If $|x-1|+\left|x^2+x+1\right|=\left|x^2+2 x\right|$, then x belongs to:

$(2, \infty)$ $[1, \infty)$ $(-1, \infty)$ $(-2, \infty)$

$[1, \infty)$

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.