Question

(理)设a∈R,函数f(x)=(ax²+a+1)(e为自然对数的底数).

(1)判断f(x)的单调性;

(2)若f(x)＞在x∈［1,2］上恒成立,求a的取值范围.

(文)已知函数f(x)=x³+bx²+cx+1在区间(-∞,-2］上单调递增,在区间［-2,2］上单调递减,且b≥0.

(1)求f(x)的解析式;

(2)设0＜m≤2,若对任意的x₁、x₂∈［m-2,m］,不等式|f(x₁)-f(x₂)|≤16m恒成立,求实数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当a=0时,g(x)=-1＜0,∴f′(x)＜0.

∴f(x)在R上为减函数.

②当a＞0时,g(x)=0的判别式Δ=4a²-4(a²+a)=-4a＜0,

∴g(x)＜0,即f′(x)＜0.

∴f(x)在R上为减函数.                                                        

③当a＜0时,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(-∞,),(,+∞)上为增函数;

f(x)在()上为减函数.                                       

(2)①当a≥0时,f(x)在［1,2］上为减函数.

∴f(x)_min=f(2)=.

由,得a＞.                                                      

②当a＜0时,f(2)=,

∴f(x)＞在［1,2］上不恒成立,

∴a的取值范围是(,+∞).                                                    

(文)解：(1)f′(x)=3x²+2bx+c,

∵f(x)在(-∞,-2］上单调递增,在［-2,2］上单调递减,

∴f′(x)=3x²+2bx+c=0有两个根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)已知条件等价于在［m-2,m］上f(x)_max-f(x)_min≤16m.                             

∵f(x)在［-2,2］上为减函数,

且0＜m≤2,

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

∴f(x)在［m-2,m］上为减函数.

∴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=.