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

To determine the domain of the function \( f(x) = \frac{\sin^{-1}(3 - x)}{\log(|x| - 2)} \), we need to identify the permissible values of \( x \) for which both the numerator and the denominator of the function are defined. 

Condition for the inverse sine function: \(\sin^{-1}(3-x)\) is defined if \(-1 \leq 3-x \leq 1\).

Solving the inequality:

• \(-1 \leq 3-x\) implies \(x \leq 4\).
• \(3-x \leq 1\) implies \(x \geq 2\).

Condition for the logarithmic function: \(\log(|x| - 2)\) is defined if \( |x| - 2 > 0\), i.e., \( |x| > 2 \).

• This gives two cases:
  • Case 1: \(x > 2\) (for positive \(x\)).
  • Case 2: \(x < -2\) (for negative \(x\)).

Intersection of both conditions: We need to find the common values of \(x\) that satisfy both conditions.

From \( \sin^{-1} \) condition: \( x \in [2, 4] \).

From the logarithm condition for \( x > 2 \): \( x \in (2, \infty) \).

So, combining these, we get the domain from \( x \in (2, 4] \), but it can't include \( x = 2 \) since the logarithm would be zero, which is undefined.

The common domain is \((3, 4)\).

Therefore, the domain of the function \( f(x) = \frac{\sin^{-1}(3 - x)}{\log(|x| - 2)} \) is \( (3, 4) \).