ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1023

\[1)\ f(x) = \frac{\sin^{4}x + \cos^{4}x}{\sin^{6}x + \cos^{6}x};\ a = \frac{10}{7}:\]

\[= \frac{1 - \frac{1}{2}\sin^{2}{2x}}{1 - \frac{3}{4}\sin^{2}{2x}} =\]

\[= \frac{4 - 2\sin^{2}{2x}}{4 - 3\sin^{2}{2x}} = \frac{10}{7}.\]

\[28 - 14\sin^{2}{2x} = 40 - 30\sin^{2}{2x}\]

\[16\sin^{2}{2x} = 12\]

\[\sin^{2}{2x} = \frac{3}{4}\]

\[\frac{1 - \cos{4x}}{2} = \frac{3}{4}\]

\[4 - 4\cos{4x} = 6\]

\[4\cos{4x} = - 2\]

\[\cos{4x} = - \frac{1}{2}\]

\[4x = \pm \frac{2\pi}{3} + 2\pi n\]

\[x = \pm \frac{\pi}{6} + \frac{\text{πn}}{2}.\]

\[Точки\ экстремума:\]

\[0 \leq \sin^{2}{2x} \leq 1\]

\[- 2 \leq - 2\sin^{2}{2x} \leq 0\ \ \ \]

\[2 \leq 4 - 2\sin^{2}{2x} \leq 4\]

\[1 \leq \frac{4 - 2\sin^{2}{2x}}{4 - 3\sin^{2}{2x}} \leq 2.\]

\[- 3 \leq - 3\sin^{2}{2x} \leq 0\]

\[1 \leq 4 - 3\sin^{2}{2x} \leq 4.\]

\[Ответ:\ \ x = \pm \frac{\pi}{6} + \frac{\text{πn}}{2};\ \]

\[\text{\ \ \ \ }f_{\max} = 2;\ f_{\min} = 1.\]

\[2)\ f(x) = \frac{2\sin^{4}x + 3\cos^{2}x}{2\cos^{4}x + \sin^{2}x};\]

\[\ a = \frac{15}{7}:\]

\[f(x) = \frac{2\sin^{4}x + 3\left( 1 - \sin^{2}x \right)}{2\left( 1 - \sin^{2}x \right)^{2} + \sin^{2}x} =\]

\[= \frac{2\sin^{4}x + 3 - 3\sin^{2}x}{2 - 4\sin^{2}x + 2\sin^{4}x + \sin^{2}x} =\]

\[= \frac{2\sin^{4}x - 3\sin^{2}x + 3}{2\sin^{4}x - 3\sin^{2}x + 2} = \frac{15}{7};\]

\[14\sin^{4}x - 21\sin^{2}x + 21 =\]

\[= 30\sin^{4}x - 45\sin^{2}x + 30\]

\[16\sin^{4}x - 24\sin^{2}x + 9 = 0\]

\[\left( 4\sin^{2}x - 3 \right)^{2} = 0\]

\[4\sin^{2}x - 3 = 0\]

\[4\sin^{2}x = 3\]

\[\sin^{2}x = \frac{3}{4}\]

\[\frac{1 - \cos{2x}}{2} = \frac{3}{4}\]

\[4 - 4\cos{2x} = 6\]

\[4\cos{2x} = - 2\]

\[\cos{2x} = - \frac{1}{2}\]

\[2x = \pm \frac{2\pi}{3} + 2\pi n\]

\[x = \pm \frac{\pi}{3} + \pi n.\]

\[Точки\ экстремума:\]

\[\sin^{2}x_{0} = - \frac{b}{2a} = - \frac{- 3}{2 \bullet 2} = \frac{3}{4};\]

\[f\left( x_{0} \right) = 2 \bullet \frac{9}{16} - 3 \bullet \frac{3}{4} =\]

\[= \frac{9 - 18}{8} = - \frac{9}{8};\]

\[f(0) = 2 \bullet 0 - 3 \bullet 0 = 0;\]

\[f(1) = 2 \bullet 1 - 3 \bullet 1 = - 1;\]

\[\frac{15}{8} \leq 2\sin^{4}x - 3\sin^{2}x + 3 \leq 3\]

\[\ \frac{7}{8} \leq 2\sin^{4}x - 3\sin^{2}x + 2 \leq 2\]

\[\frac{3}{2} \leq \frac{2\sin^{4}x - 3\sin^{2}x + 3}{2\sin^{4}x - 3\sin^{2}x + 2} \leq \frac{15}{7}.\]

\[Ответ:\ \ x = \pm \frac{\pi}{3} + \pi n;\ \]

\[\text{\ \ }f_{\max} = \frac{15}{7};\ f_{\min} = \frac{3}{2}.\]