$(D): A = [-\frac{3}{2}, \frac{3}{2}], B = [2\sqrt{2}, 4\sqrt{2}]$
Step-by-Step Solution
Key Concept: A function is bijective if and only if it is both injective (one-to-one) and surjective (onto) its codomain.
$f(x) = \sqrt{2}(\frac{2}{3}\sin x - \frac{1}{3}\cos x) + 3\sqrt{2} = \sqrt{2}\sin(x - \frac{\pi}{6}) + 3\sqrt{2}$. The range of $\sin(x - \frac{\pi}{6})$ is $[-1, 1]$, so the range of $f(x)$ is $[3\sqrt{2} - \sqrt{2}, 3\sqrt{2} + \sqrt{2}] = [2\sqrt{2}, 4\sqrt{2}]$. For injectivity on domain $[-\frac{3}{2}, \frac{3}{2}]$, the function must be strictly monotonic. Since $f'(x) = \sqrt{2}\cos(x - \frac{\pi}{6})$ changes sign on this interval, option D provides a domain where $f$ is bijective.
Correct Answer: 4