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For the function $$f(x) = \sqrt{x^2 - 3x + 2} + \sqrt{3x - x^2 - 2}$$
We need $$x^2 - 3x + 2 \ge 0$$ and $$-(x^2 - 3x + 2) \ge 0$$ together, so $$x^2 - 3x + 2 = 0$$, i.e. $$x = 1$$ or $$x = 2$$. The domain is $$\{1, 2\}$$, and at both points $$f(x) = 0$$, so the range is $$\{0\}$$. Hence options b and c are both correct.
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