Question

已知函數(shù)f(x)的定義域?yàn)椋?,1］,且滿足下列條件:

①對(duì)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(x₂)-3.

(1)求f(0)的值;

(2)求證:f(x)≤4;

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

(文)如圖,設(shè)拋物線y²=2px(p＞0)的焦點(diǎn)為F,經(jīng)過點(diǎn)F的直線交拋物線于A、B兩點(diǎn),且A、B兩點(diǎn)坐標(biāo)為(x₁,y₁)、(x₂,y₂),y₁＞0,y₂＜0,P是此拋物線的準(zhǔn)線上的一點(diǎn),O是坐標(biāo)原點(diǎn).

(1)求證:y₁y₂=-p²;

(2)直線PA、PF、PB的方向向量為(1,a)、(1,b)、(1,c),求證:實(shí)數(shù)a、b、c成等差數(shù)列;

(3)若=0,∠APF=α,∠BPF=β,∠PFO=θ,求證:θ=|α-β|.

Answer 1

(1)解:令x₁=x₂=0,

由①對(duì)于任意x∈［0,1］,總有f(x)≥3,

∴f(0)≥3.                                                              

又由②得f(0)≥2f(0)-3,

即f(0)≤3;                                                            

∴f(0)=3.                                                               

(2)證明:任取x₁、x₂∈［0,1］,且設(shè)x₁＜x₂,

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

∵0＜x₂-x₁≤1,∴f(x₂-x₁)≥3,

即f(x₂-x₁)-3≥0.

∴f(x₁)≤f(x₂).                                                           

∴當(dāng)x∈［0,1］時(shí),f(x)≤f(1)=4.                                            

(3)證明:先用數(shù)學(xué)歸納法證明

f()≤+3(n∈N^*).

(1)當(dāng)n=1時(shí),f()=f(1)=4=1+3=+3,不等式成立;

(2)假設(shè)當(dāng)n=k時(shí),f()≤+3(k∈N^*),

由f()=f［+(+)］

≥f()+f(+)-3

≥f()+f()+f()-6

得3f()≤f()+6≤+9.

∴f()≤+3,

即當(dāng)n=k+1時(shí),不等式成立.

由(1)(2),可知不等式f()≤+3對(duì)一切正整數(shù)都成立.

于是,當(dāng)x∈(,］(n=1,2,3,…)時(shí),3x+3＞3×+3=+3≥f(),

而x∈［0,1］,f(x)單調(diào)遞增,

∴f()＜f().

∴f(x)＜f()＜3x+3.                                                     

(文)證明:(1)①當(dāng)直線AB的斜率不存在時(shí),設(shè)直線AB的方程為x=,則A(,p),B(,-p),

∴y₁y₂=-p².                                                               

②當(dāng)直線AB的斜率存在且不為0時(shí),設(shè)直線AB的方程為y=k(x),

則由可得ky²-2py-kp²=0(k≠0).∴y₁y₂=-p².                        

(2)由已知a=k_PA,b=k_PF,c=k_PB,

設(shè)P(,t),F(,0),

∴a=,b=,c=;

且x₁=,x₂=.

故a+c=

=

=

=

=.

∴a、b、c成等差數(shù)列.                                                  

(3)證法一:∵=0,

∴PA⊥PB,故a·c=-1.

由(2)可知a+c=2b,即a-b=b-c.

①若AB⊥x軸,則α=β=45°,θ=0°,

∴θ=α-β.

②若k_AB＞0,則

tanα=.

同理可得tanβ=a,

∴tan(α-β)=,

即|tan(α-β)|=|b|=tanθ.

易知∠PFO、∠BPF、∠APF都是銳角.

∴θ=|α-β|.

③若k_AB＜0,類似地,也可證明θ=|α-β|.

綜上所述,θ=|α-β|.                                                       

證法二:∵=0,∴PA⊥PB.

故a·c=-1.

①如圖,若AB⊥x軸,則α=β=45°,θ=0°,

∴θ=α-β.

②若k_AB＞0,∵A、B在拋物線上,

∴|AF|=|AC|,|BF|=|BD|.

設(shè)AB中點(diǎn)為M,則

|PM|=,

∴PM是梯形ABDC的中位線,

故P是CD中點(diǎn).

∴P(),t=,

F(,0),

=(p,),

又=(x₂-x₁,y₂-y₁),

∴=p(x₂-x₁)

=p(x₂-x₁)=0.

∴.

∴△PDB≌△PBF.

∴∠BPF=∠DPB=β.

∴θ+2β=90°=α+β.∴θ=α-β.

③若k_AB＜0,類似②可證θ=β-α,

∴θ=|α-β|.