The value of $\alpha$ for which $4 \alpha \int_{-1}^{2} e^{-\alpha|x|} d x=5$, is

(A) $\log _{e}\left(\frac{3}{2}\right)$

(B) $\log _{e}\left(\frac{4}{3}\right)$

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Definite Integrals

The value of

α

for which

4 α \int_{- 1}^{2} e^{- α ∣ x ∣} d x = 5

, is

Answer: (D)

Split the integral at

x = 0

to handle the absolute value:

\int_{- 1}^{2} e^{- α ∣ x ∣} d x = \frac{1}{α} (1 - e^{- α}) + \frac{1}{α} (1 - e^{- 2 α})

. Then

4 α

times this equals

4 (2 - e^{- α} - e^{- 2 α}) = 5

. Letting

t = e^{- α}

turns it into

4 t^{2} + 4 t - 3 = 0

, i.e.

(2 t - 1) (2 t + 3) = 0

. Only

t = \frac{1}{2}

is admissible, so

e^{- α} = \frac{1}{2}

and

α = ln 2

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option

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