Limits, Continuity & Differentiability
Differentiability and Continuity
Grade 12
Question:
<p>Let \(f(x) = e^{\frac{-1}{x^2}} + \int_0^{\frac{\pi x}{2}} \sqrt{1+\sin t}\, dt \; \forall\, x \in (0, \infty)\), then:</p>
<p>(a) \(f'(x)\) exist and is continuous \(\forall\, x \in (0, \infty)\)</p>
<p>(b) \(f''(x)\) exist \(\forall\, x \in (0, \infty)\)</p>
<p>(c) \(f'(x)\) is bounded</p>
<p>(d) there exist \(\alpha > 0\) such that \(|f(x)| > |f'(x)|\) \(\forall\, x \in (\alpha, \infty)\)</p>
Step-by-Step Solution
Key Concept: The function f(x) combines a smooth exponential term (which approaches 0 as x→0⁺) with an integral term whose behavior near x=0 is determined by substitution u=πx/2. As x→0⁺, the upper limit→0, so the integral→0, making f(x)→0. Differentiability requires checking if f'(0⁺) exists using the definition and properties of both components.
<p><strong>Step 1: Analyze behavior as x→0⁺</strong></p><p>As x→0⁺: e^(-1/x²)→0 (exponential decay dominates)</p><p>As x→0⁺: ∫₀^(πx/2) √(1+sin t) dt→0 (upper limit→0, integrand bounded)</p><p>Therefore: lim(x→0⁺) f(x) = 0</p><p><strong>Step 2: Check continuity at x=0</strong></p><p>Define f(0)=0. Since both component terms→0 as x→0⁺, and f is continuous for x>0, we have f continuous on [0,∞). ✓</p><p><strong>Step 3: Check differentiability at x=0</strong></p><p>For x>0: f'(x) = -2/x³·e^(-1/x²) + √(1+sin(πx/2))·(π/2)</p><p>Right derivative: f'₊(0) = lim(h→0⁺) [f(h)-f(0)]/h</p><p>= lim(h→0⁺) [e^(-1/h²) + ∫₀^(πh/2) √(1+sin t) dt]/h</p><p>The first term: e^(-1/h²)/h→0 (exponential decay beats polynomial growth)</p><p>The second term: [∫₀^(πh/2) √(1+sin t) dt]/h ≈ √(1+0)·(π/2) = π/2 (by L'Hôpital or integral mean value)</p><p>Therefore: f'₊(0) = π/2, and f'(0) exists.</p><p><strong>Step 4: Verify smoothness properties</strong></p><p>f is continuous on [0,∞): ✓<br/>f is differentiable on [0,∞): ✓<br/>f(0)=0 and f(x)>0 for x>0: ✓<br/>f'(0)=π/2>0: ✓</p><p>∴ Answer: A, B, C, D (all properties hold)</p>
Correct Answer: A,B,C,D