Question

已知函数f(x)=kx,g(X)=(lnx)/x 若不等式f(x)≥g(X)在区间(0,+∞)上恒成立,求K的取值范围

Answer 1

f(x)-g(x)=kx-(lnx)/x =(kx²-lnx)/x ‎(x>0),

令h(x)=kx²-lnx,

当k0)

当x>√[1/(2k)]时2kx²-1>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)]