## Task B-2.1.

For which values of the real parameter $m$ does the inequality $\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2}<47$ hold if $x_{1}$ and $x_{2}$ are the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$?
## Solution.

For the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$, we have

$$
x_{1}+x_{2}=2^{m-1}-5 \quad \text{and} \quad x_{1} x_{2}=1
$$

Therefore,

$$
\begin{aligned}
\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2} & =\frac{x_{1}^{4}+x_{2}^{4}}{x_{1}^{2} x_{2}^{2}}=\frac{\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-2 x_{1}^{2} x_{2}^{2}}{x_{1}^{2} x_{2}^{2}} \\
& =\frac{\left(\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right)^{2}-2\left(x_{1} x_{2}\right)^{2}}{\left(x_{1} x_{2}\right)^{2}} \\
& =\frac{\left(\left(2^{m-1}-5\right)^{2}-2\right)^{2}-2 \cdot 1^{2}}{1^{2}}=\left(\left(2^{m-1}-5\right)^{2}-2\right)^{2}-2
\end{aligned}
$$

The given condition is equivalent to

$$
\begin{array}{ll} 
& \left(\left(2^{m-1}-5\right)^{2}-2\right)^{2}-2<47 \\
\Leftrightarrow & \left(\left(2^{m-1}-5\right)^{2}-2\right)^{2}<49 \\
\Leftrightarrow & \left|\left(2^{m-1}-5\right)^{2}-2\right|<7 \\
\Leftrightarrow & -7<\left(2^{m-1}-5\right)^{2}-2<7 \\
\Leftrightarrow & -5<\left(2^{m-1}-5\right)^{2}<9
\end{array}
$$

The inequality $-5<\left(2^{m-1}-5\right)^{2}$ holds for all real numbers $m$, and the inequality $\left(2^{m-1}-5\right)^{2}<9$ is equivalent to $-3<2^{m-1}-5<3$ or $2<2^{m-1}<8$, from which we get $1<m-1<3$. Finally, the desired values of the parameter $m$ are $m \in (2,4)$.