Question

設(shè)x₁、x₂是函數(shù)f(x)=x³+x²-a²x(其中a＞0)的兩個(gè)極值點(diǎn),且|x₁|+|x₂|=2.

(1)求a的取值范圍;

(2)設(shè)g(a)=e^ka(k∈R),求g(a)的單調(diào)區(qū)間;

(3)若函數(shù)h(x)=f′(x)-2a(x-x₁),求證:當(dāng)x₁＜x＜2且x₁＜0時(shí),有|h(x)|≤4a.

Answer 1

解:(1)f′(x)=ax²+bx-a²,∵x₁,x₂是f(x)的兩個(gè)極值點(diǎn),∴x₁、x₂是方程f′(x)=0的兩個(gè)實(shí)數(shù)根.

∵a＞0,x₁·x₂=-a＜0,x₁+x₂=,

∴|x₁|+|x₂|=|x₁-x₂|=.∵|x₁|+|x₂|=2,∴+4a=4,即b²=4a²-4a³.

由b²≥0得4a²-4a³≥0,∴0＜a≤1.

(2)由(1)知b²=4a²-4a³,∴g(a)=(1-a)e^ka,0＜a≤1,g′(a)=e^ka(k-1-ak),

k=0時(shí),g′(a)=-1,∴g(a)=(1-a)e^ka在(0,1］上為單調(diào)遞減的,即單調(diào)遞減區(qū)間為(0,1］;

當(dāng)k≠0時(shí),令g′(a)=e^ka(k-1-ak)=0,得a=1;

當(dāng)k＜0時(shí),a=1＞1,g′(a)在(0,1］上恒小于零,

∴g(a)=(1-a)e^ka在(0,1］上單調(diào)遞減,即單調(diào)遞減區(qū)間為(0,1］;當(dāng)k＞1時(shí),a=1在(0,1］之間,在(0,1)上g′(a)恒大于零,在(1,1］上g′(a)恒小于零,∴單調(diào)遞減區(qū)間為(1,1］,單調(diào)遞增區(qū)間為(0,1);

當(dāng)0＜k≤1時(shí),a=1＜0,在(0,1］上g′(a)恒小于零,∴單調(diào)遞減區(qū)間為(0,1］.

(3)∵x₁、x₂是方程f′(x)=0的兩根,

∴f′(x)=a(x-x₁)(x-x₂).∴h(x)=a(x-x₁)(x-x₂)-2a(x-x₁)=a(x-x₁)(x-x₂-2).

∴|h(x)|=a|x-x₁||x-x₂-2|≤a()².

∵x＞x₁,∴|x-x₁|=x-x₁.又x₁＜0,x₁x₂＜0,∴x₂＞0.∵x＜2,故x-x₂-2＜0,∴|x-x₁|+|x-x₂-2|=x-x₁+x₂+2-x=x₂-x₁+2=|x₁|+|x₂|+2=4.

∴|h(x)|≤4a.