Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 15

\[\boxed{\mathbf{15.}}\]

\[\textbf{а)}\ 2cos^{2}\left( x \cdot \sin\text{πx} \right) =\]

\[= 2 + \log_{2}\left( x^{2} - 4x + 5 \right)\]

\[2cos^{2}\left( x \cdot \sin\text{πx} \right) =\]

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

\[0 \leq 2cos^{2}\left( x \cdot \sin\text{πx} \right) \leq 2;\]

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

\[\left\{ \begin{matrix} 2cos^{2}\left( x \cdot \sin\text{πx} \right) = 2\ \ \ \ \ \ \ \ \ \ \ \\ 2 + \log_{2}\left( (x - 2)^{2} + 1 \right) = 2 \\ \end{matrix} \right.\ \]

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

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

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

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

\[x = 2.\]

\[Проверим:\]

\[2cos^{2}\left( 2 \cdot \sin{2\pi} \right) = 2\]

\[\text{co}s^{2}(2 \cdot 0) = 1\]

\[\text{co}s^{2}0 = 1\]

\[1 = 1.\]

\[Ответ:x = 2.\]

\[\textbf{б)}\ 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) =\]

\[= 3 + \log_{3}\left( x^{2} - 6x + 10 \right)\]

\[3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) =\]

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

\[0 \leq 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) \leq 3;\]

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

\[\left\{ \begin{matrix} 3sin^{2}\left( \frac{\text{πx}}{2} \cdot \sin\frac{\text{πx}}{2} \right) = 3\ \ \ \ \ \ \ \\ 3 + \log_{3}\left( (x - 3)^{2} + 1 \right) = 3 \\ \end{matrix} \right.\ \]

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

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

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

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

\[x = 3.\]

\[Проверим:\]

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

\[\text{si}n^{2}\left( \frac{3\pi}{2} \cdot ( - 1) \right) = 1\]

\[\text{si}n^{2}\left( - \frac{3\pi}{2} \right) = 1\]

\[( - 1)^{2} = 1\]

\[1 = 1.\]

\[Ответ:x = 3.\]