Limits, Continuity & Differentiability
Differentiability using Squeeze theorem
Grade 12
Question:
<p>If \(|f(x)| \leq x^2\), then which of the following is true?</p>
<p>\(f\) is differentiable at \(x = 0\) and \(f'(0) = 1\)</p>
<p>\(f\) is differentiable at \(x = 0\) and \(f'(0) = 0\)</p>
<p>\(f\) is not differentiable at \(x = 0\)</p>
<p>\(f\) is continuous but not differentiable at \(x = 0\)</p>
Step-by-Step Solution
Key Concept: Since |f(x)| ≤ x² and x² → 0 as x → 0, the Squeeze Theorem forces f(x) → 0. The bounded constraint around x = 0 guarantees both continuity and differentiability at origin with f'(0) = 0.
<p><strong>Step 1:</strong> From |f(x)| ≤ x², we have -x² ≤ f(x) ≤ x² for all x in the domain.</p><p><strong>Step 2:</strong> As x → 0: -x² → 0 and x² → 0. By Squeeze Theorem, f(x) → 0, so f is continuous at x = 0 (assuming f(0) exists or defining f(0) = 0).</p><p><strong>Step 3:</strong> For differentiability at x = 0: Consider |f(x) - f(0)|/x = |f(x)|/|x| ≤ x²/|x| = |x| → 0 as x → 0.</p><p><strong>Step 4:</strong> Therefore f'(0) = 0 exists, and f is differentiable at x = 0.</p><p><strong>Conclusion:</strong> f is continuous and differentiable at x = 0 with f(0) = 0 and f'(0) = 0.</p><p>∴ Answer: B</p>
Correct Answer: B