If `f:R to R` such that` f (x + y) + f (x - y) =2f (x) f (y) AA x,y in R` and `f (0) ne 0`, then `f (x)` is an :

Options

A. even function
B. odd function
C. periodic function
D. none

Answer

To solve the functional equation \( f(x + y) + f(x - y) = 2f(x)f(y) \) for all \( x, y \in \mathbb{R} \) given that \( f(0) \neq 0 \), we can follow these steps:

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

Let's start by substituting \( x = 0 \) and \( y = 0 \) into the equation.

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

This simplifies to:

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

or

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

### Step 2: Simplify the equation

We can divide both sides by 2 (since \( f(0) \neq 0 \)):

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

### Step 3: Solve for \( f(0) \)

This gives us the equation:

\[ f(0)^2 - f(0) = 0 \]

Factoring out \( f(0) \):

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

Since \( f(0) \neq 0 \), we conclude that:

\[ f(0) = 1 \]

### Step 4: Substitute \( x = 0 \) again

Now, let's substitute \( x = 0 \) into the original equation:

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

This simplifies to:

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

or

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

### Step 5: Rearranging the equation

Rearranging gives us:

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

This implies that:

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

### Step 6: Conclusion about the function

Since \( f(-y) = f(y) \) for all \( y \), we conclude that \( f(x) \) is an even function.

### Final Result

Thus, the function \( f(x) \) is an even function.

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