Question

If a function \( f \) satisfies \( f(x+1) + f(x-1) = \sqrt{2}f(x) \), then \( f(x+2) + f(x-2) = \)

• A. \( 2f(x) \)
• B. \( f(x+1) - f(x-1) \)
• C. \( 4f(x) \)
• D. \( 0 \)

Hint:

Functional Equations}
Shifting variables (\(x \to x \pm 1\)) often reveals patterns
Linear combinations of shifted equations can simplify problems
Trigonometric functions often satisfy such recurrence relations

Answer 1

The Correct Option is D

Step 1: Given the functional equation: \[ f(x+1) + f(x-1) = \sqrt{2}f(x). \] Step 2: Shift the equation by \( +1 \): \[ f(x+2) + f(x) = \sqrt{2}f(x+1). \quad \text{(1)} \] Step 3: Shift the equation by \( -1 \): \[ f(x) + f(x-2) = \sqrt{2}f(x-1). \quad \text{(2)} \] Step 4: Add equations (1) and (2): \[ f(x+2) + 2f(x) + f(x-2) = \sqrt{2}(f(x+1) + f(x-1)) = \sqrt{2} \cdot \sqrt{2}f(x) = 2f(x). \] Step 5: Simplify: \[ f(x+2) + f(x-2) = 0. \] Thus, the correct answer is \(\boxed{0}\).