题目

已知函数f(x)=kx^2+(k-1)x,k为常数
若k>0,且对任意x∈[1,正无穷),总有g(x)=f(x)+1/x>=1成立,求k的取值范围
详解谢谢

提问时间：2021-12-13

答案

f(x)=kx^2+(k-1)x
h(x)=1/x g(1)=f(1)+h(1)=2k-1+1>=1,k>=1/2f'(x)=2kx+k-1>=x-1/2>=1/2,f(x)递增|f'(x)|=2kx+k-1>=1/2 并递增h'(x)=-1/x^2=|h'(1)|时可满足题意k>=1/2,3k-1>=1k>=3/2或者(2)当|f'(x)|=|h'(x)|时,g(x)=1可满足题意2kx+k-1=1/x^2,kx^2+(k-1)x+1/x=12kx^3+(k-1)x^2-1=0,kx^3+(k-1)x^2-x+1=0k(2x^3+x^2)-x^2-1=0,k(x^3+x^2)-x^2-x+1=0k=(x^2+1)/(2x^3+x^2)=(x^2+x-1)/(x^3+x^2)(x^2+1)(x^3+x^2)=(x^2+x-1)(2x^3+x^2)x^3+2x^2-2x-2=0x只有一个大于的解1.1701,另外2个解应该小于0此时k=(x^2+1)/(2x^3+x^2)=0.1856=3/2