## Task A-3.2.

Determine the real numbers $a$ and $b$ such that the range of the function $f(x)=2022^{-2(x-2 a)(x+a)}+1$ is the interval $\langle b, 2023]$.
## Solution.

Since the quadratic function $g(x)=-2(x-2a)(x+a)$ reaches its maximum value at the vertex, the maximum value of this quadratic function is equal to the ordinate of the vertex and is $\frac{9a^2}{2}$.

1 point

The image of the function $g(x)=-2(x-2a)(x+a)$ is the interval $\left\langle-\infty, \frac{9a^2}{2}\right]$.

1 point

Since the function $h(x)=2022^{-2(x-2a)(x+a)}$ is increasing, its image is the interval $\left\langle 0, 2022^{\frac{9a^2}{2}}\right]$.

The image of the function $f(x)=2022^{-2(x-2a)(x+a)}+1$ is the interval $\left\langle 1, 2022^{\frac{9a^2}{2}}+1\right]$.

By comparing the obtained and desired intervals, we conclude

$$
\begin{aligned}
b & =1 \\
2022^{\frac{9a^2}{2}}+1 & =2023
\end{aligned}
$$

From the second equation, we get $\frac{9a^2}{2}=1$, so $a= \pm \frac{\sqrt{2}}{3}$.