Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 31

\[\boxed{\mathbf{31.}}\]

\[\textbf{а)}\ x^{2} - 1 = 2\ln x\]

\[x^{2} - 1 - 2\ln x = 0\]

\[x > 0.\]

\[M = (0; + \infty).\]

\[f^{'}(x) = 2x - \frac{2}{x}.\]

\[2x - \frac{2}{x} = 0\]

\[\frac{2x^{2} - 2}{x} = 0\]

\[2x^{2} - 2 = 0\]

\[2\left( x^{2} - 1 \right) = 0\]

\[x^{2} = 1\]

\[x = - 1\ (не\ принадлежит\ M);\]

\[x = 1.\]

\[f^{'}(x) < 0:\]

\[2x - \frac{2}{x} < 0\]

\[x < 1.\]

\[x \in (0;1).\]

\[f^{'}(x) > 0:\]

\[2x - \frac{2}{x} > 0\]

\[x > 1.\]

\[x \in (1; + \infty).\]

\[x = 1 - точка\ максимума.\]

\[Ответ:x = 1.\]

\[\textbf{б)}\ x^{\frac{3}{2}}(1 - x) = \frac{6}{25}\sqrt{\frac{3}{5}}\]

\[x^{\frac{3}{2}}(1 - x) - \frac{6}{25}\sqrt{\frac{3}{5}} = 0\]

\[x > 0.\]

\[M = (0; + \infty).\]

\[f^{'}(x) = \frac{3}{2}x^{\frac{1}{2}}(1 - x) - x^{\frac{3}{2}} =\]

\[= x^{\frac{1}{2}}\left( \frac{3}{2} - \frac{3}{2}x - x \right) =\]

\[= x^{\frac{1}{2}}\left( \frac{3}{2} - \frac{5}{2}x \right).\]

\[x^{\frac{1}{2}}\left( \frac{3}{2} - \frac{5}{2}x \right) = 0\]

\[x^{\frac{1}{2}} = 0\]

\[x = 0\ (не\ принадлежит\ M).\]

\[\frac{3}{2} - \frac{5}{2}x = 0\]

\[\frac{5}{2}x = \frac{3}{2}\]

\[x = \frac{3}{5}.\]

\[f^{'}(x) < 0:\]

\[x > \frac{3}{5}.\]

\[x \in \left( \frac{3}{5}; + \infty \right).\]

\[f^{'}(x) > 0:\]

\[x < \frac{3}{5}.\]

\[x \in \left( 0;\frac{3}{5} \right).\]

\[x = \frac{3}{5} - точка\ максимума.\]

\[Ответ:x = \frac{3}{5}.\]

\[\textbf{в)}\ x - \frac{x^{2}}{2} = \ln(x + 1)\]

\[x - \frac{x^{2}}{2} - \ln(x + 1) = 0\]

\[x + 1 > 0\]

\[x > - 1.\]

\[M = ( - 1; + \infty).\]

\[f^{'}(x) = 1 - x - \frac{1}{x + 1}.\]

\[1 - x - \frac{1}{x + 1} = 0\]

\[x + \frac{1}{x + 1} = 1\]

\[\frac{x(x + 1) + 1}{x + 1} = 1\]

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

\[x^{2} = 0\]

\[x = 0.\]

\[f^{'}(x) > 0:\]

\[x > 0.\]

\[x \in (0; + \infty).\]

\[f^{'}(x) < 0:\]

\[x < 0.\]

\[x \in ( - 1;0).\]

\[x = 0 - точка\ максимума.\]

\[Ответ:x = 0.\]