\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]
\( [1,\infty) \cup (-2,0) \)
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: \[ [-2,-1] \cup [1,2]. \]
Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]
\[ D = \begin{vmatrix} -\frac{bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & -\frac{ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & -\frac{ab}{c^2} \end{vmatrix} \]