If the area of the bounded region `R= {(x,y): "max" {0, log_(e)x} le y le 2^(x), (1)/(2) le x le 2}` is `alpha (log_(e)2)^(-1) + beta (log_(e )2) +gamma`, then the value of `(alpha +beta-2gamma)^(2)` is equal to

Options

A. 8
B. 2
C. 4
D. 1

Answer

To solve the problem, we need to find the area of the bounded region \( R = \{(x,y): \max\{0, \log_e x\} \leq y \leq 2^x, \frac{1}{2} \leq x \leq 2\} \).

### Step 1: Understand the boundaries of the region
The region is bounded by:
1. The curve \( y = 2^x \)
2. The curve \( y = \log_e x \) (or \( y = \ln x \))
3. The vertical lines \( x = \frac{1}{2} \) and \( x = 2 \)

### Step 2: Determine the points of intersection
We need to find the points where \( y = \log_e x \) intersects \( y = 2^x \) within the interval \( \left[\frac{1}{2}, 2\right] \).

1. Set \( \log_e x = 2^x \).
2. This is a transcendental equation, but we can evaluate it at specific points:
- At \( x = 1 \): \( \log_e 1 = 0 \) and \( 2^1 = 2 \) (not an intersection).
- At \( x = 2 \): \( \log_e 2 \) and \( 2^2 = 4 \) (not an intersection).
- At \( x = \frac{1}{2} \): \( \log_e \frac{1}{2} = -\log_e 2 \) and \( 2^{\frac{1}{2}} = \sqrt{2} \) (not an intersection).

Since \( \log_e x \) starts below \( 2^x \) and eventually grows slower than \( 2^x \), we conclude that \( \log_e x \) does not intersect \( 2^x \) within the interval.

### Step 3: Set up the area integral
The area \( A \) can be computed as:
\[ A = \int_{\frac{1}{2}}^{2} (2^x - \log_e x) \, dx \]

### Step 4: Calculate the integral
1. Compute \( \int 2^x \, dx \):
\[ \int 2^x \, dx = \frac{2^x}{\log_e 2} + C \]
2. Compute \( \int \log_e x \, dx \):
\[ \int \log_e x \, dx = x \log_e x - x + C \]

### Step 5: Evaluate the definite integrals
Now, evaluate:
\[ A = \left[ \frac{2^x}{\log_e 2} \right]_{\frac{1}{2}}^{2} - \left[ x \log_e x - x \right]_{\frac{1}{2}}^{2} \]

Calculating \( \frac{2^x}{\log_e 2} \):
- At \( x = 2 \): \( \frac{2^2}{\log_e 2} = \frac{4}{\log_e 2} \)
- At \( x = \frac{1}{2} \): \( \frac{2^{\frac{1}{2}}}{\log_e 2} = \frac{\sqrt{2}}{\log_e 2} \)

Thus,
\[ \left[ \frac{2^x}{\log_e 2} \right]_{\frac{1}{2}}^{2} = \frac{4}{\log_e 2} - \frac{\sqrt{2}}{\log_e 2} = \frac{4 - \sqrt{2}}{\log_e 2} \]

Calculating \( x \log_e x - x \):
- At \( x = 2 \): \( 2 \log_e 2 - 2 \)
- At \( x = \frac{1}{2} \): \( \frac{1}{2} \log_e \frac{1}{2} - \frac{1}{2} = -\frac{1}{2} \log_e 2 - \frac{1}{2} \)

Thus,
\[ \left[ x \log_e x - x \right]_{\frac{1}{2}}^{2} = (2 \log_e 2 - 2) - \left(-\frac{1}{2} \log_e 2 - \frac{1}{2}\right) \]
\[ = 2 \log_e 2 - 2 + \frac{1}{2} \log_e 2 + \frac{1}{2} = \frac{5}{2} \log_e 2 - \frac{3}{2} \]

### Step 6: Combine the results
Putting it all together:
\[ A = \frac{4 - \sqrt{2}}{\log_e 2} - \left(\frac{5}{2} \log_e 2 - \frac{3}{2}\right) \]
\[ = \frac{4 - \sqrt{2}}{\log_e 2} - \frac{5}{2} \log_e 2 + \frac{3}{2} \]

### Step 7: Compare with the given expression
We have:
\[ A = \alpha (\log_e 2)^{-1} + \beta \log_e 2 + \gamma \]
From our calculations, we can identify \( \alpha, \beta, \gamma \) based on the coefficients.

### Final Calculation
Finally, compute \( \alpha + \beta - 2\gamma \) and square it to find the answer.