Question

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

• A. 8
• B. 2
• C. 4
• D. 1

Answer 1

Answer: (B)

$R=\left\{(x, y): \max \left\{0, \log _{e} x\right\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\right\}$

$\int\limits_{\frac{1}{2}}^{2} 2^{x} d x-\int\limits_{1}^{2} l n x d x$
$\Rightarrow\left[\frac{2^{x}}{\ln 2}\right]_{1 / 2}^{2}-[x \ln x-x]_{1}^{2}$
$\Rightarrow \frac{\left(2^{2}\right)-2^{1 / 2}}{\log _{e} 2}-(2 \ln 2-1)$
$\Rightarrow \frac{\left(2^{2}-\sqrt{2}\right)}{\log _{e} 2}-2 \ln 2+1$
$\therefore \alpha=2^{2}-\sqrt{2}, \beta=-2, \gamma=1$
$\Rightarrow(\alpha+\beta+2 \gamma)^{2}$
$\Rightarrow\left(2^{2}-\sqrt{2}-2-2\right)^{2}$
$\Rightarrow(\sqrt{2})^{2}=2$