问题
解答题
记f(x)=lg(3-|x-1|)的定义域为A,集合B={x|x2-(a+5)x+5a<0}.
(1)当a=1时,求A∩B;
(2)若A∩B=A,求a的取值范围.
答案
(1)当a=1时,A=(-2,4),B=(1,5),∴A∩B=(1,4)
(2)A=(-2,4),B={x|(x-5)(x-a)<0}.
若A∩B=A,则A⊆B
∴a≤-2
记f(x)=lg(3-|x-1|)的定义域为A,集合B={x|x2-(a+5)x+5a<0}.
(1)当a=1时,求A∩B;
(2)若A∩B=A,求a的取值范围.
(1)当a=1时,A=(-2,4),B=(1,5),∴A∩B=(1,4)
(2)A=(-2,4),B={x|(x-5)(x-a)<0}.
若A∩B=A,则A⊆B
∴a≤-2
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