What is the domain of the function \(f(x)=\sqrt{1-|x-2|}\)?

Correct answer: \([1,3]\)

For a real square root, the expression inside the root must be non-negative. Thus, \(1-|x-2|\ge 0\), so \(|x-2|\le 1\). This gives \(-1\le x-2\le 1\), hence \(1\le x\le 3\). Therefore, the domain is \([1,3]\). Option C excludes the endpoints, but at \(x=1\) and \(x=3\), the radicand is 0, so both values are allowed. Exam tip: For a square-root function, require the radicand to be \(\ge 0\).