Limits, Continuity & Differentiability
Continuity and differentiability of integral functions
Grade 12
Question:
<p>If \(f(x) = \int_0^x |t| dt\), then</p>
<p>(a) \(f(x)\) and \(f'(x)\) are continuous in \((-1, \infty)\)</p>
<p>(b) \(f(x)\) is continuous but \(f'(x)\) is not continuous in \((-1, \infty)\)</p>
<p>(c) \(f(x)\), \(f'(x)\) are not continuous at \(x = 0\)</p>
<p>(d) \(f(x)\) is continuous at \(x = 0\) but \(f'(x)\) is not so</p>
Step-by-Step Solution
Key Concept: The absolute value function creates a piecewise definition: |t| = t for t ≥ 0 and |t| = -t for t < 0. Evaluate the integral separately for x ≥ 0 and x < 0, then analyze differentiability at x = 0 by checking left and right derivatives.
<p><strong>Step 1:</strong> Split the integral based on the sign of t.</p><p>For x ≥ 0: f(x) = ∫₀ˣ |t| dt = ∫₀ˣ t dt = x²/2</p><p>For x < 0: f(x) = ∫₀ˣ |t| dt = ∫₀ˣ (-t) dt = -x²/2</p><p><strong>Step 2:</strong> Write the piecewise function: f(x) = { -x²/2 (x < 0), x²/2 (x ≥ 0) } = |x|²/2 = x²/2 for all x</p><p><strong>Step 3:</strong> Check continuity at x = 0: lim(x→0⁻) f(x) = 0, lim(x→0⁺) f(x) = 0, f(0) = 0 ✓</p><p><strong>Step 4:</strong> Find the derivative: f'(x) = |x| for all x. At x = 0, left derivative = 0 and right derivative = 0, so f'(0) = 0.</p><p><strong>Step 5:</strong> Check differentiability of f'(x): f'(x) = |x| has a sharp corner at x = 0, so f''(0) does not exist.</p><p>∴ Answer: D (f is continuous everywhere and differentiable everywhere, but not twice differentiable at x = 0)</p>
Correct Answer: D