Let `f: R->R` satisfying `f((x+y)/k)=(f(x)+f(y))/k( k != 0,2)`.Let `f(x)` be differentiable on `R and f'(0) = a`, then determine `f(x)`.

Options

A. A. even function
B. B. neither even nor odd function
C. C. either zero or odd function
D. D. either zero or even function

Answer

To solve the problem, we need to determine the function \( f(x) \) given the functional equation:

\[ f\left(\frac{x+y}{k}\right) = \frac{f(x) + f(y)}{k} \]

where \( k \neq 0, 2 \), and that \( f(x) \) is differentiable on \( \mathbb{R} \) with \( f'(0) = a \).

### Step 1: Substitute Values into the Functional Equation

Let's set \( k = 1 \) and \( y = h \). This gives us:

\[ f(x + h) = f(x) + f(h) \]

This is a Cauchy-type functional equation.

### Step 2: Analyze the Functional Equation

The equation \( f(x + h) = f(x) + f(h) \) suggests that \( f \) is linear. To see this, we can set \( x = 0 \):

\[ f(h) = f(0) + f(h) \]

This implies \( f(0) = 0 \).

### Step 3: Differentiate the Function

Since \( f(x) \) is differentiable, we can express the derivative:

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

Using the functional equation, we have:

\[ f(x + h) - f(x) = f(h) \]

Thus,

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

### Step 4: Evaluate the Limit

As \( h \to 0 \), we know from the problem statement that \( f'(0) = a \). Therefore,

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

This indicates that \( f(h) \) behaves like \( ah \) as \( h \) approaches 0.

### Step 5: General Form of the Function

Since \( f(x) \) is linear and \( f(0) = 0 \), we can express \( f(x) \) in the form:

\[ f(x) = ax + c \]

Given that \( f(0) = 0 \), we find \( c = 0 \). Thus, we have:

\[ f(x) = ax \]

### Step 6: Check for Odd Function

To verify if \( f(x) \) is an odd function, we check:

\[ f(-x) = a(-x) = -ax = -f(x) \]

This confirms that \( f(x) \) is indeed an odd function.

### Conclusion

The function \( f(x) \) is given by:

\[ f(x) = ax \]

### Final Answer

The correct option is (C): either \( 0 \) or an odd function.

---