f (x) = \(\sqrt{\dfrac{{x-2}}{3-x}} \)
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 and 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 is null set. Thus, this case is not possible.
Hence, x ∈ [2, 3] – {3}
x ∈ [2, 3]
Domain (f) = [2, 3]
Answered by Aaryan | 1 year agoLet R = {(a, b) : a, b, ϵ N and a < b}.Show that R is a binary relation on N, which is neither reflexive nor symmetric. Show that R is transitive.
Let A = {3, 4, 5, 6} and R = {(a, b) : a, b ϵ A and a
(i) Write R in roster form.
(ii) Find: dom (R) and range (R)
(iii) Write R–1 in roster form
Let A = (1, 2, 3} and B = {4} How many relations can be defined from A to B.
Let R = {(x, x2) : x is a prime number less than 10}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
If A = {5} and B = {5, 6}, write down all possible subsets of A × B.