To determine the domain of the function [tex]\( f(x) = \log |x| + \frac{1}{\sqrt{|x|}} + \frac{1}{\log |x|} \)[/tex], we need to consider the restrictions on [tex]\( x \)[/tex] for each component of the function to be defined and valid.
Let's analyze each term in detail:
1. [tex]\(\log |x|\)[/tex]:
- The logarithmic function [tex]\(\log |x|\)[/tex] is defined when its argument [tex]\(|x|\)[/tex] is positive:
[tex]\[
|x| > 0 \quad \Rightarrow \quad x \neq 0
\][/tex]
2. [tex]\(\frac{1}{\sqrt{|x|}}\)[/tex]:
- The square root function [tex]\(\sqrt{|x|}\)[/tex] is defined when its argument [tex]\(|x|\)[/tex] is non-negative, but since it is in the denominator, it must be positive:
[tex]\[
|x| > 0 \quad \Rightarrow \quad x \neq 0
\][/tex]
3. [tex]\(\frac{1}{\log |x|}\)[/tex]:
- The term [tex]\(\log |x|\)[/tex] must be defined (which we already addressed), and it must not be zero because we have [tex]\(\frac{1}{\log |x|}\)[/tex]. Therefore:
[tex]\[
\log |x| \neq 0 \quad \Rightarrow \quad |x| \neq 1
\][/tex]
From the above analysis, we derive that:
- [tex]\(|x| > 0\)[/tex], so [tex]\(x \neq 0\)[/tex]
- [tex]\(|x| \neq 1\)[/tex], so [tex]\(x \neq 1\)[/tex] and [tex]\(x \neq -1\)[/tex]
Thus, the values of [tex]\( x \)[/tex] we must exclude from the domain are [tex]\( x = 0 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = -1 \)[/tex].
Consequently, the set [tex]\( A \)[/tex] is:
[tex]\[ A = \{-1, 0, 1\} \][/tex]
Therefore, the domain of the function [tex]\( f(x) \)[/tex] is [tex]\( \mathbb{R} - A \)[/tex].
The correct answer is:
[tex]\[ \boxed{\{-1, 0, 1\}} \][/tex]
This corresponds to option (1).
