Question

设函数f(x)=log2x-logx2（0<x<1），数列{an}满足f(2^an)=2n(n=1,2，……)

（1）求数列{an}的通项公式；
（2）判断数列{an}的单调性

Answer 1

(1)
f(x)=log2X-logx2=log2X-log2(2)/log2X
=log2X-1/log2X

f(2的an次方)=2n
即log2(2的an次方)-1/log2(2的an次方)=an-1/an=2n
an^2-2n·an-1＝0 （1）

此时2的an次方应该满足函数的定义域，即0<2的an次方<1
an<0

根据(1)解得：an=n-√(n^2+1)

(2)an=n-√(n^2+1)=-1/[n+√(n^2+1)] 分子有理化！
a(n+1))=-1/[n+1+√[(n+1)^2+1]>-1/[n+√(n^2+1)]=an
即：a(n+1)>an

{an}为单调递增！

后话：an<0,但不断接近0

Answer 2

f(x)=log2x-logx2=log2x-1/log2x
f(2^an)=an-1/an=2n an^2-2n*an-1=0 an=n±根号(n^2+1)
0<x<1 0<2^an<1 an<0 an=n-根号(n^2+1)

an=n-根号(n^2+1) =-(n+根号(n^2+1)),增函数

Answer 3

(1)
f(2^an) = 2n
f(2^an) = 2n = log2^(an+1)- log2^(2an)
= log2^(1-an)
= (1-an) log 2
an = 1- (2n/log2)
(2)
an = 1-(2n/log2)
{an} is decreasing

Answer 4

f(2^an)=log2(2^an)-log2^an(2), 2^an是底数。=an-1/an=2n