If |x-1|+≤ft|x^2+x+1|=≤ft|x^2+2 x|, then x belongs to:

Above is possible if

$(x-1)\left(x^2+x+1\right) \geq 0$                ......(1)

Now for $x^2+x+1$

Discriminant = 1 - 4 = -3 < 0

∴ $x^2+x+1>0 ~\forall ~ x \in R$            .......(2)

From (1) and (2), $(x-1) \geq 0 \Rightarrow x \in[1, \infty)$

Hence (2) is the correct answer.