18А. Найдите все значения параметра a, при каждом из которых уравнение \(2\left( {a + 5} \right)\,{x^2} + 2\left( {a-7} \right)\,x-a + 4 = 0\) имеет два различных корня, модуль каждого из которых меньше 1.
Если \(a = -5\), то уравнение будет линейным, которое не может иметь два различных корня. Поэтому \(a = -5\) не подходит. Первый случай (см. рис. 1): \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4{{\left( {a-7} \right)}^2}-8\left( {a + 5} \right)\left( {4-a} \right) > 0,\,\,\,\,\,}\end{array}}\\{-1 < \frac{{7-a}}{{2\left( {a + 5} \right)}} < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2a + 10-2a + 14-a + 4 > 0,\,\,\,\,\,\,\,\,\,\,\,}\\{2a + 10 + 2a-14-a + 4 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{12{a^2}-48a + 36 > 0,}\end{array}}\\{\frac{{-3a-3}}{{2\left( {a + 5} \right)}} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{a + 17}}{{2\left( {a + 5} \right)}} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{28-a > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3a > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {-5;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right) \cup \left( {3;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-5} \right) \cup \left( {-1;\infty } \right),\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;-17} \right) \cup \left( {-5;\infty } \right),\,\,\,\,}\\{a \in \left( {-\infty ;28} \right)\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a \in \left( {0;\,\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\, \Leftrightarrow \,\,\,\,\,a\, \in \,\left( {0;1} \right) \cup \left( {3;28} \right).\) \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{4{{\left( {a-7} \right)}^2}-8\left( {a + 5} \right)\left( {4-a} \right) > 0,\,\,\,\,\,}\end{array}}\\{-1 < \frac{{7-a}}{{2\left( {a + 5} \right)}} < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{2a + 10-2a + 14-a + 4 < 0,\,\,\,\,\,\,\,\,\,\,\,}\\{2a + 10 + 2a-14-a + 4 < 0\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < -5,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{12{a^2}-48a + 36 > 0,}\end{array}}\\{\frac{{-3a-3}}{{2\left( {a + 5} \right)}} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{a + 17}}{{2\left( {a + 5} \right)}} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{28-a < 0,\,\,\,\,\,\,\,\,\,\,\,\,}\\{3a < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {-\infty ;-5} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right) \cup \left( {3;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-6} \right) \cup \left( {-2;\infty } \right),\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;-17} \right) \cup \left( {-5;\infty } \right),\,}\\{a \in \left( {28;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a \in \left( {-\infty ;0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\emptyset .\) Таким образом, уравнение будет иметь два различных корня модуль каждого из которых меньше 1 при \(a\, \in \,\left( {0;1} \right) \cup \left( {3;28} \right).\) ОТВЕТ: \(\left( {0;\,1} \right) \cup \left( {3;\,28} \right).\)
Если \(a \ne -5\), то графиком функции \(f\left( x \right) = 2\left( {a + 5} \right){x^2} + 2\left( {a-7} \right)x-a + 4\) является парабола. Уравнение будет иметь два различных корня, модуль каждого из которых меньше 1, то есть принадлежат интервалу \(\,\left( {-1;1} \right)\), в следующих случаях:
Второй случай (см. рис. 2):