The curves `y = e^(-ax) sin bx` and `y = e^(- ax)` touch at the points for which bx=

Options

A. `2npi+pi/2`
B. `2npi+pi/3`
C. `2npi+pi/4`
D. none

Answer

To solve the problem of finding the values of \( bx \) for which the curves \( y = e^{-ax} \sin(bx) \) and \( y = e^{-ax} \) touch each other, we can follow these steps:

### Step 1: Set the equations equal to each other
Since the curves touch at certain points, we set the two equations equal to each other:
\[ e^{-ax} \sin(bx) = e^{-ax} \]

### Step 2: Simplify the equation
We can simplify the equation by dividing both sides by \( e^{-ax} \) (assuming \( e^{-ax} \neq 0 \)):
\[ \sin(bx) = 1 \]

### Step 3: Solve for \( bx \)
The sine function equals 1 at specific angles. The general solution for \( \sin(bx) = 1 \) is given by:
\[ bx = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
where \( n \) is any integer.

### Step 4: Express \( bx \) in terms of \( n \)
Rearranging gives:
\[ bx = 2n\pi + \frac{\pi}{2} \]

### Step 5: Factor out \( b \)
To express \( x \) in terms of \( n \):
\[ x = \frac{2n\pi + \frac{\pi}{2}}{b} \]

### Step 6: Identify specific values of \( bx \)
For \( n = 0 \):
\[ bx = \frac{\pi}{2} \]
For \( n = 1 \):
\[ bx = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2} \]
For \( n = -1 \):
\[ bx = -2\pi + \frac{\pi}{2} = -\frac{3\pi}{2} \]

### Conclusion
The values of \( bx \) for which the curves touch are:
\[ bx = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
The specific values of \( bx \) can be expressed as:
\[ bx = 2n\pi + \frac{\pi}{2} \]