Задача 10. Решите неравенство при всех значениях параметра а: \(\left( {1-{a^2}} \right){x^2} + 2\,a\,x + 1 \geqslant 0.\)
ОТВЕТ: \(x \in \left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\,\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right]\) при \(a \in \left( {-\infty ;\,-1} \right) \cup \left( {1;\,\infty } \right);\) \(x \in \left( {-\infty ;\,\dfrac{1}{2}} \right]\) при \(a = -1;\) \(x \in \left[ {-\dfrac{1}{2};\,\infty } \right)\) при \(a = 1;\) \(x \in \left( {-\infty ;\,\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right] \cup \left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\,\infty } \right)\) при \(a \in \left( {-1;\,-\dfrac{{\sqrt 2 }}{2}} \right) \cup \left( {\dfrac{{\sqrt 2 }}{2};\,1} \right);\) \(x \in R\) при \(a \in \left[ {-\dfrac{{\sqrt 2 }}{2};\,\dfrac{{\sqrt 2 }}{2}} \right].\)
Если \(a = 1\), то неравенство примет вид: \(2x + 1 \ge 0\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x\, \in \,\left[ {-\dfrac{1}{2};\infty } \right)\). Если \(a = -1\), то неравенство примет вид: \(-2x + 1 \ge 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;\dfrac{1}{2}} \right]\). Если \(a \ne \pm 1\), то: \(D = 4{a^2}-4 + 4{a^2} = 8{a^2}-4\). Если \(1-{a^2} < 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,a\, \in \,\left( {-\infty ;-1} \right) \cup \left( {1;\infty } \right)\), то: \(D > 0;\,\,\,\,\,\,\,\,{x_1} = \dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}};\,\,\,\,\,\,\,\,\,{x_2} = \dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}}\) и тогда решением неравенства будет являться: \(x\, \in \,\left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right].\) Если \(\left\{ {\begin{array}{*{20}{c}}{1-{a^2} > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{D = 8{a^2}-4 > 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{a\, \in \,\left( {-1;1} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{a\, \in \,\left( {-\infty ;-\dfrac{{\sqrt 2 }}{2}} \right) \cup \left( {\dfrac{{\sqrt 2 }}{2};\infty } \right)}\end{array}\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\,\,\,\,\,\,\,a \in \left( {-1;-\dfrac{{\sqrt 2 }}{2}} \right) \cup \left( {\dfrac{{\sqrt 2 }}{2};1} \right),\) то решением неравенства будет являться: \(x\, \in \,\left( {-\infty ;\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right] \cup \left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\infty } \right).\) Если \(D = 8{a^2}-4 \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,a\, \in \,\left[ {-\dfrac{{\sqrt 2 }}{2};\dfrac{{\sqrt 2 }}{2}} \right]\), то решением неравенства является: \(x\, \in \,R\). ОТВЕТ: \(x \in \left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\,\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right]\) при \(a \in \left( {-\infty ;\,-1} \right) \cup \left( {1;\,\infty } \right);\) \(x \in \left( {-\infty ;\,\dfrac{1}{2}} \right]\) при \(a = -1;\) \(x \in \left[ {-\dfrac{1}{2};\,\infty } \right)\) при \(a = 1;\) \(x \in \left( {-\infty ;\,\dfrac{{-a-\sqrt {2{a^2}-1} }}{{1-{a^2}}}} \right] \cup \left[ {\dfrac{{-a + \sqrt {2{a^2}-1} }}{{1-{a^2}}};\,\infty } \right)\) при \(a \in \left( {-1;\,-\dfrac{{\sqrt 2 }}{2}} \right) \cup \left( {\dfrac{{\sqrt 2 }}{2};\,1} \right);\) \(x \in R\) при \(a \in \left[ {-\dfrac{{\sqrt 2 }}{2};\,\dfrac{{\sqrt 2 }}{2}} \right].\)