Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1041

\[y = - 5;\]

\[y = 4x^{2} + 8ax - a;\]

\[y = 4ax^{2} - 8x + a - 2.\]

\[1)\ y = 4x^{2} + 8ax - a:\]

\[x_{0} = - \frac{b}{2a} = - \frac{8a}{2 \bullet 4} = - \frac{8a}{8} = - a;\]

\[y_{0} = 4a^{2} - 8a^{2} - a =\]

\[= - 4a^{2} - a > - 5;\]

\[4a^{2} + a - 5 < 0\]

\[D = 1 + 80 = 81:\]

\[a_{1} = \frac{- 1 - 9}{2 \bullet 4} = - \frac{5}{4};\ \]

\[a_{2} = \frac{- 1 + 9}{2 \bullet 4} = 1;\]

\[\left( a + \frac{5}{4} \right)(a - 1) < 0\]

\[- \frac{5}{4} < a < 1.\]

\[2)\ y = 4ax^{2} - 8x + a - 2:\]

\[x_{0} = - \frac{b}{2a} = - \frac{- 8}{2 \bullet 4a} = \frac{8}{8a} = \frac{1}{a};\]

\[y_{0} = \frac{4}{a} - \frac{8}{a} + a - 2 =\]

\[= a - \frac{4}{a} - 2 > - 5;\]

\[a + 3 - \frac{4}{a} > 0\]

\[\frac{a^{2} + 3a - 4}{a} > 0\]

\[D = 9 + 16 = 25\]

\[a_{1} = \frac{- 3 - 5}{2} = - 4;\]

\[a_{2} = \frac{- 3 + 5}{2} = 1;\]

\[\frac{(a + 4)(a - 1)}{a} > 0\]

\[- 4 < a < 0;\ \ \ a > 1.\]

\[Ответ:\ \ \]

\[a \in ( - \infty;\ - 4) \cup \left( - \frac{5}{4};\ 0 \right).\]