Question

20．若f（x）=$\left\{\begin{array}{l}{{x}^{2}-4ax+2（x＜1）}\\{lo{g}_{a}x（x≥1）}\end{array}\right.$，滿足對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}{-x}_{2}}$＜0成立，則實(shí)數(shù)a的取值范圍是[$\frac{1}{2}$，$\frac{3}{4}$]．

Answer 1

分析 判斷得出f（x）在R上單調(diào)遞減，根據(jù)函數(shù)性質(zhì)得出不等式組$\left\{\begin{array}{l}{0＜a＜1}\\{2a≥1}\\{3-4a≥0}\end{array}\right.$求解即可．

解答 解：∵滿足對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}{-x}_{2}}$＜0成立，
∴f（x）在R上單調(diào)遞減，
∴$\left\{\begin{array}{l}{0＜a＜1}\\{2a≥1}\\{3-4a≥0}\end{array}\right.$
即$\frac{1}{2}≤a≤\frac{3}{4}$
故答案為：[$\frac{1}{2}$，$\frac{3}{4}$]

點(diǎn)評(píng) 本題綜合考查了函數(shù)單調(diào)性的定義，性質(zhì)，結(jié)合二次函數(shù)，對(duì)數(shù)函數(shù)性質(zhì)，運(yùn)用不等式求解即可，難度不大，屬于中檔題．