Relations & Functions
Functional equations
Grade 12
Question:
<p>If \(f(x) + 2f\left(\dfrac{1}{x}\right) = 3x\), \(x \neq 0\) and \(S = \{x \in \mathbb{R} : f(x) = f(-x)\}\), then \(S\)</p>
<p>contains more than two elements.</p>
<p>is an empty set.</p>
<p>contains exactly one element.</p>
<p>contains exactly two elements.</p>
Step-by-Step Solution
Key Concept: Substitute x with 1/x in the functional equation to create a system of two equations, then solve for f(x) explicitly. Check which values satisfy the even function condition f(x) = f(-x).
<p><strong>Step 1:</strong> Given: f(x) + 2f(1/x) = 3x ... (1)</p><p><strong>Step 2:</strong> Replace x with 1/x in equation (1):<br>f(1/x) + 2f(x) = 3/x ... (2)</p><p><strong>Step 3:</strong> Multiply equation (1) by 2:<br>2f(x) + 4f(1/x) = 6x ... (3)</p><p><strong>Step 4:</strong> Subtract equation (2) from equation (3):<br>3f(1/x) = 6x - 3/x<br>f(1/x) = 2x - 1/x</p><p><strong>Step 5:</strong> Substitute back into equation (1):<br>f(x) + 2(2x - 1/x) = 3x<br>f(x) = 3x - 4x + 2/x<br>f(x) = -x + 2/x</p><p><strong>Step 6:</strong> For f(x) = f(-x):<br>-x + 2/x = -(-x) + 2/(-x)<br>-x + 2/x = x - 2/x<br>-2x + 4/x = 0<br>4/x = 2x<br>4 = 2x²<br>x² = 2<br>x = ±√2</p><p><strong>Step 7:</strong> Verify: f(√2) = -√2 + 2/√2 = -√2 + √2 = 0<br>f(-√2) = √2 - 2/√2 = √2 - √2 = 0 ✓</p><p>∴ Answer: S = {-√2, √2} or {x ∈ ℝ : x = ±√2}</p>
Correct Answer: D