## Task A-1.2.

Determine all real numbers $a$ for which the equation

$$
|| x-2|-x+a|=x+3
$$

has exactly two real solutions.
## First Solution.

We consider the graph of the function $f(x)=|| x-2|-x+a|-x-3$, i.e., the curve $y=|| x-2|-x+a|-x-3$, and examine how many points it can intersect the $x$-axis.

For $x \geqslant 2$, the curve has the equation $y=|a-2|-3-x$, which is a line with a slope of -1, and at the point $x=2$ it has the value $y=|a-2|-5$.

For $x<2$, the curve has the equation $y=|2+a-2 x|-3-x$, which is a broken line or a line. Specifically, the graph of the function $g(x)=|2+a-2 x|-3-x$ generally consists of two half-lines lying on lines with equations $y=a-1-3 x$ and $y=x-5-a$. It is only necessary to determine whether the cusp where the half-lines intersect is in the region where $x<2$. The mentioned lines intersect at the point which is the solution to the system, i.e., where the expression $2+a-2 x$ changes sign. This is $x=1+\frac{a}{2}$. The cusp is in the region $x<2$ for $1+\frac{a}{2}<2$, i.e., for $a<2$.

If $a \geqslant 2$, then the cusp is to the right of $x=2$ and in the region $x<2$ we have $2+a-2 x \geqslant 0$, so the graph of the function $g(x)$ in that region is just the line $y=a-1-3 x$.

Thus, we arrive at the following conclusions about the graph of the function $f$ depending on the parameter $a$.

For $a \geqslant 2$, we have two segments with slopes -3 and -1, with a transition at $x=2$. For such $a$, the function $f$ is always decreasing and we have only one zero point regardless of $a$.

For $a<2$, we have three segments with slopes $-3,1,-1$ and transitions at $x=1+\frac{a}{2}$ and $x=2$ where the values are $y=0-3-1-\frac{a}{2}=-4-\frac{a}{2}$ and $y=-a-3$. Note that $-4-\frac{a}{2}<-a-3$ since this is equivalent to $a<2$. We have exactly two zero points if one of these two values is equal to 0, i.e., if $a=-8$ or $a=-3$.