Question

19．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{a}^{x}，x＞2}\\{（3a-5）（x-2）^{2}+2，x≤2}\end{array}\right.$是R上的單調(diào)遞增函數(shù)，則實數(shù)a的取值范圍為[$\sqrt{2}$，$\frac{5}{3}$）．

Answer 1

分析 根據(jù)分段函數(shù)單調(diào)性的性質(zhì)建立不等式關(guān)系進行求解即可．

解答 解：若函數(shù)f（x）是增函數(shù)，
則滿足$\left\{\begin{array}{l}{a＞1}\\{3a-5＜0}\\{{a}^{2}≥2}\end{array}\right.$，即$\left\{\begin{array}{l}{a＞1}\\{a＜\frac{5}{3}}\\{a≥\sqrt{2}或a≤-\sqrt{2}}\end{array}\right.$，
即$\sqrt{2}$≤a＜$\frac{5}{3}$，
故答案為：[$\sqrt{2}$，$\frac{5}{3}$）．

點評 本題主要考查分段函數(shù)單調(diào)性的性質(zhì)的應(yīng)用，根據(jù)函數(shù)單調(diào)性的關(guān)系建立不等式組是解決本題的關(guān)鍵．