Question

13．已知函數(shù)f（x）=|x+a|．
（1）若a=2，解關(guān)于x的不等式f（x）+f（x-3）≥5；
（2）若關(guān)于x的不等式f（x）-f（x+2）+4≥|1-3m|恒成立，求實(shí)數(shù)m的取值范圍．

Answer 1

分析 （1）不等式f（x）+f（x-3）≥5可化為；|x+2|+|x-1|≥5．⇒$\left\{\begin{array}{l}{x＜-2}\\{-2x-1≥5}\end{array}\right.$或$\left\{\begin{array}{l}{-2≤x≤1}\\{3≥5}\end{array}\right.$或$\left\{\begin{array}{l}{x＞1}\\{2x+1≥5}\end{array}\right.$，即可求解；
（2）關(guān)于x的不等式f（x）-f（x+2）+4≥|1-3m|恒成立?|x+a|-|x+2+a|+4≥|1-3m|恒成立．即2+4≥|1-3m|⇒$\left\{\begin{array}{l}{1-3m≥-2}\\{1-3m≤2}\end{array}\right.$⇒-$\frac{1}{3}$≤m≤1．

解答 解：（1）當(dāng)a=2，關(guān)于x的不等式f（x）+f（x-3）≥5可化為；|x+2|+|x-1|≥5．
⇒$\left\{\begin{array}{l}{x＜-2}\\{-2x-1≥5}\end{array}\right.$或$\left\{\begin{array}{l}{-2≤x≤1}\\{3≥5}\end{array}\right.$或$\left\{\begin{array}{l}{x＞1}\\{2x+1≥5}\end{array}\right.$，
⇒x≤-3或∅或x≥2
故不等式的解集為[2，+∞）∪（-∞，-3]．
（2）關(guān)于x的不等式f（x）-f（x+2）+4≥|1-3m|恒成立?|x+a|-|x+2+a|+4≥|1-3m|恒成立．
因?yàn)?2≤|x+a|-|x+2+a|≤2，
∴-2+4≥|1-3m|⇒$\left\{\begin{array}{l}{1-3m≥-2}\\{1-3m≤2}\end{array}\right.$⇒-$\frac{1}{3}$≤m≤1．
∴實(shí)數(shù)m的取值范圍為[-$\frac{1}{3}$，1]

點(diǎn)評(píng) 本題考查了絕對(duì)值不等式的解法，考查了含參不等式的恒成立問題的處理方法，屬于中檔題，