Task 11.6. (14 points)

Real numbers $a$ and $b$ are such that $\left(a+\sqrt{1+a^{2}}\right)\left(b+\sqrt{1+b^{2}}\right)=1$. Find the sum $a+b$.

#
# Solution.

$$
\begin{aligned}
& b+\sqrt{1+b^{2}}=\frac{1}{a+\sqrt{1+a^{2}}}=-a+\sqrt{1+a^{2}} \\
& a+\sqrt{1+a^{2}}=\frac{1}{b+\sqrt{1+b^{2}}}=-b+\sqrt{1+b^{2}}
\end{aligned}
$$

By adding the obtained equations, we get

$$
a+b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}}=-a-b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}} \Rightarrow a+b=0
$$

## Answer. 0.

| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 14 |
| All main logical steps of the solution are presented. Some justifications are missing in the solution. The answer is correct. | $\pm$ | 11 |
| The idea of the solution is found. The main logical steps of the solution are presented. The answer is incorrect or missing. | $+/ 2$ | 7 |
| The answer is correct. The solution is missing or incorrect. | $\mp$ | 3 |
| The solution does not meet any of the criteria described |  |  |
| above. |  |  |