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Step-by-Step Solution
Step 1: Express $f(x)$ in terms of even and odd components
We are given:
$$f(x) = a^x \quad \text{(where } a > 0\text{)}.$$
We decompose $f(x)$ into an even part $f_1(x)$ and an odd part $f_2(x)$:
The even part $f_1(x)$ is defined by
$$f_1(x) = \frac{f(x) + f(-x)}{2}.$$
The odd part $f_2(x)$ is defined by
$$f_2(x) = \frac{f(x) - f(-x)}{2}.$$
Step 2: Simplify each component
Substitute $f(x) = a^x$ to find explicit forms of $f_1(x)$ and $f_2(x)$:
Even part:
$$f_1(x) = \frac{a^x + a^{-x}}{2}.$$
Odd part:
$$f_2(x) = \frac{a^x - a^{-x}}{2}.$$
Step 3: Write down $f_1(x+y) + f_1(x-y)$
We need to find the expression for
$$f_1(x+y) + f_1(x-y).$$
By definition,
$$f_1(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2}, \quad
f_1(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2}.$$
Add them together:
$$f_1(x+y) + f_1(x-y)
= \frac{a^{x+y} + a^{-(x+y)}}{2}
+ \frac{a^{x-y} + a^{-(x-y)}}{2}.$$
Combine terms under a common factor of $1/2$:
$$f_1(x+y) + f_1(x-y)
= \frac{1}{2}\bigl(a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}\bigr).$$
Step 4: Compute $f_1(x) \, f_1(y)$
Now consider:
$$f_1(x)\,f_1(y)
= \left(\frac{a^x + a^{-x}}{2}\right)\left(\frac{a^y + a^{-y}}{2}\right).$$
Multiply out the terms:
$$f_1(x)\,f_1(y)
= \frac{1}{4}\bigl(a^{x}a^{y} + a^{x}a^{-y} + a^{-x}a^{y} + a^{-x}a^{-y}\bigr).$$
Simplify exponents:
$$f_1(x)\,f_1(y)
= \frac{1}{4}\bigl(a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y}\bigr).$$
Step 5: Relate $f_1(x+y) + f_1(x-y)$ to $2 f_1(x) f_1(y)$
Observe that the expression within the parentheses of
$f_1(x+y) + f_1(x-y)$ is twice the expression inside $4 \times f_1(x) f_1(y)$. Specifically,
$$f_1(x+y) + f_1(x-y)
= \frac{1}{2}\bigl(a^{x+y} + a^{-(x+y)} + a^{x-y} + a^{-(x-y)}\bigr).$$
Meanwhile,
$$4\,f_1(x)\,f_1(y)
= 4 \times \frac{1}{4}\bigl(a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y}\bigr)
= a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y}.$$
Notice that the set of exponents $a^{x+y}, a^{x-y}, a^{-x+y}, a^{-x-y}$ matches those in the sum for $f_1(x+y)+f_1(x-y)$. Hence,
$$f_1(x+y) + f_1(x-y) = 2\,f_1(x)\,f_1(y).$$
Step 6: Conclude the result
Therefore, we have established that:
$$f_1(x + y) + f_1(x - y) = 2\,f_1(x)\,f_1(y),$$
which matches the correct answer.
