Question

設函數F(x)=ax³+bx²+cx(a＜b＜c),其圖象在x=1, x=M處的切線的斜率分別為0,-a,

(1)求證:0≤＜1;

(2)若函數F(x)的遞增區(qū)間為［s,t］,求|s-t|的取值范圍;?

(3)若當x≥k時,(k是與a,b,c無關的常數),恒有F′(x)+a＜0,試求k的最小值.

Answer 1

(1)證明:f′(x)=ax²+bx+c,由題意,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.                                                                                            ?

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

由③有實根,得Δ=b²+4AB≥0,?

即()²+4≥0.?

解得≤-4或≥0.                                                                                                 ?

綜上,0≤＜1.                                                                                                        ?

(2)解:由f′(x)=ax²+bx+c的判別式Δ=b²-4AC＞0得f′(x)=ax²+bx+c=0有兩個不等實根,?

設為x₁,x₂,?

又f′(1)=0知x₁=1是方程的根,?

∴x₂=--1＜0＜x₁.?

當x＜x₂或x＞x₁時,f′(x)＜0;當x₂＜x＜x₁時,f′(x)＞0.                                                 ?

∴函數f(x)的遞增區(qū)間為［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.?

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

∴即

解之,得x≤或x≥.                                                                 ?

∴k≥.因此k的最小值為.