Question

If the area of the bounded region
$R=\left\{(x,y):\max \left\{0,{\log }_{e}x\right\}\le y\le 2^x,\frac{1}{2}\le x\le 2\right\}$ 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)

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

$\underset{\frac{1}{2}}{\overset{2}{\int }}{2}^{x}dx-\underset{1}{\overset{2}{\int }}lnxdx$
$\Rightarrow {\left[\frac{2^x}{\ln 2}\right]}_{1/2}^2-[x\ln x-x{]}_1^2$
$\Rightarrow \frac{\left(2^2\right)-2^{1/2}}{{\log }_{e}2}-(2\ln 2-1)$
$\Rightarrow \frac{\left(2^2-\sqrt{2}\right)}{{\log }_{e}2}-2\ln 2+1$
$\therefore \alpha =2^2-\sqrt{2},\beta =-2,\gamma =1$
$\Rightarrow (\alpha +\beta +2\gamma {)}^2$
$\Rightarrow {\left(2^2-\sqrt{2}-2-2\right)}^2$
$\Rightarrow (\sqrt{2}{)}^2=2$