Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 744

\[\boxed{\mathbf{744}\mathbf{.}}\]

\[1)\ y = tg\left( x + \frac{\pi}{4} \right);\]

\[\textbf{а)}\ x + \frac{\pi}{4} \neq \frac{\pi}{2} + \pi n\]

\[x \neq \frac{\pi}{2} - \frac{\pi}{4} + \pi n = \frac{\pi}{4} + \pi n.\]

\[\textbf{б)}\ E(y) = ( - \infty;\ + \infty).\]

\[\textbf{в)}\ y(x + T) = y(x)\]

\[\text{tg}\left( x + T + \frac{\pi}{4} \right) = tg\left( x + \frac{\pi}{4} \right)\]

\[T = \pi.\]

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

\[y( - x) = tg\left( - x + \frac{\pi}{4} \right) =\]

\[= - tg\left( x - \frac{\pi}{4} \right).\]

\[\textbf{д)}\ \text{tg}\left( x + \frac{\pi}{4} \right) = 0\]

\[x + \frac{\pi}{4} = arctg\ 0 + \pi n = \pi n\]

\[x = - \frac{\pi}{4} = \pi n.\]

\[\textbf{е)}\ Возрастает:\]

\[- \frac{3\pi}{4} + \pi n < x < \frac{\pi}{4} + \pi n.\]

\[Положительна:\ \]

\[- \frac{\pi}{4} + \pi n < x < \frac{\pi}{4} + \pi n.\]

\[Отрицательна:\]

\[\frac{\pi}{4} + \pi n < x < \frac{3\pi}{4} + \pi n.\]

\[2)\ y = tg\frac{x}{2}\]

\[\textbf{а)}\ \frac{x}{2} \neq \frac{\pi}{2} + \pi n\]

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

\[\textbf{б)}\ E(y) = ( - \infty;\ + \infty).\]

\[\textbf{в)}\ y(x + T) = y(x)\]

\[\text{tg}\left( \frac{1}{2} \bullet (x + T) \right) = tg\frac{x}{2}\]

\[\text{tg}\left( \frac{x}{2} + \frac{T}{2} \right) = tg\frac{x}{2}\]

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

\[T = 2\pi.\]

\[\textbf{г)}\ Функция\ нечетная:\]

\[y( - x) = tg\left( - \frac{x}{2} \right) =\]

\[= - tg\frac{x}{2} = - y(x).\]

\[\textbf{д)}\ \text{tg}\frac{x}{2} = 0\]

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

\[x = 2\pi n.\]

\[\textbf{е)}\ Возрастает:\]

\[- \pi + 2\pi n < x < \pi + 2\pi n.\]

\[Положительна:\]

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

\[Отрицательна:\]

\[- \pi + 2\pi n < x < 2\pi n.\]