Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1622

\[\boxed{\mathbf{1622}\mathbf{.}}\]

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

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

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

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

\[= 4 \bullet 2x - 4a + 0 = 8x - 4a.\]

\[Промежуток\ возрастания:\]

\[8x - 4a > 0\]

\[2x - a > 0\]

\[2x > a\]

\[x > \frac{a}{2}.\]

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

\[0 \leq \frac{a}{2} \leq 2\]

\[0 \leq a \leq 4.\]

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

\[3 = 4 \bullet \frac{a^{2}}{4} - 4a \bullet \frac{a}{2} + a^{2} - 2a + 2\]

\[a^{2} - 2a^{2} + a^{2} - 2a + 2 - 3 = 0\]

\[- 2a - 1 = 0\]

\[2a = - 1\]

\[a = - \frac{1}{2}.\]

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

\[x = 2\ \ и\ \ a \geq 4;\]

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

\[16 - 8a + a^{2} - 2a + 2 - 3 = 0\]

\[a^{2} - 10a + 15 = 0\]

\[D = 100 - 60 = 40 = 4 \bullet 10\]

\[a = \frac{10 \pm \sqrt{40}}{2} = \frac{10 \pm 2\sqrt{10}}{2} =\]

\[= 5 \pm \sqrt{10}.\]

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

\[x = 0\ \ и\ \ a \leq 0;\]

\[3 = 4 \bullet 0^{2} + 8a \bullet 0 + a^{2} - 2a + 2\]

\[a^{2} - 2a + 2 - 3 = 0\]

\[a^{2} - 2a - 1 = 0\]

\[D = 4 + 4 = 8 = 4 \bullet 2\]

\[a = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} =\]

\[= 1 \pm \sqrt{2}.\]

\[Ответ:\ \ \]

\[a_{1} = 1 - \sqrt{2};\ \ a_{2} = 5 + \sqrt{10}.\]