The value of a for which exactly one root of the equation `e^(a)x^(2)-e^(2a)x+e^(a)-1=0` lies between 1 and 2 are given by

Options

A. `ln((5-sqrt(13))/(4))lt a lt ln ((5+sqrt(13))/(4))`
B. `0 lt a lt 100`
C. `ln.(5)/(4)lt a lt ln.(10)/(3)`
D. none of these

Answer

To solve the problem of finding the value of \( a \) for which exactly one root of the equation

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

lies between 1 and 2, we will follow these steps:

### Step 1: Rewrite the quadratic equation
The given quadratic equation can be rewritten as:

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

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

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

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

### Step 3: Set conditions for exactly one root between 1 and 2
For the quadratic equation to have exactly one root between 1 and 2, the following conditions must hold:

1. \( f(1) < 0 \) (the function must be negative at \( x = 1 \))
2. \( f(2) > 0 \) (the function must be positive at \( x = 2 \))

### Step 4: Solve the inequalities
1. **Inequality 1**: \( f(1) < 0 \)
\[ 2e^{a} - e^{2a} - 1 < 0 \]
Rearranging gives:
\[ e^{2a} - 2e^{a} + 1 > 0 \]
This can be factored as:
\[ (e^{a} - 1)^{2} > 0 \]
This inequality holds for \( e^{a} \neq 1 \), which means \( a \neq 0 \).

2. **Inequality 2**: \( f(2) > 0 \)
\[ 5e^{a} - 2e^{2a} - 1 > 0 \]
Rearranging gives:
\[ 2e^{2a} - 5e^{a} + 1 < 0 \]
Let \( t = e^{a} \):
\[ 2t^{2} - 5t + 1 < 0 \]
To find the roots, we calculate the discriminant:
\[ D = (-5)^{2} - 4 \cdot 2 \cdot 1 = 25 - 8 = 17 \]
The roots are:
\[ t_{1} = \frac{5 - \sqrt{17}}{4}, \quad t_{2} = \frac{5 + \sqrt{17}}{4} \]

### Step 5: Determine the interval for \( t \)
The quadratic \( 2t^{2} - 5t + 1 < 0 \) is negative between its roots:
\[ \frac{5 - \sqrt{17}}{4} < t < \frac{5 + \sqrt{17}}{4} \]

### Step 6: Convert back to \( a \)
Since \( t = e^{a} \), we take the natural logarithm:
\[ \ln\left(\frac{5 - \sqrt{17}}{4}\right) < a < \ln\left(\frac{5 + \sqrt{17}}{4}\right) \]

### Final Answer
Thus, the values of \( a \) for which exactly one root of the equation lies between 1 and 2 are given by:
\[ \boxed{\left( \ln\left(\frac{5 - \sqrt{17}}{4}\right), \ln\left(\frac{5 + \sqrt{17}}{4}\right) \right)} \]