Limits, Continuity & Differentiability
Differentiability
Grade 12
Question:
<p>If \( f \) is a real-valued differentiable function satisfying \( |f(x) - f(y)| \leq (x-y)^2 \), \( x, y \in R \) and \( f(0) = 0 \), then \( f(1) \) equals</p>
<p>\(-1\)</p>
<p>\(0\)</p>
<p>\(2\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Use the Lipschitz-like condition |f(x) - f(y)| ≤ (x-y)² to show f'(x) = 0 everywhere by examining the derivative definition: |f(x+h) - f(x)|/|h| ≤ |h| → 0 as h → 0, forcing f'(x) = 0 for all x.
<p><strong>Step 1:</strong> Apply the definition of derivative at any point x:</p><p>f'(x) = lim(h→0) [f(x+h) - f(x)]/h</p><p><strong>Step 2:</strong> Use the given constraint |f(x) - f(y)| ≤ (x-y)² with y = x+h:</p><p>|f(x+h) - f(x)| ≤ (x+h-x)² = h²</p><p><strong>Step 3:</strong> Divide by |h|:</p><p>|[f(x+h) - f(x)]/h| ≤ |h|</p><p><strong>Step 4:</strong> Take limit as h → 0:</p><p>|f'(x)| ≤ lim(h→0) |h| = 0</p><p>Therefore f'(x) = 0 for all x ∈ ℝ</p><p><strong>Step 5:</strong> A function with zero derivative everywhere is constant:</p><p>f(x) = c for all x</p><p><strong>Step 6:</strong> Use initial condition f(0) = 0:</p><p>c = 0, so f(x) = 0 for all x</p><p><strong>Step 7:</strong> Therefore f(1) = 0</p><p>∴ Answer: B (which is 0)</p>
Correct Answer: B