Question

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

• A. \( [-2,0) \cup (1,2] \)
• B. \( [-2,-1] \cup [1,2] \)
• C. \( [-1,0] \cup [1,2] \)
• D. \( [1,\infty) \cup (-2,0) \)

Hint:

For logarithmic functions, ensure that the argument is positive. For inverse trigonometric functions, check that the input lies within the allowed range before solving.

Answer 1

The Correct Option is B

We need to find the domain of the function: \[ f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right). \] 

Step 1: Condition for Logarithm 
For the logarithmic function \( \log_2 \left( \frac{x^2}{2} \right) \) to be defined, the argument must be positive: \[ \frac{x^2}{2}>0. \] Since \( x^2 \) is always non-negative and is only zero at \( x=0 \), we have: \[ x \neq 0. \] 

Step 2: Condition for Inverse Sine 
The function \( \sin^{-1}(y) \) is defined for \( y \in [-1,1] \). Thus, we require: \[ -1 \leq \log_2 \left( \frac{x^2}{2} \right) \leq 1. \] 

Step 3: Solve for \( x \) 
Rewriting the inequality: \[ -1 \leq \log_2 \left( \frac{x^2}{2} \right) \leq 1. \] Converting to exponential form: \[ 2^{-1} \leq \frac{x^2}{2} \leq 2^1. \] \[ \frac{1}{2} \leq \frac{x^2}{2} \leq 2. \] Multiplying by 2: \[ 1 \leq x^2 \leq 4. \] 

Step 4: Solve for \( x \) 
Taking square roots: \[ -2 \leq x \leq -1 \quad \text{or} \quad 1 \leq x \leq 2. \] Thus, the domain of \( f(x) \) is: \[ [-2,-1] \cup [1,2]. \]