Question

8．已知函數(shù)f（x）=$\left\{\begin{array}{l}{ax+2，x＜1}\\{{x}^{2}-ax+2，x≥1}\end{array}\right.$在R上單調(diào)遞增，則實數(shù)a的取值范圍為（0，$\frac{1}{2}$]．

Answer 1

分析 若f（x）=$\left\{\begin{array}{l}{ax+2，x＜1}\\{{x}^{2}-ax+2，x≥1}\end{array}\right.$在R上單調(diào)遞增，則每段函數(shù)均為增函數(shù)，且當x=1時，前一段函數(shù)的函數(shù)值不大于后一段函數(shù)的函數(shù)值，由此可構(gòu)造滿足條件的不等式組，解出實數(shù)a的取值范圍．

解答 解：∵函數(shù)f（x）=$\left\{\begin{array}{l}{ax+2，x＜1}\\{{x}^{2}-ax+2，x≥1}\end{array}\right.$在R上單調(diào)遞增，
則$\left\{\begin{array}{l}a＞0\\ \frac{a}{2}≤1\\ a+2≤3-a\end{array}\right.$，
解得：a∈（0，$\frac{1}{2}$]，
故答案為：（0，$\frac{1}{2}$]

點評 本題考查的知識點是函數(shù)單調(diào)性的性質(zhì)，熟練掌握分段函數(shù)的單調(diào)性是解答的關鍵．