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(B) The function is defined when $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$. Since $\log_2\left(\frac{x^2}{2}\right) = \log_2(x^2) - \log_2(2)

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$[-2, -1] \cup [1, 2]$

Vedclass · Answer

(B) The function is defined when $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$. Since $\log_2\left(\frac{x^2}{2}\right) = \log_2(x^2) - \log_2(2) = \log_2(x^2) - 1$, we have: $-1 \leq \log_2(x^2) - 1 \leq 1$. Adding $1$ to all parts: $0 \leq \log_2(x^2) \leq 2$. Converting from logarithmic to exponential form: $2^0 \leq x^2 \leq 2^2 \Rightarrow 1 \leq x^2 \leq 4$. Taking the square root, we get $|x| \in [1, 2]$, which implies $x \in [-2, -1] \cup [1, 2]$.

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

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