Question

5．函數(shù)f（x）=$\left\{\begin{array}{l}{a{x}^{2}+\frac{1}{5}（x≥0）}\\{（\frac{1}{2}-a）x+{a}^{2}（x＜0）}\end{array}\right.$是增函數(shù)，則實數(shù)a的取值范圍是（0，$\frac{\sqrt{5}}{5}$）．

Answer 1

分析 根據(jù)分段函數(shù)的單調(diào)性的性質(zhì)進(jìn)行求解即可．

解答 解：若函數(shù)在R上是增函數(shù)，
則在x≥0和x＜0上分別遞增，
且滿足$\left\{\begin{array}{l}{a＞0}\\{\frac{1}{2}-a＞0}\\{{a}^{2}≤\frac{1}{5}}\end{array}\right.$，
即$\left\{\begin{array}{l}{a＞0}\\{a＜\frac{1}{2}}\\{-\frac{\sqrt{5}}{5}＜a＜\frac{\sqrt{5}}{5}}\end{array}\right.$．解得0＜a＜$\frac{\sqrt{5}}{5}$，
故答案為：（0，$\frac{\sqrt{5}}{5}$）．

點評 本題主要考查函數(shù)單調(diào)性的性質(zhì)，利用分段函數(shù)的單調(diào)性的性質(zhì)是解決本題的關(guān)鍵．注意在端點處，函數(shù)值的大小關(guān)系．