Question

(理)設a∈R,函數(shù)f(x)=(ax²+a+1)(e為自然對數(shù)的底數(shù)).

(1)判斷f(x)的單調(diào)性;

(2)若f(x)＞在x∈［1,2］上恒成立,求a的取值范圍.

(文)已知函數(shù)f(x)=x³+bx²+cx+1在區(qū)間(-∞,-2］上單調(diào)遞增,在區(qū)間［-2,2］上單調(diào)遞減,且b≥0.

(1)求f(x)的解析式;

(2)設0＜m≤2,若對任意的x₁、x₂∈［m-2,m］,不等式|f(x₁)-f(x₂)|≤16m恒成立,求實數(shù)m的最小值.

Answer 1

(理)解：(1)由已知f′(x)=e^-x(ax²+a+1)+e^-x·(2ax)=e^-x(-ax²+2ax-a-1),          

令g(x)=-ax²+2ax-a-1.

①當a=0時,g(x)=-1＜0,∴f′(x)＜0.

∴f(x)在R上為減函數(shù).

②當a＞0時,g(x)=0的判別式Δ=4a²-4(a²+a)=-4a＜0,

∴g(x)＜0,即f′(x)＜0.

∴f(x)在R上為減函數(shù).                                                        

③當a＜0時,由-ax²+2ax-a-1＞0,得x＜1-或x＞1+;

由-ax²+2ax-a-1＜0,得1-＜x＜1+.

∴f(x)在(-∞,),(,+∞)上為增函數(shù);

f(x)在()上為減函數(shù).                                       

(2)①當a≥0時,f(x)在［1,2］上為減函數(shù).

∴f(x)_min=f(2)=.

由,得a＞.                                                      

②當a＜0時,f(2)=,

∴f(x)＞在［1,2］上不恒成立,

∴a的取值范圍是(,+∞).                                                    

(文)解：(1)f′(x)=3x²+2bx+c,

∵f(x)在(-∞,-2］上單調(diào)遞增,在［-2,2］上單調(diào)遞減,

∴f′(x)=3x²+2bx+c=0有兩個根x₁、x₂,且x₁=-2,x₂≥2,                                

∵x₁+x₂=,x₁x₂=,

∴x₂=+2.∴+2≥2.∴b≤0.                                             

又b≥0,∴b=0.

∴x₂=2,c=-12.

∴f(x)=x³-12x+1.6分

(2)已知條件等價于在［m-2,m］上f(x)_max-f(x)_min≤16m.                             

∵f(x)在［-2,2］上為減函數(shù),

且0＜m≤2,

∴［m-2,m］［-2,2］.                                                     

∴f(x)在［m-2,m］上為減函數(shù).

∴f(x)_max=f(m-2)=(m-2)³-12(m-2)+1,

f(x)_min=f(m)=m³-12m+1.

∴f(x)_max-f(x)_min=-6m²+12m+16≤16m,

得m≤-2或m≥.

又∵0＜m≤2,∴m_min=.