Question

11．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{a}^{x}，x＜0}\\{（a-3）x+4a，x≥0}\end{array}\right.$滿足對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，則函數(shù)f（x）是單調(diào)減函數(shù)，a的取值范圍是0＜a≤$\frac{1}{4}$．

Answer 1

分析 若對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，則函數(shù)f（x）是單調(diào)減函數(shù)；故$\left\{\begin{array}{l}0＜a＜1\\ a-3＜0\\ 1≥4a\end{array}\right.$，解得a的取值范圍．

解答 解：若對(duì)任意x₁≠x₂，都有$\frac{f（{x}_{1}）-f（{x}_{2}）}{{x}_{1}-{x}_{2}}$＜0成立，
則函數(shù)f（x）是單調(diào)減函數(shù)；
故$\left\{\begin{array}{l}0＜a＜1\\ a-3＜0\\ 1≥4a\end{array}\right.$，
解得：0＜a≤$\frac{1}{4}$
故答案為：減，0＜a≤$\frac{1}{4}$

點(diǎn)評(píng) 本題考查的知識(shí)點(diǎn)是分段函數(shù)的應(yīng)用，正確理解分段函數(shù)的單調(diào)性是解答的關(guān)鍵．