12А. Найдите все значения параметра a, при каждом из которых корни уравнения \(\left( {a-1} \right){x^2}-\left( {a + 1} \right)\,x + a = 0\) принадлежат интервалу \(\left( {0;\,3} \right)\).
Если \(a = 1\), то уравнение является линейным и оно примет вид \(-2x + 1 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = \frac{1}{2}\, \in \,\left( {0;3} \right)\), то есть \(a = 1\) подходит. Первый случай (см. рис. 1): \(\left\{ {\begin{array}{*{20}{c}}{a-1 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{0 < {x_B} = -\frac{b}{{2a}} < 3,}\\{\begin{array}{*{20}{c}}{f\left( 0 \right) > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 3 \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {a + 1} \right)}^2}-4a\left( {a-1} \right) \ge 0,}\\{0 < \frac{{a + 1}}{{2\left( {a-1} \right)}} < 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{a > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{9a-9-3a-3 + a > 0}\end{array}}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\left\{ {\begin{array}{*{20}{c}}{a > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{-3{a^2} + 6a + 1 \ge 0,}\\{\frac{{a + 1}}{{2\left( {a-1} \right)}} > 0,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{\frac{{-5a + 7}}{{2\left( {a-1} \right)}} < 0,\,\,\,\,\,\,\,\,\,\,}\\{7a-12 > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a \in \left( {1;\infty } \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left[ {\frac{{3-2\sqrt 3 }}{3};\frac{{3 + 2\sqrt 3 }}{3}} \right],}\end{array}}\\{a\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;\infty } \right),\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right) \cup \left( {\frac{7}{5};\infty } \right),\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {\frac{{12}}{7};\infty } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,a \in \,\left( {\frac{{12}}{7};\frac{{3 + 2\sqrt 3 }}{3}} \right].\) \(\left\{ {\begin{array}{*{20}{c}}{a-1 < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{0 < {x_B} = -\frac{b}{{2a}} < 3,}\\{\begin{array}{*{20}{c}}{f\left( 0 \right) < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( 3 \right) < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{\left( {a + 1} \right)}^2}-4a\left( {a-1} \right) \ge 0,}\\{0 < \frac{{a + 1}}{{2\left( {a-1} \right)}} < 3,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{9a-9-3a-3 + a < 0}\end{array}}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\left\{ {\begin{array}{*{20}{c}}{a < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{-3{a^2} + 6a + 1 \ge 0,}\\{\frac{{a + 1}}{{2\left( {a-1} \right)}} > 0,\,\,\,\,\,\,\,\,\,\,\,}\\{\begin{array}{*{20}{c}}{\frac{{-5a + 7}}{{2\left( {a-1} \right)}} < 0,\,\,\,\,\,\,\,\,\,\,}\\{7a-12 < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{a \in \left( {\infty ;\,0} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left[ {\frac{{3-2\sqrt 3 }}{3};\frac{{3 + 2\sqrt 3 }}{3}} \right],}\end{array}}\\{a\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;\infty } \right),\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;1} \right) \cup \left( {\frac{7}{5};\infty } \right),\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;\frac{{12}}{7}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\emptyset .\) Таким образом, корни уравнения будут принадлежать интервалу \(\,\left( {0;3} \right)\) при \(\,a \in \left( {\frac{{12}}{7};\frac{{3 + 2\sqrt 3 }}{3}} \right] \cup \left\{ 1 \right\}.\) ОТВЕТ: \(\left( {\frac{{12}}{7};\,\frac{{3 + 2\sqrt 3 }}{3}} \right] \cup \left\{ 1 \right\}.\)
Если \(a \ne 1\), то графиком функции \(f\left( x \right) = \left( {a-1} \right){x^2}-\left( {a + 1} \right)x + a\) является парабола. Корни уравнения будут принадлежать интервалу \(\,\left( {0;3} \right)\), в следующих случаях:
Второй случай (см. рис. 2):