题目内容
已知函数f(x)=2x﹣1,对于满足0<x1<x2的任意x1,x2,给出下列结论:
(1)(x2﹣x1)[f(x2)﹣f(x1)]<0
(2)x2f(x1)<x1f(x2)
(3)f(x2)﹣f(x1)>x2﹣x1
(4)
>f(
)其中正确结论的序号是
(1)(x2﹣x1)[f(x2)﹣f(x1)]<0
(2)x2f(x1)<x1f(x2)
(3)f(x2)﹣f(x1)>x2﹣x1
(4)
[ ]
A.(1)(2)
B.(1)(3)
C.(2)(4)
D.(3)(4)
B.(1)(3)
C.(2)(4)
D.(3)(4)
C
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