Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1624

\[\boxed{\mathbf{16}\mathbf{24}\mathbf{.}}\]

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

\[2\cos^{4}x + \sin^{2}x =\]

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

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

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

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

\[= \cos^{2}x \bullet \cos{2x} + 1.\]

\[= \cos^{2}x \bullet \cos{2x} + 2.\]

\[Получим:\]

\[y = \frac{\cos^{2}x \bullet \cos{2x} + 1}{\cos^{2}x \bullet \cos{2x} + 2}\]

\[u = \cos^{2}x \bullet \cos{2x} + 1:\]

\[= - 2\cos x \bullet \sin(x + 2x) =\]

\[= - 2\cos x \bullet \sin{3x}.\]

\[z = \cos^{2}x \bullet \cos{2x} + 2:\]

\[z^{'}(x) = - 2\cos x \bullet \sin{3x};\]

\[y^{'}(x) = \frac{u^{'} \bullet z - u \bullet z^{'}}{z^{2}} =\]

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

\[Точки\ экстремума:\]

\[- 2\cos x \bullet \sin{3x} \bullet 1 = 0\]

\[\cos x \bullet \sin{3x} = 0.\]

\[1)\ \cos x = 0\]

\[x = \arccos 0 + \pi n = \frac{\pi}{2} + \pi n.\]

\[2)\ \sin{3x} = 0\]

\[3x = \arcsin 0 + \pi n = \pi n\]

\[x = \frac{\text{πn}}{3}.\]

\[Период\ равен\ \pi:\]

\[y\left( \frac{\pi}{2} \right) = \frac{\cos^{2}\frac{\pi}{2} \bullet \cos\pi + 1}{\cos^{2}\frac{\pi}{2} \bullet \cos\pi + 2} =\]

\[= \frac{0^{2} \bullet ( - 1) + 1}{0^{2} \bullet ( - 1) + 2} = \frac{1}{2};\]

\[y(0) = \frac{\cos^{2}0 \bullet \cos 0 + 1}{\cos^{2}0 \bullet \cos 0 + 2} =\]

\[= \frac{1^{2} \bullet 1 + 1}{1^{2} \bullet 1 + 2} = \frac{1 + 1}{1 + 2} = \frac{2}{3};\]

\[y\left( \frac{\pi}{3} \right) = \frac{\cos^{2}\frac{\pi}{3} \bullet \cos\frac{2\pi}{3} + 1}{\cos^{2}\frac{\pi}{3} \bullet \cos\frac{2\pi}{3} + 2} =\]

\[= \frac{\left( \frac{1}{2} \right)^{2} \bullet \left( - \frac{1}{2} \right) + 1}{\left( \frac{1}{2} \right)^{2} \bullet \left( - \frac{1}{2} \right) + 2} = \frac{7}{8}\ :\frac{15}{8} = \frac{7}{15};\]

\[y\left( \frac{2\pi}{3} \right) = \frac{\cos^{2}\frac{2\pi}{3} \bullet \cos\frac{4\pi}{3} + 1}{\cos^{2}\frac{2\pi}{3} \bullet \cos\frac{4\pi}{3} + 2} =\]

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

\[= \frac{7}{8}:\frac{15}{8} = \frac{7}{15}.\]

\[Ответ:\ \ y_{\max} = \frac{2}{3};\ \ y_{\min} = \frac{7}{15}.\]