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[多选题]
若函数f(x)=㏒ a (x³-ax)(a>0,a≠1),在区间(-½,0)内单调递增,则a的取值范围是()
A.[1/4,1)
B.[3/4,1)
C.(9/4,+∞)
D.(1,9/4)
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[341)
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A.[1/4,1)
B.[3/4,1)
C.(9/4,+∞)
D.(1,9/4)
[341)
函数f(x)=ax在R上是减函数,则()
A.a<1
B.0<a<1
C.|a|<1
D.0<|a|<1
证明:若函数f(x)在a连续,且f(a)<0,则有
f(x)<0.
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B.#图片3$#
C.#图片4$#
D.#图片5$#