Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 773

Авторы:
Год:2020-2021-2022-2023
Тип:учебник
Серия:Алгебра и начала математического анализа, геометрия

Задание 773

\[\boxed{\mathbf{77}\mathbf{3}\mathbf{.}}\]

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

\[\textbf{а)}\ D(x) = ( - \infty\ + \infty)\]

\[\textbf{б)}\ - 1 \leq \sin\left( \frac{x}{2} + \frac{\pi}{3} \right) \leq 1\]

\[- 2 \leq 2\sin\left( \frac{x}{2} + \frac{\pi}{3} \right) \leq 2\]

\[- 4 \leq 2\sin\left( \frac{x}{2} + \frac{\pi}{3} \right) \leq 0\]

\[E(y) = \lbrack - 4\ 0\rbrack\]

\[\textbf{в)}\ y(x + T) = y(x)\]

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

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

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

\[\frac{T}{2} = 2\pi\]

\[T = 2 \bullet 2\pi = 4\pi.\]

\[\textbf{г)}\ Ни\ четная,\ ни\ нечетная:\]

\[y( - x) = 2\sin{\left( - \frac{x}{2} + \frac{\pi}{3} \right) - 2}.\]

\[\textbf{д)}\ 2\sin\left( \frac{x}{2} + \frac{\pi}{3} \right) - 2 = 0\]

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

\[\frac{x}{2} + \frac{\pi}{3} = \arcsin 1 + 2\pi n\]

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

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

\[\frac{x}{2} = \frac{3\pi}{6} - \frac{2\pi}{6} + 2\pi n\]

\[\frac{x}{2} = \frac{\pi}{6} + 2\pi n\]

\[x = 2 \bullet \left( \frac{\pi}{6} + 2\pi n \right)\]

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

\[\textbf{е)}\ Максимальные\ значения:\]

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

\[\textbf{ж)}\ Минимальные\ значения:\]

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

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

\[\frac{x}{2} + \frac{\pi}{3} = - \arcsin 1 + 2\pi n\]

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

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

\[\frac{x}{2} = - \frac{3\pi}{6} - \frac{2\pi}{6} + 2\pi n\]

\[\frac{x}{2} = - \frac{5\pi}{6} + 2\pi n\]

\[x = 2 \bullet \left( - \frac{5\pi}{6} + 2\pi n \right)\]

\[x = - \frac{5\pi}{3} + \pi n.\]

\[2)\ y = \cos x - \sqrt{\cos^{2}x} =\]

\[= \cos x - \left| \cos x \right|\]

\[\textbf{а)}\cos x \geq 0:\]

\[y = \cos x - \cos x = 0\]

\[\cos x \geq 0\]

\[- \arccos 0 + 2\pi n \leq x \leq \arccos 0 + 2\pi n\]

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

\[\textbf{б)}\ \cos x < 0:\]

\[y = \cos x + \cos x = 2\cos x\]

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

\[- 1 \leq \cos x \leq 0\]

\[- 2 \leq 2\cos x \leq 0\]

\[E(y) = \lbrack - 2\ 0\rbrack.\]

\[Нули\ функции:\]

\[2\cos x = 0\]

\[\cos x = 0\]

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

\[Минимальные\ значения:\]

\[2\cos x = - 2\]

\[\cos x = - 1\]

\[x = \pi - \arccos 0 + 2\pi n =\]

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

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