已知函數(shù)f(x)=5x-5x2,記函數(shù)f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…,fn(x)=f[fn-1(x)],…,考察區(qū)間A=(-∞,0),對(duì)任意實(shí)數(shù)x∈A,有f1(x)=f(x)=a<0,f2(x)=f[f1(x)]=f(a)<0,且n≥2時(shí),fn(x)<0,問(wèn):是否還有其它區(qū)間,對(duì)于該區(qū)間的任意實(shí)數(shù)x,只要n≥2,都有fn(x)<0?
分析:由函數(shù)f1(x)=f(x),f2(x)=f[f1(x)],f3(x)=f[f2(x)],…,fn(x)=f[fn-1(x)]…知若使fn(x)<0等價(jià)于f[fn-1(x)]<0,由f(x)<0,得x<0或x>1,則有fn-1(x)<0或fn-1(x)>1,依此類推,要使一切n∈N+,n≥2,都有fn(x)<0,必須有f1(x)<0或f1(x)>1即f(x)<0或f(x)>1的解集即為所求.
解答:解:f(x)<0,即5x
2-5x>0,故x<0或x>1.
∴f
n(x)<0?f[f
n-1(x)]<0?f
n-1(x)<0或f
n-1(x)>1.
要使一切n∈N
+,n≥2,都有f
n(x)<0,必須使f
1(x)<0或f
1(x)>1,
∴f(x)<0或f(x)>1,即5x-5x
2<0或5x-5x
2>1.
解得x<0或x>1或
<x<
,
∴還有區(qū)間(
,
)和(1,+∞)
使得對(duì)于這些區(qū)間內(nèi)任意實(shí)數(shù)x,只要n≥2,都有f
n(x)<0
點(diǎn)評(píng):本題一道方案類型題,做題時(shí)要用即定的規(guī)律進(jìn)行遞推,轉(zhuǎn)化,還涉及到不等式的法和恒成立問(wèn)題.