If the area covered by `y=(2)/(x)` and `y=(2)/(2x-1)` from `x=1 or x=e` is ln (a) sq. units, then `(2e-1)^(2)a^(2)` is equal to

Options

A. `(e)/(2)`
B. `e^(4)`
C. 1
D. e

Answer

To solve the problem, we need to find the area between the curves \( y = \frac{2}{x} \) and \( y = \frac{2}{2x - 1} \) from \( x = 1 \) to \( x = e \). We will follow these steps:

### Step 1: Set up the area between the curves
The area \( A \) between the two curves from \( x = 1 \) to \( x = e \) can be expressed as:
\[ A = \int_{1}^{e} \left( \frac{2}{x} - \frac{2}{2x - 1} \right) \, dx \]

### Step 2: Calculate the integral
We will break this integral into two parts:
\[ A = \int_{1}^{e} \frac{2}{x} \, dx - \int_{1}^{e} \frac{2}{2x - 1} \, dx \]

#### Part 1: Calculate \( \int_{1}^{e} \frac{2}{x} \, dx \)
\[ \int \frac{2}{x} \, dx = 2 \ln |x| \]
Evaluating from \( 1 \) to \( e \):
\[ \left[ 2 \ln x \right]_{1}^{e} = 2 \ln e - 2 \ln 1 = 2 \cdot 1 - 0 = 2 \]

#### Part 2: Calculate \( \int_{1}^{e} \frac{2}{2x - 1} \, dx \)
Let \( u = 2x - 1 \), then \( du = 2 \, dx \) or \( dx = \frac{du}{2} \).
When \( x = 1 \), \( u = 1 \) and when \( x = e \), \( u = 2e - 1 \).

Now, substituting:
\[ \int_{1}^{e} \frac{2}{2x - 1} \, dx = \int_{1}^{2e - 1} \frac{2}{u} \cdot \frac{du}{2} = \int_{1}^{2e - 1} \frac{1}{u} \, du \]
\[ = \left[ \ln |u| \right]_{1}^{2e - 1} = \ln(2e - 1) - \ln(1) = \ln(2e - 1) \]

### Step 3: Combine the results
Now, substituting back into the area formula:
\[ A = 2 - \ln(2e - 1) \]

### Step 4: Set equal to \( \ln(a) \)
Given that \( A = \ln(a) \), we can equate:
\[ \ln(a) = 2 - \ln(2e - 1) \]
Using properties of logarithms:
\[ \ln(a) = \ln(e^2) - \ln(2e - 1) = \ln\left(\frac{e^2}{2e - 1}\right) \]
Thus, we have:
\[ a = \frac{e^2}{2e - 1} \]

### Step 5: Calculate \( (2e - 1)^2 a^2 \)
Now we need to find \( (2e - 1)^2 a^2 \):
\[ a^2 = \left(\frac{e^2}{2e - 1}\right)^2 = \frac{e^4}{(2e - 1)^2} \]
Thus:
\[ (2e - 1)^2 a^2 = (2e - 1)^2 \cdot \frac{e^4}{(2e - 1)^2} = e^4 \]

### Final Answer
The value of \( (2e - 1)^2 a^2 \) is:
\[ \boxed{e^4} \]