Answer: f: R → R ∈ f(x) = x2 g : R → C ∈ g(x) = x2 Both the functions are equal only when the domain and codomain of both the functions are equal. So, the domain of f ≠ domain of g. ∴ f and g are...
Write the following relations as sets of ordered pairs and find which of them are functions: {(x, y): x + y = 3, x, y ∈ {0, 1, 2, 3}}
Answer: If x = 0, 0 + y = 3 y = 3 If x = 1, 1 + y = 3 y = 2 If x = 2, 2 + y = 3 y = 1 If x = 3, 3 + y = 3 y = 0 ∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)} Hence, R is a...
Write the following relations as sets of ordered pairs and find which of them are functions: (i) {(x, y): y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}} (ii) {(x, y): y > x + 1, x = 1, 2 and y = 2, 4, 6}
Answers: (i) If x = 1, y = 3(1) = 3 If x = 2, y = 3(2) = 6 If x = 3, y = 3(3) = 9 ∴ R = {(1, 3), (2, 6), (3, 9)} Hence, R is a function. (ii) If x = 1, y > 1 + 1 or y > 2 y = {4, 6} If x = 2,...
Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = loge x. Determine whether f (xy) = f (x) + f (y) holds.
Answer: Given, f (x) = loge x ⇒ f (y) = loge y Consider, f (xy) F (xy) = loge (xy) f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c] f (xy) = loge x + loge y ∴ f (xy) = f (x) + f...
Let f: R+→ R, where R+ is the set of all positive real numbers, be such that f(x) = loge x. Determine (i) the image set of the domain of f (ii) {x: f (x) = –2}
Answers: (i) Domain of f = R+ Value of logarithm to the base e (natural logarithm) can take all possible real values. ∴ The image set of f = R (ii) f(x) = –2 loge x = –2 ∴ x = e-2 [logb a = c ⇒ a =...
A function f: R → R is defined by f(x) = x2. Determine {y: f(y) = –1}
Answer: f(y) = –1 y2 = –1 Domain of f - R, Value of y2 i- non-negative There is no real y for which y2 = –1. ∴{y: f(y) = –1} = ∅
A function f: R → R is defined by f(x) = x2. Determine (i) range of f (ii) {x: f(x) = 4}
Answers: (i) Domain of f = R (set of real numbers) Square of a real number is always positive or zero. ∴ range of f = R+∪ {0} (ii) f(x) = 4 x2 = 4 x2 – 4 = 0 (x – 2)(x + 2) = 0 ∴ x = ± 2 ∴ {x: f(x)...
If a function f: R → R be defined by
Find: f (1), f (–1), f (0), f (2). Answers: If, x > 0, f (x) = 4x + 1 Substitute, x = 1 f (1) = 4(1) + 1 f (1) = 4 + 1 f (1) = 5 If x < 0, f(x) = 3x – 2 Substitute, x = –1 f (–1) =...
Let A = {–2, –1, 0, 1, 2} and f: A → Z be a function defined by f(x) = x2 – 2x – 3. Find: (i) range of f i.e. f (A) (ii) pre-images of 6, –3 and 5
Answers: A = {–2, –1, 0, 1, 2} f : A → Z f(x) = x2 – 2x – 3 (i) A - Domain of the function f. The range is the set of elements f(x) for all x ∈ A. Substitute, x = –2 in f(x) f(–2) = (–2)2 – 2(–2) –...
What is the fundamental difference between a relation and a function? Is every relation a function?
Consider ‘f’ as a function and R as a relation defined from set X to set Y. Domain of the relation R might be a subset of the set X, but the domain of the function f must be equal to X. This is...
Define a function as a correspondence between two sets.
Function as a correspondence between two sets: Let A and B be two non-empty sets. Then a function ‘f’ from set A to B is a rule or method or correspondence which associates elements of set A to...
Define a function as a set of ordered pairs.
Function as a set of ordered pairs: Let A and B be two non-empty sets. A relation from A to B, i.e., a subset of A×B, is called a function (or a mapping) from A to B. 1. if for each a ∈ A there...