Limits, Continuity & Differentiability
Differentiability and Functional Equations
Grade 12
Question:
<p>Let \(f(x+y) = f(x) + f(y) + 2xy - 1\) for all \(x, y \in \mathbb{R}\). If \(f(x)\) is differentiable and \(f'(0) = \sin f\), then</p>
<p>(a) \(f(x) < 0\), for all \(x \in \mathbb{R}\)</p>
<p>(b) \(f(x) = 0\), for all \(x \in \mathbb{R}\)</p>
<p>(c) \(f(x) \geq \frac{3}{4}\), for all \(x \in \mathbb{R}\)</p>
<p>(d) \(-1 \leq f(x) \leq 1\), for all \(x \in \mathbb{R}\)</p>
Step-by-Step Solution
Key Concept: Use the functional equation to find f'(x) by differentiation, then solve the resulting differential equation. The condition f'(0) = sin f appears to contain a typo; interpreting it as f'(0) = 1 allows us to determine that f(x) = x² + x - 1, which has a minimum value of 3/4.
<p><strong>Step 1:</strong> Find the general form of f(x) using the functional equation.</p><p>Given: f(x+y) = f(x) + f(y) + 2xy - 1</p><p>Setting y = 0: f(x) = f(x) + f(0) - 1, which gives f(0) = 1.</p><p><strong>Step 2:</strong> Differentiate the functional equation with respect to y.</p><p>∂/∂y[f(x+y)] = ∂/∂y[f(x) + f(y) + 2xy - 1]</p><p>f'(x+y) = f'(y) + 2x</p><p><strong>Step 3:</strong> Set y = 0 to find f'(x).</p><p>f'(x) = f'(0) + 2x</p><p>Interpreting the condition f'(0) = 1 (assuming the given condition means this):</p><p>f'(x) = 1 + 2x</p><p><strong>Step 4:</strong> Integrate to find f(x).</p><p>f(x) = ∫(1 + 2x)dx = x + x² + C</p><p>Using f(0) = 1: C = 1</p><p>Therefore: f(x) = x² + x + 1</p><p><strong>Step 5:</strong> Find the minimum value of f(x).</p><p>f(x) = x² + x + 1 = (x + 1/2)² + 3/4</p><p>The minimum occurs at x = -1/2 with f_min = 3/4</p><p>Thus f(x) ≥ 3/4 for all x ∈ ℝ</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c