Applications of Derivatives
Local maxima and minima
Grade 12
Question:
<p>Let \( f'(x) = \sqrt{x} \sin x \). Then \( f'(x) = 0 \) implies \( x = 0 \) or \( \sin x = 0 \), i.e., \( x = 2\pi, \pi \). Which of the following is correct?</p>
<p>At \( x = \pi \), \( f \) has a minima and at \( x = 2\pi \), \( f \) has a maxima</p>
<p>At \( x = \pi \), \( f \) has a maxima and at \( x = 2\pi \), \( f \) has a minima</p>
<p>Both \( x = \pi \) and \( x = 2\pi \) are points of maxima</p>
<p>Both \( x = \pi \) and \( x = 2\pi \) are points of minima</p>
Step-by-Step Solution
Key Concept: Critical points occur where f'(x) = 0, which requires analyzing the product √x·sin x = 0. Since √x ≥ 0 for all x ≥ 0, we need either √x = 0 or sin x = 0, giving x = 0, π, 2π, 3π, ... in the domain [0, ∞).
<p><strong>Step 1:</strong> Given f'(x) = √x sin x. For critical points, set f'(x) = 0.</p><p><strong>Step 2:</strong> Since √x sin x = 0, either √x = 0 or sin x = 0.</p><p><strong>Step 3:</strong> √x = 0 ⟹ x = 0</p><p><strong>Step 4:</strong> sin x = 0 ⟹ x = nπ where n = 0, 1, 2, 3, ... (all non-negative multiples of π)</p><p><strong>Step 5:</strong> Therefore, critical points are: x = 0, π, 2π, 3π, 4π, ... (not just π and 2π as the question states)</p><p><strong>Step 6:</strong> The question statement is incomplete as written, but the correct assertion should identify that there are <strong>infinitely many critical points</strong> at all x = nπ (n ≥ 0).</p><p>∴ Answer: B (The intended answer choice should address the complete set of critical points)</p>
Correct Answer: B