Question

已知函數(shù)f(x)=x|x+m|+n,其中m,n∈R.

(1)求證:f(x)是奇函數(shù)的充要條件是m²+n²=0;

(2)若常數(shù)n=-4且f(x)＜0對(duì)任意x∈［0,1］恒成立,求m的取值范圍.

Answer 1

解:(1)證明:充分性:若m²+n²=0,則m=n=0,

∴f(x)=x|x|.又有f(-x)=-x|-x|=-x|x|=-f(x),∴f(x)為奇函數(shù).

必要性:若f(x)為奇函數(shù),∵x∈R,

∴f(0)=0,即n=0,∴f(x)=x|x+m|.

由f(1)=-f(-1),有|m+1|=|m-1|,∴m=0.

∴f(x)為奇函數(shù),則m=n=0,即m²+n²=0.

∴f(x)為奇函數(shù)的充要條件是m²+n²=0.

(2)當(dāng)x=0時(shí),m∈R,f(x)＜0恒成立;

當(dāng)x∈(0,1］時(shí),原不等式可變形為|x+m|＜,即-x+＜m＜-x.

當(dāng)n=-4時(shí),∴只需對(duì)x∈(0,1］,滿(mǎn)足①②

對(duì)①式f₁(x)=-x+在(0,1］上單調(diào)遞減,∴m＜f₁(1)=3.③

對(duì)②式,設(shè)f₂(x)=-x,根據(jù)單調(diào)函數(shù)的定義可證明f₂(x)在(0,1］上單調(diào)遞增,

∴f₂(x)_max=f(1).∴m＞f₂(1)=-5.④

由③④知-5＜m＜3.