Let f(x+y) + f(x-y) = 2f(x)f(y) for x, `y in R` and `f(0) != 0`. Then f(x) must be

Options

A. One-one function
B. Onto function
C. Even function
D. Odd function

Answer

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

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

for all \( x, y \in \mathbb{R} \) and given that \( f(0) \neq 0 \).

### Step 1: Substitute \( x = 0 \) and \( y = 0 \)
Let's substitute \( x = 0 \) 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: Rearranging the equation
Rearranging gives us:

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

Factoring out \( f(0) \):

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

### Step 4: Analyze the roots
The possible solutions are:

1. \( f(0) = 0 \) (not valid since \( f(0) \neq 0 \))
2. \( f(0) = 1 \)

Thus, we conclude:

\[ f(0) = 1 \]

### Step 5: Substitute \( x = 0 \) in the original equation
Now, substitute \( x = 0 \) in 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 6: Rearranging the equation
Rearranging gives us:

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

### Step 7: Conclusion
This shows that:

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

Thus, \( f(y) \) is an even function. Therefore, we conclude that:

\[ f(x) \text{ must be an even function.} \]

### Final Answer
The function \( f(x) \) must be an even function.

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