Question

If the real-valued function

\[ f(x) = \sin^{-1}(x^2 - 1) - 3\log_3(3^x - 2) \]

is not defined for all \( x \in (-\infty, a] \cup (b, \infty) \), then what is \( 3^a + b^2 \)?

• A. \( 5 \)
• B. \( 6 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

When dealing with functions involving square roots and logarithms, remember to find the domain by considering the individual constraints for each term.

Answer 1

The Correct Option is D

Given the function: \[ f(x) = \sin^{-1}(x^2 - 1) - 3\log_3(3^x - 2) \] For \( f(x) \) to be defined, both terms must be defined. Step 1: Domain of \( \sin^{-1}(x^2 - 1) \) The inverse sine function \( \sin^{-1}(y) \) is defined for \( y \in [-1,1] \). Thus, we require: \[ -1 \leq x^2 - 1 \leq 1 \] Adding 1 to all sides: \[ 0 \leq x^2 \leq 2 \] Since \( x^2 \geq 0 \) always, the valid range is: \[ - \sqrt{2} \leq x \leq \sqrt{2} \] Step 2: Domain of \( 3\log_3(3^x - 2) \) For the logarithmic function to be defined, its argument must be positive: \[ 3^x - 2>0 \] \[ 3^x>2 \] Taking \( \log_3 \) on both sides: \[ x>\log_3 2 \] Step 3: Finding the Interval The function is not defined where at least one condition fails, i.e., \[ x \in (-\infty, \log_3 2] \cup (\sqrt{2}, \infty) \] Comparing with \( (-\infty, a] \cup (b, \infty) \), we identify: \[ a = \log_3 2, \quad b = \sqrt{2} \] Step 4: Compute \( 3^a + b^2 \) \[ 3^a = 3^{\log_3 2} = 2, \quad b^2 = (\sqrt{2})^2 = 2 \] \[ 3^a + b^2 = 2 + 2 = 4 \] Final Answer: \(\boxed{4}\) \bigskip