ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1042

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

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

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

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

\[1)\ \sin^{2}x_{0} = - \frac{b}{2a} = - \frac{- 3}{2 \bullet 2} = \frac{3}{4};\]

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

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

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

\[f(1) = 2 \bullet 1^{2} - 3 \bullet 1 = 2 - 3 = - 1.\]

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

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

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

\[Ответ:\ \ \frac{7}{15};\ \frac{2}{3}.\]