The area bounded by `y=(1)/(x) and y=(1)/(2x-1)` from x = 1 to x = 2 is ln (a) sq. units, then `3a^(2)` is equal to

Options

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

Answer

To find the area bounded by the curves \( y = \frac{1}{x} \) and \( y = \frac{1}{2x - 1} \) from \( x = 1 \) to \( x = 2 \), we will follow these steps:

### Step 1: Find the points of intersection
We need to set the two equations equal to each other to find the points of intersection:

\[ \frac{1}{x} = \frac{1}{2x - 1} \]

Cross-multiplying gives:

\[ 2x - 1 = x \]

Solving for \( x \):

\[ 2x - x = 1 \implies x = 1 \]

Now, substituting \( x = 1 \) back into either equation to find \( y \):

\[ y = \frac{1}{1} = 1 \]

So, one point of intersection is \( (1, 1) \). Now, we check if there are any other intersections in the interval \( x = 1 \) to \( x = 2 \):

Substituting \( x = 2 \):

\[ y = \frac{1}{2} \quad \text{and} \quad y = \frac{1}{3} \]

Since \( y = \frac{1}{x} \) is decreasing and \( y = \frac{1}{2x - 1} \) is also decreasing, there are no other intersections in this interval.

### Step 2: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 2 \) can be found using the integral:

\[ A = \int_{1}^{2} \left( \frac{1}{x} - \frac{1}{2x - 1} \right) dx \]

### Step 3: Calculate the integral
Now we compute the integral:

\[ A = \int_{1}^{2} \left( \frac{1}{x} - \frac{1}{2x - 1} \right) dx \]

This can be split into two separate integrals:

\[ A = \int_{1}^{2} \frac{1}{x} \, dx - \int_{1}^{2} \frac{1}{2x - 1} \, dx \]

Calculating the first integral:

\[ \int \frac{1}{x} \, dx = \ln |x| \quad \Rightarrow \quad \left[ \ln x \right]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \]

Calculating the second integral:

\[ \int \frac{1}{2x - 1} \, dx = \frac{1}{2} \ln |2x - 1| \quad \Rightarrow \quad \left[ \frac{1}{2} \ln |2x - 1| \right]_{1}^{2} \]

Calculating at the bounds:

\[ = \frac{1}{2} \left( \ln(3) - \ln(1) \right) = \frac{1}{2} \ln(3) \]

### Step 4: Combine the results
Now substituting back into the area formula:

\[ A = \ln 2 - \frac{1}{2} \ln 3 \]

Using properties of logarithms:

\[ A = \ln \left( \frac{2}{\sqrt{3}} \right) \]

### Step 5: Relate to the given area
We know from the problem statement that the area is \( \ln(a) \). Thus:

\[ \ln(a) = \ln \left( \frac{2}{\sqrt{3}} \right) \implies a = \frac{2}{\sqrt{3}} \]

### Step 6: Calculate \( 3a^2 \)
Now we need to find \( 3a^2 \):

\[ a^2 = \left( \frac{2}{\sqrt{3}} \right)^2 = \frac{4}{3} \]

Thus,

\[ 3a^2 = 3 \cdot \frac{4}{3} = 4 \]

### Final Answer
The value of \( 3a^2 \) is \( \boxed{4} \).