Задание №89455 ЕГЭ Профильной математике

Уравнение вида $\left|\mathrm{f}\right(\mathrm{x}\left)\right|=\mathrm{g}\left(\mathrm{x}\right)$, равносильно совокупности:

$\left\{\begin{array}{l}\mathrm{g}\left(\mathrm{x}\right)\ge 0\\ \left[\begin{array}{c}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right); \\ \mathrm{f}\left(\mathrm{x}\right)=-\mathrm{g}\left(\mathrm{x}\right).\end{array}\right.\end{array}\right.$

$$\begin{gathered} \left\{\begin{array}{l}1-\mathrm{x}\ge 0\\ \left[\begin{array}{c}{\mathrm{x}}^2-2\mathrm{x}-1=1-\mathrm{x} \\ {\mathrm{x}}^2-2\mathrm{x}-1=-1+\mathrm{x}\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\le 1\\ \left[\begin{array}{c}{\mathrm{x}}^2-\mathrm{x}-2=0\\ {\mathrm{x}}^2-3\mathrm{x}=0\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\le 1\\ \left[\begin{array}{c}(\mathrm{x}-2)(\mathrm{x}+1)=0\\ \mathrm{x}(\mathrm{x}-3)=0\end{array}\right.\end{array}\right. \\ \left\{\begin{array}{l}\mathrm{x}\le 1\\ \left[\begin{array}{c}\mathrm{x}=2\\ \mathrm{x}=-1\\ \mathrm{х}=0\\ \mathrm{х}=3\end{array}\right.\end{array}\right. \\ \left[\begin{array}{c}\mathrm{x}=-1\\ \mathrm{x}=0.\end{array}\right. \end{gathered}$$

б) Отберём корни, принадлежащие отрезку  $[-\sqrt{2};-\frac{\mathrm{\pi }}{4}].$

Применим значение $\mathrm{\pi }=3,14.$

Тогда промежуток принимает вид $[-\sqrt{2};-\frac{3,14}{4}].$

$$\begin{gathered} -\sqrt{2}\le -1\le -\frac{3,14}{4} \Rightarrow -1\in \left[\begin{array}{cc}-\sqrt{2};& -\frac{\mathrm{\pi }}{4}\end{array}\right]. \\ 0\notin \left[\begin{array}{cc}-\sqrt{2};& -\frac{\mathrm{\pi }}{4}\end{array}\right]. \end{gathered}$$

Ответ: а) $-1; 0;$ б) $-1.$