Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1507

\[\boxed{\mathbf{1507}\mathbf{.}}\]

\[y = 2\sin x + \cos{2x};\ на\left\lbrack 0;\ \frac{\pi}{2} \right\rbrack:\]

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

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

\[Стационарные\ точки:\]

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

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

\[\cos x - 2\sin x \bullet \cos x = 0\]

\[\cos x \bullet \left( 1 - 2\sin x \right) = 0.\]

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

\[x = \arccos 0 + \pi n\]

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

\[2)\ 1 - 2\sin x = 0\]

\[1 = 2\sin x\]

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

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

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

\[y(0) = 2\sin 0 + \cos(2 \bullet 0) =\]

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

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

\[= 2 \bullet \frac{1}{2} + \cos\frac{\pi}{3} = 1 + \frac{1}{2} = 1,5;\]

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

\[= 2 \bullet 1 + \cos\pi = 2 - 1 = 1.\]

\[Ответ:\ \ y_{\min} = 1;\ \ y_{\max} = 1,5.\]