Question

已知實(shí)數(shù)c≥0,曲線C：y=與直線l：y=x-c的交點(diǎn)為P(異于原點(diǎn)O)，在曲線C上取一點(diǎn)P₁(x₁,y₁)，過點(diǎn)P₁作P₁Q₁平行于x軸，交直線l于點(diǎn)Q₁，過點(diǎn)Q₁作Q₁P₂平行于y軸，交曲線C于點(diǎn)P₂(x₂,y₂)，接著過點(diǎn)P₂作P₂Q₂平行于x軸，交直線l于點(diǎn)Q₂，過點(diǎn)Q₂作直線Q₂P₃平行于y軸，交曲線C于點(diǎn)P₃(x₃,y₃)，如此下去，可以得到點(diǎn)P₄(x₄,y₄),P₅(x₅,y₅),…, P_n(x_n,

x_N),….設(shè)點(diǎn)P的坐標(biāo)為(a,),x₁=b,0＜b＜a.

(1)試用c表示a，并證明a≥1;

(2)試證明x₂＞x₁,且x_n＜a(N∈N^*);

(3)當(dāng)c=0,b≥時(shí)，求證: ＜ (k,N∈N^*).

Answer 1

(1)解:點(diǎn)P的坐標(biāo)(a,)滿足方程組所以=a-c.?

解a--c=0,得=,所以a=(1+2c+).                         ?

因?yàn)?I >c≥0,所以1+2c+≥2.所以a≥1.                                                       

(2)證明:由已知P₁(b,),Q₁(+c,),P₂(+c,),?

即x₁=b,x₂=+c,                                                                                            ?

x₂-x₁=+c-b,?

由(1)c=a-,所以x₂-x₁=+a--b=(-)(+-1).?

因?yàn)?＜b＜a,a≥1,所以x₂＞x₁.                                                                                 ?

下面用數(shù)學(xué)歸納法證明x_n＜a(n∈N^*).?

當(dāng)n=1時(shí)，x₁=b＜a;假設(shè)當(dāng)n=k時(shí)，x_k＜a,?

由已知，x_k+1=y_k+c,x_k＞0,所以x_k+1=+c=+a-＜a.?

綜上，x_n＜a(n∈N^*).                                                                                               ?

(3)解:當(dāng)c=0時(shí)，≤b＜a=1,x_n+1=y_n=(n∈N^*),??

所以x_n=x_n-1=x_n-2=…=x₁=b.                                                              ?

因?yàn)?I >b≥,所以當(dāng)k≥1時(shí)，x_k+2≥x₃≥().?

所以.                                                                                                     ?

又x_k+1-x_k=b-b＞0,?

所以≤b=x₁＜x_n＜a=1,x_n-x₁＜1-=.                                                                 

所以≤=(x_n+1-x₁)＜.