## Task B-4.3.

For which real parameters $b$ and $c$ is the natural domain of the function $f(x)=\frac{1}{\sqrt{-10 x^{2}+b x+c}}$ equal to the set $A=\left\{x \in \mathbb{R} \left\lvert\, \log _{\frac{1}{2}} \frac{2 x-1}{2-3 x}>0\right.\right\}$?
## Solution.

First, let's solve the inequality $\log _{\frac{1}{2}} \frac{2 x-1}{2-3 x}>0$.

The logarithm is defined if $\frac{2 x-1}{2-3 x}>0$, which holds for $x \in\left\langle\frac{1}{2}, \frac{2}{3}\right\rangle$.

Then we proceed as follows:

$$
\begin{gathered}
\log _{\frac{1}{2}} \frac{2 x-1}{2-3 x}>0 \\
\frac{2 x-1}{2-3 x} < 1 \quad \text{(since the base is less than 1, the argument must be less than 1 for the logarithm to be positive)} \\
2 x - 1 < 2 - 3 x \\
5 x < 3 \\
x < \frac{3}{5}
\end{gathered}
$$

Combining this with the domain of the logarithm, we get $x \in \left\langle \frac{1}{2}, \frac{3}{5} \right\rangle$.

If $x_{1}$ and $x_{2}$ are the solutions to the equation $-10 x^{2}+b x+c=0$, then the solution to the inequality $-10 x^{2}+b x+c>0$ is the interval $\left\langle x_{1}, x_{2}\right\rangle$, so it must be $x_{1}=\frac{1}{2}, x_{2}=\frac{3}{5}$.

Using Vieta's formulas, we will calculate the desired parameters:

$$
\begin{aligned}
& x_{1}+x_{2}=\frac{b}{10} \Rightarrow b=10 \cdot\left(\frac{1}{2}+\frac{3}{5}\right)=11 \\
& x_{1} \cdot x_{2}=-\frac{c}{10} \Rightarrow c=-10 \cdot \frac{1}{2} \cdot \frac{3}{5}=-3
\end{aligned}
$$

Thus, the desired parameters are $b=11, c=-3$.

Note: The desired parameters can also be obtained at the end by substituting $x_{1}=\frac{1}{2}, x_{2}=\frac{3}{5}$ into $-10 x^{2}+b x+c=0$:

$$
\begin{aligned}
& x_{1}=\frac{1}{2} \Rightarrow-10\left(\frac{1}{2}\right)^{2}+b \cdot \frac{1}{2}+c=0 \\
& x_{2}=\frac{3}{5} \Rightarrow-10\left(\frac{3}{5}\right)^{2}+b \cdot \frac{3}{5}+c=0
\end{aligned}
$$

Solving the system $3 b+5 c=18, b+2 c=5$ yields $b=11, c=-3$.