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: $$\boxed{2f(x)f(y)}$$