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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 17
Задание 17
\[\boxed{\mathbf{17.}}\]
\[\textbf{а)}\ 2cos^{2}\left( x\sin x \right) =\]
\[= 2 + \left| \log_{2}\left( x^{2} - 4x + 1 \right) \right|\]
\[- 1 \leq \cos\left( x\sin x \right) \leq 1\]
\[0 \leq cos^{2}\left( x\sin x \right) \leq 1\]
\[0 \leq 2cos^{2}\left( x\sin x \right) \leq 2.\]
\[\left| \log_{2}\left( x^{2} - 4x + 1 \right) \right| \geq 0\]
\[2 + \left| \log_{2}\left( x^{2} - 4x + 1 \right) \right| \geq 2.\]
\[\left\{ \begin{matrix} 2cos^{2}\left( x\sin x \right) = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 + \left| \log_{2}\left( x^{2} - 4x + 1 \right) \right| = 2 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \text{co}s^{2}\left( x\sin x \right) = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \left| \log_{2}\left( x^{2} - 4x + 1 \right) \right| = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x + 1 = 1\]
\[x^{2} - 4x = 0\]
\[x(x - 4) = 0\]
\[x = 0;\ \ \ x = 4.\]
\[Проверим:\]
\[\text{co}s^{2}\left( 0\sin 0 \right) = 1^{2} = 1 - верно.\]
\[\text{co}s^{2}\left( 4\sin 4 \right) \neq 1 - неверно.\]
\[Ответ:x = 0.\]
\[\textbf{б)}\ 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) =\]
\[= 3 + \log_{3}\left( x^{2} - 6x + 10 \right)\]
\[0 \leq \ 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) \leq 3.\]
\[\log_{3}\left( x^{2} - 6x + 10 \right) \geq 0\]
\[{3 + \log_{3}}\left( x^{2} - 6x + 10 \right) \geq 3.\]
\[\left\{ \begin{matrix} 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) = 3\ \ \ \ \ \ \ \ \\ 3 + \log_{3}\left( x^{2} - 6x + 10 \right) = 3 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \text{si}n^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) = 1\ \ \ \\ \log_{3}\left( x^{2} - 6x + 10 \right) = 0 \\ \end{matrix} \right.\ \]
\[\log_{3}\left( x^{2} - 6x + 10 \right) = 0\]
\[x^{2} - 6x + 10 = 1\]
\[x^{2} - 6x + 9 = 0\]
\[(x - 3)^{2} = 0\]
\[x = 3.\]
\[Проверим:\ \]
\[\text{si}n^{2}\left( \frac{3\pi}{2} \cdot \sin\frac{3\pi}{2} \right) =\]
\[= \text{si}n^{2}\left( \frac{3\pi}{2} \cdot ( - 1) \right) =\]
\[= \text{si}n^{2}\left( - \frac{3\pi}{2} \right) = ( - 1)^{2} = 1.\]
\[Ответ:x = 3.\]