ГДЗ по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 18

\[\boxed{\mathbf{18.}}\]

\[\textbf{а)}\ x^{2} - \pi x + \frac{\pi^{2}}{4} \leq 3\sin x - 3\ \]

\[\left( x - \frac{\pi}{2} \right)^{2} \leq 3 \cdot \left( \sin x - 1 \right)\]

\[\left( x - \frac{\pi}{2} \right)^{2} \geq 0;\]

\[3 \cdot \left( \sin x - 1 \right) \leq 0.\]

\[Равносильная\ система\ \]

\[уравнений:\]

\[\left\{ \begin{matrix} \left( x - \frac{\pi}{2} \right)^{2} = 0\ \ \ \ \ \ \ \ \ \\ 3 \cdot \left( \sin x - 1 \right) = 0 \\ \end{matrix} \right.\ \]

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

\[x = \frac{\pi}{2}.\]

\[Проверим:\]

\[3 \cdot \left( \sin\frac{\pi}{2} - 1 \right) = 3 \cdot (1 - 1) =\]

\[= 3 \cdot 0 = 0.\]

\[Ответ:x = \frac{\pi}{2}.\]

\[- (x - \pi)^{2} \geq 2\left( \cos x + 1 \right)\]

\[(x - \pi)^{2} \leq - 2\left( \cos x + 1 \right)\]

\[(x - \pi)^{2} \geq 0;\]

\[- 2\left( \cos x + 1 \right) \leq 0.\]

\[Равносильная\ система\ \]

\[уравнений:\]

\[\left\{ \begin{matrix} (x - \pi)^{2} = 0\ \ \ \ \ \ \ \ \ \ \ \\ - 2\left( \cos x + 1 \right) = 0 \\ \end{matrix} \right.\ \]

\[x - \pi = 0\]

\[x = \pi.\]

\[Проверим:\]

\[- 2\left( \cos\pi + 1 \right) = 0\]

\[\cos\pi + 1 = 0\]

\[- 1 + 1 = 0\]

\[0 = 0.\]

\[Ответ:x = \text{π.}\]

\[\log_{2}\left( (x + 2)^{2} + 1 \right) \leq - (x + 2)^{2}\]

\[\log_{2}\left( (x + 2)^{2} + 1 \right) \geq 0;\]

\[- (x + 2)^{2} \leq 0.\]

\[Равносильная\ система\ \]

\[уравнений:\]

\[\left\{ \begin{matrix} \log_{2}\left( (x + 2)^{2} + 1 \right) = 0 \\ - (x + 2)^{2} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

\[x = - 2.\]

\[Проверим:\]

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

\[= \log_{2}1 = 0.\]

\[Ответ:x = - 2.\]

\[\log_{0,6}\left( (x - 3)^{2} + 1 \right) \geq (x - 3)^{2}\]

\[\log_{0,6}\left( (x - 3)^{2} + 1 \right) \leq 0;\]

\[(x - 3)^{2} \geq 0.\]

\[Равносильная\ система\ \]

\[уравнений:\]

\[\left\{ \begin{matrix} \log_{0,6}\left( (x - 3)^{2} + 1 \right) = 0 \\ (x - 3)^{2} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x - 3 = 0\]

\[x = 3.\]

\[Проверим:\]

\[\log_{0,6}\left( (3 - 3)^{2} + 1 \right) =\]

\[= \log_{0,6}1 = 0.\]

\[Ответ:x = 3.\]