Question

(理)如果f(x)在某個區(qū)間I內(nèi)滿足:

對任意的x₁、x₂∈I,都有［f(x₁)+f(x₂)］≥f(),則稱f(x)在I上為下凸函數(shù).

已知函數(shù)f(x)=-alnx.

(1)證明當(dāng)a＞0時,f(x)在(0,+∞)上為下凸函數(shù);

(2)若f′(x)為f(x)的導(dǎo)函數(shù),且x∈［,2］時,|f′(x)|＜1,求實數(shù)a的取值范圍.

(文)如果f(x)在某個區(qū)間I內(nèi)滿足:

對任意的x₁、x₂∈I,都有［f(x₁)+f(x₂)］≥f(),則稱f(x)在I上為下凸函數(shù),已知函數(shù)f(x)=ax²+x.

(1)證明當(dāng)a＞0時,f(x)在R上為下凸函數(shù);

(2)若x∈(0,1)時,|f(x)|≤1,求實數(shù)a的取值范圍.

Answer 1

(理)(1)證明:任取x₁、x₂∈(0,+∞),

則［f(x₁)+f(x₂)］

=［-alnx₁+-alnx₂］

=-aln,                                                         

,                                          

∵x₁²+x₂²≥2x₁x₂,∴(x₁+x₂)²≥4x₁x₂.

又x₁＞0,x₂＞0,∴.                                            

又≥,a＞0,

∴-aln≥-aln,                                                  

即［f(x₁)+f(x₂)］≥f().                                              

∴f(x)為(0,+∞)上的下凸函數(shù).

(2)解：f′(x)=,                                                    

∵|f′(x)|＜1,即||＜1,

∴-(x+)＜a＜x-.                                                        

∵x∈［,2］時,|f′(x)|＜1恒成立,

∴a∈(-2,).                                                            

(文)(1)證明：f(x₁)+f(x₂)-2f()

=ax₁²+x₁+ax₂²+x₂-2［a()²-］

=,                                                           

∵a＞0,∴［f(x₁)+f(x₂)］≥f().

∴當(dāng)a＞0時,f(x)為R上的下凸函數(shù).                                         

(2)解：∵|f(x)|≤1,

∴-1≤ax²+x≤1,

≤a≤.                                                       

∵x∈(0,1),

∴-2≤a≤0.