Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1042

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

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

\[x^{2} + px - kx - 1 = 0\]

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

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

\[x_{1} = \frac{k - p - \sqrt{D}}{2};\text{\ \ }\]

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

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

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

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

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

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

\[= \sqrt{D}.\]

\[5)\ x_{2}^{3} - x_{1}^{3} =\]

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

\[= \sqrt{D} \bullet \left( \frac{1}{4}(k - p)^{2} - \frac{1}{12}D + 1 \right);\]

\[Пусть\ u = (p - k)^{2} + 4;\ \ \ \ \]

\[S(u) = \frac{1}{6}u\sqrt{u} = \frac{1}{6}u^{\frac{3}{2}}:\]

\[S^{'}(k) =\]

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

\[= \left( - 2 \bullet (p - k) + 0 \right) \bullet \frac{1}{6} \bullet \frac{3}{2}u^{\frac{1}{2}}\]

\[S^{'}(k) = ( - 2p + 2k) \bullet \frac{1}{4}\sqrt{u} =\]

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

\[Промежуток\ возрастания:\]

\[2k - 2p > 0\]

\[2k > 2p\ \]

\[k > p.\]

\[k = p - точка\ минимума.\]

\[Ответ:\ \ k = p.\]