Question

11．函數(shù)y=a|2x-b|+2在[$\frac{1}{2}$，+∞）上單調(diào)遞增，則a，b應(yīng)滿足的條件為a＞0，b≤1．

Answer 1

分析 先去絕對(duì)值把函數(shù)變成：y=$\left\{\begin{array}{l}2ax-ab+2，x≥\frac{2}\\-2ax+ab+2，x＜\frac{2}\end{array}\right.$，當(dāng)a＞0，該函數(shù)在[$\frac{1}{2}$，+∞）上單調(diào)遞增，所以該函數(shù)在[$\frac{2}$，+∞）上應(yīng)單調(diào)遞增，所以便得到a＞0，b≤1．

解答 解：函數(shù)y=a|2x-b|+2=$\left\{\begin{array}{l}2ax-ab+2，x≥\frac{2}\\-2ax+ab+2，x＜\frac{2}\end{array}\right.$，
當(dāng)a＞0，該函數(shù)在[$\frac{2}$，+∞）上單調(diào)遞增，
又由函數(shù)y=a|2x-b|+2在[$\frac{1}{2}$，+∞）上單調(diào)遞增，
∴a＞0，且$\frac{2}$≤$\frac{1}{2}$，
即a＞0，b≤1，
故答案為：a＞0，b≤1

點(diǎn)評(píng) 本題考查分段函數(shù)的應(yīng)用，含絕對(duì)值函數(shù)的單調(diào)性及處理辦法：去絕對(duì)值號(hào)，難度中檔．