If `e^(ax) +e^(-bx) = p_0 +p_1x +p_2x^2 + ....oo` then `(p_0 , p_1,p_2)` =

Options

A. `(2,a -b , (a^2+b^2)/2) `
B. `(2,a + b , (a^2+b^2)/2) `
C. `(2, a-b,a^2+b^2)`
D. `(2,a - b , (a^2- b^2)/2) `

Answer

To solve the problem, we need to find the coefficients \( p_0, p_1, \) and \( p_2 \) from the expression \( e^{ax} + e^{-bx} \).

### Step-by-Step Solution:

1. **Write the Series Expansion**:
The exponential function can be expanded using its Taylor series:
\[ e^{ax} = 1 + \frac{(ax)^1}{1!} + \frac{(ax)^2}{2!} + \frac{(ax)^3}{3!} + \ldots \]
\[ e^{-bx} = 1 - \frac{(bx)^1}{1!} + \frac{(bx)^2}{2!} - \frac{(bx)^3}{3!} + \ldots \]

2. **Combine the Two Expansions**:
Now, we add the two series:
\[ e^{ax} + e^{-bx} = \left(1 + \frac{(ax)^1}{1!} + \frac{(ax)^2}{2!} + \ldots\right) + \left(1 - \frac{(bx)^1}{1!} + \frac{(bx)^2}{2!} - \ldots\right) \]
This simplifies to:
\[ = 2 + (a - b)x + \left(\frac{a^2}{2!} + \frac{b^2}{2!}\right)x^2 + \ldots \]

3. **Identify Coefficients**:
From the combined series, we can identify the coefficients \( p_0, p_1, \) and \( p_2 \):
- \( p_0 = 2 \)
- \( p_1 = a - b \)
- \( p_2 = \frac{a^2 + b^2}{2} \)

4. **Final Answer**:
Therefore, the values of \( p_0, p_1, \) and \( p_2 \) are:
\[ (p_0, p_1, p_2) = \left(2, a - b, \frac{a^2 + b^2}{2}\right) \]

### Summary:
The final answer is:
\[ (p_0, p_1, p_2) = \left(2, a - b, \frac{a^2 + b^2}{2}\right) \]