Question

11．已知函數(shù)f（x）=$\left\{\begin{array}{l}{\frac{a+1}{x}，x＞1}\\{（-2a-1）x+1，x≤1}\end{array}\right.$是R上的單調(diào)遞減函數(shù)，則實(shí)數(shù)a的取值范圍是（-$\frac{1}{2}$，$-\frac{1}{3}$]．

Answer 1

分析 若函數(shù)f（x）=$\left\{\begin{array}{l}{\frac{a+1}{x}，x＞1}\\{（-2a-1）x+1，x≤1}\end{array}\right.$是R上的單調(diào)遞減函數(shù)，則$\left\{\begin{array}{l}a+1＞0\\-2a-1＜0\\ a+1≤-2a-1+1\end{array}\right.$，解得答案．

解答 解：函數(shù)f（x）=$\left\{\begin{array}{l}{\frac{a+1}{x}，x＞1}\\{（-2a-1）x+1，x≤1}\end{array}\right.$是R上的單調(diào)遞減函數(shù)，
∴$\left\{\begin{array}{l}a+1＞0\\-2a-1＜0\\ a+1≤-2a-1+1\end{array}\right.$，
解得：a∈（-$\frac{1}{2}$，$-\frac{1}{3}$]，
故答案為：（-$\frac{1}{2}$，$-\frac{1}{3}$]．

點(diǎn)評 本題考查的知識點(diǎn)是分段函數(shù)的單調(diào)性，正確理解分段函數(shù)單調(diào)性的意義是解答的關(guān)鍵．