已知函数f(x)=ax2-(1+5a)x+3满足f(2)>f(1)>f(3)>f(0),则实数a的取值范围为_.
问题描述:
已知函数f(x)=ax2-(1+5a)x+3满足f(2)>f(1)>f(3)>f(0),则实数a的取值范围为______.
答
∵函数f(x)=ax2-(1+5a)x+3满足f(2)>f(1)>f(3)>f(0),
∴4a-2-10a+3>a-1-5a+3>9a-3-15a+3>3,
∴-1<a<-
.1 2
故答案为:-1<a<-
.1 2