15А. Найдите все значения параметра a, при каждом из которых корни уравнения \(a\,{x^2} + \left( {4-2a} \right)\,x + 1 = 0\) по модулю меньше 1.
Если \(a = 0\), то уравнение будет являться линейным и примет вид \(4x + 1 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = -\frac{1}{4}\, \in \,\left( {-1;1} \right)\), то есть \(a = 0\) подходит. Первый случай (см. рис. 1): \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {4-2a} \right)}^2}-4a \ge 0,}\end{array}}\\{-1 < \frac{{a-2}}{a} < 1,\,\,\,\,\,\,\,\,}\\{a-4 + 2a + 1 > 0,}\\{a + 4-2a + 1 > 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{a^2}-5a + 4 \ge 0,}\\{\frac{{-2}}{a} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{\frac{{2a-2}}{a} > 0,\,\,\,\,\,\,\,\,\,}\\{3a-3 > 0,\,\,\,\,\,\,\,\,\,\,\,}\\{5-a > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a\, \in \,\left( {0;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right] \cup \left[ {4;\infty } \right),}\\{a\, \in \,\left( {0;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{a\, \in \,\left( {-\infty ;0} \right) \cup \left( {1;\infty } \right),}\\{a\, \in \,\left( {1;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;\,5} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a \in \left[ {4;\,5} \right).\) \(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {4-2a} \right)}^2}-4a \ge 0,}\end{array}}\\{-1 < \frac{{a-2}}{a} < 1,\,\,\,\,\,\,\,\,}\\{a-4 + 2a + 1 < 0,}\\{a + 4-2a + 1 < 0\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{a^2}-5a + 4 \ge 0,}\\{\frac{{-2}}{a} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{\frac{{2a-2}}{a} > 0,\,\,\,\,\,\,\,\,\,}\\{3a-3 < 0,\,\,\,\,\,\,\,\,\,\,\,}\\{5-a < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;0} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right] \cup \left[ {4;\infty } \right),}\\{a\, \in \,\left( {0;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{a\, \in \,\left( {-\infty ;0} \right) \cup \left( {1;\infty } \right),}\\{a\, \in \,\left( {-\infty ;1} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {\,5;\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .\) Таким образом, корни уравнения будут по модулю меньше 1 при \(a\, \in \,\left\{ 0 \right\} \cup \left[ {4;\,5} \right).\) ОТВЕТ: \(\left\{ 0 \right\} \cup \left[ {4;\,5} \right).\)
Если \(a \ne 0\), то графиком функции \(f\left( x \right) = a\,{x^2} + \left( {4-2a} \right)x + 1\) является парабола. Корни уравнения по модулю будут меньше 1, то есть принадлежали интервалу \(\,\left( {-1;1} \right)\), в следующих случаях:
Второй случай (см. рис. 2):