## Solution.

For the function $f_{1}(x)$ to have a maximum, it must be $k-8<0$, i.e., $k<8$,

The maximum value of the function $f_{1}(x)$ is $\quad \frac{4(k-8)(k-9)-4(k-5)^{2}}{4(k-8)}$,

i.e., $\max _{x \in \mathbb{R}} f_{1}(x)=\frac{47-7 k}{k-8}$.

The minimum value of the function $f_{2}(x)$ is $\frac{4(k-4)(k+7)-4(k-1)^{2}}{4(k-4)}$,

i.e., $\min _{x \in \mathbb{R}} f_{2}(x)=\frac{5 k-29}{k-4}$.

The extreme values of these functions coincide, so we have:

$$
\frac{47-7 k}{k-8}=\frac{5 k-29}{k-4}
$$

By rearranging, we obtain the quadratic equation $k^{2}-12 k+35=0$

whose solutions are $k=5, k=7$.

Both numbers satisfy the condition $(4<k<8)$, so these are the desired parameters.

Note: A student who does not set the condition $4<k<8$ but verifies at the end that the extreme values of the functions coincide should receive full marks.

For $k=5, \max f_{1}(x)=-4=\min f_{2}(x) ; \quad$ for $k=7, \quad \max f_{1}(x)=2=\min f_{2}(x)$.
## Task A-2.1.

Determine, if it exists, the real parameter $k$ such that the maximum value of the function

$$
f_{1}(x)=(k-8) x^{2}-2(k-5) x+k-9
$$

is equal to the minimum value of the function

$$
f_{2}(x)=(k-4) x^{2}-2(k-1) x+k+7
$$