Question

已知函數(shù)f(x)=ax³+x²-a²x(a＞0)，存在實(shí)數(shù)x₁,x₂滿足下列條件：①x₁＜x₂;②f′(x₁)?=f′(x₂)=0;③|x₁|+|x₂|=2.

(1)證明0＜a≤3;

(2)求b的取值范圍；

(3)若函數(shù)h(x)=f′(x)-6a(x-x₁),證明當(dāng)x₁＜x＜2時(shí)，|h(x)|≤12a.

Answer 1

解：設(shè)f′(x)=3ax²+2x-a²=3a(x-x₁)(x-x₂),?

∴x₁+x₂=-,x₁x₂=-.?

由a＞0，得x₁＜0＜x₂.?

∵|x₁|+|x₂|=2,?

∴x₂-x₁=2.?

(1)證明:∵-x₁和x₂是方程T²-2T+=0的兩個(gè)實(shí)根，?

∵方程有解，∴Δ=4-≥0,?

得0＜a≤3.                                                                                                              ?

(2)由(x₁+x₂)²-4x₁x₂=4,?

得+=4.?

∴b=3a²(3-a)=-3a³+9a².?

∴b′=-9a²+18a.?

由b′=0,得a=0或a=2,?

∵0＜a≤3,?

∴0＜a≤2時(shí)，b′≥0,b在(0,2］上單調(diào)遞增；?

2≤a≤3時(shí)，b′≤0,b在［2,3］上單調(diào)遞減.?

∴a=2時(shí)，b取最大值12,?

a=3時(shí)，b=0,?

a=0時(shí),b=0.                                                                                                             ?

∴0≤b≤12.?

(也可用下面方法?

∴b=3a²(3-a)≤12()³=12.?

∴0≤b≤12)?

(3)證明:h(x)=3a(x-x₁)(x-x₂)-6a(x-x₁)?

=3a(x-x₁)［(x-x₂)-2］.?

其圖象是開(kāi)口向上的拋物線.?

∵x₁＜x＜2且x₁＜x₂,x₂-x₁=2,?

對(duì)稱軸為x===x₂.?

由x₁＜x＜2,?

若x₂≤2,則0=h(x₁)＞h(x)≥h(x)=3a(x₂-x₁)(-2)=-12a,?

∴|h(x)|≤12a.?

若2＜x₂,則0=h(x₁)＞h(x)＞h(2)＞h(x₂)=-12a,??

∴|h(x)|＜12a.?

由0＜a≤3,綜上,得|h(x)|≤12a.