Question

已知函数f(x)=kx,g(x)=(lnx)/x.若不等式f(x)>=g(x)在区间（0,正无穷大）恒成立正,求K取值范

Answer 1

f(x)-g(x)=kx-(lnx)/x =(kx?-lnx)/x ?(x>0),

令h(x)=kx?-lnx,

当k0)

当x>√[1/(2k)]时2kx?-1>0,

当x0)恒成立

即h(x)>=0恒成立,则须h(x)最小值)(1/2){1-ln[1/(2k)]}>=0,即1-ln[1/(2k)]>=0,ln[1/(2k)]