ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 259

\[1)\ y = 2\sqrt{x};\ \ \ y = 2\sqrt{6 - x};\]

\[2\sqrt{x} = 2\sqrt{6 - x}\]

\[\sqrt{x} = \sqrt{6 - x}\]

\[x = 6 - x\]

\[2x = 6\]

\[x = 3.\]

\[6 - x > 0;\ \ \ x < 6;\]

\[x > 0.\]

\[Угол\ между\ кривыми:\]

\[y^{'}(3) = \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{3}};\]

\[y^{'}(3) = \frac{- 2}{2\sqrt{6 - x}} = - \frac{1}{\sqrt{3}};\]

\[a = \frac{\pi}{6};\text{\ \ \ }\]

\[b = - \frac{\pi}{6};\]

\[c = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}.\]

\[Ответ:\ \ \frac{\pi}{3}.\]

\[2)\ y = \sqrt{2x + 1},\ \ \ y = 1;\]

\[\sqrt{2x + 1} = 1;\]

\[2x + 1 = 1;\]

\[2x = 0;\]

\[x = 0.\]

\[2x + 1 > 0\]

\[x > - \frac{1}{2}.\]

\[Угол\ между\ кривыми:\]

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

\[y^{'}(0) = 0;\]

\[a = \frac{\pi}{4};\text{\ \ \ }\]

\[b = 0;\]

\[c = \frac{\pi}{4} - 0 = \frac{\pi}{4}.\]

\[Ответ:\ \ \frac{\pi}{4}.\]