Let f be a real vlaued fuction with domain R such that `f(x+1)+f(x-1)=sqrt(2)f(x)` for all ` x in R`, then ,

Options

A. f(x) is a periodic function with period 8
B. f(x) is a periodic function with period 12
C. f(x) is a non-periodic function
D. f(x) is a periodic function with indeterminate period

Answer

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

\[ f(x+1) + f(x-1) = \sqrt{2} f(x) \]

for all \( x \in \mathbb{R} \).

### Step 1: Substitute \( x \) with \( x + 1 \)

We replace \( x \) with \( x + 1 \) in the original equation.

\[ f((x+1)+1) + f((x+1)-1) = \sqrt{2} f(x+1) \]

This simplifies to:

\[ f(x+2) + f(x) = \sqrt{2} f(x+1) \tag{1} \]

### Step 2: Substitute \( x \) with \( x - 1 \)

Next, we replace \( x \) with \( x - 1 \) in the original equation.

\[ f((x-1)+1) + f((x-1)-1) = \sqrt{2} f(x-1) \]

This simplifies to:

\[ f(x) + f(x-2) = \sqrt{2} f(x-1) \tag{2} \]

### Step 3: Add Equations (1) and (2)

Now, we add equations (1) and (2):

\[ (f(x+2) + f(x)) + (f(x) + f(x-2)) = \sqrt{2} f(x+1) + \sqrt{2} f(x-1) \]

This simplifies to:

\[ f(x+2) + 2f(x) + f(x-2) = \sqrt{2} (f(x+1) + f(x-1)) \]

From the original equation, we know that:

\[ f(x+1) + f(x-1) = \sqrt{2} f(x) \]

Substituting this into our equation gives:

\[ f(x+2) + 2f(x) + f(x-2) = \sqrt{2} \cdot \sqrt{2} f(x) = 2f(x) \]

### Step 4: Simplify the Equation

This leads to:

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

which implies:

\[ f(x+2) = -f(x-2) \tag{3} \]

### Step 5: Substitute \( x \) with \( x + 2 \) in Equation (3)

Now, we replace \( x \) with \( x + 2 \):

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

This simplifies to:

\[ f(x+4) = -f(x) \tag{4} \]

### Step 6: Substitute \( x \) with \( x + 4 \) in Equation (4)

Next, we replace \( x \) with \( x + 4 \):

\[ f((x+4)+4) = -f((x+4)) \]

This simplifies to:

\[ f(x+8) = -f(x+4) \]

From equation (4), we know \( f(x+4) = -f(x) \), so substituting gives:

\[ f(x+8) = -(-f(x)) = f(x) \]

### Conclusion

Thus, we have shown that:

\[ f(x+8) = f(x) \]

This indicates that \( f(x) \) is periodic with a period of 8.

### Final Answer

The function \( f(x) \) is periodic with a period of 8.

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