问题
选择题
设f-1(x)是函数f(x)=
|
答案
由题意设y=
(2x-2-x)整理化简得22x-2y2x-1=0,1 2
解得:2x=y±y2+1
∵2x>0,∴2x=y+
,y2+1
∴x=log2(y+
)y2+1
∴f-1(x)=log2(x+
)x2+1
由使f-1(x)>1得log2(x+
)>1x2+1
∵2>1,∴x+
>2x2+1
由此解得:x>3 4
故选A.
设f-1(x)是函数f(x)=
|
由题意设y=
(2x-2-x)整理化简得22x-2y2x-1=0,1 2
解得:2x=y±y2+1
∵2x>0,∴2x=y+
,y2+1
∴x=log2(y+
)y2+1
∴f-1(x)=log2(x+
)x2+1
由使f-1(x)>1得log2(x+
)>1x2+1
∵2>1,∴x+
>2x2+1
由此解得:x>3 4
故选A.