已知f(x)=x2+(lga+2)x+lgb,f(-1)=-2,当x∈R时f(x)≥2x恒成立,求实数a的值,并求此时f(x)的最小值?
已知f(x)=x2+(lga+2)x+lgb,f(-1)=-2,当x∈R时f(x)≥2x恒成立,求实数a的值,并求此时f(x)的最小值?
数学人气:756 ℃时间:2019-08-20 13:52:10
优质解答
由f(-1)=-2,得:f(-1)=1-(lga+2)+lgb=-2,解之得:lga-lgb=1,∴ab=10,a=10b.又由x∈R,f(x)≥2x恒成立.知:x2+(lga+2)x+lgb≥2x,即x2+xlga+lgb≥0,对x∈R恒成立,由△=lg2a-4lgb≤0,故得(1+lgb)...
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