Question

(理)如果f(x)在某个区间I内满足:

对任意的x₁、x₂∈I,都有［f(x₁)+f(x₂)］≥f(),则称f(x)在I上为下凸函数.

已知函数f(x)=-alnx.

(1)证明当a＞0时,f(x)在(0,+∞)上为下凸函数;

(2)若f′(x)为f(x)的导函数,且x∈［,2］时,|f′(x)|＜1,求实数a的取值范围.

(文)如果f(x)在某个区间I内满足:

对任意的x₁、x₂∈I,都有［f(x₁)+f(x₂)］≥f(),则称f(x)在I上为下凸函数,已知函数f(x)=ax²+x.

(1)证明当a＞0时,f(x)在R上为下凸函数;

(2)若x∈(0,1)时,|f(x)|≤1,求实数a的取值范围.

Answer 1

(理)(1)证明:任取x₁、x₂∈(0,+∞),

则［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)为(0,+∞)上的下凸函数.

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

∵|f′(x)|＜1,即||＜1,

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

∵x∈［,2］时,|f′(x)|＜1恒成立,

∴a∈(-2,).                                                            

(文)(1)证明：f(x₁)+f(x₂)-2f()

=ax₁²+x₁+ax₂²+x₂-2［a()²-］

=,                                                           

∵a＞0,∴［f(x₁)+f(x₂)］≥f().

∴当a＞0时,f(x)为R上的下凸函数.                                         

(2)解：∵|f(x)|≤1,

∴-1≤ax²+x≤1,

≤a≤.                                                       

∵x∈(0,1),

∴-2≤a≤0.