The set of all a ∈ ℝ for which the equation x | x - 1 | + | x + 2 | + a = 0 has exactly one real…

Given,

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

Now taking, Case I: $x<-2$, we get

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

$\Rightarrow a=x^2+2$

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

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

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

$\Rightarrow a=x^2-2x-2$

And $y=x^2-2x-2$ 

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

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

Now taking Case III: $x\ge 1$ we get,

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

$\Rightarrow a=-\left(x^2+2\right)$

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,

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