Question

已知函数f(x)=(4x方+k*2x方+1)/4x方+2x方+1

Answer 1

由f(x1)+f(x2)>f(x3),记f(x)最小值为m,最大值为M

则有2m>M

令t=2^x>0,则f(x)=(t^2+kt+1)/(t^2+t+1)=1+t(k-1)/(t^2+t+1)=1+(k-1)/(t+1/t+1)

因t>0,t+1/t>=2,当t=1时取等号

故当k>1时,f(x)有最大值M=1+(k-1)/3,最小值m=1(f(x)>1),故有2>=1+(k-1)/3,得:1