题目内容
11.已知函数f(x)=$\left\{\begin{array}{l}{\frac{a+1}{x},x>1}\\{(-2a-1)x+1,x≤1}\end{array}\right.$是R上的单调递减函数,则实数a的取值范围是(-$\frac{1}{2}$,$-\frac{1}{3}$].分析 若函数f(x)=$\left\{\begin{array}{l}{\frac{a+1}{x},x>1}\\{(-2a-1)x+1,x≤1}\end{array}\right.$是R上的单调递减函数,则$\left\{\begin{array}{l}a+1>0\\-2a-1<0\\ a+1≤-2a-1+1\end{array}\right.$,解得答案.
解答 解:函数f(x)=$\left\{\begin{array}{l}{\frac{a+1}{x},x>1}\\{(-2a-1)x+1,x≤1}\end{array}\right.$是R上的单调递减函数,
∴$\left\{\begin{array}{l}a+1>0\\-2a-1<0\\ a+1≤-2a-1+1\end{array}\right.$,
解得:a∈(-$\frac{1}{2}$,$-\frac{1}{3}$],
故答案为:(-$\frac{1}{2}$,$-\frac{1}{3}$].
点评 本题考查的知识点是分段函数的单调性,正确理解分段函数单调性的意义是解答的关键.
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