Question

已知函数f(x)=x2+(2+lga)x+lgb,且f(-1)=-2.若方程f(x)=2x有两个相等的实数根,求a ,b

Answer 1

f(-1)=1-2-lga+lgb=-2

lgb=lga-1

f(x)=2x

则x^2+(2+lga)x+lgb=2x

x^2+lga*x+lgb=0

有两个相等的实数根

(lga)^2-4lgb=0

令m=lga,则lgb=m-1

m^2-4(m-1)=m^2-4m+4=(m-2)^2=0

m=lga=2

a=100,

lgb=lga-1=1,b=10