Slope of tangent to the curve `y=2e^(x)sin((pi)/(4)-(x)/(2))cos((pi)/(4)-(x)/(2))`, where `0le xle2pi` is minimum at x =

Options

A. 0
B. `pi`
C. `2pi`
D. none of these

Explanation

To find the minimum slope of the tangent to the curve given by the function \( y = 2e^x \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) \) for \( 0 \leq x \leq 2\pi \), we will follow these steps:

### Step 1: Differentiate the function
To find the slope of the tangent, we need to compute the derivative \( \frac{dy}{dx} \).

Given:
\[ y = 2e^x \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) \]

Using the product rule and the chain rule, we differentiate:
\[ \frac{dy}{dx} = 2e^x \left( \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) + e^x \left( \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) \cdot \left(-\frac{1}{2}\sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) + \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cdot \left(-\frac{1}{2}\cos\left(\frac{\pi}{4} - \frac{x}{2}\right)\right) \right) \right) \]

### Step 2: Set the derivative to zero
To find the critical points where the slope is minimum, we set \( \frac{dy}{dx} = 0 \):
\[ 2e^x \left( \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) + e^x \left( -\frac{1}{2}\sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \cos\left(\frac{\pi}{4} - \frac{x}{2}\right) - \frac{1}{2}\cos\left(\frac{\pi}{4} - \frac{x}{2}\right) \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) \right) \right) = 0 \]

This simplifies to:
\[ -2e^x \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) = 0 \]

### Step 3: Solve for x
From the equation \( -2e^x \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) = 0 \), we find:
\[ \sin\left(\frac{\pi}{4} - \frac{x}{2}\right) = 0 \]

This occurs when:
\[ \frac{\pi}{4} - \frac{x}{2} = n\pi \quad \text{for integer } n \]

Solving for \( x \):
\[ x = \frac{\pi}{2} - 2n\pi \]

### Step 4: Evaluate critical points within the interval
We need to check the values of \( x \) within the interval \( [0, 2\pi] \):
- For \( n = 0 \): \( x = \frac{\pi}{2} \)
- For \( n = 1 \): \( x = \frac{\pi}{2} - 2\pi \) (not in the interval)
- For \( n = -1 \): \( x = \frac{\pi}{2} + 2\pi \) (not in the interval)

### Step 5: Check endpoints and critical points
Now we evaluate \( \frac{dy}{dx} \) at the endpoints \( x = 0 \) and \( x = 2\pi \) and at the critical point \( x = \pi \):
- At \( x = 0 \): \( \frac{dy}{dx} = 2e^0 \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right) = 2 \cdot 1 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1 \)
- At \( x = \pi \): \( \frac{dy}{dx} = 0 \)
- At \( x = 2\pi \): \( \frac{dy}{dx} = 2e^{2\pi} \sin\left(-\frac{\pi}{4}\right)\cos\left(-\frac{\pi}{4}\right) = 2e^{2\pi} \cdot -\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = -e^{2\pi} \)

### Conclusion
The minimum slope of the tangent occurs at \( x = \pi \).