ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1618

\[\boxed{\mathbf{1618}\mathbf{.}}\]

\[f(x) = x^{2} + 2x - 3;y = kx + 1:\]

\[x^{2} + 2x - 3 = kx + 1\]

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

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

\[= 4 - 4k + k^{2} + 16 =\]

\[= k^{2} - 4k + 20\]

\[x = \frac{k - 2 \pm \sqrt{k^{2} - 4k + 20}}{2} =\]

\[= \frac{k - 2 \pm \sqrt{D}}{2}.\]

\[S = \int_{x_{1}}^{x_{2}}\left( kx + 1 - x^{2} - 2x + 3 \right) =\]

\[= \int_{x_{1}}^{x_{2}}\left( kx + 4 - x^{2} - 2x \right) =\]

\[= \left. \ \left( k \bullet \frac{x^{2}}{2} + 4 \bullet \frac{x^{1}}{1} - \frac{x^{3}}{3} - 2 \bullet \frac{x^{2}}{2} \right) \right|_{x_{1}}^{x_{2}} =\]

\[= \left. \ \left( \frac{x^{2}(k - 2)}{2} + 4x - \frac{x^{3}}{3} \right) \right|_{x_{1}}^{x_{2}} =\]

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

\[= \frac{k - 2 + \sqrt{D}}{2} - \frac{k - 2 - \sqrt{D}}{2} =\]

\[= \frac{2\sqrt{D}}{2} = \sqrt{D};\]

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

\[= \frac{1}{4} \bullet 4(k - 2)\sqrt{D} = (k - 2)\sqrt{D}.\]

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

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

\[= \frac{1}{4}\sqrt{D}\left( D + 3(k - 2)^{2} \right).\]

\[Подставим:\]

\[= \frac{\sqrt{D}}{6} \bullet \left( \frac{3}{2}k^{2} - 6k + 6 + 24 - \frac{D}{2} \right) =\]

\[= \frac{\sqrt{D}}{6} \bullet \left( \frac{3}{2}k^{2} - 6k + 30 - \frac{D}{2} \right) =\]

\[= \frac{\sqrt{D}}{6} \bullet \left( k^{2} + 20 - 4k \right) =\]

\[= \frac{1}{6} \bullet \sqrt{k^{2} - 4k + 20} \bullet \left( k^{2} + 20 - 4k \right) =\]

\[= \frac{1}{6} \bullet \sqrt{\left( k^{2} + 20 - 4k \right)^{3}};\]

\[S^{'}(x) = \frac{1}{6} \bullet \left( k^{2} + 20 - 4k \right)^{\frac{3}{2}} =\]

\[= \frac{1}{6} \bullet \frac{3}{2} \bullet (2k - 4) \bullet \left( k^{2} + 20 - 4k \right)^{\frac{1}{2}} =\]

\[= \frac{1}{4} \bullet 2(k - 2) \bullet \sqrt{k^{2} + 20 - 4k}.\]

\[Промежуток\ возрастания:\]

\[k - 2 > 0\]

\[k > 2.\]

\[k = 2 - точка\ минимума.\]

\[Ответ:\ \ k = 2.\]