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

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
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

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
f(2)=4a-2(1+5a)+3=-6a+1
f(1)=a-1-5a+3=-4a+2
f(3)=9a-3-15a+3=-6a
f(0)=3
由题意：
-6a+1>-4a+2>-6a>3
分别解得：a-1,且a