Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 692

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\[1)\ y = 1 + \sin x\]

\[- 1 \leq \sin x \leq 1\]

\[0 \leq 1 + \sin x \leq 2.\]

\[Ответ:\ \ E(y) = \lbrack 0;\ 2\rbrack.\]

\[2)\ y = 1 - \cos x\]

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

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

\[0 \leq 1 - \cos x \leq 2.\]

\[Ответ:\ \ E(y) = \lbrack 0;\ 2\rbrack.\]

\[3)\ y = 2\sin x + 3\]

\[- 1 \leq \sin x \leq 1\]

\[- 2 \leq 2\sin x \leq 2\]

\[1 \leq 2\sin x + 3 \leq 5.\]

\[Ответ:\ \ E(y) = \lbrack 1;\ 5\rbrack.\]

\[4)\ y = 1 - 4\cos{2x}\]

\[- 1 \leq \cos{2x} \leq 1\]

\[- 4 \leq - 4\cos{2x} \leq 4\]

\[- 3 \leq 1 - 4\cos{2x} \leq 5.\]

\[Ответ:\ \ E(y) = \lbrack - 3;\ 5\rbrack.\]

\[5)\ y = \sin{2x} \bullet \cos{2x} + 2 =\]

\[= \frac{1}{2}\sin{4x} + 2;\]

\[- 1 \leq \sin{4x} \leq 1\]

\[- \frac{1}{2} \leq \frac{1}{2}\sin{4x} \leq \frac{1}{2}\]

\[\frac{3}{2} \leq \frac{1}{2}\sin{4x} + 2 \leq \frac{5}{2}.\]

\[Ответ:\ \ E(y) = \lbrack 1,5;\ 2,5\rbrack.\]

\[6)\ y = \frac{1}{2}\sin x \bullet \cos x - 1 =\]

\[= \frac{1}{4}\sin{2x} - 1;\]

\[- 1 \leq \sin{2x} \leq 1\]

\[- \frac{1}{4} \leq \frac{1}{4}\sin{2x} \leq \frac{1}{4}\]

\[- \frac{5}{4} \leq \frac{1}{4}\sin{2x} - 1 \leq - \frac{3}{4}.\]

\[Ответ:\ \ E(y) = \lbrack - 1,25;\ - 0,75\rbrack.\]