(1)对于函数f(x)=
1
x ,D=(-∞,0)∪(0,+∞),若f(x)∈M,
则存在非零实数x 0,使得
1
x 0 +1 =
1
x 0 +1,即 x 0 2 +x 0+1=0,显然此方程无实数解,
∴f(x)∉M;
函数g(x)=x 2,D=R,若g(x)∈M成立,
则有 (x 0 +1) 2 = x 0 2 +1,解得x 0=0,
∴g(x)∈M;
(2)由条件得:D=R,a>0,由f(x)∈M知,
存在实数x 0,使得lg
a
(x 0 +1) 2 +1 =lg
a
x 0 2 +1 +lg
a
2 ,
∴
a
(x 0 +1) 2 +1 =
a
x 0 2 +1 •
a
2 ,
化简得:(a-2) x 0 2 +2ax 0+2a-2=0,
当a=2时,x 0=-
1
2 ,符号题意;
当a≠2时,由△≥0得:4a 2-4(a-2)(2a-2)≥0,
即3-
5 ≤a≤3+
5 (a≠2),
综上所述,a的取值范围是[3-
5 ,3+
5 ].