Limits, Continuity & Differentiability
Differentiability and derivatives
Grade 12
Question:
<p>We have \(|f(x) - f(y)| \leq (x-y)^2\) for all \(x, y \in \mathbb{R}\) and \(f(0) = 0\). Find \(f(1)\).</p>
<p>1</p>
<p>2</p>
<p>0</p>
<p>Cannot be determined</p>
Step-by-Step Solution
Key Concept: The condition |f(x) - f(y)| ≤ (x-y)² forces f to be constant, since the Lipschitz-type bound becomes arbitrarily tight as points approach each other, making f'(x) = 0 everywhere.
<p><strong>Step 1:</strong> Set y = 0 in the given inequality.</p><p>|f(x) - f(0)| ≤ (x - 0)²</p><p>|f(x) - 0| ≤ x²</p><p>|f(x)| ≤ x²</p><p><strong>Step 2:</strong> For any fixed x ≠ 0, set y = x/2 and apply the condition repeatedly.</p><p>|f(x/2) - f(0)| ≤ (x/2)² ⟹ |f(x/2)| ≤ x²/4</p><p><strong>Step 3:</strong> Use the condition on the interval [0, x]:</p><p>|f(x) - f(0)| ≤ x², so |f(x)| ≤ x²</p><p>This holds for all x. Now apply the inequality to consecutive intervals of length h:</p><p>|f(x+h) - f(x)| ≤ h²</p><p><strong>Step 4:</strong> The derivative condition: if f is differentiable,</p><p>|f'(x)| = lim_{h→0} |f(x+h) - f(x)|/h ≤ lim_{h→0} h²/h = lim_{h→0} h = 0</p><p>Therefore f'(x) = 0 for all x where the derivative exists.</p><p><strong>Step 5:</strong> A function with zero derivative everywhere must be constant.</p><p>Since f(0) = 0, we have f(x) = 0 for all x ∈ ℝ.</p><p><strong>Step 6:</strong> Verify: |f(x) - f(y)| = |0 - 0| = 0 ≤ (x-y)² ✓</p><p>∴ f(1) = <strong>0</strong></p>
Correct Answer: C