题目

已知f=x^2+x+lgb,f=-2且f大于等于2x恒成立,求a和b的值.
各位请给个完整的步骤,不然不行,

提问时间：2021-01-12

答案

f(x)=x^2+(2+lga)x+lgb
f(-1)=(-1)^2+(2+lga)*(-1)+lgb=-2
1-2-lga+lgb=-2
lga-lgb=lga/b=1
a/b=10
a=10b
f(x)=x^2+(2+lga)x+lgb>=2x
x^2+lgax+lgb>=0
△=(lga)^2-4lgb<=0
[lg(10b)]^2-4lgb<=0
(lg10+lgb)^2-4lgb<=0
1+2lgb+(lgb)^2-4lgb<=0
1-2lgb+(lgb)^2<=0
(1-lgb)^2<=0
lgb=1
b=10
a=10b=100