ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 101

\[1)\ y = \arcsin\frac{x - 3}{2}\]

\[- 1 \leq \frac{x - 3}{2} \leq 1\]

\[- 2 \leq x - 3 \leq 2\]

\[1 \leq x \leq 5.\]

\[D(x) = \lbrack 1;\ 5\rbrack.\]

\[2)\ y = \arccos(2 - 3x)\]

\[- 1 \leq 2 - 3x \leq 1\]

\[- 3 \leq - 3x \leq - 1\]

\[1 \leq 3x \leq 3\]

\[\frac{1}{3} \leq x \leq 1.\]

\[D(x) = \left\lbrack \frac{1}{3};\ 1 \right\rbrack.\]

\[3)\ y = \arccos\left( 2\sqrt{x} - 3 \right)\]

\[- 1 \leq 2\sqrt{x} - 3 \leq 1;\]

\[2 \leq 2\sqrt{x} \leq 4;\]

\[1 \leq \sqrt{x} \leq 2;\]

\[1 \leq x \leq 4;\]

\[D(x) = \lbrack 1;\ 4\rbrack.\]

\[4)\ y = \arcsin\frac{2x^{2} - 5}{3}\]

\[- 1 \leq \frac{2x^{2} - 5}{3} \leq 1\]

\[- 3 \leq 2x^{2} - 5 \leq 3\]

\[2 \leq 2x^{2} \leq 8\]

\[1 \leq x^{2} \leq 4\]

\[- 2 \leq x \leq - 1\]

\[1 \leq x \leq 2\]

\[D(x) = \lbrack - 2;\ - 1\rbrack \cup \lbrack 1;\ 2\rbrack.\]

\[5)\ y = \arccos\frac{2 - \sqrt{x}}{3}\]

\[- 1 \leq \frac{2 - \sqrt{x}}{3} \leq 1\]

\[- 3 \leq 2 - \sqrt{x} \leq 3\]

\[- 5 \leq - \sqrt{x} \leq 1\]

\[- 1 \leq \sqrt{x} \leq 5\]

\[0 \leq x \leq 25\]

\[D(x) = \lbrack 0;\ 25\rbrack.\]

\[6)\ y = \arcsin\left( 3\sqrt{x} - 2 \right)\]

\[- 1 \leq 3\sqrt{x} - 2 \leq 1\]

\[1 \leq 3\sqrt{x} \leq 3\]

\[\frac{1}{3} \leq \sqrt{x} \leq 1\]

\[\frac{1}{9} \leq x \leq 1\]

\[D(x) = \left\lbrack \frac{1}{9};\ 1 \right\rbrack.\]

\[7)\ y = \arcsin\left( x^{2} - 2 \right)\]

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

\[1 \leq x^{2} \leq 3\]

\[- \sqrt{3} \leq x \leq - 1\]

\[1 \leq x \leq \sqrt{3}\]

\[D(x) = \left\lbrack - \sqrt{3};\ - 1 \right\rbrack \cup \left\lbrack 1;\ \sqrt{3} \right\rbrack.\]

\[8)\ y = \arccos\left( x^{2} - x \right)\]

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

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

\[x^{2} - x - 1 \leq 0\]

\[D = 1^{2} + 4 \bullet 1 = 1 + 4 = 5\]

\[x = \frac{1 \pm \sqrt{5}}{2};\]

\[\frac{1 - \sqrt{5}}{2} \leq x \leq \frac{1 + \sqrt{5}}{2}.\]

\[D(x) = \left\lbrack \frac{1 - \sqrt{5}}{2};\ \frac{1 + \sqrt{5}}{2} \right\rbrack.\]