Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>In \((0, x)\), \(f(x) = \sin^m x + \csc^m x\) attains</p>
<p>a maximum independent of m</p>
<p>a minimum value which is a function of m</p>
<p>a minimum value which is independent of m</p>
<p>the minimum value at the same point independent of m</p>
Step-by-Step Solution
Key Concept: Find critical points by differentiating f(x) = sin^m(x) + csc^m(x) and analyze the behavior of f'(x). The minimum occurs when sin(x) = csc(x), which happens at x = π/2 where both terms are minimized simultaneously.
<p><strong>Step 1:</strong> Recognize that f(x) = sin^m(x) + csc^m(x) = sin^m(x) + 1/sin^m(x). Let u = sin(x) where u ∈ (0,1] for x ∈ (0, π).</p><p><strong>Step 2:</strong> The function becomes g(u) = u^m + u^(-m). Find dg/du = m·u^(m-1) - m·u^(-m-1) = m·u^(-m-1)[u^(2m) - 1].</p><p><strong>Step 3:</strong> Setting dg/du = 0 gives u^(2m) = 1, so u = 1. This occurs at sin(x) = 1, which means x = π/2.</p><p><strong>Step 4:</strong> At x = π/2: f(π/2) = 1^m + 1^m = 2 (minimum value). As x → 0⁺ or x → π⁻, sin(x) → 0⁺, so csc^m(x) → ∞, making f(x) → ∞.</p><p><strong>Step 5:</strong> For x ∈ (0, π), the function attains a <strong>minimum value of 2 at x = π/2</strong> and has <strong>no maximum</strong> (approaches infinity at boundaries).</p><p>∴ Answer: CD (Minimum at interior point with specific value)</p>
Correct Answer: CD