Question

對(duì)于任意函數(shù)f(x),x∈D,可按圖示構(gòu)造一個(gè)數(shù)列發(fā)生器,其工作原理如下:

①輸入數(shù)據(jù)x₀∈D,經(jīng)數(shù)列發(fā)生器輸出x₁=f(x₀)；

②若x₁D,則數(shù)列發(fā)生器結(jié)束工作；若x₁∈D,則將x₁反饋回輸入端,再輸出x₂=f(x₁),并依此規(guī)律繼續(xù)下去.現(xiàn)在定義f(x)=.

(1)若輸入x₀=,則由數(shù)列發(fā)生器產(chǎn)生數(shù)列{x_n}.請(qǐng)寫(xiě)出數(shù)列{x_n}的所有項(xiàng).

(2)若要數(shù)列發(fā)生器產(chǎn)生一個(gè)無(wú)窮的常數(shù)數(shù)列,試求輸入的初始數(shù)據(jù)x₀的值.

(3)若輸入x₀時(shí),產(chǎn)生的無(wú)窮數(shù)列{x_n}滿足:對(duì)任意正整數(shù)n,均有x_n＜x_n+1,求x₀的取值范圍.

Answer 1

解析：(1)∵f(x)的定義域D=(-∞,-1)∪(-1,+∞),

∴ 數(shù)列x_n只有三項(xiàng):x₁=,x₂=,x₃=-1 .

(2)∵f(x)==x,即x²-3x+2=0,∴x=1或x=2,即當(dāng)x₀=1或2時(shí),x_n+1==x_n.故當(dāng)x₀=1時(shí),x_n=1；當(dāng)x₀=2時(shí),x_n=2(n∈N^*).

(3)解不等式x＜,得x＜-1或1＜x＜2.要使x₁＜x₂,則x₁＜-1或1＜x₁＜2.

對(duì)于函數(shù)f(x)==4-,

若x₁＜-1,則x₂=f(x₁)＞4,x₃=f(x₂)＜x₂.

當(dāng)1＜x₁＜2時(shí),x₂=f(x₁)＞x₁,且1＜x₂＜2,依次類(lèi)推可得數(shù)列{x_n}的所有項(xiàng)均滿足x_n+1＞x_n(n∈N^*).

綜上所述,x₁∈(1,2).由x₁=f(x₀),得x₀∈(1,2).