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Question

The value of α for which 4α21eα|x|dx=5, is :

A
loge2
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B
loge2
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C
loge(43)
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D
loge(32)
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Solution

The correct option is A loge2
4α21eα|x|dx=5
4α[01eα|x|dx+20eα|x|dx]=5

4α[01eαxdx+20eαxdx]=5

4α[(1eαα)+(e2α1α)]=5

4[1eαe2α+1]=5

Let eα=t
4t2+4t3=0
t=12=eαα=loge2

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