RELATION AND FUNCTION

 

In mathematics, relations and functions are fundamental concepts that describe how elements from one set are related to elements in another set. They are used to model and analyze various mathematical and real-world situations.

1.Relation:

   - A relation is a set of ordered pairs (x, y) where x is from a set called the domain, and y is from a set called the codomain.

   - Relations can be represented using tables, lists of ordered pairs, or graphs.

   - Relations can be classified based on their properties, such as being reflexive, symmetric, or transitive.

   **Example**: Consider a set A = {1, 2, 3} and a set B = {a, b, c}. The relation R could be {(1, a), (2, b), (3, c)}.

2. Function :

   - A function is a specific type of relation where each element in the domain is related to exactly one element in the codomain.

   - In other words, for every x in the domain, there exists a unique y in the codomain such that (x, y) belongs to the function.

   - Functions are often represented using function notation, such as f(x) = y, where f is the function, x is the input, and y is the output.

   **Example**: The relation R = {(1, a), (2, b), (3, c)} can be a function if each element in the domain (1, 2, 3) maps to a unique element in the codomain (a, b, c).

Types of Functions:

- **One-to-One Function**: Also called an injective function, where each element in the domain maps to a unique element in the codomain, and no two elements in the domain map to the same element in the codomain.

- **Onto Function**: Also called a surjective function, where every element in the codomain is the image of at least one element in the domain.

- **Many-to-One Function**: A function where multiple elements in the domain can map to the same element in the codomain.

- **Inverse Function**: A function that "undoes" the action of another function.

Functions are fundamental in various fields of mathematics and science, as they provide a way to model relationships between quantities and analyze how they behave. They are used extensively in calculus, linear algebra, statistics, and many other areas of mathematics and its applications.

1. A relation R in a set A is called _______, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.

(a) symmetric

(b) transitive

(c) equivalence

(d) non-symmetric

Correct option: (a) symmetric

Solution:

A relation R in a set A is called symmetric, if (a1, a2) ∈ R implies (a2, a1) ∈ R, for all a1, a2 ∈ A.

2. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(a) Reflexive and symmetric

(b) Transitive and symmetric

(c) Equivalence

(d) Reflexive, transitive but not symmetric

Correct option: (d) Reflexive, transitive but not symmetric

Solution:

Given that n divides m, ∀ n ∈ N, R is reflexive.

Let n = 3 and m = 6

R is not symmetric since for 3, 6 ∈ N, 3 R 6 ≠ 6 R 3.

R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will also divide r.

So, the given relation is reflexive, transitive, but not symmetric.

3. The maximum number of equivalence relations on the set A = {1, 2, 3} are

(a) 1

(b) 2

(c) 3

(d) 5

Correct option: (d) 5

Solution:

Given, set A = {1, 2, 3}

The equivalence relations for the given set are:

R1 = {(1, 1), (2, 2), (3, 3)}

R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

R3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R5 = {(1, 2, 3) ⇔ A x A = A2}

Therefore, the maximum number of an equivalence relation is ‘5’.

4. If set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720

(b) 120

(c) 0

(d) none of these

Correct option: (c) 0

Solution:

Given,

n(A) = 5

n(B) = 6

Each element in set B is assigned to only one element in set A for the one-one function.

Here, only ‘5’ elements of set B are assigned to ‘5’ elements of set ‘A’ and one element will be left in set B.

The range of the function must be equal to B. However, for the given sets, it is not possible.

Thus, the range of functions does not contain all ‘6’ elements of set ‘ B’. Therefore, if the function is one-one it cannot be onto.

Hence, the number of one-one and onto mappings from A to B is 0.

5. Let f : [2, ∞) → R be the function defined by f(x) = x2 – 4x + 5, then the range of f is

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

Correct option: (b) [1, ∞)

Solution:

Given,

f : [2, ∞) → R

f(x) = x2 – 4x + 5

Let f(x) = y

x2 – 4x + 5 = y

x2 – 4x + 4 + 1 = y

(x – 2)2 + 1 = y

(x – 2)2 = y – 1

x – 2 = √(y – 1)

x = 2 + √(y – 1)

If the function is a real-valued function, then y – 1 ≥ 0

So, y ≥ 1.

Therefore, the range is [1, ∞).

6. Let f : R → R be defined by f(x) = 1/x ∀ x ∈ R. Then f is

(a) one-one

(b) onto

(c) bijective

(d) f is not defined

Correct option: (d) f is not defined

Solution:

f(x) = 1/x ∀ x ∈ R

Suppose x = 0, then f is not defined.

i.e., f (0) = 1/0 = undefined

So, the function f is not defined.

8. If f : R → R be defined by f(x) = 3x2 – 5 and g : R → R by g(x) = x/(x2 + 1), then g o f is

(a) (3x2 – 5)/(9x4 – 30x2 + 26)

(b) (3x2 – 5)/(9x4 – 6x2 + 26)

(c) 3x2/(x4 + 2x2 – 4)

(d) 3x2/(9x4 + 30x2 – 2)

Correct option: (a) (3x2 – 5)/(9x4 – 30x2 + 26)

Solution:

Given,

f(x) = 3x2 – 5

g(x) = x/(x2 + 1)

g o f = g[f(x)]

= g(3x2 – 5)

= (3x2 – 5)/ [(3x2 – 5)2 + 1]

= (3x2 – 5)/(9x4 – 30x2 + 25 + 1)

= (3x2 – 5)/(9x4 – 30x2 + 26)

9. Let f : R → R be given by f (x) = tan x. Then f–1(1) is

(a) π/4

(b) {n π + π/4 : n ∈ Z}

(c) does not exist

(d) none of these

Correct option: (a) π/4

Solution:

Given,

f(x) = tan x

Let f(x) = y

⇒ y = tan x

⇒ x = tan-1y

⇒ f-1(y) = tan-1y [since f(x) = y ⇒ x = f-1(y) ]

⇒ f-1(x) = tan-1x

Similarly,

f-1(1) = tan-1(1)

= tan-1(tan π/4)

= π/4

10. If f: R → R be given by f(x) = (3 – x3)1/3, then fof(x) is

(a) x1/3

(b) x3

(c) x

(d) (3 – x3)

Correct option: (c) x

Solution:

Given,

f(x) = (3 – x3)1/3, then

fof(x) = f[(3 – x3)1/3]

= [3 – {(3 – x3)1/3)}3]1/3

= [3 – (3 – x3)]1/3

= [3 – 3 + x3]1/3

= (x3)1/3

= x

11. Let f : A → B and g : B → C be the bijective functions. Then (g ∘ f)–1 is

(a) f–1 ∘ g–1

(b) f ∘ g

(c) g–1 ∘ f–1

(d) g ∘ f

Correct option: (a) f–1 ∘ g–1

Solution:

Given that, f : A → B and g : B → C be the bijective functions.

Let us consider three sets such as A = {1,3,4}, B ={2,5,1} and C = {3,4,2}.

f : A → B is a bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)}

f-1 = {(2,1),(5,3),(1,4)}

g : B → C is a bijective function.

∴ g = {(2, 3), (5, 4), (1, 2)}

g-1 ={(3,2),(4, 5),(2, 1)}

Now,

g∘f (1) = g[f(1)] = g(2) = 3

g∘f (3) = g[f(3)] = g(5) = 4

g∘f (4) = g[f(4)] = g(1) = 2

∴ g∘f = {(1,3),(3,4),(4,2)} ….(1)

Then (g∘f)-1 = {(3,1),(4, 3),(2, 4)} ….(2)

Now, f-1∘g-1(3) = f-1[g-1(3)] = f-1(2) = 1
f-1∘g-1(4) = f-1[g-1(4)] = f-1(5) = 3
f-1∘g-1(2) = f-1[g-1(2)] = f-1(1) = 4

∴ f-1∘g-1 = {(3, 1), (4, 3), (2, 4)} ….(3)

From the (2) and (3), we can conclude that (g ∘ f)–1 = f-1∘g-1.