The set of all a ∈ ℝ for which the equation x | x - 1 | + | x + 2 |...

Given,

$x | x - 1 | + | x + 2 | + a = 0$

Now taking, Case I: $x < - 2$, we get

$- {x}^{2} + x - x - 2 + a = 0$

$⇒ a = {x}^{2} + 2$

And $y = {x}^{2} + 2$ is decreasing $\forall x \in ( - \infty , - 2 )$

Now taking Case II: $- 2 \le x < 1$ we get,

$- {x}^{2} + x + x + 2 + a = 0$

$⇒ a = {x}^{2} - 2x - 2$

And $y = {x}^{2} - 2x - 2$

So, $\dfrac{dy}{dx} = 2x - 1 \le 0 \forall x \in [ - 2, 1 )$

Hence, $y$ is decreasing $\forall x \in [ - 2, 1 )$

Now taking Case III: $x \ge 1$ we get,

${x}^{2} - x + x + 2 + a = 0$

$⇒ a = - \left({x}^{2} + 2\right)$

And $y = - \left({x}^{2} + 2\right)$ is decreasing $\forall x \in [ 1, \infty )$

Hence, from all the cases we can say that nature of function is continuously decreasing so, it will cut $x -$axis only one time,

$∴$ Exactly one real root $\forall a \in R$