Question

已知f(x)=|x+1|+|x-1|,不等式f(x)<4的解集為M.

(1)求M.

(2)當(dāng)a,b∈M時(shí),證明:2|a+b|<|4+ab|.

 

Answer 1

(1) M=(-2,2) (2)見(jiàn)解析

【解析】(1)f(x)=|x+1|+|x-1|=

當(dāng)x<-1時(shí),由-2x<4,得-2<x<-1.

當(dāng)-1≤x≤1時(shí),f(x)=2<4;

當(dāng)x>1時(shí),由2x<4,得1<x<2.

所以M=(-2,2).

(2)a,b∈M,即-2<a<2,-2<b<2,

∴4(a+b)2-(4+ab)2

=4(a2+2ab+b2)-(16+8ab+a2b2)

=(a2-4)(4-b2)<0.

∴4(a+b)2<(4+ab)2.

∴2|a+b|<|4+ab|.