16А. Найдите все значения параметра a, при каждом из которых корни уравнения  \(a\,{x^2}-\left( {a + 1} \right)\,x + 2 = 0\)  по модулю меньше 1.

Ответ

ОТВЕТ: \(\left[ {3 + 2\sqrt 2 ;\,\infty } \right).\)

Решение

Если \(a = 0\), то уравнение будет являться линейным и примет вид \(-x + 2 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = 2 \notin \,\left( {-1;1} \right)\), то есть \(a = 0\) не подходит.

Если \(a \ne 0\), то графиком функции  \(f\left( x \right) = a\,{x^2}-\left( {a + 1} \right)x + 2\)  является парабола. Корни уравнения по модулю будут меньше 1, то есть принадлежали интервалу \(\,\left( {-1;1} \right)\), в следующих случаях:

Первый случай (см. рис. 1):

\(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {a + 1} \right)}^2}-8a \ge 0,\,}\end{array}}\\{-1 < \frac{{a + 1}}{{2a}} < 1,\,\,\,\,\,\,\,}\\{a + a + 1 + 2 > 0,}\\{a-a-1 + 2 > 0}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{a^2}-6a + 1 \ge 0,}\end{array}}\\{\frac{{1-a}}{{2a}} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{3a + 1}}{{2a}} > 0,\,\,\,\,\,\,\,\,\,}\\{2a + 3 > 0\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a \in \left( {0;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{a\, \in \,\left( {-\infty ;3-2\sqrt 2 } \right] \cup \left[ {3 + 2\sqrt 2 ;\infty } \right),}\\{a\, \in \,\left( {-\infty ;0} \right) \cup \left( {1;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-\frac{1}{3}} \right) \cup \left( {0;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\,a\, \in \,\left( {-\frac{3}{2};\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a\, \in \,\left[ {3 + 2\sqrt 2 ;\infty } \right).\)

Второй случай (см. рис. 2):

\(\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\\{-1 < {x_B} = -\frac{b}{{2a}} < 1,}\\{f\left( {-1} \right) < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 1 \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {a + 1} \right)}^2}-8a \ge 0,\,}\end{array}}\\{-1 < \frac{{a + 1}}{{2a}} < 1,\,\,\,\,\,\,\,}\\{a + a + 1 + 2 < 0,}\\{a-a-1 + 2 < 0}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{a^2}-6a + 1 \ge 0,}\end{array}}\\{\frac{{1-a}}{{2a}} < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\frac{{3a + 1}}{{2a}} > 0,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{1 < 0,\,\,\,\,\,\,\,\,\,\,\,}\\{2a + 3 < 0}\end{array}\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\emptyset .\)

Таким образом, корни уравнения будут по модулю меньше 1 при \(a\, \in \,\left[ {3 + 2\sqrt 2 ;\infty } \right).\)

ОТВЕТ: \(\left[ {3 + 2\sqrt 2 ;\,\infty } \right).\)