Задание №18 — Системы с параметром

Решение:
$\left\{\begin{matrix} y = - \left|x - a\right| + 1 , \\ \left|y\right| = - + x^{2} - 2 x \end{matrix}\right.$
$\left\{\begin{matrix} y = - \left|x - a\right| + 1 \\ \left\{\begin{matrix} \begin{matrix}y \geq 0 \\ y = - x^{2} - 2 x\end{matrix} \end{matrix}\right. \\ \left\{\begin{matrix} \begin{matrix}y < 0 \\ y = x^{2} + 2 x\end{matrix} \end{matrix}\right. \end{matrix}\right.$


$a \in \left(\left(3\right) ; \left(2\right)\right) \cup \left(\left(2\right) ; \left(1\right)\right)$
$x_{0} = \frac{- 2}{2} = - 1$
$- x^{2} - 2 x = 0 \Rightarrow x_{1} = 0 ; x_{2} = - 2$
$y_{0} \left(1\right) = - 1 + 2 = + 1$
$x_{0} = - \frac{2}{2} = - 1$
$x^{2} + 2 x = 0 \Rightarrow x_{1} = 0 ; x_{2} = - 2$
$y_{0} \left(1\right) = 1 - 2 = - 1 .$
1) $y = - x + a + 1$
$y = - x^{2} - 2 x$
$- x^{2} - 2 x + x - a - 1 = 0$
$- x^{2} - x - a - 1 = 0$
$x^{2} + x + a + 1 = 0$
$D = 1 - 4 \left(a + 1\right) = 1 - 4 a - 4 = - 3 - 4 a$
$- 3 - 4 a = 0$
$a = - \frac{3}{4} .$
2) $a = 1 .$
3) $y = a - x + 1$
$y = x^{2} + 2 x$
$x^{2} + 2 x = a - x + 1 \Rightarrow$$x^{2} + 3 x - a + 1 = 0$
$D = 9 - 4 \left(- a + 1\right) =$$9 + 4 a - 4 = 5 + 4 a$
$5 + 4 a = 0$
$a = - \frac{5}{4} .$

Ответ: $a \in \left(- \frac{5}{4} ; 1\right) \cup \left(- 1 ; - \frac{3}{4}\right)$

Источник: ФИПИ