Official solution
\[4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5\]
\[ \Rightarrow \] \[4\alpha \int\limits_{ - 1}^0 {{e^{\alpha x}}dx} + 4\alpha \int\limits_0^2 {{e^{ - \alpha x}}dx} \] = 5
\[ \Rightarrow \] \[4\alpha \left[ {{{{e^{\alpha x}}} \over \alpha }} \right]_{ - 1}^0 + 4\alpha \left[ {{{{e^{ - \alpha x}}} \over { - \alpha }}} \right]_0^2\] = 5
\[ \Rightarrow \] \[4{e^{ - 2\alpha }} + 4{e^{ - \alpha }} - 3\] = 0
Let e-\[\alpha \] = t
\[ \therefore \] 4t2 + 4t - 3 = 0
\[ \Rightarrow \] t = \[{1 \over 2}\]
\[ \therefore \] e-\[\alpha \] = \[{1 \over 2}\]
\[ \Rightarrow \] \[\alpha \] = \[{\log _e}2\]