Question

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

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

Answer 1

The correct option is A

${\log }_e\left(2\right)$

Finding the value of :

Given, $4\alpha {\int }_{-1}^2{e}^{-\alpha \left|x\right|}dx=5$

$\Rightarrow$$4\alpha \left[{\int }_{-1}^0{e}^{-\alpha \left|x\right|}dx+{\int }_0^2{e}^{-\alpha \left|x\right|}dx\right]=5$

$\Rightarrow$ $4\alpha \left[{\int }_{-1}^0{e}^{-\alpha x}dx+{\int }_0^2{e}^{-\alpha x}dx\right]=5$

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

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

$\Rightarrow$ $(2-e^{-\alpha }-e^{-2\alpha }-\frac{5}{4})=0$

$\Rightarrow$ $4e^{-2\alpha }+4e^{-\alpha }-3=0$

$\Rightarrow$ $4e^{-2\alpha }+6e^{-\alpha }-2e^{-\alpha }-3=0$

$\Rightarrow$ $(2e^{-\alpha }+3)(2e^{-\alpha }-1)=0$

$\Rightarrow$ $e^{-a}=\frac{1}{2}$ $\left[\mathbf{\therefore }{\mathit{e}}^{\mathbf{-}\mathbf{a}}\mathbf{}\mathbf{\ne }\mathbf{}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\right]$

$\mathbf{\therefore }\alpha =lo{g}_e\left(2\right)$

Hence, Option (A) is the correct answer.