Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If \(u = \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} + \sqrt{a^2\sin^2\theta + b^2\cos^2\theta}\), then the difference between the maximum and minimum values of \(u^2\) is given by</p>
<p>\(2(a^2 + b^2)\)</p>
<p>\(2\sqrt{a^2 + b^2}\)</p>
<p>\((a + b)^2\)</p>
<p>\((a - b)^2\)</p>
Step-by-Step Solution
Key Concept: Express u² in terms of cos²θ and sin²θ, then use the constraint cos²θ + sin²θ = 1 to reduce it to a single-variable function. Find extrema by taking derivatives or recognizing it as a quadratic in one variable.
<p><strong>Step 1:</strong> Let x = cos²θ, so sin²θ = 1 - x, where 0 ≤ x ≤ 1.</p><p>u = √(a²x + b²(1-x)) + √(a²(1-x) + b²x)</p><p>u = √(b² + (a² - b²)x) + √(a² - (a² - b²)x)</p><p><strong>Step 2:</strong> Square both sides:<br/>u² = [b² + (a² - b²)x] + [a² - (a² - b²)x] + 2√{[b² + (a² - b²)x][a² - (a² - b²)x]}</p><p>u² = a² + b² + 2√{[b² + (a² - b²)x][a² - (a² - b²)x]}</p><p><strong>Step 3:</strong> Let p = (a² - b²)x. The product under the square root becomes:<br/>[b² + p][a² - p] = a²b² + a²p - b²p - p² = a²b² + p(a² - b²) - p²</p><p><strong>Step 4:</strong> To maximize/minimize u², we need to extremize f(p) = a²b² + p(a² - b²) - p²<br/>df/dp = (a² - b²) - 2p = 0 ⟹ p = (a² - b²)/2</p><p><strong>Step 5:</strong> Maximum of f(p): f_max = a²b² + (a² - b²)²/4 = (a² + b²)²/4<br/>So u²_max = a² + b² + 2·(a² + b²)/2 = 2(a² + b²)</p><p><strong>Step 6:</strong> At boundaries (x = 0 or x = 1):<br/>f(0) = a²b² and f((a²-b²)) = a²b² - (a² - b²)²<br/>Minimum of f(p) = |a² - b²|²/4<br/>So u²_min = a² + b² + |a² - b²| = (a + b)²</p><p><strong>Step 7:</strong> Difference = 2(a² + b²) - (a + b)² = 2a² + 2b² - a² - 2ab - b² = (a - b)² = a² + b² - 2ab</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D