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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 e2−54eBoth the curves are defined for x>0Both are positive when x>1 and negative when 0<x<1We know limx→0+(logx)→−∞Hence, limx→0+(logx)→−∞ .Thus y-axis is asymptote of second curve.And limx→0+exlogx[(0)×∞form]=limx→0+elogx1x(−∞∞form)=limx→0+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, thereforeexlogx=logxex⇒(e2x2−1)logx=0⇒x=1,1e(sincex>0)Therefore required area =∫11e((logx)ex−exlogx)dx=1e[(logx)22]11e−e[x24(2logx−1)]11e=e2−54e

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