Question

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

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

Hint:

Split the integral at 0 to remove the modulus, evaluate each part, and solve a quadratic in \(t=e^{-\alpha}\).

Answer 1

The Correct Option is A

Step 1: \(\int_{-1}^{2}e^{-\alpha|x|}dx=\int_{-1}^{0}e^{\alpha x}dx+\int_0^2e^{-\alpha x}dx=\dfrac{1-e^{-\alpha}}\alpha+\dfrac{1-e^{-2\alpha}}\alpha\).
Step 2: Multiply by \(4\alpha\): \(4\left(2-e^{-\alpha}-e^{-2\alpha}\right)=5\).
Step 3: Let \(t=e^{-\alpha}\): \(4t^2+4t-3=0\Rightarrow(2t-1)(2t+3)=0\Rightarrow t=\tfrac12\) (t is positive).
Step 4: \(e^{-\alpha}=\tfrac12\Rightarrow\alpha=\ln2\).
Answer: (A) \(\log_e2\)