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
Step-by-Step Solution
Key Concept: Analyse $g(x)=-x|x-1|-|x+2|$ case by case on $(-\infty,-2)$, $[-2,1)$, $[1,\infty)$ and determine its monotonicity.
Case I ($x<-2$): $a=x^2+2$, decreasing. Case II ($-2\leq x<1$): $a=x^2-2x-2$, decreasing (derivative $\leq 0$). Case III ($x\geq 1$): $a=-(x^2+2)$, decreasing. Combined, $g$ is strictly decreasing on $\mathbb{R}$, so exactly one root for all $a\in(-\infty,\infty)$.
Correct Answer: 2