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

We are given the function: \[ f(x) = \frac{a^x + a^{-x}}{2}. \] Now, let's calculate \( f(x+y) \) and \( f(x-y) \): \[ f(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2}, \quad f(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2}. \] By adding these two expressions together: \[ f(x+y) + f(x-y) = \frac{a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}}{2}. \] Next, simplify this using the properties of exponents: \[ f(x+y) + f(x-y) = \frac{a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})}{2}. \] Factor the expression: \[ f(x+y) + f(x-y) = \frac{(a^x + a^{-x})(a^y + a^{-y})}{2}. \] Now, recognizing that \( f(x) = \frac{a^x + a^{-x}}{2} \) and \( f(y) = \frac{a^y + a^{-y}}{2} \), we substitute these back in: \[ f(x+y) + f(x-y) = 2f(x)f(y). \] Final Answer: \[ \boxed{2f(x)f(y)} \]