4α∫−10eαxdx+∫02e−αxdx=5
⇒4α(αeαx)−10+−αe−αx∣02=5
⇒4α(α1−e−α)−(αe−2α−1)=5
⇒4(2−e−α−e−2α)=5
Put e−α=t
⇒4t2+4t−3=0⇒(2t+3)(2t−1)=0
⇒e−α=21⇒α=ln2
The value of α for which 4α∫−12e−α∣x∣dx=5 , is
Held on 7 Jan 2020 · Verified 6 Jul 2026.
loge2
loge(23)
loge2
loge(34)
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