Question

16．已知f（x）=$\left\{\begin{array}{l}{\frac{x-3a-1}{x-2}，x＜1}\\{-{x}^{2}-2（a-1）x-\frac{1}{6}，x≥1}\end{array}\right.$是定義在（-∞，+∞）上是減函數(shù)，則a的取值范圍是（　�。�

• A． [$\frac{1}{6}$，$\frac{1}{3}$]
• B． [0，$\frac{1}{3}$]
• C． [0，$\frac{1}{3}$）
• D． [$\frac{1}{6}$，$\frac{1}{3}$）

Answer 1

分析 由題意根據(jù)函數(shù)的單調(diào)性可得$\left\{\begin{array}{l}{1-3a＞0}\\{1-a≤1}\end{array}\right.$，解得即可．

解答 解：f（x）=$\left\{\begin{array}{l}{\frac{x-3a-1}{x-2}，x＜1}\\{-{x}^{2}-2（a-1）x-\frac{1}{6}，x≥1}\end{array}\right.$=$\left\{\begin{array}{l}1+\frac{1-3a}{x-2}，x＜1\\-{x}^{2}-2（a-1）x-\frac{1}{6}，x≥1\end{array}\right.$是定義在（-∞，+∞）上是減函數(shù)，
∴$\left\{\begin{array}{l}{1-3a＞0}\\{1-a≤1}\end{array}\right.$，
解得0≤a＜$\frac{1}{3}$，
故選：C．

點(diǎn)評(píng) 本題主要考查函數(shù)的單調(diào)性的性質(zhì)，屬于基礎(chǔ)題．