ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1621

\[\boxed{\mathbf{1621}\mathbf{.}}\]

\[\mathbf{Н}а\ отрезке\ \lbrack 0;\ 2\rbrack:\]

\[y = x^{2} + (a + 4)x + 2a + 3.\]

\[y^{'}(x) =\]

\[= \left( x^{2} \right)^{'} + (a + 4)(x)^{'} + (2a + 3)^{'} =\]

\[= 2x + (a + 4) + 0 = 2x + a + 4.\]

\[Промежуток\ возрастания:\]

\[2x + a + 4 > 0\]

\[2x > - a - 4\]

\[x > - \frac{a + 4}{2}.\]

\[Вершина\ лежит\ на\ отрезке\ \lbrack 0;\ 2\rbrack:\]

\[0 \leq - \frac{a + 4}{2} \leq 2\]

\[- 2 \leq \frac{a + 4}{2} \leq 0\]

\[- 4 \leq a + 4 \leq 0\]

\[- 8 \leq a \leq - 4.\]

\[На\ отрезке\ \lbrack 0;\ 2\rbrack:\]

\[\frac{1}{4}(a + 4)^{2} - \frac{1}{2}(a + 4)^{2} + 2a + 7 = 0\]

\[- \frac{1}{4}(a + 4)^{2} + 2a + 7 = 0\]

\[- \frac{1}{4}\left( a^{2} + 8a + 16 \right) + 2a + 7 = 0\]

\[- \frac{1}{4}a^{2} - 2a - 4 + 2a + 7 = 0\]

\[- \frac{1}{4}a^{2} + 3 = 0\ \ \ \ \ | \bullet ( - 4)\]

\[a^{2} - 12 = 0\]

\[a^{2} = 12\]

\[a = \pm \sqrt{12} = \pm 2\sqrt{3}.\]

\[Правее\ отрезка\ \lbrack 0;\ 2\rbrack:\]

\[x = 2\ \ и\ \ a \leq - 8;\]

\[- 4 = 2 \bullet 2 + 2(a + 4) + 2a + 3\]

\[- 4 = 4 + 2a + 8 + 2a + 3\]

\[- 4 = 4a + 15\]

\[4a = - 19\]

\[a = - \frac{19}{4} = - 4\frac{3}{4}.\]

\[Левее\ отрезка\ \lbrack 0;\ 2\rbrack:\]

\[x = 0\ \ и\ \ a \geq - 4;\]

\[- 4 = 0^{2} + (a + 4) \bullet 0 + 2a + 3\]

\[- 4 = 2a + 3\]

\[2a = - 7\]

\[a = - 3,5.\]

\[Ответ:\ \ a = - 3,5.\]