(1)由题意知f(x)=sinx,要f(x 0+1)=f(x 0)+f(1),即需sin(x 0+1)=sinx 0+sin1
显然当x 0=0时等式成立,即f(x)=sinx∈M.
(2)∵函数f(x)= lg
2k
x 2 +1 ∈M ,∴f(x+1)=f(x)+f(1)有解,即 lg
2k
(x+1) 2 +1 =lg
2k
x 2 +1 +lg
2k
2 lg
2k
(x+1) 2 +1 =lg
2k
x 2 +1 •
2k
2
2k
(x+1) 2 +1 =
2k
x 2 +1 •
2k
2 ,
∴x 2+1=k(x 2+2x+2),∴(k-1)x 2+2kx+2k-1=0有解,
①k=1时, x=-
1
2 有解,符合;
②k≠1时,△=4k 2-4(k-1)(2k-1)≥0,∴
3-
5
2 ≤k≤
3+
5
2 ,k≠1 ,
综上:
3-
5
2 ≤k≤
3+
5
2 .
(3)∵函数f(x)=2 x+x 2∈M,要证f(x)∈M,
∴f(x+1)=f(x)+f(1)有解,∴2 x+1+(x+1) 2=2 x+x 2+3有解,即2 x+2x-2=0有解,
设h(x)=2 x+2x-2,∵h(0)=-1,h(1)=2,
根据函数的零点存在性判定理得,存在x 0∈(0,1),h(x 0)=0,
即f(x 0+1)=f(x 0)+f(1)成立,∴f(x)∈M.