∵f(x)=
1 |
x |
∴
1 |
x+1 |
1 |
x |
∵方程x2+x+1=0无实数解,
∴f(x)=
1 |
x |
(2)∵函数f(x)=lg
t |
x2+1 |
∴lg
t |
(x+1)2+1 |
t |
x2+1 |
t |
2 |
∴(t-2)x2+2tx+2(t-1)=0有实数解,
t=2时,x=−
1 |
2 |
t≠2时,由△=4t2-4(t-2)×2(t-1)≥0,
得t2−6t+4≤0⇒t∈[3−
5 |
5 |
∴t∈[3−
5 |
5 |
1 |
x |
t |
x2+1 |
1 |
x |
1 |
x+1 |
1 |
x |
1 |
x |
t |
x2+1 |
t |
(x+1)2+1 |
t |
x2+1 |
t |
2 |
1 |
2 |
5 |
5 |
5 |
5 |