Sets, Relations & Functions
Functional Equations
Grade 11
Question:
<p>If \(f(x)\) be a function satisfying \(f(x) \cdot f\!\left(\dfrac{1}{x}\right) = f(x) + f\!\left(\dfrac{1}{x}\right)\) and \(f(4) = 65\), then find \(f(6)\).</p>
Step-by-Step Solution
Key Concept: Use the functional equation to establish that f(x) = 1 + g(x) where g(x)·g(1/x) = 1, making g(x) multiplicative. This transforms the problem into finding the form of f and using the given condition f(4) = 65 to determine constants.
<p><strong>Step 1:</strong> Rewrite the functional equation f(x)·f(1/x) = f(x) + f(1/x).</p><p>Rearrange: f(x)·f(1/x) - f(x) - f(1/x) = 0</p><p>Add 1 to both sides: f(x)·f(1/x) - f(x) - f(1/x) + 1 = 1</p><p>Factor: [f(x) - 1]·[f(1/x) - 1] = 1</p><p><strong>Step 2:</strong> Let g(x) = f(x) - 1. Then g(x)·g(1/x) = 1, so g(1/x) = 1/g(x).</p><p>This suggests g(x) = x^k for some constant k (a multiplicative function).</p><p>Therefore: f(x) = 1 + x^k</p><p><strong>Step 3:</strong> Use f(4) = 65 to find k.</p><p>1 + 4^k = 65</p><p>4^k = 64 = 4³</p><p>k = 3</p><p><strong>Step 4:</strong> So f(x) = 1 + x³</p><p>Verify: f(x)·f(1/x) = (1 + x³)(1 + 1/x³) = 1 + x³ + 1/x³ + 1 = f(x) + f(1/x) ✓</p><p>∴ f(6) = 1 + 6³ = 1 + 216 = <strong>217</strong></p>
Correct Answer: 217