Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>For \(x \in \left(0, \dfrac{5\pi}{2}\right)\), define \(f(x) = \int_0^x \sqrt{t} \sin t \, dt\). Then \(f\) has</p>
<p>local minimum at \(\pi\) and \(2\pi\).</p>
<p>local minimum at \(\pi\) and local maximum at \(2\pi\).</p>
<p>local maximum at \(\pi\) and local minimum at \(2\pi\).</p>
<p>local maximum at \(\pi\) and \(2\pi\).</p>
Step-by-Step Solution
Key Concept: Use Leibniz rule for derivatives of integrals: f'(x) = √x sin x. Analyze the sign of f'(x) by examining when √x sin x is positive or negative on the given interval to find local extrema.
<p><strong>Step 1:</strong> Apply Leibniz rule: f'(x) = √x sin x for x ∈ (0, 5π/2)</p><p><strong>Step 2:</strong> Since √x > 0 for x > 0, the sign of f'(x) depends entirely on sin x:</p><ul><li>f'(x) > 0 when sin x > 0, i.e., x ∈ (0, π) ∪ (2π, 5π/2)</li><li>f'(x) < 0 when sin x < 0, i.e., x ∈ (π, 2π)</li><li>f'(x) = 0 at x = π, 2π</li></ul><p><strong>Step 3:</strong> Analyze extrema:</p><ul><li>f is increasing on (0, π) → local maximum at x = π</li><li>f is decreasing on (π, 2π) → local minimum at x = 2π</li><li>f is increasing on (2π, 5π/2) → local maximum at x = 5π/2 (endpoint)</li></ul><p><strong>Step 4:</strong> Therefore, f has exactly <strong>one local maximum</strong> at x = π and <strong>one local minimum</strong> at x = 2π (assuming option C states this).</p><p>∴ Answer: C</p>
Correct Answer: C