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Sagot :
Let's find the domain of the function [tex]\( f(x) = \sqrt{\log_e \frac{1}{|\sin x - 1|}} \)[/tex].
### Step-by-Step Solution
1. Understanding the Function:
The given function is
[tex]\[ f(x) = \sqrt{\log_e \frac{1}{|\sin x - 1|}} \][/tex]
To determine the domain, we must ensure that the expression inside the square root is non-negative and defined.
2. Condition for the Logarithmic Function:
For [tex]\(\log_e \left( \frac{1}{|\sin x - 1|} \right)\)[/tex] to be defined, the argument [tex]\(\frac{1}{|\sin x - 1|}\)[/tex] must be positive.
[tex]\[ \frac{1}{|\sin x - 1|} > 0 \][/tex]
Since the absolute value of [tex]\(\sin x - 1\)[/tex] is always non-negative, the denominator will always be positive except when [tex]\(\sin x = 1\)[/tex]. Thus, we must exclude [tex]\( \sin x = 1 \)[/tex] because it would make the denominator zero, leading to an undefined logarithm.
3. Condition for the Logarithm Argument:
Now we need:
[tex]\[ \log_e \left( \frac{1}{|\sin x - 1|} \right) \geq 0 \][/tex]
This inequality holds if:
[tex]\[ \frac{1}{|\sin x - 1|} \geq 1 \][/tex]
4. Solving the Inequality:
Simplifying the above inequality, we get:
[tex]\[ \frac{1}{|\sin x - 1|} \geq 1 \implies 1 \geq |\sin x - 1| \][/tex]
So,
[tex]\[ |\sin x - 1| \leq 1 \][/tex]
The absolute value inequality can be broken down into:
[tex]\[ -1 \leq \sin x - 1 \leq 1 \][/tex]
5. Solving the Compound Inequality:
Split it into two parts:
- [tex]\(-1 \leq \sin x - 1\)[/tex]
- [tex]\(\sin x - 1 \leq 1\)[/tex]
Solving each part separately:
- [tex]\(-1 \leq \sin x - 1 \implies 0 \leq \sin x\)[/tex]
- [tex]\(\sin x - 1 \leq 1 \implies \sin x \leq 2\)[/tex]
Since [tex]\(\sin x\)[/tex] ranges between [tex]\(-1\)[/tex] and [tex]\(1\)[/tex], [tex]\(\sin x \leq 2\)[/tex] is always true. So, we only need to satisfy [tex]\(0 \leq \sin x \leq 1\)[/tex].
6. Excluding [tex]\(\sin x = 1\)[/tex]:
Recall the earlier point that [tex]\(\sin x = 1\)[/tex] makes the logarithm undefined, so we cannot include [tex]\( x \)[/tex] values where [tex]\(\sin x = 1\)[/tex].
7. Determining the Valid [tex]\(x\)[/tex] Values:
[tex]\(\sin x = 1\)[/tex] occurs at [tex]\(x = \frac{\pi}{2} + 2n\pi\)[/tex], where [tex]\(n\)[/tex] is any integer. Hence, we exclude these points:
[tex]\[ x \neq \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \][/tex]
8. Final Domain:
Therefore, the domain of the function [tex]\(f(x)\)[/tex] is:
[tex]\[ x \in \{ x \in \mathbb{R} \mid 0 \leq \sin x < 1 \} \][/tex]
excluding points where [tex]\(\sin x = 1\)[/tex].
In interval notation, the domain can be expressed as:
[tex]\[ x \in (0, \pi/2) \cup (\pi/2, \pi) \cup (2\pi, 5\pi/2) \cup (5\pi/2, 3\pi) \cup \ldots \][/tex]
excluding [tex]\(\frac{\pi}{2} + 2k\pi\)[/tex] where [tex]\(k \in \mathbb{Z}\)[/tex].
### Step-by-Step Solution
1. Understanding the Function:
The given function is
[tex]\[ f(x) = \sqrt{\log_e \frac{1}{|\sin x - 1|}} \][/tex]
To determine the domain, we must ensure that the expression inside the square root is non-negative and defined.
2. Condition for the Logarithmic Function:
For [tex]\(\log_e \left( \frac{1}{|\sin x - 1|} \right)\)[/tex] to be defined, the argument [tex]\(\frac{1}{|\sin x - 1|}\)[/tex] must be positive.
[tex]\[ \frac{1}{|\sin x - 1|} > 0 \][/tex]
Since the absolute value of [tex]\(\sin x - 1\)[/tex] is always non-negative, the denominator will always be positive except when [tex]\(\sin x = 1\)[/tex]. Thus, we must exclude [tex]\( \sin x = 1 \)[/tex] because it would make the denominator zero, leading to an undefined logarithm.
3. Condition for the Logarithm Argument:
Now we need:
[tex]\[ \log_e \left( \frac{1}{|\sin x - 1|} \right) \geq 0 \][/tex]
This inequality holds if:
[tex]\[ \frac{1}{|\sin x - 1|} \geq 1 \][/tex]
4. Solving the Inequality:
Simplifying the above inequality, we get:
[tex]\[ \frac{1}{|\sin x - 1|} \geq 1 \implies 1 \geq |\sin x - 1| \][/tex]
So,
[tex]\[ |\sin x - 1| \leq 1 \][/tex]
The absolute value inequality can be broken down into:
[tex]\[ -1 \leq \sin x - 1 \leq 1 \][/tex]
5. Solving the Compound Inequality:
Split it into two parts:
- [tex]\(-1 \leq \sin x - 1\)[/tex]
- [tex]\(\sin x - 1 \leq 1\)[/tex]
Solving each part separately:
- [tex]\(-1 \leq \sin x - 1 \implies 0 \leq \sin x\)[/tex]
- [tex]\(\sin x - 1 \leq 1 \implies \sin x \leq 2\)[/tex]
Since [tex]\(\sin x\)[/tex] ranges between [tex]\(-1\)[/tex] and [tex]\(1\)[/tex], [tex]\(\sin x \leq 2\)[/tex] is always true. So, we only need to satisfy [tex]\(0 \leq \sin x \leq 1\)[/tex].
6. Excluding [tex]\(\sin x = 1\)[/tex]:
Recall the earlier point that [tex]\(\sin x = 1\)[/tex] makes the logarithm undefined, so we cannot include [tex]\( x \)[/tex] values where [tex]\(\sin x = 1\)[/tex].
7. Determining the Valid [tex]\(x\)[/tex] Values:
[tex]\(\sin x = 1\)[/tex] occurs at [tex]\(x = \frac{\pi}{2} + 2n\pi\)[/tex], where [tex]\(n\)[/tex] is any integer. Hence, we exclude these points:
[tex]\[ x \neq \frac{\pi}{2} + 2n\pi, \quad n \in \mathbb{Z} \][/tex]
8. Final Domain:
Therefore, the domain of the function [tex]\(f(x)\)[/tex] is:
[tex]\[ x \in \{ x \in \mathbb{R} \mid 0 \leq \sin x < 1 \} \][/tex]
excluding points where [tex]\(\sin x = 1\)[/tex].
In interval notation, the domain can be expressed as:
[tex]\[ x \in (0, \pi/2) \cup (\pi/2, \pi) \cup (2\pi, 5\pi/2) \cup (5\pi/2, 3\pi) \cup \ldots \][/tex]
excluding [tex]\(\frac{\pi}{2} + 2k\pi\)[/tex] where [tex]\(k \in \mathbb{Z}\)[/tex].
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