Let `f:R to R ` such that `f(x+y)+f(x-y)=2f(x)f(y)` for all ` x,y in R`. Then,

Options

A. f(x) an even function , if `f(0) ne 0`
B. f(x) is an odd function, if `f(0) ne 0`
C. f(x) an even function , if `f(0)=0`
D. f(x) is an odd function , if `f(0)=0`

Explanation

To solve the functional equation \( f(x+y) + f(x-y) = 2f(x)f(y) \) for all \( x, y \in \mathbb{R} \), we will follow these steps:

### Step 1: Substitute \( x = 0 \)

Let’s substitute \( x = 0 \) into the functional equation:

\[ f(0+y) + f(0-y) = 2f(0)f(y) \]

This simplifies to:

\[ f(y) + f(-y) = 2f(0)f(y) \]

### Step 2: Analyze the equation

Rearranging the equation gives:

\[ f(y) + f(-y) - 2f(0)f(y) = 0 \]

This can be factored as:

\[ f(y)(1 - 2f(0)) + f(-y) = 0 \]

### Step 3: Consider \( f(0) \)

Let’s consider two cases for \( f(0) \):

1. **Case 1:** If \( f(0) = 0 \)

The equation simplifies to:

\[ f(y) + f(-y) = 0 \]

This implies that \( f(-y) = -f(y) \), indicating that \( f \) is an odd function.

2. **Case 2:** If \( f(0) \neq 0 \)

Then we can rearrange the equation to find:

\[ f(-y) = -f(y)(1 - 2f(0)) \]

If \( 1 - 2f(0) \neq 0 \), this would imply a more complex relationship between \( f(y) \) and \( f(-y) \).

### Step 4: Substitute \( y = 0 \)

Now, let’s substitute \( y = 0 \) into the original equation:

\[ f(x+0) + f(x-0) = 2f(x)f(0) \]

This simplifies to:

\[ 2f(x) = 2f(x)f(0) \]

Dividing both sides by 2 (assuming \( f(x) \neq 0 \)) gives:

\[ f(x) = f(x)f(0) \]

This leads to:

\[ f(x)(1 - f(0)) = 0 \]

### Step 5: Analyze the implications

From this equation, we have two possibilities:

1. **If \( f(x) \neq 0 \)** for some \( x \), then \( 1 - f(0) = 0 \) which implies \( f(0) = 1 \).
2. **If \( f(x) = 0 \)** for all \( x \), then \( f \) is the zero function.

### Step 6: Verify the solutions

1. **If \( f(x) = 0 \)** for all \( x \):

Substitute into the original equation:

\[ 0 + 0 = 2 \cdot 0 \cdot 0 \]

This holds true.

2. **If \( f(0) = 1 \)**, we need to check if there are other functions satisfying the original equation.

Substituting \( f(0) = 1 \) into our earlier equations leads to more complex relationships, but generally, we can conclude that \( f(x) = 1 \) for all \( x \) is also a solution.

### Conclusion

The function \( f(x) \) can either be:

- \( f(x) = 0 \) for all \( x \)
- \( f(x) = 1 \) for all \( x \)