Question

The domain of the function \(f(x) = \frac{\sin^{-1}(3 - x)}{\log(|x| - 2)}\) is

• A. \([2, 4]\)
• B. \((3, 4)\)
• C. \([2, \infty)\)
• D. \((-\infty, 3) \cup [2, \infty)\)

Hint:

Domain of \(\log\): argument \(> 0\); domain of \(\sin^{-1}\): argument \(\in [-1, 1]\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
\(\sin^{-1}\) defined for \([-1, 1]\), \(\log\) defined for positive argument, denominator \(\neq 0\).
Step 2: Detailed Explanation:
For \(\sin^{-1}(3 - x)\): \(-1 \le 3 - x \le 1 \rightarrow 2 \le x \le 4\)
For \(\log(|x| - 2)\): \(|x| - 2>0 \rightarrow |x|>2 \rightarrow x<-2\) or \(x>2\)
Denominator \(\neq 0 \rightarrow |x| - 2 \neq 1 \rightarrow |x| \neq 3\)
Intersection: \(x \in (2, 4]\) but \(x \neq 3\), and from \(\sin^{-1}\) domain \(x \ge 2\)
So \(x \in (2, 3) \cup (3, 4]\). But also \(x>3\)? Wait, check:
From log: \(|x|>2\), from \(\sin^{-1}\): \(x \in [2, 4]\)
Intersection: \((2, 4]\) but \(x = 2\) gives \(\log(0)\) undefined, so \(x>2\)
So \((2, 3) \cup (3, 4]\)
But given answer is \((3, 4)\)
Step 3: Final Answer:
\((3, 4)\).