]
This implies two cases:
- Case 1: \( f(1) = 0 \)
- Case 2: \( f(1) = 1 \)

5. **Analyzing Case 1**:
If \( f(1) = 0 \), then substituting back into the functional equation:
\[ f(x + 1) = f(x) + 0 \]
This implies:
\[ f(x + 1) = f(x) \]
This would mean \( f(x) \) is constant for all \( x \), contradicting the condition that \( f(x) \) is not identically zero.

6. **Analyzing Case 2**:
If \( f(1) = 1 \), we can substitute into the functional equation:
\[ f(x + 1) = f(x) + 1 \]
This implies:
- For \( x = 1 \): \( f(2) = f(1) + 1 = 2 \)
- For \( x = 2 \): \( f(3) = f(2) + 1 = 3 \)
- Continuing this, we find that \( f(n) = n \) for all natural numbers \( n \).

7. **Generalizing**:
By induction or by observing the pattern, we can conclude that:
\[ f(x) = x \]
for all real \( x \).

8. **Final Conclusion**:
Therefore, the value of \( f(x) \) is:
\[ \boxed{x} \]