ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1023

\[\boxed{\mathbf{1023}\mathbf{.}}\]

\[1)\ y = x^{2} + 10;\ \ \ \ \ (0;\ 1);\]

\[Уравнениие\ касательной:\]

\[y^{'}(a) = \left( x^{2} \right)^{'} + (10)^{'} =\]

\[= 2x + 0 = 2x = 2a;\]

\[y(a) = a^{2} + 10;\]

\[y = a^{2} + 10 + 2a \bullet (x - a) =\]

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

\[= - a^{2} + 2ax + 10.\]

\[- a^{2} + 2a \bullet 0 + 10 = 1\]

\[- a^{2} = - 9\]

\[a^{2} = 9\ \]

\[a = \pm 3.\]

\[1)\ y =\]

\[= - ( - 3)^{2} + 2x \bullet ( - 3) + 10 =\]

\[= - 9 - 6x + 10 = 1 - 6x\]

\[x^{2} + 10 = 1 - 6x\]

\[x^{2} + 6x + 9 = 0\]

\[(x + 3)^{2} = 0\ \]

\[x = - 3.\]

\[2)\ y = - 3^{2} + 2 \bullet 3x + 10 =\]

\[= - 9 + 6x + 10 = 1 + 6x\]

\[x^{2} + 10 = 1 + 6x\]

\[x^{2} - 6x + 9 = 0\]

\[(x - 3)^{2} = 0\]

\[x = 3.\]

\[Пересечение\ касательных:\]

\[1 - 6x = 1 + 6x\]

\[1 - 1 = 6x + 6x\]

\[0 = 12x\ \]

\[x = 0.\]

\[= 9 - 27 + 27 + 9 - 27 + 27 =\]

\[= 9 + 9 = 18;\]

\[Ответ:\ \ 18.\]

\[2)\ y = \frac{1}{x};\ \ \ x = 1;\ \ x = 2;\]

\[Уравнениие\ касательной:\]

\[y^{'}(x) = \left( \frac{1}{x} \right)^{'} = - \frac{1}{x^{2}}\]

\[y^{'}(2) = - \frac{1}{2^{2}} = - \frac{1}{4}\]

\[y(2) = \frac{1}{2}\]

\[y = \frac{1}{2} - \frac{1}{4} \bullet (x - 2) =\]

\[= \frac{1}{2} - \frac{x}{4} + \frac{1}{2} = 1 - \frac{x}{4}.\]

\[Точки\ пересечения\ функций:\]

\[\frac{1}{x} = 1 - \frac{x}{4}\]

\[1 = x - \frac{x^{2}}{4}\]

\[4 = 4x - x^{2}\]

\[x^{2} - 4x + 4 = 0\]

\[(x - 2)^{2} = 0\ \]

\[x = 2.\]

\[S = \int_{1}^{2}{\left( \frac{1}{x} - \left( 1 - \frac{x}{4} \right) \right)\text{\ dx}} =\]

\[= \left. \ \left( \ln x - x + \frac{x^{2}}{2}\ :4 \right) \right|_{1}^{2} =\]

\[= \left. \ \left( \ln x - x + \frac{x^{2}}{8} \right) \right|_{1}^{2} =\]

\[= \ln 2 - 2 + \frac{2^{2}}{8} - \ln 1 + 1 - \frac{1^{2}}{8} =\]

\[= \ln 2 - 2 + \frac{4}{8} - 0 + 1 - \frac{1}{8} =\]

\[= \ln 2 - 1 + \frac{3}{8} = \ln 2 - \frac{5}{8}.\]

\[Ответ:\ \ \ln 2 - \frac{5}{8}.\]