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设函数f(x)=x3-4x+a(0<a<2)有三个零点x1、x2、x3,且x1<...
设函数f(x)=x3-4x+a(0<a<2)有三个零点x1、x2、x3,且x1<x2<x3,则下列结论正确的是( )
A.x1>-1
B.x2<0
C.0<x2<1
D.x3>2
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