Question

The domain of the function
\[ f(x) = \sin^{-1} \!\left( \log_2 \left( \frac{1}{2} x^2 \right) \right) \]
is

• A. \([-2,-1)\cup[1,2]\)
• B. \((-2,-1]\cup[1,2)\)
• C. \([-2,-1]\cup[1,2]\)
• D. (-2,-1)∪(1,2)

Hint:

Always check both logarithmic and inverse–trigonometric restrictions.

Answer 1

The Correct Option is A

Step 1: Argument of \(\sin^{-1}\) lies in \([-1, 1]\):
\[ -1 \le \log_2 \left( \frac{x^2}{2} \right) \le 1 \]
Step 2: Solving:
\[ \frac{1}{2} \le \frac{x^2}{2} \le 2 \implies 1 \le x^2 \le 4 \]
Step 3:
\[ x \in [-2, -1] \cup [1, 2] \]