Key Concepts and Formulas

• Decomposition of a Function into Even and Odd Parts: Any real-valued function $f(x)$ can be uniquely expressed as the sum of an even function $f_1(x)$ and an odd function $f_2(x)$ using the formulas:
  • Even part: $f_1(x) = \frac{f(x) + f(-x)}{2}$
  • Odd part: $f_2(x) = \frac{f(x) - f(-x)}{2}$
• Properties of Exponents: For any $a > 0$ and real numbers $m, n$:
  • $a^m \cdot a^n = a^{m+n}$
  • $a^{-m} = \frac{1}{a^m}$

Step-by-Step Solution

Step 1: Identify the given function and its components. We are given the function $f(x) = a^x$, where $a > 0$. We are also told that $f(x)$ can be written as $f(x) = f_1(x) + f_2(x)$, where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. We need to find the expression for $f_1(x+y) + f_1(x-y)$.

Step 2: Determine the even component $f_1(x)$. Using the formula for the even part of a function, we have: $f_1(x) = \frac{f(x) + f(-x)}{2}$ Substitute $f(x) = a^x$. Then, $f(-x) = a^{-x}$. $f_1(x) = \frac{a^x + a^{-x}}{2}$ This is the specific form of the even component for the given function $f(x) = a^x$.

Step 3: Determine the odd component $f_2(x)$ (for completeness). Using the formula for the odd part of a function, we have: $f_2(x) = \frac{f(x) - f(-x)}{2}$ Substitute $f(x) = a^x$ and $f(-x) = a^{-x}$: $f_2(x) = \frac{a^x - a^{-x}}{2}$ While $f_2(x)$ is not directly used in the final calculation, understanding its derivation confirms the decomposition.

Step 4: Evaluate $f_1(x+y)$ and $f_1(x-y)$. We need to find the sum $f_1(x+y) + f_1(x-y)$. First, we express $f_1$ at the required points using the formula derived in Step 2: $f_1(x+y) = \frac{a^{x+y} + a^{-(x+y)}}{2} = \frac{a^{x+y} + a^{-x-y}}{2}$ $f_1(x-y) = \frac{a^{x-y} + a^{-(x-y)}}{2} = \frac{a^{x-y} + a^{-x+y}}{2}$

Step 5: Calculate the sum $f_1(x+y) + f_1(x-y)$. Now, we add the expressions from Step 4: $f_1(x+y) + f_1(x-y) = \frac{a^{x+y} + a^{-x-y}}{2} + \frac{a^{x-y} + a^{-x+y}}{2}$ Combine the terms over the common denominator: $f_1(x+y) + f_1(x-y) = \frac{a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y}}{2}$

Step 6: Evaluate option (A): $2f_1(x)f_1(y)$. We need to calculate the value of $2f_1(x)f_1(y)$ and compare it with the result from Step 5. First, find the product $f_1(x)f_1(y)$: $f_1(x)f_1(y) = \left(\frac{a^x + a^{-x}}{2}\right) \left(\frac{a^y + a^{-y}}{2}\right)$ $f_1(x)f_1(y) = \frac{1}{4} (a^x \cdot a^y + a^x \cdot a^{-y} + a^{-x} \cdot a^y + a^{-x} \cdot a^{-y})$ Using the exponent rule $a^m \cdot a^n = a^{m+n}$: $f_1(x)f_1(y) = \frac{1}{4} (a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y})$ Now, multiply by 2: $2f_1(x)f_1(y) = 2 \cdot \frac{1}{4} (a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y})$ $2f_1(x)f_1(y) = \frac{1}{2} (a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y})$

Step 7: Compare the results from Step 5 and Step 6. By comparing the expression for $f_1(x+y) + f_1(x-y)$ from Step 5 and the expression for $2f_1(x)f_1(y)$ from Step 6, we see that they are identical: $f_1(x+y) + f_1(x-y) = \frac{a^{x+y} + a^{-x-y} + a^{x-y} + a^{-x+y}}{2}$ $2f_1(x)f_1(y) = \frac{a^{x+y} + a^{x-y} + a^{-x+y} + a^{-x-y}}{2}$ Therefore, $f_1(x+y) + f_1(x-y) = 2f_1(x)f_1(y)$.

Common Mistakes & Tips

• Correctly Applying Exponent Rules: Ensure accurate use of $a^m \cdot a^n = a^{m+n}$ and the meaning of negative exponents like $a^{-(x+y)} = a^{-x-y}$.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when combining fractions and expanding products.
• Understanding Even/Odd Definitions: The core of the problem relies on correctly identifying the even and odd parts of the function using the provided formulas.

Summary

The problem requires us to decompose the given function $f(x) = a^x$ into its even and odd components. The even component, $f_1(x)$, is found to be $\frac{a^x + a^{-x}}{2}$. We then evaluate the expression $f_1(x+y) + f_1(x-y)$ and compare it with the given options. By carefully applying the definitions and exponent rules, we find that $f_1(x+y) + f_1(x-y)$ is equal to $2f_1(x)f_1(y)$.

The final answer is $\boxed{2f_1(x)f_1(y)}$, which corresponds to option (A).