4. Find all values of the parameter $a$ for which the inequality $\log _{\sqrt{a / 10}}\left(\frac{a+8 x-x^{2}}{20}\right) \geq 2$

has at least one solution $x \geq 2$, and every solution of the inequality is also a solution

of the equation $|a+2 x-16|+|a-2 x+9|=|2 a-7|$.

(16 points)
# Solution:

$\log _{\sqrt{a / 10}}\left(\frac{a+8 x-x^{2}}{20}\right) \geq 2 \Leftrightarrow$

$\left\{\begin{array}{c}a+8 x-x^{2}>0, \\ a>0, \\ a \neq 10, \\ (\sqrt{a / 10}-1)\left(\frac{a+8 x-x^{2}}{20}-\frac{a}{10}\right) \geq 0\end{array} \Leftrightarrow\right.$

$\left\{\begin{array}{cl}a>x^{2}-8 x, & \\ a>0, & \\ a \neq 10, & \text { in the system } \\ (a-10)\left(8 x-x^{2}-a\right) \geq 0 . & \end{array}\right.$

Let's represent the set of solutions to this system. These are the areas on the diagram, colored blue and green. Consider the equation $|a+2 x-16|+|a-2 x+9|=|2 a-7|$.

Let $u=a+2 x-16, v=a-2 x+9$. Then the original equation will have the form $|u|+|v|=|u+v|$. The solutions to the latter equation are all $u$ and $v$ such that $u v \geq 0$, or $(a+2 x-16)(a-2 x+9) \geq 0$. In the system $O x a$

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-10.jpg?height=1447&width=848&top_left_y=533&top_left_x=1112)

the set of solutions to the last inequality is the area between the two lines $a+2 x-16=0$ and $a-2 x+9=0$, colored yellow on the diagram.

Each solution $x \geq 2$ of the inequality $\log _{\sqrt{a / 10}}\left(\frac{a+8 x-x^{2}}{20}\right) \geq 2$ is a solution to the equation $|a+2 x-16|+|a-2 x+9|=|2 a-7|$ for $a \in[9 ; 10) \cup[12 ; 16]$.

Answer: $a \in[9 ; 10) \cup[12 ; 16]$.