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(1), is

Options

A. (a)0
B. (b)1
C. (c)2
D. (d)Not defined

Answer

To solve the problem, we need to analyze the given functional equation and the derivative condition step by step.

### Step 1: Analyze the Functional Equation
We start with the functional equation provided:

\[ f(x + y^{2n+1}) = f(x) + (f(y))^{2n+1} \]

for any real numbers \(x\) and \(y\), and \(n\) belonging to the natural numbers.

### Step 2: Substitute Values
Let's substitute \(x = 0\) and \(y = 0\):

\[ f(0 + 0^{2n+1}) = f(0) + (f(0))^{2n+1} \]

This simplifies to:

\[ f(0) = f(0) + (f(0))^{2n+1} \]

This implies:

\[ 0 = (f(0))^{2n+1} \]

Since \(f(x)\) is not identically zero, we conclude that:

\[ f(0) = 0 \]

### Step 3: Differentiate the Function
Next, we consider the derivative condition given in the problem:

\[ f'(0) \geq 0 \]

Using the definition of the derivative, we have:

\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{f(x)}{x} \]

Since \(f(0) = 0\), we can rewrite this as:

\[ f'(0) = \lim_{x \to 0} \frac{f(x)}{x} \]

### Step 4: Analyze the Behavior of \(f(x)\)
From the above limit, if \(f'(0) \geq 0\), then as \(x\) approaches \(0\), \(f(x)\) must behave such that:

- If \(x > 0\), \(f(x) \geq 0\).
- If \(x < 0\), we need to analyze further.

### Step 5: Substitute \(y = 1\)
Now, we substitute \(y = 1\) into the original functional equation:

\[ f(x + 1^{2n+1}) = f(x) + (f(1))^{2n+1} \]

This simplifies to:

\[ f(x + 1) = f(x) + (f(1))^{2n+1} \]

### Step 6: Evaluate \(f(1)\)
Let’s evaluate \(f(1)\) when \(x = 0\):

\[ f(1) = f(0) + (f(1))^{2n+1} \]

Since \(f(0) = 0\), we have:

\[ f(1) = (f(1))^{2n+1} \]

### Step 7: Solve the Equation
This implies two possibilities:

1. \(f(1) = 0\)
2. \(f(1) = 1\) (since \(f(1)^{2n+1} = f(1)\) implies \(f(1) = 1\) if \(f(1) \neq 0\))

### Conclusion
Thus, the possible values of \(f(1)\) are:

\[ f(1) = 0 \quad \text{or} \quad f(1) = 1 \]

### Final Answer
The value of \(f(1)\) is either \(0\) or \(1\).

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