已知函数f(x)=lg(ax-1)-lg(x-1) 在[10,+∞]上为单调增函数,求实数a的取值范围.

f(x)=lg((ax-1)/(x-1)) (*)
首先满足ax-1>0(x>=10),ax>1,a>1/x,1/x在10到正无穷上是递减,故a>1/10;
其次,对于(*)式,满足了上述ax-1>0之后,要使得整个函数在10到无穷上递增,必须使得(ax-1)/(x-1)在10到无穷上是递增函数(因为对于函数lgx,在0到无穷上递增);(ax-1)/(x-1)=a+(a-1)/(x-1),而影响f(x)递增性的就在于(a-1)/(x-1),注意到在10到正无穷上,1/(x-1)递减,故要使得f(x)递增,a-1

f(x)=lg((ax-1)/(x-1))
m=(ax-1)/(x-1)
=(ax-a+a-1)/(x-1)
=a+(a-1)/(x-1)
m'=-3/2*(a-1)*(x-1)^(-3/2)
x>=10
m>0
(a-1)/(x-1)^(-3/2)<0
10a-1>0,a>1/10
1/10

1年前

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