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Question

Statement-1: eπ>πe
Statement-2: πeπ>eeπ

A
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1
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B
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
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C
Statement-1 is true, Statement-2 is false
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D
Statement-1 is false, Statement-2 is true
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Solution

The correct option is B Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1
eπ=(e1/e)eπ=(x1/x11)eπ (1)πe=(π1/π)eπ=(x1/x22)eπ (2)
Now, let y=x1/x [x>0]
Taking loge both the sides,
logey=1xlogex
Differentiating w.r.t x
1ydydx=1x.1xlogex×1x2
dydx=x1/xx2(1logex)
For critical point, dydx=0
x1/xx2(1logex)=01logex=0 (x1/x,x2 never be zero)
x=e
dydxx=e=+ve, dydxx=e+=ve
Since, first derivative changes its sign from +ve to ve, therefore, e is the maximum point.
e1/e is the maximum value of x1/x.
From eqn(1) and (2), we have
(e1/e)eπ>(π1/π)eπ
eπ>πe

Now, πeπ>eeπ [π>e]
Both Statement-1 and 2 are true.

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