Question -
Answer -
(i) f(x) = √(x-2)
We know the square of a real number is never negative.
f (x) takes real values only when x – 2 ≥ 0
x ≥ 2
∴ x ∈ [2,∞)
∴Domain (f) = [2, ∞)
(ii) f(x) = 1/(√(x2-1))
We know the square of a real number is never negative.
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, ∞)
In addition, f (x) is also undefined when x2 – 1 = 0 because denominator will be zero and theresult will be indeterminate.
x2 – 1 = 0 ⇒ x= ± 1
So, x ∈ (–∞, –1] ∪ [1, ∞) – {–1, 1}
x ∈ (–∞, –1) ∪ (1, ∞)
∴Domain (f) = (–∞, –1) ∪ (1, ∞)
(iii) f(x) = √(9-x2)
We know the square of a real number is never negative.
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]
∴Domain (f) = [–3, 3]
(iv) f(x) = √(x-2)/(3-x)
We know the square root of a real number is never negative.
f (x) takes real values only when x – 2 and 3 – x are both positive andnegative.
(a) Bothx – 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) Bothx – 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 is null set. Thus, this case isnot possible.
Hence, x ∈ [2, 3] – {3}
x ∈ [2, 3]
∴Domain (f) = [2, 3]