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

Find the set of all $$a \in \mathbb{R}$$ for which $$x|x-1| + |x+2| + a = 0$$ has exactly one real root.

Let $$f(x) = x|x-1| + |x+2|$$. The equation becomes $$a = -f(x)$$. We analyze $$g(x) = -f(x)$$ in three intervals based on the critical points $$x = -2$$ and $$x = 1$$.

Case 1: $$x \geq 1$$

$$f(x) = x(x-1) + (x+2) = x^2 + 2$$, so $$g(x) = -(x^2 + 2)$$.

This is strictly decreasing on $$[1, \infty)$$ with $$g(1) = -3$$ and $$g(x) \to -\infty$$.

Case 2: $$-2 \leq x < 1$$

$$f(x) = x(1-x) + (x+2) = -x^2 + 2x + 2$$, so $$g(x) = x^2 - 2x - 2 = (x-1)^2 - 3$$.

On $$[-2, 1)$$: $$g(-2) = 6$$ and $$g(x) \to -3$$ as $$x \to 1^-$$. Since $$(x-1)^2$$ is decreasing for $$x < 1$$, $$g$$ is strictly decreasing on $$[-2, 1)$$.

Case 3: $$x < -2$$

$$f(x) = x(1-x) - (x+2) = -x^2 - 2$$, so $$g(x) = x^2 + 2$$.

As $$x$$ increases from $$-\infty$$ to $$-2^-$$: $$g$$ decreases from $$+\infty$$ to $$6^+$$.

Combining all cases:

The function $$g(x)$$ is strictly decreasing over all of $$\mathbb{R}$$ and continuous, with range $$(-\infty, +\infty)$$. At the boundaries: $$g(-2) = 6$$ from both Cases 2 and 3, and $$g(1) = -3$$ from both Cases 1 and 2. Therefore, for every real value of $$a$$, the equation $$a = g(x)$$ has exactly one solution.

The answer is Option B: $$(-\infty, \infty)$$.