Question

已知函數(shù)f(x)的定義域為[0，1]，且滿足下列條件：①對于任意x∈[0，1]，總有f(x)≥3且f(1)=4.②若x₁≥0,x₂≥0,x₁+x₂≤1.則有f(x₁+x₂)≥f(x₁)+f(x₃)-3.

(1)求f(0)的值；

(2)求證:f(x)≤4；

(3)當(dāng)x∈(,)(n=1,2,3…)時，試證明f(x)＜3x+3.

Answer 1

答案：(1)解：∵當(dāng)x∈［0,1］時，f(x)≥3,

∴f(0)≥3,又在f(x₁+x₂)≥f(x₁)+f(x₂)-3中

令x₁=x₂=0,得f(0)≤3,∴f(0)=3.                                                  

(2)證明：設(shè)0≤x₁＜x₂≤1.∴0＜x₂-x₁≤1,

∴f(x₂)-f(x₁)=f［(x₂-x₁)+x₁］-f(x₁)≥f(x₂-x₁)+f(x₁)-3-f(x₁)=f(x₂-x₁)-3≥0,

若f(x₂-x₁)-3=0,即f(x₂-x₁)=3,則f(x)在[0，1]上恒為3，這與f(1)=4矛盾.

∴f(x₂-x₁)＞3,即f(x)在[0，1]上為單調(diào)遞增函數(shù),∴f(x)≤f(1)=4.                      

(3)證明：由f(x₁+x₂)≥f(x₁)+f(x₂)-3,令x₁=x₂=x,得f(x)≤.

取x=,得f()≤≤+3,x=時,f()≤.

由數(shù)學(xué)歸納法得f()≤+3.∴取x∈()時，則有f(x)＜2x+2,

而x∈()(),∴f(x)＜2x+2.

而在()上2x+2＜3x+3,∴當(dāng)x∈()時,f(x)＜3x+3成立.