Question:

Step 1: Determine the domain constraint for the inverse sine function. For $f(x) = \sin^{-1}\left(\log_2 \dfrac{x^2}{2}\right)$ to be defined, the argument of $\sin^{-1}$ must lie in $[-1, 1]$. Therefore, we require: $$-1 \leq \log_2 \dfrac{x^2}{2} \leq 1$$ Step 2: Convert the logarithmic inequality to exponential form. Taking exponentials with base 2 on all parts: $$2^{-1} \leq \dfrac{x^2}{2} \leq 2^{1}$$ $$\dfrac{1}{2} \leq \dfrac{x^2}{2} \leq 2$$ Step 3: Solve for $x$ by isolating $x^2$. Multiplying all parts by 2: $$1 \leq x^2 \leq 4$$ Taking square roots (considering both positive and negative values): $$1 \leq |x| \leq 2$$ This means: $$x \in [-2, -1] \cup [1, 2]$$ Step 4: Find the range of the logarithmic expression. For $x \in [-2, -1] \cup [1, 2]$, the expression $\dfrac{x^2}{2}$ takes values in the interval $\left[\dfrac{1}{2}, 2\right]$. Therefore, $\log_2 \dfrac{x^2}{2}$ ranges over: $$\left[\log_2 \dfrac{1}{2}, \log_2 2\right] = [-1, 1]$$ Step 5: Apply the inverse sine function to determine the final range. The inverse sine function $\sin^{-1}(t)$ has range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ for all $t \in [-1, 1]$. Since the argument $\log_2 \dfrac{x^2}{2}$ covers the entire interval $[-1, 1]$, the range of $f(x)$ is: $$\boxed{\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]}$$ **Answer: Option 1**