Sets, Relations & Functions
Even and odd functions
Grade 11
Question:
<p>Let \(f(x) = a^x\) (\(a > 0\)) 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. Then \(f_1(x+y) + f_1(x-y)\) equals:</p>
<p>\(2f_1(x)f_1(y)\)</p>
<p>\(2f_1(x+y)f_1(x-y)\)</p>
<p>\(2f_1(x)f_2(y)\)</p>
<p>\(2f_1(x+y)f_2(x-y)\)</p>
Step-by-Step Solution
Key Concept: Decompose f(x) = a^x into even and odd components using f(x) = f₁(x) + f₂(x), then use the property that f₁(x) = [f(x) + f(-x)]/2 for the even part, and apply functional equations for exponential functions.
<p><strong>Step 1:</strong> Decompose f(x) = a^x into even and odd parts:</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) + f₁(x-y):</p><p>f₁(x+y) = (a^(x+y) + a^(-(x+y)))/2 = (a^x·a^y + a^(-x)·a^(-y))/2</p><p>f₁(x-y) = (a^(x-y) + a^(-(x-y)))/2 = (a^x·a^(-y) + a^(-x)·a^y)/2</p><p><strong>Step 3:</strong> Add the two expressions:</p><p>f₁(x+y) + f₁(x-y) = [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^y + a^(-y))(a^x + a^(-x))]/2</p><p>= [(a^y + a^(-y))/2]·[(a^x + a^(-x))/2]·2</p><p>= 2f₁(x)·f₁(y)</p><p>∴ Answer: <strong>2f₁(x)f₁(y)</strong></p>
Correct Answer: A