<p>Given \( f(x) = a^x \) (\( a > 0 \)) and \( f(x) = f_1(x) + f_2(x) \), where \( f_1(x) \) is an even function and \( f_2(x) \) is an odd function. Then \( f_1(x+y) + f_1(x-y) \) equals</p>
Step-by-Step Solution
Key Concept: Decompose f(x) = a^x into even and odd parts using f(x) = f₁(x) + f₂(x), where f₁(x) = [f(x) + f(-x)]/2 and f₂(x) = [f(x) - f(-x)]/2. Then use the resulting expressions to evaluate the given sum.
<p><strong>Step 1:</strong> Decompose f(x) = a^x into even and odd functions.</p><p>Since f(x) = f₁(x) + f₂(x), where f₁ is even and f₂ is odd:</p><p>f₁(x) = [f(x) + f(-x)]/2 = (a^x + a^(-x))/2</p><p>f₂(x) = [f(x) - f(-x)]/2 = (a^x - a^(-x))/2</p><p><strong>Step 2:</strong> Calculate f₁(x+y) and f₁(x-y).</p><p>f₁(x+y) = (a^(x+y) + a^(-(x+y)))/2 = (a^(x+y) + a^(-x-y))/2</p><p>f₁(x-y) = (a^(x-y) + a^(-(x-y)))/2 = (a^(x-y) + a^(-x+y))/2</p><p><strong>Step 3:</strong> Add f₁(x+y) and f₁(x-y).</p><p>f₁(x+y) + f₁(x-y) = [a^(x+y) + a^(-x-y) + a^(x-y) + a^(-x+y)]/2</p><p>= [a^(x+y) + a^(x-y) + a^(-(x-y)) + a^(-(x+y))]/2</p><p>= [a^x·a^y + a^x·a^(-y) + a^(-x)·a^(-y) + a^(-x)·a^y]/2</p><p>= [a^x(a^y + a^(-y)) + a^(-x)(a^y + a^(-y))]/2</p><p>= [(a^x + a^(-x))(a^y + a^(-y))]/2</p><p>= 2f₁(x)·f₁(y)</p><p>∴ Answer: <strong>2f₁(x)f₁(y)</strong></p>
Correct Answer: A