The function xx is increasing, when
x>1e
x<1e
x<0
∀x
Explanation for the correct option
Given function is xx
Let y=xxy=elnxx∵elna=adydx=ddxelnxx=ddxexlnx=exlnxdxdx·lnx+x·ddxlnx∵ddxfgx=f'gx.g'x,ddxU·V=U·dVdx+V·dUdx=elnxxlnx+x·1x=xxlnx+1
A function is increasing if f'x>0. So,xxlnx+1>0⇒lnx+1>0∵xx>0⇒lnx>-1⇒x>1e∵lnx=y⇒x=ey
Thus, the function is increasing if x>1e.
Hence, option(A) i.e. x>1e is correct.