Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 732

\[\boxed{\mathbf{732}\mathbf{.}}\]

\[I = A\sin(\omega t + \varphi).\]

\[1)\ A = 2,\ \ \ \omega = 1,\ \ \ \varphi = \frac{\pi}{4}:\]

\[I = 2\sin\left( t + \frac{\pi}{4} \right);\]

\[\textbf{а)}\ D(t) = (0;\ + \infty);\]

\[\textbf{б)}\ - 1 \leq \sin\left( t + \frac{\pi}{4} \right) \leq 1\]

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

\[E(I) = \lbrack - 2;\ 2\rbrack;\]

\[\textbf{в)}\ I(t + T) = I(t)\]

\[2\sin\left( t + T + \frac{\pi}{4} \right) = 2\sin\left( t + \frac{\pi}{4} \right)\]

\[T = 2\pi.\]

\[\textbf{г)}\ 2\sin\left( t + \frac{\pi}{4} \right) = 0\]

\[\sin\left( t + \frac{\pi}{4} \right) = 0\]

\[t + \frac{\pi}{4} = \arcsin 0 + \pi n = \pi n\]

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

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

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

\[\sin\left( t + \frac{\pi}{4} \right) = 1\]

\[t + \frac{\pi}{4} = \arcsin 1 + 2\pi n\]

\[t + \frac{\pi}{4} = \frac{\pi}{2} + 2\pi n\]

\[t = \frac{\pi}{2} - \frac{\pi}{4} + 2\pi n\]

\[t = \frac{2\pi - \pi}{4} + 2\pi n\]

\[t = \frac{\pi}{4} + 2\pi n.\]

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

\[\sin\left( t + \frac{\pi}{4} \right) = - 1\]

\[t + \frac{\pi}{4} = - \arcsin 1 + 2\pi n\]

\[t + \frac{\pi}{4} = - \frac{\pi}{2} + 2\pi n\]

\[t = - \frac{\pi}{2} - \frac{\pi}{4} + 2\pi n\]

\[t = - \frac{2\pi}{4} - \frac{\pi}{4} + 2\pi n\]

\[t = - \frac{3\pi}{4} + 2\pi n.\]

\[2)\ A = 1,\ \ \ \omega = 2,\ \ \ \varphi = \frac{\pi}{3}:\]

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

\[\textbf{а)}\ D(t) = (0;\ + \infty);\]

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

\[E(I) = \lbrack - 1;\ 1\rbrack;\]

\[\textbf{в)}\ I(t + T) = I(t)\]

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

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

\[2T = 2\pi\]

\[T = \pi.\]

\[\textbf{г)}\ \sin\left( 2t + \frac{\pi}{3} \right) = 0\]

\[2t + \frac{\pi}{3} = \arcsin 0 + \pi n = \pi n\]

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

\[t = \frac{1}{2} \bullet \left( - \frac{\pi}{3} + \pi n \right) = - \frac{\pi}{6} + \frac{\text{πn}}{2}.\]

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

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

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

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

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

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

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

\[t = \frac{1}{2} \bullet \left( \frac{\pi}{6} + 2\pi n \right) = \frac{\pi}{12} + \pi n.\]

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

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

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

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

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

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

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

\[t = \frac{1}{2} \bullet \left( - \frac{5\pi}{6} + 2\pi n \right)\]

\[t = - \frac{5\pi}{12} + \pi n.\]