Applications of Derivatives
Inequalities using derivatives
Grade 12
Question:
<p>Let \( f(x) \) be a polynomial function satisfying \( 0 < xf(y) < yf(x) \) \( \forall \, x, y \) such that \( 0 < x < y < 1 \) and \( f(0) = 0 \) then:</p>
<p>(a) \( f'(x) < f(1) \)</p>
<p>(b) \( f(1) < 2\displaystyle\int_0^1 f(x)\,dx \)</p>
<p>(c) \( 3f\!\left(\dfrac{1}{3}\right) > 2f\!\left(\dfrac{1}{2}\right) \)</p>
<p>(d) \( 6f\!\left(\dfrac{1}{6}\right) < 5f\!\left(\dfrac{1}{5}\right) \)</p>
Step-by-Step Solution
Key Concept: Use the functional equation f(x)f(1/x) = f(x) + f(1/x) to determine the polynomial form, then apply calculus to find extrema. The symmetry condition severely constrains the polynomial structure.
<p><strong>Step 1:</strong> Analyze 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: 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.</p><p>For polynomial f(x), we need g(x) = x^n for some integer n (to satisfy the reciprocal property).</p><p>Therefore: f(x) = 1 + x^n</p><p><strong>Step 3:</strong> Use 0 < f(x) for all x > 0. This requires n to be even, so f(x) = 1 + x^{2m}.</p><p><strong>Step 4:</strong> Find extrema: f'(x) = 2mx^{2m-1}</p><p>For x > 0: f'(x) = 0 has no solution (f is strictly increasing on (0,∞)).</p><p>The minimum occurs as x → 0⁺, giving f_min → 1.</p><p>∴ Answer: B</p>
Correct Answer: B