Find the domain of each of the following real valued functions of real variable:
(i) f(x) = \( \dfrac{1}{x}\)
(ii) f(x) = \( \dfrac{1}{(x-7)}\)
(iii) f(x) = \(\dfrac{ (3x-2)}{(x+1)}\)
(iv) f(x) = \(\dfrac{ (2x+1)}{(x^2-9)}\)
(v) f(x) = \(\dfrac{ (x^2+2x+1)}{(x^2-8x+12)}\)
(i) We know, f (x) is defined for all real values of x, except for the case when x = 0.
Domain of f = R – {0}
(ii) We know, f (x) is defined for all real values of x, except for the case when x – 7 = 0 or x = 7.
Domain of f = R – {7}
(iii) We know, f(x) is defined for all real values of x, except for the case when x + 1 = 0 or x = –1.
Domain of f = R – {–1}
(iv) We know, f (x) is defined for all real values of x, except for the case when x2 – 9 = 0.
x2 – 9 = 0
x2 – 32 = 0
(x + 3)(x – 3) = 0
x + 3 = 0 or x – 3 = 0
x = ± 3
Domain of f = R – {–3, 3}
(v) We know, f(x) is defined for all real values of x, except for the case when x2 – 8x + 12 = 0.
x2 – 8x + 12 = 0
x2 – 2x – 6x + 12 = 0
x(x – 2) – 6(x – 2) = 0
(x – 2)(x – 6) = 0
x – 2 = 0 or x – 6 = 0
x = 2 or 6
Domain of f = R – {2, 6}
Answered by Sakshi | 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.