Question

Given, the function \( f(x) = \frac{a^x + a^{-x}}{2} \) (\( a > 2 \)), then \( f(x+y) + f(x-y) \) is equal to

• A. \( f(x) - f(y) \)
• B. \( f(y) \)
• C. \( 2f(x)f(y) \)
• D. \( f(x)f(y) \)

Hint:

To simplify expressions involving exponential terms like \( a^{x+y} \) and \( a^{x-y} \), use exponent rules to group terms efficiently.

Answer 1

The Correct Option is C

The function given is \( f(x) = \frac{a^x + a^{-x}}{2} \). We compute \( f(x+y) \) and \( f(x-y) \) as follows: \[ f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2}, \quad f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2}. \] Adding these two expressions yields: \[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2}. \] Using exponent properties, this simplifies to: \[ f(x+y) + f(x-y) = \frac{a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})}{2}. \] Factoring the expression gives: \[ f(x+y) + f(x-y) = \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}. \] Recognizing that \( f(x) = \frac{a^x + a^{-x}}{2} \) and \( f(y) = \frac{a^y + a^{-y}}{2} \), we substitute these into the equation: \[ f(x+y) + f(x-y) = 2f(x)f(y). \] Final Answer: \[ \boxed{2f(x)f(y)} \]