## Task A-2.3.

Determine all values of the parameter $a$ for which the system

$$
\begin{aligned}
2^{|x|}+|x| & =x^{2}+y+a \\
x^{2}+y^{2} & =1
\end{aligned}
$$

has exactly one solution $(x, y) \in \mathbb{R}^{2}$.
## Solution.

If the pair $(x, y)$ is a solution to the given system, then the pair $(-x, y)$ is also a solution.

We conclude that the unique solution of this system must be of the form $(0, y)$.

Substituting $x=0$ into the given system, we get

$$
\begin{aligned}
1 & =y+a \\
y^{2} & =1
\end{aligned}
$$

so $y=1$ or $y=-1$, and therefore $a=0$ or $a=2$.

Let's first consider the case $a=0$.

The initial system reduces to

$$
\begin{aligned}
2^{|x|}+|x| & =x^{2}+y \\
x^{2}+y^{2} & =1
\end{aligned}
$$

It is easy to see that $(0,1)$ is a solution. Let's show that there are no other solutions.

From the second equation, it follows that $|x| \leqslant 1$ and $|y| \leqslant 1$. Since $0 \leqslant|x| \leqslant 1$, we have $|x| \geqslant x^{2}$.

We also have

$$
2^{|x|} \geqslant 2^{0}=1 \geqslant|y| \geqslant y
$$

so

$$
2^{|x|}+|x| \geqslant x^{2}+y
$$

For equality to hold, we must have $|x|=0$ and $|y|=y$, i.e., $x=0$ and $y \geqslant 0$.

Due to the second equation, we then have $(x, y)=(0,1)$, so this system has no other solutions.

In the case $a=2$, we can observe that in addition to $(0,-1)$, the solutions of the system

$$
\begin{aligned}
2^{|x|}+|x| & =x^{2}+y+2 \\
x^{2}+y^{2} & =1
\end{aligned}
$$

are also the pairs $(1,0)$ and $(-1,0)$.

We conclude that the system has a unique solution if and only if $a=0$.