What is the domain of f(x)=√(x-2)/√(8-x)?

Correct answer: [2,8)

The numerator √(x−2) is real only when x−2≥0, which gives x≥2. The denominator √(8−x) must satisfy two conditions: its radicand must be non-negative and the denominator itself must not be zero. Because it is in the denominator, we need √(8−x)>0, equivalently 8−x>0, so x<8. Combining x≥2 with x<8 gives 2≤x<8, or [2,8). The endpoint x=2 is allowed because the numerator is zero while the denominator is √6, which is nonzero. The endpoint x=8 is forbidden because the denominator becomes zero. Therefore option A is correct. B unnecessarily removes 2, C includes an undefined value, and D reverses the relevant inequalities.