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.1x−logex×1x2
⇒dydx=x1/xx2(1−logex)
For critical point, dydx=0
x1/xx2(1−logex)=0⇒1−logex=0 (x1/x,x2 never be zero)
⇒x=e
dydx∣∣x=e−=+ve, dydx∣∣x=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.