We know the square of a real number is never negative.

Clearly, f(x) takes real values only when x – 2 and 3 – x are both positive or negative.

(a) Both x – 2 and 3 – x are positive

x – 2 ≥ 0 ⇒ x ≥ 2

3 – x ≥ 0 ⇒ x ≤ 3

Hence, x ≥ 2 and x ≤ 3

∴ x ∈ [2, 3]

(b) Both x – 2 and 3 – x are negative

x – 2 ≤ 0 ⇒ x ≤ 2

3 – x ≤ 0 ⇒ x ≥ 3

Hence, x ≤ 2 and x ≥ 3

However, the intersection of these sets in null set. Thus, this case is not possible.

In addition, f(x) is also undefined when 3 – x = 0 because the denominator will be zero and the result will be indeterminate.

3 – x = 0 ⇒ x = 3

Hence, x ∈ [2, 3] – {3}

∴ x ∈ [2, 3)

Thus, domain of f = [2, 3)