Задача 54810 Найдите область определения функции: 1)...

5f7e961376b7cf17294b59d5

22.10.2020 08:06:27

Найдите область определения функции:

1) [m] y = \arcsin \frac{1}{x} [/m];
2) [m] y = \arcsin \frac{1}{x - 2} [/m];
3) [m] y = 2 \arccos \frac{2}{x + 2} [/m];
4) [m] y = 2 - \arccos \frac{1}{x - 1} [/m].

5f3ea7e3faf909182968ddd9

22.10.2020 09:43:11

★

[m] D( arcsin)=D(arccos)=[-1;1][/m]

1) [m] -1 ≤ \frac{1}{x} ≤ 1[/m] ⇔ [m]\left\{\begin{matrix}
\frac{1}{x} ≤ 1\\\frac{1}{x} ≥- 1 \end{matrix}\right.[/m] ; [m]\left\{\begin{matrix}
\frac{1-x}{x} ≤ 0\\\frac{1+x}{x} ≥0 \end{matrix}\right.[/m]

Решаем каждое методом интервалов:

__________-____ (0) __-__ [1]_____

__+__[-1] _____ (0) ___+___

О т в е т. (0;1)

Остальные аналогично.

2)
[m] -1 ≤ \frac{1}{x-2} ≤ 1[/m] ⇔ [m]\left\{\begin{matrix}
\frac{1}{x-2} ≤ 1\\\frac{1}{x-2} ≥- 1 \end{matrix}\right.[/m] ;[m]\left\{\begin{matrix}
\frac{1-x+2}{x-2} ≤ 0\\\frac{1+x-2}{x-2} ≥0 \end{matrix}\right.[/m] ;

3)[m] -1 ≤ \frac{2}{x+2} ≤ 1[/m] ⇔ [m]\left\{\begin{matrix}
\frac{1-x-2}{x+2} ≤ 1\\\frac{2+x+2}{x+2} ≥- 1 \end{matrix}\right.[/m]

4) [m] -1 ≤ \frac{1}{x-1} ≤ 1[/m] ⇔ [m]\left\{\begin{matrix}
\frac{1-x+1}{x-1} ≤ 1\\\frac{1+x-1}{x-1} ≥- 1 \end{matrix}\right.[/m]