ГДЗ по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 10

\[\boxed{\mathbf{10.}}\]

\[\textbf{а)}\ y = \sqrt{1 - x^{2}}\]

\[Область\ существования:\]

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

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

\[(x + 1)(x - 1) \leq 0\]

\[- 1 \leq x \leq 1.\]

\[Область\ изменения:\]

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

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

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

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

\[y \in \lbrack 0;1\rbrack.\]

\[\textbf{б)}\ y = \sqrt{1 - x^{2}};X = \left\lbrack 0;\ \frac{\sqrt{3}}{2} \right\rbrack.\]

\[Область\ изменения:\]

\[0 \leq x \leq \frac{\sqrt{3}}{2}\]

\[0 \leq x^{2} \leq \frac{3}{4}\]

\[- \frac{3}{4} \leq - x^{2} \leq 0\]

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

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

\[Y = \left\lbrack \frac{1}{2};1 \right\rbrack.\]

\[\textbf{в)}\ y = \sqrt{1 - x^{2}};X = \left\lbrack - \frac{\sqrt{3}}{2};1 \right).\]

\[Область\ изменения:\]

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

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

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

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

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

\[Y = (0;1\rbrack.\]

\[\textbf{г)}\ y = \frac{300}{\sqrt{100 - x^{2}}};\ \ X = \lbrack - 8;1\rbrack.\]

\[Область\ изменения:\]

\[- 8 \leq x \leq 0\]

\[0 \leq x^{2} \leq 64\]

\[- 64 \leq - x^{2} \leq 0\]

\[36 \leq 100 - x^{2} \leq 100\]

\[6 < \sqrt{100 - x^{2}} \leq 10\]

\[3 \leq \frac{30}{\sqrt{100 - x^{2}}} \leq 5.\]

\[Y = \lbrack 3;5\rbrack.\]

\[\textbf{д)}\ y = \frac{1}{\sqrt{1 - x^{2}}};\ \ X = \left( - \frac{\sqrt{2}}{2};0 \right).\]

\[Область\ изменения.\]

\[- \frac{\sqrt{2}}{2} < x < 0\]

\[0 < x^{2} < \frac{1}{2}\]

\[- \frac{1}{2} < - x^{2} < 0\]

\[\frac{1}{2} < 1 - x^{2} < 1\]

\[\frac{1}{\sqrt{2}} < \sqrt{1 - x^{2}} < 1\]

\[1 < \frac{1}{\sqrt{1 - x^{2}}} < \sqrt{2}.\]

\[Y = \left( 1;\sqrt{2} \right).\]

\[\textbf{е)}\ y = \log_{2}\left( \sqrt{x^{2} - 1} \right).\]

\[Область\ определения:\]

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

\[x^{2} > 1\]

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

\[x > 1;\ \ x < - 1\]

\[X = ( - \infty; - 1) \cup (1; + \infty).\]

\[Y = R.\]

\[\textbf{ж)}\ y = \frac{1}{\sqrt{1 - x^{2}}};X = \left\lbrack - \frac{\sqrt{3}}{2};\frac{\sqrt{3}}{2} \right\rbrack.\]

\[Область\ определения:\]

\[- \frac{\sqrt{3}}{2} \leq x \leq 0;0 \leq x \leq \frac{\sqrt{3}}{2}\]

\[0 < x^{2} \leq \frac{3}{4}\]

\[- \frac{1}{2} < - x^{2} < 0.\]

\[Область\ изменения:\]

\[0 < x^{2} \leq \frac{3}{4}\]

\[- \frac{3}{4} \leq - x^{2} \leq 1\]

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

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

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

\[Y = \lbrack 1;2\rbrack.\]

\[\textbf{з)}\ y = \log_{2}\left( \sqrt{1 - x^{2}} \right).\]

\[Область\ значения:\]

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

\[x^{2} < 1\]

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

\[- 1 < x < 1.\]

\[y_{\max} = 0\ при\ x = 0.\]

\[Область\ изменения:\]

\[\log_{2}\left( \sqrt{1 - x^{2}} \right) \leq 0\]

\[Y = ( - \infty;0\rbrack.\]