1. Consider two quadratic equations

$$
x^{2}-a x-b=0, \quad x^{2}-b x-a=0
$$

with real parameters $a, b$. Determine the smallest and the largest values that the sum $a+b$ can take, given that there exists exactly one real number $x$ that satisfies both equations simultaneously. Also, determine all pairs $(a, b)$ of real parameters for which the considered sum of these values is achieved.
1. By subtracting the given equations, we obtain the equality $(b-a) x+a-b=0$ or $(b-a)(x-1)=0$, from which it follows that $b=a$ or $x=1$.

If $b=a$, both equations take the form $x^{2}-a x-a=0$. Exactly one solution exists if and only if the discriminant $a^{2}+4 a$ is zero. This holds for $a=0$ and for $a=-4$. Since $b=a$, the sum $a+b$ is 0 in the first case and -8 in the second case.

If $x=1$, we get from the given equations $a+b=1$, thus $b=1-a$. The equations then take the form

$$
x^{2}-a x+a-1=0 \quad \text { and } \quad x^{2}+(a-1) x-a=0 .
$$

The first has roots 1 and $a-1$, the second has roots $1$ and $-a$. Exactly one common solution is obtained in all cases except when $a-1=-a$ or $a=\frac{1}{2}$, in which case there are two common solutions.

Conclusion. The smallest value of the sum $a+b$ is -8 and is achieved for $a=b=-4$. The largest value of the sum $a+b$ is 1; this value is achieved for all pairs $(a, 1-a)$, where $a \neq \frac{1}{2}$ is any real number.

For a complete solution, award 6 points. Award one point for deriving the condition $(b=a) \vee(x=1)$, two points for solving the case $b=a$, two points for solving the case $x=1$, and one point for the correct conclusion.