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,證明:2|a+b|<|4+ab|.

Answer 1

(1) M=(-2,2)    (2)見解析

(1)f(x)=|x+1|+|x-1|=
當(dāng)x<-1時,由-2x<4,得-2<x<-1.
當(dāng)-1≤x≤1時,f(x)=2<4;
當(dāng)x>1時,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|.