Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let \(f(x) = \sin^3 x + \lambda \sin^2 x\), \(\frac{\pi}{2} < x < \frac{\pi}{2}\). Find the interval in which \(\lambda\) should lie so that \(f(x)\) has exactly one minimum and exactly one maximum.</p>
<p>\(\lambda \in (-3/2, 0) \cup (0, 3/2)\)</p>
<p>\(\lambda \in (-3/2, 3/2)\)</p>
<p>\(\lambda \in (-\infty, -3/2) \cup (3/2, \infty)\)</p>
<p>\(\lambda \in (-3/2, 0)\)</p>
Step-by-Step Solution
Key Concept: For f to have a local extremum in the open interval (π/2, π), f'(x) = 0 must have a solution there, and the critical point must satisfy the second derivative test or sign change condition for derivatives.
<p><strong>Step 1:</strong> Find f'(x).</p><p>f(x) = sin³x + λsin²x</p><p>f'(x) = 3sin²x·cosx + 2λsinx·cosx = sinx·cosx(3sinx + 2λ)</p><p><strong>Step 2:</strong> Set f'(x) = 0 for critical points in (π/2, π).</p><p>Since x ∈ (π/2, π): sinx > 0 and cosx < 0, so sinx·cosx ≠ 0</p><p>Therefore: 3sinx + 2λ = 0 ⟹ sinx = -2λ/3</p><p><strong>Step 3:</strong> For a critical point to exist in (π/2, π), we need:</p><p>0 < -2λ/3 < 1 (since sinx ∈ (0,1) on this interval)</p><p>This gives: -3/2 < λ < 0</p><p><strong>Step 4:</strong> Verify local extremum using second derivative test or first derivative sign change.</p><p>f''(x) = cos²x - sin²x + 2λ(cos²x - sin²x) = (1 + 2λ)(cos2x)</p><p>At the critical point where sinx = -2λ/3 ∈ (0,1), we have x ∈ (π/2, π), so cos2x < 0.</p><p>For a local extremum: f''(x) ≠ 0 ⟹ 1 + 2λ ≠ 0 ⟹ λ ≠ -1/2</p><p>Combined with -3/2 < λ < 0, the range is: <strong>λ ∈ (-3/2, -1/2) ∪ (-1/2, 0)</strong></p><p>∴ Answer: A</p>
Correct Answer: A