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

The region is bounded by $y=2^x$ (upper curve) and $y=\max\{0, \ln x\}$ (lower curve) for $x \in [\frac{1}{2}, 2]$.
We need to analyze the lower boundary function, $y = \max\{0, \ln x\}$.
- For $x \in [\frac{1}{2}, 1)$, $\ln x<0$, so $\max\{0, \ln x\} = 0$.
- For $x \in [1, 2]$, $\ln x \ge 0$, so $\max\{0, \ln x\} = \ln x$.
This means we must split the integral for the area into two parts at $x=1$.
Area $A = \int_{1/2}^{1} (2^x - 0) dx + \int_{1}^{2} (2^x - \ln x) dx$.
Let's evaluate the integrals.
$\int 2^x dx = \frac{2^x}{\ln 2}$.
$\int \ln x dx = x\ln x - x$ (using integration by parts).
First integral:
$\int_{1/2}^{1} 2^x dx = \left[\frac{2^x}{\ln 2}\right]_{1/2}^{1} = \frac{2^1}{\ln 2} - \frac{2^{1/2}}{\ln 2} = \frac{2 - \sqrt{2}}{\ln 2}$.
Second integral:
$\int_{1}^{2} (2^x - \ln x) dx = \left[\frac{2^x}{\ln 2} - (x\ln x - x)\right]_{1}^{2}$
$= \left(\frac{2^2}{\ln 2} - (2\ln 2 - 2)\right) - \left(\frac{2^1}{\ln 2} - (1\ln 1 - 1)\right)$
$= \left(\frac{4}{\ln 2} - 2\ln 2 + 2\right) - \left(\frac{2}{\ln 2} - (-1)\right)$
$= \frac{4}{\ln 2} - 2\ln 2 + 2 - \frac{2}{\ln 2} - 1 = \frac{2}{\ln 2} - 2\ln 2 + 1$.
Total Area $A = \left(\frac{2 - \sqrt{2}}{\ln 2}\right) + \left(\frac{2}{\ln 2} - 2\ln 2 + 1\right)$
$A = \frac{2 - \sqrt{2} + 2}{\ln 2} - 2\ln 2 + 1 = \frac{4 - \sqrt{2}}{\ln 2} - 2\ln 2 + 1$.
We are given the area is $\alpha(\log_e 2)^{-1} + \beta(\log_e 2) + \gamma$.
$A = (4-\sqrt{2})(\ln 2)^{-1} + (-2)(\ln 2) + 1$.
Comparing the forms, we get:
$\alpha = 4 - \sqrt{2}$
$\beta = -2$
$\gamma = 1$
Now, we compute the required value: $(\alpha + \beta - 2\gamma)^2$.
$\alpha + \beta - 2\gamma = (4 - \sqrt{2}) + (-2) - 2(1) = 4 - \sqrt{2} - 2 - 2 = -\sqrt{2}$.
$(\alpha + \beta - 2\gamma)^2 = (-\sqrt{2})^2 = 2$.