问题
选择题
设f-1(x)是函数f(x)=
|
答案
由题意设y=
(ax-a-x)整理化简得a2x-2yax-1=0,1 2
解得:ax=y± y2+1
∵ax>0,∴ax=y+
,y2+1
∴x=loga(y+
)y2+1
∴f-1(x)=loga(x+
)x2+1
由使f-1(x)>1得loga(x+
)>1x2+1
∵a>1,∴x+
>ax2+1
由此解得:x>a2-1 2a
故选A
设f-1(x)是函数f(x)=
|
由题意设y=
(ax-a-x)整理化简得a2x-2yax-1=0,1 2
解得:ax=y± y2+1
∵ax>0,∴ax=y+
,y2+1
∴x=loga(y+
)y2+1
∴f-1(x)=loga(x+
)x2+1
由使f-1(x)>1得loga(x+
)>1x2+1
∵a>1,∴x+
>ax2+1
由此解得:x>a2-1 2a
故选A