(1)∵1为f(x)的一个零点,
∴f(1)=0,解得c=1.
(2)由(1)知: f(x)=
x-1
x+1 ,
所以 f( 4 a )+f( 4 b )=
4 a -1
4 a +1 +
4 b -1
4 b +1 =
2• 4 a+b -2
( 4 a +1)•( 4 b +1) =0 .
(3)先证f(x)的单调性.
设0≤x 1<x 2≤2,则 f( x 2 )-f( x 1 )=
c x 2 -1
x 2 +1 -
c x 1 -1
x 1 +1 =
( x 2 - x 1 )•(c+1)
( x 2 +1)•( x 1 +1) .
∵0≤x 1<x 2≤2,∴当c>-1时,f(x 2)>f(x 1),即函数f(x)在[0,2]上单调递增,
所以f(x) max=f(2)=3,即
2c-1
2+1 =3,解得c=5;
当c=-1时,f(x 1)=f(x 2),即f(x)在[0,2]上是常函数,
所以f(x)=-1,不合题意;
当c<-1时,f(x 1)<f(x 2),即函数f(x)在[0,2]上单调递减,
所以f(x) max=f(0)=3,即-1=3,显然不成立,
综上所述,c=5.