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МатематикаГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1113
Задание 1113
\[f(x) = x^{2} + 2x - 3;\ \]
\[y = kx + 1.\]
\[1)\ x^{2} + 2x - 3 = kx + 1\]
\[x^{2} + (2 - k)x - 4 = 0\]
\[D = (2 - k)^{2} + 4 \bullet 4 =\]
\[= 4 - 4k + k^{2} + 16 =\]
\[= k^{2} - 4k + 20\]
\[x = \frac{k - 2 \pm \sqrt{k^{2} - 4k + 20}}{2} =\]
\[= \frac{k - 2 \pm \sqrt{D}}{2}.\]
\[2)\ S =\]
\[= \int_{x_{1}}^{x_{2}}{\left( kx + 1 - x^{2} - 2x + 3 \right)\text{dx}} =\]
\[= \int_{x_{1}}^{x_{2}}{\left( kx + 4 - x^{2} - 2x \right)\text{dx}} =\]
\[= \left. \ \left( k \bullet \frac{x^{2}}{2} + 4 \bullet \frac{x^{1}}{1} - \frac{x^{3}}{3} - 2 \bullet \frac{x^{2}}{2} \right) \right|_{x_{1}}^{x_{2}} =\]
\[= \left. \ \left( \frac{x^{2}(k - 2)}{2} + 4x - \frac{x^{3}}{3} \right) \right|_{x_{1}}^{x_{2}} =\]
\[3)\ x_{2} - x_{1} =\]
\[= \frac{k - 2 + \sqrt{D}}{2} - \frac{k - 2 - \sqrt{D}}{2} =\]
\[= \frac{2\sqrt{D}}{2} = \sqrt{D};\]
\[x_{2} + x_{1} =\]
\[= \frac{k - 2 + \sqrt{D}}{2} + \frac{k - 2 - \sqrt{D}}{2} =\]
\[= \frac{2(k - 2)}{2} = k - 2;\]
\[x_{2}^{2} - x_{1}^{2} = \left( x_{2} - x_{1} \right)\left( x_{2} + x_{1} \right) =\]
\[= (k - 2)\sqrt{D};\]
\[x_{2}x_{1} = \frac{c}{a} = - 4,\ \ \ x_{2}^{2} + x_{1}^{2} =\]
\[= \left( x_{2} + x_{1} \right)^{2} - 2x_{2}x_{1} =\]
\[= (k - 2)^{2} + 8;\]
\[x_{2}^{3} - x_{1}^{3} =\]
\[= \left( x_{2} - x_{1} \right)\left( x_{2}^{2} + x_{2}x_{1} + x_{1}^{2} \right) =\]
\[= \sqrt{D}\left( (k - 2)^{2} + 4 \right).\]
\[4)\ Подставим:\]
\[S(k) =\]
\[= \frac{\sqrt{D}(k - 2)^{2} + 16\sqrt{D}}{6} =\]
\[= \frac{\sqrt{D}\left( k^{2} - 4k + 4 + 16 \right)}{6} =\]
\[= \frac{1}{6}\sqrt{D} \bullet \left( k^{2} - 4k + 20 \right) =\]
\[= \frac{1}{6}\sqrt{D} \bullet D = \frac{1}{6}\sqrt{D^{3}}.\]
\[5)\ S^{'}(k) = \frac{1}{6}{\left( k^{2} - 4k + 20 \right)^{\frac{3}{2}}}^{'} =\]
\[= \frac{1}{6} \bullet \frac{3}{2}\left( k^{2} - 4k + 20 \right)^{\frac{1}{2}} \bullet (2k - 4) =\]
\[= \frac{\sqrt{k^{2} - 4k + 20} \bullet (k - 2)}{2}.\]
\[6)\ Промежуток\ возрастания:\]
\[k - 2 \geq 0\]
\[k \geq 2.\]
\[7)\ Точка\ минимума:\]
\[k = 2.\]
\[Ответ:\ \ k = 2.\]
