Question

The value of $\alpha$ for which $4\alpha \underset{-1}{\overset{2}{\int }}{e}^{-\alpha |x|}dx=5$, is :

• A. ${\log }_{e}2$
• B. ${\log }_e\sqrt{2}$
• C. ${\log }_e\left(\frac{4}{3}\right)$
• D. ${\log }_e\left(\frac{3}{2}\right)$

Answer 1

The correct option is A ${\log }_{e}2$

$4\alpha \underset{-1}{\overset{2}{\int }}{e}^{-\alpha |x|dx}=5$
$\Rightarrow 4\alpha [\underset{-1}{\overset{0}{\int }}{e}^{-\alpha |x|}dx+\underset{0}{\overset{2}{\int }}{e}^{-\alpha |x|}dx]=5$

$\Rightarrow 4\alpha [\underset{-1}{\overset{0}{\int }}{e}^{\alpha x}dx+\underset{0}{\overset{2}{\int }}{e}^{-\alpha x}dx]=5$

$\Rightarrow 4\alpha [\left(\frac{1-e^{-\alpha }}{\alpha }\right)+\left(\frac{e^{-2\alpha }-1}{-\alpha }\right)]=5$

$\Rightarrow 4[1-e^{-\alpha }-e^{-2\alpha }+1]=5$

Let $e^{-\alpha }=t$
$\Rightarrow 4t^2+4t-3=0$
$\Rightarrow t=\frac{1}{2}=e^{-\alpha }\Rightarrow \alpha ={\log }_{e}2$