f(x)≤g(x)
lg(x+1)≤2lg(2x+t)
x+1≤(2x+t)^2
F(x)=4x^2+(4t-1)x+t^2-1≥0
△=(4t-1)^2-4*4(t^2-1)=-8t+17≤0,t≥17/8時,F(x)≥0恒成立
△>0時
對稱軸-(4t-1)/8≥1,t≤1/4時
F(1)=4+(4t-1)+t^2-1=t^2+4t+2=(t+2)^2-2≥0
t≥-2+√2,或,t≤-2-√2
即:-2+√2≤t≤1/4,或,t≤-2-√2
對稱軸-(4t-1)/8≤0,t≥1/4時
F(0)=t^2-1≥0
t≥1,或,t≤-1
即：t≥1
所以,參數(shù)t的取值范圍:(-∞,-2-√2]U[-2+√2,1/4]U[1,+∞)