Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 399

\[y = x^{2} + px;\ \ \ y = kx + 1.\]

\[1)\ Точки\ пересечения:\]

\[x^{2} + px = kx + 1\]

\[x^{2} + (p - k)x - 1 = 0\]

\[D = (p - k)^{2} + 4\]

\[x = \frac{- (p - k) \pm \sqrt{(p - k)^{2} + 4}}{2}.\]

\[2)\ x_{2} - x_{1} = \sqrt{D};\ \ x_{2}x_{1} = - 1:\]

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

\[= (k - p)\sqrt{D};\]

\[x_{2}^{2} + x_{1}^{2} =\]

\[= \left( \frac{- b - \sqrt{D}}{2} \right)^{2} + \left( \frac{- b + \sqrt{D}}{2} \right)^{2} =\]

\[= \frac{b^{2} + 2b\sqrt{D} + D}{4} + \frac{b^{2} - 2b\sqrt{D} + D}{4} =\]

\[= \frac{b^{2} + D}{2} =\]

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

\[= (p - k)^{2} + 2 =\]

\[x_{2}^{3} - x_{1}^{3} =\]

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

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

\[3)\ S =\]

\[= \int_{x_{1}}^{x_{2}}{\left( (kx + 1) - \left( x^{2} + px \right) \right)\text{dx}} =\]

\[S = \frac{1}{6}D\sqrt{D} = \frac{1}{6}D^{\frac{3}{2}}.\]

\[4)\ Промежуток\ возрастания:\]

\[D^{'}(k) = 2(p - k) \bullet ( - 1) + 0;\]

\[D^{'}(k) = 2(k - p) \geq 0;\]

\[k \geq p.\]

\[5)\ Точка\ минимума:\]

\[k = p.\]

\[Ответ:\ \ k = p.\]