Question

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

• A. $\ln \sqrt{2}$
• B. $\ln 2$
• C. $\ln \sqrt{8}$
• D. $\ln \left(\frac{3}{2}\right)$

Answer 1

The correct option is B $\ln 2$

Let $I=4\alpha \underset{-1}{\overset{2}{\int }}{e}^{-\alpha |x|}dx$
$\Rightarrow I=4\alpha \left(\underset{-1}{\overset{0}{\int }}{e}^{\alpha x}dx+\underset{0}{\overset{2}{\int }}{e}^{-\alpha x}dx\right)$
$\Rightarrow I=4\alpha ({\frac{e^{\alpha x}}{\alpha }|}_{-1}^0-{\frac{e^{-\alpha x}}{\alpha }|}_0^2)$
$\Rightarrow I=4(2-e^{-\alpha }-e^{-2\alpha })=5$
Let $e^{-\alpha }=t$
$\Rightarrow 8-4t-4t^2=5$
$\Rightarrow 4t^2+4t-3=0$
$\Rightarrow (2t-1)(2t+3)=0$
$\therefore t=\frac{1}{2}$ and $t\ne \frac{-3}{2}(\because e^x>0)$
$\Rightarrow e^{-\alpha }=\frac{1}{2}$
$\therefore \alpha =\ln 2$