Question

已知函數(shù)f(x)=|x－a|,其中a>1.
(1)當(dāng)a=2時(shí),求不等式f(x)≥4－|x－4|的解集;
(2)已知關(guān)于x的不等式|f(2x+a)－2f(x)|≤2的解集為{x|1≤x≤2},求a的值.

Answer 1

(1){x|x≤1或x≥5}.
(2)3

(1)當(dāng)a=2時(shí), f(x)+|x－4|=
當(dāng)x≤2時(shí),由f(x)≥4－|x－4|得－2x+6≥4,解得x≤1;
當(dāng)2<x<4時(shí), f(x)≥4－|x－4|無(wú)解;
當(dāng)x≥4時(shí),由f(x)≥4－|x－4|得2x－6≥4,解得x≥5;
所以f(x)≥4－|x－4|的解集為{x|x≤1或x≥5}.
(2)記h(x)=f(2x+a)－2f(x),
則h(x)=
由|h(x)|≤2,解得≤x≤
又已知|h(x)|≤2的解集為{x|1≤x≤2}.
所以=1且=2
于是a=3.