Given the function f(x) = frac{a^x + a^{-x}}{ | vedclass

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(A) We are given $f(x) = \frac{a^x + a^{-x}}{2}$.Consider the expression $f(x + y) + f(x - y)$:$f(x + y) + f(x - y) = \frac{a^{x+y} + a^{-(x+y)}}{2} +

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$2f(x)f(y)$

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(A) We are given $f(x) = \frac{a^x + a^{-x}}{2}$.Consider the expression $f(x + y) + f(x - y)$:$f(x + y) + f(x - y) = \frac{a^{x+y} + a^{-(x+y)}}{2} + \frac{a^{x-y} + a^{-(x-y)}}{2}$$= \frac{1}{2} [a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y}]$$= \frac{1}{2} [a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})]$$= \frac{1}{2} (a^x + a^{-x})(a^y + a^{-y})$Since $f(x) = \frac{a^x + a^{-x}}{2}$,we have $(a^x + a^{-x}) = 2f(x)$ and $(a^y + a^{-y}) = 2f(y)$.Substituting these values:$= \frac{1}{2} [2f(x)] [2f(y)]$$= 2f(x)f(y)$.

Given the function $f(x) = \frac{a^x + a^{-x}}{2}$,where $a > 2$. Then $f(x + y) + f(x - y) = $

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