已知函数f(x)=ax-1-2(a>0且f(x)=ax-1-2)的反函数y=f-1(x)定义域为集合a≠1,集合B={x||x−t|≤1/2,x∈R}.若A∩B=φ,求实数t的取值范围.
已知函数f(x)=ax-1-2(a>0且f(x)=ax-1-2)的反函数y=f-1(x)定义域为集合a≠1,集合B={x||x−t|≤
,x∈R}.若A∩B=φ,求实数t的取值范围.
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数学人气:245 ℃时间:2019-10-19 21:49:47
优质解答
由题意得,函数f(x)=ax-1-2(a>0且a≠1)值域为(-2,+∞)所以,y=f-1(x)的定义域为A=(-2,+∞)(6分)又由B={x||x−t|≤12,x∈R}得t−12≤x≤t+12(8分)∵A∩B=ϕ,∴t+12≤−2,即 t≤−52(11分)所...
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