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Question

Find the area enclosed the curves : y=exlogx and y=logxex where loge=1

A
e254e
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B
e2+54e
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C
e232e
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D
e2+32e
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Solution

The correct option is A e254e
Both the curves are defined for x>0
Both are positive when x>1 and negative when 0<x<1
We know limx0+(logx)
Hence, limx0+(logx) .Thus y-axis is asymptote of second curve.
And limx0+exlogx[(0)×form]
=limx0+elogx1x(form)
=limx0+e(1x)(1x2)=0 (using L'hopital rule)
Thus, the first curve starts from (0,0) but does not include (0,0)
Now, the given curves intersect, therefore
exlogx=logxex(e2x21)logx=0
x=1,1e(sincex>0)
Therefore required area
=11e((logx)exexlogx)dx=1e[(logx)22]11ee[x24(2logx1)]11e
=e254e

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