Task 2. (10 points) Find the greatest value of the parameter $b$ for which the inequality $b \sqrt{b}\left(x^{2}-10 x+25\right)+\frac{\sqrt{b}}{\left(x^{2}-10 x+25\right)} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right|$ has at least one solution.
Solution.

$$
\begin{aligned}
& b \sqrt{b}\left(x^{2}-10 x+25\right)+\frac{\sqrt{b}}{\left(x^{2}-10 x+25\right)} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right| \\
& b \sqrt{b}(x-5)^{2}+\frac{\sqrt{b}}{(x-5)^{2}} \leq \frac{1}{5} \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right| \\
& \sqrt[4]{b^{3}}(x-5)^{2}+\frac{1}{\sqrt[4]{b}(x-5)^{2}} \leq \frac{1}{5}\left|\sin \frac{\pi x}{10}\right|
\end{aligned}
$$

Applying the Cauchy inequality about the arithmetic mean and geometric mean, we transition to the inequality:

$$
2 \sqrt{\sqrt[4]{b^{3}}(x-5)^{2} \cdot \frac{1}{\sqrt[4]{b}(x-5)^{2}}} \leq \sqrt[4]{b^{3}}(x-5)^{2}+\frac{1}{\sqrt[4]{b}(x-5)^{2}} \leq \frac{1}{5} \cdot\left|\sin \frac{\pi x}{10}\right| \leq \frac{1}{5}
$$

Solving this inequality, we get

$$
2 \sqrt[4]{b} \leq \frac{1}{5}, \sqrt[4]{b} \leq \frac{1}{10}, b \leq \frac{1}{10000}
$$

If $b=\frac{1}{10000}$, then the inequality $\frac{1}{1000}(x-5)^{2}+\frac{10}{(x-5)^{2}} \leq \frac{1}{5} \cdot\left|\sin \frac{\pi x}{10}\right|$ has a solution at $x=15$:

$$
\frac{100}{1000}+\frac{10}{100} \leq \frac{1}{5} \cdot\left|\sin \frac{\pi 15}{10}\right|, 1 \leq\left|\sin \frac{3 \pi}{2}\right|
$$

## 2 points.

This is the only value for which the inequality has a solution.

$b \sqrt{b}\left(x^{2}-10 x+25\right)+\frac{\sqrt{b}}{\left(x^{2}-10 x+25\right)} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right|$,

$b \sqrt{b}(x-5)^{2}+\frac{\sqrt{b}}{(x-5)^{2}} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right| \leq \frac{1}{5} \sqrt[4]{b^{3}}$,

$b\left(\sqrt{b}(x-5)^{2}+\frac{1}{\sqrt{b}(x-5)^{2}}\right) \leq \frac{1}{5} \sqrt[4]{b^{3}}$.

In the parentheses, we obtained the sum of reciprocals, which have the property $a+\frac{1}{a} \geq 2$.

Therefore, we get the system of inequalities:

$$
\left\{\begin{array}{l}
\sqrt{b}(x-5)^{2}+\frac{1}{\sqrt{b}(x-5)^{2}} \geq 2 \\
\sqrt{b}(x-5)^{2}+\frac{1}{\sqrt{b}(x-5)^{2}} \leq \frac{1}{5} \sqrt[4]{b}
\end{array}\right.
$$

This system has a solution only if $\frac{1}{5} \sqrt[4]{b}=2, \sqrt[4]{b}=0.1, b=0.001$.

Answer. 0.0001.