Question

The area (in sq. units) defined by $|y|\le e^{-|x|}-\frac{1}{2}$ in Cartesian co-ordinate system, is

• A. $4-2\ln 2$
• B. $4-\ln 2$
• C. $1-\ln 2$
• D. $2-2\ln 2$

Answer 1

The correct option is D $2-2\ln 2$

$|y|\le e^{-|x|}-\frac{1}{2}$
For $y\ge 0,$ we have
$y\le e^{-|x|}-\frac{1}{2}$
$\Rightarrow y\le \{\begin{array}{cc}{e}^x-\frac{1}{2},& x<0\\ e^{-x}-\frac{1}{2},& x\ge 0\end{array}\cdots \left(1\right)$


For $y<0,$ we have
$-y\le e^{-|x|}-\frac{1}{2}$
$\Rightarrow y\ge \frac{1}{2}-e^{-|x|}$
$\Rightarrow y\ge \{\begin{array}{cc}\frac{1}{2}-e^x,& x<0\\ \frac{1}{2}-e^{-x},& x\ge 0\end{array}\cdots \left(2\right)$


Combining $\left(1\right)$ and $\left(2\right),$ we have the required area given as


Now, Area $=4\times$ Area in first quadrant
$\Rightarrow$ Area $=4\underset{0}{\overset{\ln 2}{\int }}(e^{-x}-\frac{1}{2})dx=4(\frac{1}{2}-\frac{1}{2}\ln 2)$
$\therefore$ Area $=2-2\ln 2$