The set of all $a \in \mathbb{R}$ for which the equation $x|x-1|+|x+2|+a=0$ has exactly one real root, is :

$x|x-1|+|x+2|+a=0$ <br/><br/>Case I : If $x<-2$ then <br/><br/>$-x^2+x-x-2+a=0$ <br/><br/>$a=x^2+2$ <br/><br/>$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$ <br/><br/>Case II : If $-2 \leq x<1$ then <br/><br/>$$ \begin{aligned} & -x^2+x+x+2+a=0 \\\\ & a=x^2-2 x-2 \end{aligned} $$ <br/><br/>$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$. <br/><br/>Case III: If $x \geq 1$ then <br/><br/>$$ \begin{aligned} & x^2-x+x+2+a=0 \\\\ & a=-\left(x^2+2\right) \end{aligned} $$ <br/><br/>$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$ <br/><br/>$\therefore$ Exactly one real root $\forall x \in R$