Applications of Derivatives
Inequalities using derivatives
Grade 12
Question:
<p>Let \( f(x) \) be a polynomial function satisfying \( 0 < xf(y) < yf(x) \) for all \( x, y \) 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 given functional equation f(x)·f(1/x) = f(x) + f(1/x) to determine the form of f(x), then apply calculus constraints (f'(0) exists, f is increasing) to find specific coefficients.
<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 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>For a polynomial f(x), we need g(x) = x^n for some integer n. Thus f(x) = x^n + 1.</p><p><strong>Step 3:</strong> Since f'(0) exists and must be finite, f(x) cannot have negative powers. Since f is increasing on (0,∞), we need f'(x) > 0 for x > 0.</p><p>For f(x) = x^n + 1: f'(x) = nx^(n-1) > 0 requires n > 0 and n ≥ 1.</p><p><strong>Step 4:</strong> Test the boundary condition 0 < f(x) < 2 for x ∈ (0,1): This gives 1 < x^n + 1 < 2, so 0 < x^n < 1 for all x ∈ (0,1). This works for any positive integer n.</p><p>The simplest and most constrained case is n = 1, giving f(x) = x + 1.</p><p>∴ Answer: B</p>
Correct Answer: B