Let `f(x+y)+f(x-y)=2f(x)f(y) AA x,y in R` and `f(0)=k`, then

Options

A. I,III
B. II,III
C. I,II
D. III,IV

Answer

To solve the problem, we start with the functional equation given:

\[ f(x+y) + f(x-y) = 2f(x)f(y) \quad \text{for } x, y \in \mathbb{R} \]

and the condition \( f(0) = k \).

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

Let's substitute \( y = 0 \) into the functional equation:

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

This simplifies to:

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

or

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

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

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

### Step 2: Analyze the equation

From the equation \( f(x) = f(x)k \), we can rearrange it to:

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

This implies two cases:

1. \( f(x) = 0 \) for all \( x \) (which is a trivial solution).
2. \( k = 1 \) (which allows \( f(x) \) to be non-zero).

### Step 3: Consider the case \( k = 1 \)

If \( k = 1 \), we substitute back into the original functional equation:

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

Now let’s check if \( f(x) = \cos(x) \) satisfies this equation.

### Step 4: Check if \( f(x) = \cos(x) \)

Substituting \( f(x) = \cos(x) \):

The left-hand side becomes:

\[ \cos(x+y) + \cos(x-y) \]

Using the cosine addition formula:

\[ \cos(x+y) + \cos(x-y) = 2\cos(x)\cos(y) \]

The right-hand side is:

\[ 2f(x)f(y) = 2\cos(x)\cos(y) \]

Since both sides are equal, \( f(x) = \cos(x) \) is indeed a solution when \( k = 1 \).

### Step 5: Consider the case \( k = 0 \)

If \( k = 0 \), we substitute back into the original functional equation:

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

This implies that \( f(x+y) = -f(x-y) \), suggesting that \( f \) could be an odd function.

### Step 6: Check if \( f(x) = \sin(x) \)

Substituting \( f(x) = \sin(x) \):

The left-hand side becomes:

\[ \sin(x+y) + \sin(x-y) \]

Using the sine addition formula:

\[ \sin(x+y) + \sin(x-y) = 2\sin(x)\cos(y) \]

The right-hand side is:

\[ 2f(x)f(y) = 2\sin(x)\sin(y) \]

Since both sides are not equal, \( f(x) = \sin(x) \) does not satisfy the equation.

### Conclusion

From the analysis:

- If \( k = 1 \), \( f(x) \) can be an even function (like \( \cos(x) \)).
- If \( k = 0 \), \( f(x) \) can be odd, but we need to find a specific function.

Thus, the final conclusion is:

- \( f \) is even if \( k = 1 \).
- \( f \) is odd if \( k = 0 \).
- \( f \) is always odd for \( k \neq 0 \) and \( k \neq 1 \).