Question

If the area of the bounded region $R = \left\{(x, y) : \max\{0, \log_e x\} \le y \le 2^x, \frac{1}{2} \le x \le 2\right\}$ is, $\alpha(\log_e 2)^{-1} + \beta(\log_e 2) + \gamma$, then the value of $(\alpha + \beta - 2\gamma)^2$ is equal to :

• A. 1
• B. 2
• C. 4
• D. 8

Hint:

When dealing with functions defined by `max` or `min` (or absolute values), always split the integration interval at the points where the function definition changes. Here, $\ln x$ crosses zero at $x=1$, which is the critical point for the split.

Answer 1

The Correct Option is B

Solution

We are given the region: 

\( R = \{(x, y) : \max\{0, \log_e x\} \le y \le 2^x, \frac{1}{2} \le x \le 2\} \)

Step 1: Understand the lower boundary

• For \( \frac{1}{2} \le x < 1 \), we have \( \log_e x < 0 \), so lower bound is \( y = 0 \).
• For \( 1 \le x \le 2 \), we have \( \log_e x \ge 0 \), so lower bound is \( y = \log_e x \).

Step 2: Break the integral

First interval: \( \frac{1}{2} \le x \le 1 \)

\( \displaystyle \int_{\frac{1}{2}}^{1} 2^x \, dx \)

\( \int 2^x dx = \frac{2^x}{\ln 2} \)

\( \displaystyle \left[\frac{2^x}{\ln 2}\right]_{\frac{1}{2}}^{1} = \frac{2}{\ln 2} - \frac{\sqrt{2}}{\ln 2} \)

Second interval: \( 1 \le x \le 2 \)

\( \displaystyle \int_{1}^{2} (2^x - \ln x)\,dx \)

\( \int 2^x dx = \frac{2^x}{\ln 2} \)

\( \int \ln x\,dx = x\ln x - x \)

\( \displaystyle \left[\frac{2^x}{\ln 2}\right]_{1}^{2} - \left[(x\ln x - x)\right]_{1}^{2} \)

\( = \left(\frac{4}{\ln 2} - \frac{2}{\ln 2}\right) - \left[(2\ln 2 - 2) - (-1)\right] \)

\( = \frac{2}{\ln 2} - (2\ln 2 - 1) \)

Step 3: Total Area

Area = \( \displaystyle \left(\frac{2}{\ln 2} - \frac{\sqrt{2}}{\ln 2}\right) + \left(\frac{2}{\ln 2} - 2\ln 2 - 1\right) \)

\( \displaystyle = \frac{4 - \sqrt{2}}{\ln 2} - 2\ln 2 - 1 \)

Step 4: Compare with form

\( \alpha(\ln 2)^{-1} + \beta(\ln 2) + \gamma \)

\( \alpha = 4 - \sqrt{2}, \quad \beta = -2, \quad \gamma = -1 \)

Step 5: Compute \( (\alpha + \beta - 2\gamma)^2 \)

\( \alpha + \beta - 2\gamma = (4 - \sqrt{2}) - 2 - 2(-1) = 4 - \sqrt{2} \)

\( (4 - \sqrt{2})^2 = 16 - 8\sqrt{2} + 2 = 18 - 8\sqrt{2} \)

Final Answer

\( \boxed{18 - 8\sqrt{2}} \)