(i) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when x – 2 ≥ 0

x ≥ 2

∴ x ∈ [2, ∞)

∴ The domain (f) = [2, ∞)

(ii) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when x² – 1 ≥ 0

x² – 1² ≥ 0

(x + 1) (x – 1) ≥ 0

x ≤ –1 or x ≥ 1

∴ x ∈ (–∞, –1] ∪ [1, ∞)

Here, in addition f(x) is also undefined when x² – 1 = 0 because denominator will be zero and the result will be indeterminate.

x² – 1 = 0 ⇒ x = ± 1

Therefore, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}

 x ∈ (–∞, –1) ∪ (1, ∞)

∴ The domain (f) = (–∞, –1) ∪ (1, ∞)

(iii) As we know that the square of a real number is never negative.

Now, f(x) takes real values only when 9 – x² ≥ 0

9 ≥ x²

x² ≤ 9

x² – 9 ≤ 0

x² – 3² ≤ 0

(x + 3)(x – 3) ≤ 0

x ≥ –3 and x ≤ 3

x ∈ [–3, 3]

∴ The domain (f) = [–3, 3]

(iv) As we know the square root of a real number is never negative.

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

(a) Here, both x – 2 and 3 – x are positive

x – 2 ≥ 0

x ≥ 2

3 – x ≥ 0

x ≤ 3

Thus, x ≥ 2 and x ≤ 3

∴ x ∈ [2, 3]

(b) Here, both x – 2 and 3 – x are negative

x – 2 ≤ 0 

x ≤ 2

3 – x ≤ 0 

x ≥ 3

Thus, x ≤ 2 and x ≥ 3

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

Thus, x ∈ [2, 3] – {3}

x ∈ [2, 3]

∴ The domain (f) = [2, 3]