Question

设函数F(x)=ax³+bx²+cx(a＜b＜c),其图象在x=1, x=M处的切线的斜率分别为0,-a,

(1)求证:0≤＜1;

(2)若函数F(x)的递增区间为［s,t］,求|s-t|的取值范围;?

(3)若当x≥k时,(k是与a,b,c无关的常数),恒有F′(x)+a＜0,试求k的最小值.

Answer 1

(1)证明:f′(x)=ax²+bx+c,由题意,得f′(1)=a+b+c=0,                                      ①?

f′(M)=aM²+bM+c=-a,                                                                                 ②?

又a＜b＜c,得3a＜a+b+c＜3c.∴a＜0,c＞0.?

由①得c=-a-b,代入a＜b＜c,得a＜b＜-a-b.?

由a＜0得-＜＜1.                                                                                            ?

将c=-a-b代入②得,aM²+bM-b=0.                                                                     ③?

由③有实根,得Δ=b²+4AB≥0,?

即()²+4≥0.?

解得≤-4或≥0.                                                                                                 ?

综上,0≤＜1.                                                                                                        ?

(2)解:由f′(x)=ax²+bx+c的判别式Δ=b²-4AC＞0得f′(x)=ax²+bx+c=0有两个不等实根,?

设为x₁,x₂,?

又f′(1)=0知x₁=1是方程的根,?

∴x₂=--1＜0＜x₁.?

当x＜x₂或x＞x₁时,f′(x)＜0;当x₂＜x＜x₁时,f′(x)＞0.                                                 ?

∴函数f(x)的递增区间为［x₂,x₁］.?

∴|s-T|=|x₁-x₂|=2+∈［2,3).                                                                            ?

(3)解：由f′(x)+a＜0,即ax²+bx+c+a＜0,即ax²+bx-b＜0,?

∵a＜0,∴x²+x-＞0.?

设g()=(x-1)·+x²对0≤＜1恒成立.                                                          ?

∴即

解之,得x≤或x≥.                                                                 ?

∴k≥.因此k的最小值为.