JEE Main 2021 (Online) 27th July Morning ShiftQuestion 1
If the area of the bounded region $R=\left\{(x,y):\max\{0,\log_ex\}\le y\le2^x,\ \frac12\le x\le2\right\}$ is $\alpha(\log_e2)^{-1}+\beta(\log_e2)+\gamma$, then the value of $(\alpha+\beta-2\gamma)^2$ is equal to :
Application of Integrals
Solution
$\max(0,\ln x)=0$ on $[\frac12,1]$ and $\ln x$ on $[1,2]$. $A=\int_{1/2}^2 2^xdx-\int_1^2\ln x\,dx=\frac{4-\sqrt2}{\ln2}-(2\ln2-1)$. So $\alpha=4-\sqrt2,\ \beta=-2,\ \gamma=1$. $(\alpha+\beta-2\gamma)^2=(4-\sqrt2-2-2)^2=2$. Option (B).