The value of a for which exactly one root of the equation `e^ax^2 - e^(2a)x + e^a -1` lies between 1 and 2 are given by

Options

A. In `((5-sqrt(17))/(4))lt a lt "In"((5+sqrt(17))/(4))`
B. `0 lt a lt 100`
C. In `(5)/(4) lt a lt "In"(10)/(3)`
D. none of these

Explanation

To solve the problem, we need to find the values of \( a \) for which exactly one root of the equation

\[ e^{ax^2} - e^{2a}x + e^a - 1 = 0 \]

lies between 1 and 2.

### Step 1: Define the function
Let

\[ f(x) = e^{a}x^2 - e^{2a}x + e^{a} - 1 \]

### Step 2: Evaluate the function at the endpoints
We need to evaluate \( f(1) \) and \( f(2) \):

\[ f(1) = e^{a}(1^2) - e^{2a}(1) + e^{a} - 1 = e^{a} - e^{2a} + e^{a} - 1 = 2e^{a} - e^{2a} - 1 \]

\[ f(2) = e^{a}(2^2) - e^{2a}(2) + e^{a} - 1 = 4e^{a} - 2e^{2a} + e^{a} - 1 = 5e^{a} - 2e^{2a} - 1 \]

### Step 3: Set up the condition for exactly one root
For there to be exactly one root between 1 and 2, we require that \( f(1) \) and \( f(2) \) have opposite signs:

\[ f(1) \cdot f(2) < 0 \]

### Step 4: Substitute the expressions
Substituting the expressions we found:

\[ (2e^{a} - e^{2a} - 1)(5e^{a} - 2e^{2a} - 1) < 0 \]

### Step 5: Analyze the quadratic inequalities
We will analyze the two factors separately:

1. **First factor**: \( 2e^{a} - e^{2a} - 1 \)
- This is a quadratic in terms of \( e^{a} \).
- Let \( t = e^{a} \), then we have:
\[ -t^2 + 2t - 1 = 0 \implies t^2 - 2t + 1 = 0 \implies (t-1)^2 = 0 \]
- The root is \( t = 1 \) (i.e., \( e^{a} = 1 \) or \( a = 0 \)).
- The quadratic opens downwards, so \( 2e^{a} - e^{2a} - 1 < 0 \) for \( t < 1 \) and \( t > 1 \).

2. **Second factor**: \( 5e^{a} - 2e^{2a} - 1 \)
- Again, substituting \( t = e^{a} \):
\[ -2t^2 + 5t - 1 = 0 \]
- Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 8}}{-4} = \frac{5 \pm \sqrt{17}}{4} \]
- The roots are \( t_1 = \frac{5 - \sqrt{17}}{4} \) and \( t_2 = \frac{5 + \sqrt{17}}{4} \).

### Step 6: Determine intervals
We need to find the intervals where the product \( (2e^{a} - e^{2a} - 1)(5e^{a} - 2e^{2a} - 1) < 0 \).

1. The first factor changes sign at \( t = 1 \).
2. The second factor changes sign at \( t = \frac{5 - \sqrt{17}}{4} \) and \( t = \frac{5 + \sqrt{17}}{4} \).

### Step 7: Analyze the signs
- For \( t < \frac{5 - \sqrt{17}}{4} \): both factors are positive.
- Between \( \frac{5 - \sqrt{17}}{4} < t < 1 \): first factor is negative, second is positive.
- Between \( 1 < t < \frac{5 + \sqrt{17}}{4} \): first factor is positive, second is negative.
- For \( t > \frac{5 + \sqrt{17}}{4} \): both factors are negative.

### Step 8: Final range for \( a \)
Thus, the intervals for \( t \) are:

\[ \frac{5 - \sqrt{17}}{4} < t < \frac{5 + \sqrt{17}}{4} \]

Converting back to \( a \):

\[ \ln\left(\frac{5 - \sqrt{17}}{4}\right) < a < \ln\left(\frac{5 + \sqrt{17}}{4}\right) \]

### Final Answer
The values of \( a \) for which exactly one root of the equation lies between 1 and 2 are given by:

\[ a \in \left(\ln\left(\frac{5 - \sqrt{17}}{4}\right), \ln\left(\frac{5 + \sqrt{17}}{4}\right)\right) \]