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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 34
Задание 34
\[\boxed{\mathbf{34.}}\]
\[\textbf{а)}\ y = \lbrack x\rbrack;\ \ X = R;\]
\[T = 1:\]
\[\lbrack x + T\rbrack = \lbrack x + 1\rbrack = \lbrack x\rbrack + 1\]
\[\lbrack x + T\rbrack \neq \lbrack x\rbrack.\]
\[Не\ является\ периодической.\]
\[\textbf{б)}\ y = \left\{ x \right\};\ \ X = R;\]
\[T = 1:\]
\[\left\{ x + T \right\} = \left\{ x + 1 \right\} = \left\{ x \right\}\]
\[Периодическая,\ с\ периодом\]
\[\ T = 1.\]
\[\textbf{в)}\ y = \left| \left\{ x \right\} - \frac{1}{2} \right|;\ \ X = R;\]
\[T = 1:\]
\[\left| \left\{ x + t \right\} - \frac{1}{2} \right| = \left| \left\{ x + 1 \right\} - \frac{1}{2} \right| =\]
\[= \left| \left\{ x \right\} - \frac{1}{2} \right|\]
\[Периодическая,\ с\ периодом\ \]
\[T = 1.\]
\[\textbf{г)}\ y = \left| \left\{ \frac{1}{2}x \right\} - \frac{1}{2} \right|;\ \ X = R;\]
\[T = 1:\]
\[\left| \left\{ x + T \right\} - \frac{1}{2} \right| = \left| \left\{ x + 1 \right\} - \frac{1}{2} \right| =\]
\[= \left| \left\{ x \right\} - \frac{1}{2} \right|\]
\[Периодическая,\ с\ периодом\ \]
\[T = 1.\]
\[T = 2:\]
\[\left| \left\{ \frac{1}{2}(x + T) \right\} - \frac{1}{2} \right| =\]
\[= \left| \left\{ \frac{1}{2}(x + 2) \right\} - \frac{1}{2} \right| =\]
\[= \left| \left\{ \frac{1}{2}x + 1 \right\} - \frac{1}{2} \right| = \left| \left\{ \frac{1}{2}x \right\} - \frac{1}{2} \right|\]
\[Периодическая,\ с\]
\[\ периодом\ T = 2.\]
\[\textbf{д)}\ y = \left| 2\left\{ \frac{1}{2}x \right\} - 1 \right|;\ \ X = R;\]
\[y = \left\{ x \right\};\ \ X = R;\]
\[T = 1:\]
\[\left\{ x + T \right\} = \left\{ x + 1 \right\} = \left\{ x \right\}.\]
\[Периодическая,\ с\ периодом\]
\[\ T = 1.\]
\[y = \left\{ \frac{1}{2}x \right\};\ \ X = R;\]
\[T = 2:\]
\[\left\{ \frac{1}{2}(x + T) \right\} = \left\{ \frac{1}{2}(x + 2) \right\} =\]
\[= \left\{ \frac{1}{2}x + 1 \right\} = \left\{ \frac{1}{2}x \right\}.\]
\[Периодическая,\ с\ периодом\]
\[\ T = 2.\]
\[y = \left| 2\left\{ \frac{1}{2}x \right\} - 1 \right|\]
\[T = 2:\]
\[\left| 2\left\{ \frac{1}{2}x + T \right\} - 1 \right| =\]
\[= |2\left\{ \frac{1}{2}(x + 2) - 2 \right\} =\]
\[= \left| 2\left\{ \frac{1}{2}x \right\} - - 1 \right|\]
\[Периодическая,\ с\ периодом\ \]
\[T = 2.\]
\[\textbf{е)}\ y = \left| 4\left\{ \frac{1}{4}x \right\} - 2 \right|;\ \ X = R;\]
\[y = \left\{ x \right\};\ \ X = R;\]
\[T = 1:\]
\[\left\{ x + T \right\} = \left\{ x + 1 \right\} = \left\{ x \right\}.\]
\[y = \left\{ \frac{1}{4}x \right\};\ \ X = R;\]
\[T = 4:\]
\[\left\{ \frac{1}{4}(x + t) \right\} = \left\{ \frac{1}{4}(x + 4) \right\} =\]
\[= \left\{ \frac{1}{4}x + 1 \right\} = \left\{ \frac{1}{4}x \right\}\]
\[Периодическая,\ с\ периожом\]
\[\ T = 4.\]
\[y = \left| 4\left\{ \frac{1}{4}x \right\} - 2 \right|\]
\[T = 4:\]
\[\left| 4\left\{ \frac{1}{4}(x + T) \right\} - 2 \right| =\]
\[= \left| 4\left\{ \frac{1}{4}(x + 4) \right\}\ - 2 \right| =\]
\[= \left| 4\left\{ \frac{1}{4}x \right\} - 2 \right|\]
\[Периодическая,\ с\ периодом\ \]
\[T = 4.\]