ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 971

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\[1)\ f(x) = 2\sin x + \sin{2x};\ \ \ \ \ \ \]

\[\left\lbrack 0;\ \frac{3\pi}{2} \right\rbrack\]

\[f^{'}(x) = 2 \bullet \left( \sin x \right)^{'} + \left( \sin{2x} \right)^{'};\]

\[f^{'}(x) = 2\cos x + 2\cos{2x} =\]

\[= 2 \bullet \left( \cos x + \cos{2x} \right).\]

\[Стационарные\ точки:\]

\[\cos x + \cos{2x} = 0\]

\[\cos x + \cos^{2}x - \sin^{2}x = 0\]

\[\cos x + \cos^{2}x - 1 + \cos^{2}x = 0\]

\[2\cos^{2}x + \cos x - 1 = 0\]

\[Пусть\ y = \cos x:\]

\[2y^{2} + y - 1 = 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\text{\ \ }\]

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

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

\[x = \pi - \arccos 1 + 2\pi n\]

\[x = \pi + 2\pi n.\]

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

\[x = \pm \arccos\frac{1}{2} + 2\pi n\]

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

\[f(0) = 2\sin 0 + \sin 0 =\]

\[= 2 \bullet 0 + 0 = 0;\]

\[f\left( \frac{\pi}{3} \right) = 2\sin\frac{\pi}{3} + \sin\frac{2\pi}{3} =\]

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

\[f(\pi) = 2\sin\pi + \sin{2\pi} =\]

\[= 2 \bullet 0 + 0 = 0;\]

\[f\left( \frac{3\pi}{2} \right) = 2\sin\frac{3\pi}{2} + \sin{3\pi} =\]

\[= 2 \bullet ( - 1) + 0 = - 2.\]

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

\[2)\ f(x) = 2\cos x + \sin{2x};\ \text{\ \ \ }\]

\[\lbrack 0;\ \pi\rbrack\]

\[f^{'}(x) = 2 \bullet \left( \cos x \right)^{'} + \left( \sin{2x} \right)^{'};\]

\[f^{'}(x) = - 2\sin x + 2\cos{2x} =\]

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

\[Стационарные\ точки:\]

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

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

\[1 - \sin^{2}x - \sin^{2}x - \sin x = 0\]

\[2\sin^{2}x + \sin x - 1 = 0.\]

\[Пусть\ y = \sin x:\]

\[2y^{2} + y - 1 = 0\]

\[D = 1^{2} + 4 \bullet 2 = 1 + 8 = 9\]

\[y_{1} = \frac{- 1 - 3}{2 \bullet 2} = - 1;\text{\ \ }\]

\[y_{2} = \frac{- 1 + 3}{2 \bullet 2} = \frac{1}{2}.\]

\[1)\ \sin x = - 1\]

\[x = - \arcsin 1 + 2\pi n\]

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

\[2)\ \sin x = \frac{1}{2}\]

\[x = ( - 1)^{n} \bullet \arcsin\frac{1}{2} + \pi n\]

\[x = ( - 1)^{n} \bullet \frac{\pi}{6} + \pi n.\]

\[f(0) = 2\cos 0 + \sin 0 =\]

\[= 2 \bullet 1 + 0 = 2;\]

\[f\left( \frac{\pi}{6} \right) = 2\cos\frac{\pi}{6} + \sin\frac{\pi}{3} =\]

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

\[f\left( \frac{5\pi}{6} \right) = 2\cos\frac{5\pi}{6} + \sin\frac{5\pi}{3} =\]

\[= 2 \bullet \left( - \frac{\sqrt{3}}{2} \right) - \frac{\sqrt{3}}{2} = - \frac{3\sqrt{3}}{2};\]

\[f(\pi) = 2\cos\pi + \sin{2\pi} =\]

\[= 2 \bullet ( - 1) + 0 = - 2.\]

\[Ответ:\ \ y_{\min} = - \frac{3\sqrt{3}}{2};\ \ \]

\[y_{\max} = \frac{3\sqrt{3}}{2}.\]