Let A denotes the area bounded by y=(lnx)2 and lines y = 0 and x=e2 then -
A
A<4e2−3e−1
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B
A<4(e2−1)
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C
A>5(e2−1)
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D
A>52e(e−1)
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Solution
The correct options are AA<4e2−3e−1 BA<4(e2−1) DA>52e(e−1) y=(lnx)2 dydx=2lnxx,y′′=2[1−lnxx2]=0 x=e→ Point of inflection So Area<Ar(ABFG)+Ar(BCDE) A<(e−1)×1+(e2−e)×4 A<4e2−3e−1 (Option A) A<Ar(ACDJ) A<(e2−1)×4 A<4e2−4 A>Ar(trapeziumBCDF) A>12(1+4)(e2−e) A>52e(e−1)