Let f be a differentiable function satisfying `f'(x)=f' (-x) AA x in R.` Then which of the following is correct regarding f:

Options

A. (a) If `f (1) =f (2), ` then `f (-1)=f (-2)`
B. (b) `1/2 f (x)+ 1/2 f (y) =f ((1)/(2) (x+y)))` for all ral values of x,y
C. (c) Let `f (x)` be an even function, then `f (x) =0 AA x in R`
D. (d) `f(x) +f (-x) =2f (0) AA x in R`

Answer

To solve the problem, we need to analyze the given condition \( f'(x) = f'(-x) \) for all \( x \in \mathbb{R} \) and determine the implications for the function \( f(x) \).

### Step 1: Understand the given condition
The condition \( f'(x) = f'(-x) \) implies that the derivative of \( f \) is an even function. This means that the slope of the tangent line to the graph of \( f \) at \( x \) is the same as the slope at \( -x \).

### Step 2: Integrate the condition
Since \( f'(-x) = f'(x) \), we can integrate both sides with respect to \( x \):
\[ \int f'(x) \, dx = \int f'(-x) \, dx \]
This gives us:
\[ f(x) = -f(-x) + C \]
where \( C \) is a constant of integration.

### Step 3: Analyze the implications
From the equation \( f(x) = -f(-x) + C \), we can rearrange it to find:
\[ f(x) + f(-x) = C \]
This indicates that the sum of the function values at \( x \) and \( -x \) is constant.

### Step 4: Check the options
Now, we will analyze the options provided:

**Option A:** If \( f(1) = f(2) \), then \( f(-1) = f(-2) \).

Using the derived equation \( f(x) + f(-x) = C \):
- \( f(1) + f(-1) = C \)
- \( f(2) + f(-2) = C \)

Since \( f(1) = f(2) \), we can equate:
\[ f(1) + f(-1) = f(2) + f(-2) \]
This implies \( f(-1) = f(-2) \). Thus, **Option A is correct**.

**Option B:** \( \frac{1}{2}f(x) + \frac{1}{2}f(y) = f\left(\frac{x+y}{2}\right) \).

This is a condition for linearity, which does not necessarily hold for all functions satisfying our initial condition. Therefore, **Option B is incorrect**.

**Option C:** If \( f(x) \) is even, then \( f(x) = 0 \).

This is not true because even functions can take non-zero values. For example, \( f(x) = x^2 \) is even but not zero. Thus, **Option C is incorrect**.

**Option D:** \( f(x) + f(-x) = 2f(0) \).

From our derived equation \( f(x) + f(-x) = C \) and knowing that \( C = 2f(0) \) (since \( f(0) + f(0) = 2f(0) \)), we can conclude that **Option D is correct**.

### Conclusion
The correct options are:
- **Option A**: \( f(1) = f(2) \implies f(-1) = f(-2) \)
- **Option D**: \( f(x) + f(-x) = 2f(0) \)