ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 300

\[y = kx + b - прямая:\]

\[2 = k \bullet 1 + b\]

\[b = 2 - k.\]

\[1)\ Преобразуем:\]

\[y = kx + (2 - k) = k(x - 1) + 2.\]

\[2)\ k(x - 1) + 2 = 0\]

\[k(x - 1) = - 2\]

\[x - 1 = - \frac{2}{k}\]

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

\[3)\ y = k(0 - 1) + 2 = 2 - k.\]

\[4)\ S(k) =\]

\[= \frac{1}{2}xy = \frac{1}{2}\left( 1 - \frac{2}{k} \right)(2 - k) =\]

\[= \frac{1}{2}\left( 2 - k - \frac{4}{k} + 2 \right) =\]

\[= \frac{1}{2}\left( 4 - k - \frac{4}{k} \right) =\]

\[= \frac{1}{2}\left( 0 - 1 - 4 \bullet \left( - \frac{1}{k^{2}} \right) \right) =\]

\[= \frac{4 - k^{2}}{2k^{2}}.\]

\[5)\ Промежуток\ возрастания:\]

\[4 - k^{2} \geq 0\]

\[k^{2} - 4 \leq 0;\]

\[(k + 2)(k - 2) \leq 0\]

\[- 2 \leq k \leq 2.\]

\[6)\ Точка\ минимума:\]

\[k = - 2.\]

\[Ответ:\ - 2.\]

\[\mathbf{ }\]

\[\ y = (x - 1)^{2};\ \ 0 \leq x \leq 2.\]

\[1)\ \ (1 + k) - точка\ основания:\]

\[b = 1 - k - тоже\ точка\ \]

\[основания;\]

\[y_{1} = (1 + k - 1)^{2} = k^{2};\]

\[y_{2} = (1 - k - 1)^{2} = k^{2}.\]

\[2)\ Высота\ и\ основания\ \]

\[трапеции:\]

\[x_{0} = 1;\ \ \ y_{0} = 0;\]

\[0 \leq y \leq 1;\text{\ \ \ }\]

\[k^{2} \leq 1;\]

\[h = 1 - k^{2};\]

\[a = (k + 1) - (1 - k) = 2k;\]

\[b = 5 - 3 = 2.\]

\[3)\ S(k) = \frac{1}{2}h(a + b) =\]

\[= \frac{1}{2}\left( 1 - k^{2} \right)(2 + 2k) =\]

\[= \left( 1 - k^{2} \right)(1 + k) =\]

\[= 1 + k - k^{2} - k^{3};\]

\[S^{'}(k) = 0 + 1 - 2k - 3k^{2} =\]

\[= 1 - 2k - 3k^{2}.\]

\[4)\ 1 - 2k - 3k^{2} \geq 0\]

\[3k^{2} + 2k - 1 \leq 0\]

\[D = 4 + 12 = 16\]

\[k_{1} = \frac{- 2 - 4}{2 \bullet 3} = - 1;\text{\ \ }\]

\[k_{2} = \frac{- 2 + 4}{2 \bullet 3} = \frac{1}{3};\]

\[(k + 1)\left( k - \frac{1}{3} \right) \leq 0\]

\[- 1 \leq k \leq \frac{1}{3}.\]

\[5)\ Точка\ максимума:\]

\[k = \frac{1}{3};\text{\ \ \ }\]

\[S = 1 + \frac{1}{3} - \frac{1}{9} - \frac{1}{27} = \frac{32}{27}.\]

\[Ответ:\ \ \frac{32}{27}.\]