The correct option is B e
f(x)=x(lnx−2)
⇒f′(x)=x(1x)+(lnx−2)=lnx−1
⇒f′(x)=lnx−1
f′(x)=0⇒x=e
For x∈[1,e), f′(x)<0 and for x∈(e,e2], f′(x)>0
So, the minimum value of f(x) occurs at x=e
Now, f(1)=−2
⇒f(e)=−e (least)
⇒f(e2)=0 (greatest)
Hence, |f(e2)−f(e)|=|0−(−e)|=e