The slope of the tangent to the curve `y=e^x cosx` is minimum at `x= a,0 leq a leq 2pi`, then the value of a is

Options

A. 0
B. `pi`
C. `2pi`
D. `3pi//2`

Answer

To find the value of \( a \) where the slope of the tangent to the curve \( y = e^x \cos x \) is minimum, we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)
Given the function:
\[ y = e^x \cos x \]
We apply the product rule to differentiate:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^x) \cdot \cos x + e^x \cdot \frac{d}{dx}(\cos x) \]
Calculating the derivatives:
\[ \frac{d}{dx}(e^x) = e^x \quad \text{and} \quad \frac{d}{dx}(\cos x) = -\sin x \]
Thus,
\[ \frac{dy}{dx} = e^x \cos x - e^x \sin x = e^x (\cos x - \sin x) \]

### Step 2: Set the first derivative equal to zero to find critical points
To find where the slope is minimum, we need to find the critical points by setting the first derivative to zero:
\[ e^x (\cos x - \sin x) = 0 \]
Since \( e^x \) is never zero, we have:
\[ \cos x - \sin x = 0 \implies \cos x = \sin x \]
This occurs at:
\[ x = \frac{\pi}{4} + n\pi \quad \text{for integers } n \]
Within the interval \( [0, 2\pi] \), the solutions are:
\[ x = \frac{\pi}{4}, \frac{5\pi}{4} \]

### Step 3: Find the second derivative \( \frac{d^2y}{dx^2} \)
To determine whether these points are minima or maxima, we need the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x (\cos x - \sin x)) \]
Using the product rule again:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(e^x) (\cos x - \sin x) + e^x \frac{d}{dx}(\cos x - \sin x) \]
Calculating the derivatives:
\[ \frac{d}{dx}(\cos x - \sin x) = -\sin x - \cos x \]
Thus,
\[ \frac{d^2y}{dx^2} = e^x (\cos x - \sin x) + e^x (-\sin x - \cos x) = e^x (-2\sin x) \]

### Step 4: Evaluate the second derivative at critical points
Now we evaluate \( \frac{d^2y}{dx^2} \) at \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \):
1. For \( x = \frac{\pi}{4} \):
\[ \frac{d^2y}{dx^2} = e^{\frac{\pi}{4}} (-2\sin(\frac{\pi}{4})) = e^{\frac{\pi}{4}} (-2 \cdot \frac{\sqrt{2}}{2}) = -\sqrt{2} e^{\frac{\pi}{4}} < 0 \quad (\text{local maximum}) \]
2. For \( x = \frac{5\pi}{4} \):
\[ \frac{d^2y}{dx^2} = e^{\frac{5\pi}{4}} (-2\sin(\frac{5\pi}{4})) = e^{\frac{5\pi}{4}} (-2 \cdot -\frac{\sqrt{2}}{2}) = \sqrt{2} e^{\frac{5\pi}{4}} > 0 \quad (\text{local minimum}) \]

### Conclusion
The slope of the tangent to the curve \( y = e^x \cos x \) is minimum at:
\[ a = \frac{5\pi}{4} \]