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

解:(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

于是a=3.

12.已知函數(shù)f(x)=|x+a|.

(1)當(dāng)a=-1時(shí),求不等式f(x)≥|x+1|+1的解集;

(2)若不等式f(x)+f(-x)<2存在實(shí)數(shù)解,求實(shí)數(shù)a的取值范圍.

解:(1)當(dāng)a=-1時(shí),

f(x)≥|x+1|+1可化為|x-1|-|x+1|≥1,

化簡得或或解得x≤-1,或-1<x≤-,

即所求解集為{x︱x≤-}.

(2)令g(x)=f(x)+f(-x),

則g(x)=|x+a|+|x-a|≥2|a|,所以2>2|a|,即-1<a<1.所以實(shí)數(shù)a的取值范圍是(-1,1).