What is the range of (f(x)=\sqrt{2x-x^2}+3)?

Correct answer: ( [3,4] )

The direct answer is A: [3,4]. For a square root to be real, its inside must be non-negative: 2x−x²≥0. Rewrite it as 1−(x−1)². Since (x−1)² is at least 0, the expression 1−(x−1)² has maximum 1. On its allowed domain, it has minimum 0, so 0≤2x−x²≤1. Taking square roots gives 0≤√(2x−x²)≤1. Adding 3 throughout gives 3≤f(x)≤4. Thus the range is [3,4]. Option A is correct. Option B [0,1] is the range of the square-root part before adding 3. Option C [2,4] incorrectly includes values below 3. Option D [3,∞) incorrectly allows values larger than 4, although the square-root part never exceeds 1. Memory cue: find the range inside the root first, then apply the outside operations.