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The value of  α for which 4α12eαxdx=5 , is: 

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b
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c
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d
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detailed solution

Correct option is D

4α∫−10eαxdx+∫02e−αxdx=5⇒4αeαxα−10+e−αx−α=5⇒4α1-e−αα−e−2α−1α=5⇒42−e−α−e−2α=5put e−α=t⇒4t2+4t−3=0⇒2t+32t−1=0⇒e−α=12⇒α=ln2


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