Question

12．設(shè)函數(shù)f（x）=|2x+a|+|x-$\frac{1}{a}$|．
（Ⅰ）當(dāng)a=1時(shí)，解不等式f（x）＜x+3；
（Ⅱ）當(dāng)a＞0時(shí)，證明：f（x）≥$\sqrt{2}$．

Answer 1

分析 （I）當(dāng)a=1時(shí)，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）當(dāng)a＞0時(shí)，f（x）=|2x+a|+|x-$\frac{1}{a}$|=$\left\{\begin{array}{l}{3x+a-\frac{1}{a}，x＞\frac{1}{a}}\\{x+a+\frac{1}{a}，-\frac{a}{2}≤x≤\frac{1}{a}}\\{-3x-a+\frac{1}{a}，x＜-\frac{a}{2}}\end{array}\right.$．利用單調(diào)性即可證明．

解答 解：（I）當(dāng)a=1時(shí)，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的解集為：$（-\frac{3}{4}，\frac{3}{2}）$．
證明：（II）當(dāng)a＞0時(shí)，f（x）=|2x+a|+|x-$\frac{1}{a}$|=$\left\{\begin{array}{l}{3x+a-\frac{1}{a}，x＞\frac{1}{a}}\\{x+a+\frac{1}{a}，-\frac{a}{2}≤x≤\frac{1}{a}}\\{-3x-a+\frac{1}{a}，x＜-\frac{a}{2}}\end{array}\right.$．
當(dāng)x＞$\frac{1}{a}$時(shí)，f（x）＞$\frac{2}{a}$+a．
當(dāng)x＜-$\frac{a}{2}$時(shí)，f（x）＞$\frac{a}{2}$+$\frac{1}{a}$．
當(dāng)$-\frac{a}{2}≤x≤\frac{1}{a}$時(shí)，$\frac{a}{2}$+$\frac{1}{a}$≤f（x）≤$\frac{2}{a}$+a．
∴f（x）_min=$\frac{a}{2}$+$\frac{1}{a}$≥$2\sqrt{\frac{a}{2}×\frac{1}{a}}$=$\sqrt{2}$，當(dāng)且僅當(dāng)a=$\sqrt{2}$時(shí)取等號(hào)．

點(diǎn)評(píng) 本題考查了絕對(duì)值不等式的解法、函數(shù)的單調(diào)性，考查了分類討論方法、推理能力與計(jì)算能力，屬于中檔題．