Question:
If a function \(f\) satisfies \(f(x+1)+f(x-1)=\sqrt{2}f(x)\), then \(f(x+2)+f(x-2)=\)
If a function \(f\) satisfies \(f(x+1)+f(x-1)=\sqrt{2}f(x)\), then \(f(x+2)+f(x-2)=\)
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In functional equations involving \(f(x+1)\) and \(f(x-1)\), substitute \(x+1\) and \(x-1\) separately to obtain expressions involving \(f(x+2)\) and \(f(x-2)\).
Updated On: Jun 26, 2026
- \(2f(x)\)
- \(f(x+1)-f(x-1)\)
- \(4f(x)\)
- \(0\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the given functional equation.
Given, \[ f(x+1)+f(x-1)=\sqrt{2}f(x) \]
Step 2: Replace \(x\) by \(x+1\).
\[ f(x+2)+f(x)=\sqrt{2}f(x+1) \] So, \[ f(x+2)=\sqrt{2}f(x+1)-f(x) \]
Step 3: Replace \(x\) by \(x-1\).
\[ f(x)+f(x-2)=\sqrt{2}f(x-1) \] So, \[ f(x-2)=\sqrt{2}f(x-1)-f(x) \]
Step 4: Add both equations.
\[ f(x+2)+f(x-2)=\sqrt{2}\{f(x+1)+f(x-1)\}-2f(x) \] Using the given condition, \[ f(x+1)+f(x-1)=\sqrt{2}f(x) \] Therefore, \[ f(x+2)+f(x-2)=\sqrt{2}\cdot \sqrt{2}f(x)-2f(x) \] \[ =2f(x)-2f(x) \] \[ =0 \]
Step 5: Final conclusion.
Hence, \[ \boxed{0} \]
Given, \[ f(x+1)+f(x-1)=\sqrt{2}f(x) \]
Step 2: Replace \(x\) by \(x+1\).
\[ f(x+2)+f(x)=\sqrt{2}f(x+1) \] So, \[ f(x+2)=\sqrt{2}f(x+1)-f(x) \]
Step 3: Replace \(x\) by \(x-1\).
\[ f(x)+f(x-2)=\sqrt{2}f(x-1) \] So, \[ f(x-2)=\sqrt{2}f(x-1)-f(x) \]
Step 4: Add both equations.
\[ f(x+2)+f(x-2)=\sqrt{2}\{f(x+1)+f(x-1)\}-2f(x) \] Using the given condition, \[ f(x+1)+f(x-1)=\sqrt{2}f(x) \] Therefore, \[ f(x+2)+f(x-2)=\sqrt{2}\cdot \sqrt{2}f(x)-2f(x) \] \[ =2f(x)-2f(x) \] \[ =0 \]
Step 5: Final conclusion.
Hence, \[ \boxed{0} \]
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