已知f(x)=
(1)求a的值; (2)求f(x)的反函数; (3)对任意的k∈(0,+∞)解不等式f-1(x)>log2
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(1)由题知f(0)=0,得a=1,
此时f(x)+f(-x)=
+2x-1 2x+1
=2-x-1 2-x+1
+2x-1 2x+1
=0,1-2x 1+2x
即f(x)为奇函数.
(2)∵y=
=1-2x-1 2x+1
,得2x=2 2x+1
(-1<y<1),1+y 1-y
∴f-1(x)=log2
(-1<x<1).1+x 1-x
(3)∵f-1(x)>log2
,∴1+x k
,∴
>1+x 1-x 1+x k -1<x<1
,x>1-k -1<x<1
①当0<k<2时,原不等式的解集{x|1-k<x<1},
②当k≥2时,原不等式的解集{x|-1<x<1}.