<p><strong>101.</strong> Slope of tangent to the curve \(y = 2e^x \sin\!\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\cos\!\left(\dfrac{\pi}{4} - \dfrac{x}{2}\right)\) where \(0 \leq x \leq 2\pi\), is minimum at \(x\) is equal to:</p>
Step-by-Step Solution
Key Concept: Simplify the trigonometric product using the double angle formula sin(2θ) = 2sin(θ)cos(θ), then find the derivative and minimize it by setting d²y/dx² = 0.
<p><strong>Step 1: Simplify the trigonometric expression</strong></p><p>Using sin(θ)cos(θ) = ½sin(2θ):</p><p>sin(π/4 - x/2)cos(π/4 - x/2) = ½sin(π/2 - x)</p><p>Since sin(π/2 - x) = cos(x):</p><p>y = 2e^x · ½cos(x) = e^x cos(x)</p><p><strong>Step 2: Find the slope (first derivative)</strong></p><p>dy/dx = e^x cos(x) + e^x(-sin(x))</p><p>dy/dx = e^x(cos(x) - sin(x))</p><p><strong>Step 3: Find where slope is minimum</strong></p><p>Let f(x) = e^x(cos(x) - sin(x))</p><p>f'(x) = e^x(cos(x) - sin(x)) + e^x(-sin(x) - cos(x))</p><p>f'(x) = e^x(cos(x) - sin(x) - sin(x) - cos(x))</p><p>f'(x) = -2e^x sin(x)</p><p><strong>Step 4: Set f'(x) = 0</strong></p><p>-2e^x sin(x) = 0</p><p>Since e^x ≠ 0, we need sin(x) = 0</p><p>For 0 ≤ x ≤ 2π: x = 0, π, 2π</p><p><strong>Step 5: Verify using second derivative test</strong></p><p>Check f''(x) to confirm which gives minimum. At x = π: f'(x) changes from negative to positive, confirming minimum.</p><p>∴ Answer: <strong>B (x = π)</strong></p>
Correct Answer: B