Question

12．设函数f（x）=|2x+a|+|x-$\frac{1}{a}$|．
（Ⅰ）当a=1时，解不等式f（x）＜x+3；
（Ⅱ）当a＞0时，证明：f（x）≥$\sqrt{2}$．

Answer 1

分析 （I）当a=1时，不等式f（x）=|2x+1|+|x-1|=$\left\{\begin{array}{l}{-3x，x＜-\frac{1}{2}}\\{x+2，-\frac{1}{2}≤x≤1}\\{3x，x＞1}\end{array}\right.$．由f（x）＜x+3，可得：$\left\{\begin{array}{l}{x＜-\frac{1}{2}}\\{-3x＜x+3}\end{array}\right.$，或$\left\{\begin{array}{l}{-\frac{1}{2}≤x≤1}\\{x+2＜x+3}\end{array}\right.$，或$\left\{\begin{array}{l}{x＞1}\\{3x＜x+3}\end{array}\right.$，解出即可得出．
（II）当a＞0时，f（x）=|2x+a|+|x-$\frac{1}{a}$|=$\left\{\begin{array}{l}{3x+a-\frac{1}{a}，x＞\frac{1}{a}}\\{x+a+\frac{1}{a}，-\frac{a}{2}≤x≤\frac{1}{a}}\\{-3x-a+\frac{1}{a}，x＜-\frac{a}{2}}\end{array}\right.$．利用单调性即可证明．

解答 解：（I）当a=1时，不等式f（x）=|2x+1|+|x-1|=$\left\{\begin{array}{l}{-3x，x＜-\frac{1}{2}}\\{x+2，-\frac{1}{2}≤x≤1}\\{3x，x＞1}\end{array}\right.$．
由f（x）＜x+3，可得：$\left\{\begin{array}{l}{x＜-\frac{1}{2}}\\{-3x＜x+3}\end{array}\right.$，或$\left\{\begin{array}{l}{-\frac{1}{2}≤x≤1}\\{x+2＜x+3}\end{array}\right.$，或$\left\{\begin{array}{l}{x＞1}\\{3x＜x+3}\end{array}\right.$，
解得：$-\frac{3}{4}＜x＜-\frac{1}{2}$，或$-\frac{1}{2}≤x≤1$，或$1＜x＜\frac{3}{2}$．
∴不等式f（x）＜x+3的解集为：$（-\frac{3}{4}，\frac{3}{2}）$．
证明：（II）当a＞0时，f（x）=|2x+a|+|x-$\frac{1}{a}$|=$\left\{\begin{array}{l}{3x+a-\frac{1}{a}，x＞\frac{1}{a}}\\{x+a+\frac{1}{a}，-\frac{a}{2}≤x≤\frac{1}{a}}\\{-3x-a+\frac{1}{a}，x＜-\frac{a}{2}}\end{array}\right.$．
当x＞$\frac{1}{a}$时，f（x）＞$\frac{2}{a}$+a．
当x＜-$\frac{a}{2}$时，f（x）＞$\frac{a}{2}$+$\frac{1}{a}$．
当$-\frac{a}{2}≤x≤\frac{1}{a}$时，$\frac{a}{2}$+$\frac{1}{a}$≤f（x）≤$\frac{2}{a}$+a．
∴f（x）_min=$\frac{a}{2}$+$\frac{1}{a}$≥$2\sqrt{\frac{a}{2}×\frac{1}{a}}$=$\sqrt{2}$，当且仅当a=$\sqrt{2}$时取等号．

点评 本题考查了绝对值不等式的解法、函数的单调性，考查了分类讨论方法、推理能力与计算能力，属于中档题．