Question

19．若函數(shù)f（x）=x²+a|x-$\frac{1}{2}$|在[0，+∞）上單調(diào)遞增，則實(shí)數(shù)a的取值范圍是（　�。�

• A． [-2，0]
• B． [-4，0]
• C． [-1，0]
• D． [-$\frac{1}{2}$，0]

Answer 1

分析 去絕對(duì)值，由已知條件知，函數(shù)x²+ax-a在[1，+∞）單調(diào)遞增，x²-ax+a在[0，1）單調(diào)遞增，得到關(guān)于a的不等式組，解該不等式組即得a的取值范圍．

解答 解：f（x）=x²+a|x-$\frac{1}{2}$|=$\left\{\begin{array}{l}{{x}^{2}+ax-\frac{1}{2}a，x≥\frac{1}{2}}\\{{x}^{2}-ax+\frac{1}{2}a，x＜1}\end{array}\right.$，
要使f（x）在[0，+∞）上單調(diào)遞增，則：
$\left\{\begin{array}{l}{-\frac{a}{2}≤\frac{1}{2}}\\{\frac{a}{2}≤0}\end{array}\right.$，得-1≤a≤0，
∴實(shí)數(shù)a的取值范圍是[-1，0]，
故選：C．

點(diǎn)評(píng) 考查含絕對(duì)值函數(shù)的單調(diào)性，二次函數(shù)的單調(diào)性及單調(diào)區(qū)間．