Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \sin x$. Which of the following statements is true about the function $f$?

Correct Option: D. Solution: The function $f(x) = \sin x$ is periodic with period $2\pi$, which means it is not one-one over the entire set of real numbers. Additionally, the range of $f(x) = \sin x$ is $[-1, 1]$, meaning it is not onto when considering the entire set of real numbers as the codomain. Therefore, $f(x) = \sin x$ is neither one-one nor onto.

Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \sin x$. Consider the composition of $f$ with itself, $f \circ f$. Which of the following statements is true?

Correct Option: C. Solution: The function $f(x) = \sin x$ is neither one-one nor onto as it maps $\mathbb{R}$ to $[-1, 1]$. The composition $f \circ f(x) = \sin(\sin x)$ further restricts the range to $[-\sin 1, \sin 1]$, making it neither one-one nor onto.

Let $A = \{1, 2, 3\}$. How many equivalence relations containing $(1, 2)$ exist?

Correct Option: B. Solution: The smallest equivalence relation containing $(1, 2)$ is $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. Adding $(2, 3)$ and $(3, 2)$ makes it the universal relation. Thus, there are 2 such relations.

Relations and Functions

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Summary

Chapter Summary: Relations and Functions

Key Concepts

Relations: A relation R from set A to set B is a subset of A x B.
Functions: A special type of relation where each element in the domain maps to exactly one element in the co-domain.

Types of Relations

Empty Relation: R = Φ, no elements are related.
Universal Relation: R = A x A, every element is related to every other element.
Reflexive Relation: (a, a) ∈ R for all a ∈ A.
Symmetric Relation: If (a, b) ∈ R, then (b, a) ∈ R.
Transitive Relation: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
Equivalence Relation: A relation that is reflexive, symmetric, and transitive.

Functions

One-One (Injective): f(x₁) = f(x₂) implies x₁ = x₂.
Onto (Surjective): For every y in the co-domain, there exists an x in the domain such that f(x) = y.
Bijective: A function that is both one-one and onto.

Important Definitions

Equivalence Class: The subset of X containing all elements related to a.
Composition of Functions: Combining two functions where the output of one function becomes the input of another.
Invertible Functions: Functions that have an inverse.

Examples of Relations

Symmetric but not Reflexive or Transitive: R = {(1, 2), (2, 1)}.
Equivalence Relation: R = {(x, y): x and y have the same number of pages}.
Reflexive and Transitive but not Symmetric: R = {(a, b): a ≤ b}.

Common Pitfalls

Confusing one-one and onto functions.
Misidentifying types of relations (e.g., assuming a relation is equivalence without checking all properties).

Exam Tips

Always verify the properties of relations and functions with examples.
Practice identifying and constructing equivalence classes.

Learning Objectives

Define the concept of relations and functions.
Identify different types of relations: reflexive, symmetric, transitive, and equivalence relations.
Explain the definitions of one-one (injective), onto (surjective), and bijective functions.
Describe the composition of functions and invertible functions.
Analyze examples of relations and functions to determine their properties.
Apply the definitions of relations and functions to solve mathematical problems.

Detailed Notes

Chapter 1: Relations and Functions

1.1 Introduction

The notion of relations and functions, domain, co-domain, and range were introduced in Class XI.
A relation in mathematics is defined as a recognisable connection between two objects or quantities.
Example of relations from set A (students of Class XII) to set B (students of Class XI):
- (i) {(a, b) ∈ A × B: a is brother of b}
- (ii) {(a, b) ∈ A × B: a is sister of b}
- (iii) {(a, b) ∈ A × B: age of a is greater than age of b}
- (iv) {(a, b) ∈ A × B: total marks obtained by a in the final examination is less than the total marks obtained by b}
- (v) {(a, b) ∈ A × B: a lives in the same locality as b}
A relation R from A to B is defined as an arbitrary subset of A × B.

1.2 Types of Relations

Empty Relation: R = ∅ ⊆ A × A, where no element of A is related to any element of A.
Universal Relation: R = A × A, where each element of A is related to every element of A.
Reflexive Relation: R is reflexive if (a, a) ∈ R for every a ∈ A.
Symmetric Relation: R is symmetric if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R for all a₁, a₂ ∈ A.
Transitive Relation: R is transitive if (a₁, a₂) ∈ R and (a₂, a₃) ∈ R implies (a₁, a₃) ∈ R for all a₁, a₂, a₃ ∈ A.
Equivalence Relation: A relation that is reflexive, symmetric, and transitive.

1.3 Types of Functions

One-One (Injective): A function f: X → Y is one-one if f(x₁) = f(x₂) implies x₁ = x₂.
Onto (Surjective): A function f: X → Y is onto if for every y ∈ Y, there exists an x ∈ X such that f(x) = y.
Bijective: A function is bijective if it is both one-one and onto.

Key Definitions and Properties

Equivalence Class: [a] containing a ∈ X for an equivalence relation R in X is the subset of X containing all elements b related to a.
Characteristic Property of Finite Sets: For a finite set X, a function f: X → X is one-one if and only if it is onto, and vice versa.

Historical Note

The concept of function has evolved over time, starting from R. Descartes to G. H. Hardy.
The modern set-theoretic definition of function was developed by Georg Cantor.

Examples of Relations

Example of a relation that is symmetric but neither reflexive nor transitive: R = {(1, 2), (2, 1)}.
Example of an equivalence relation: R = {(x, y) : x and y have the same number of pages in a library}.

Exercises

Determine the properties of various relations defined in different sets.
Show that specific functions are one-one, onto, or neither.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Relations: Students often confuse the definitions of reflexive, symmetric, and transitive relations. Ensure you understand each definition clearly:
- Reflexive: For every element a in set A, (a, a) must be in relation R.
- Symmetric: If (a₁, a₂) is in R, then (a₂, a₁) must also be in R.
- Transitive: If (a₁, a₂) and (a₂, a₃) are in R, then (a₁, a₃) must also be in R.
Equivalence Relations: Many students fail to check all three properties (reflexive, symmetric, transitive) when determining if a relation is an equivalence relation. Always verify all conditions.
Function Types: Confusion often arises between one-one (injective), onto (surjective), and bijective functions. Remember:
- One-one: Distinct elements in the domain map to distinct elements in the codomain.
- Onto: Every element in the codomain is an image of some element in the domain.
- Bijective: A function that is both one-one and onto.

Exam Tips

Practice with Examples: Work through examples of each type of relation and function to solidify your understanding. For instance, show that the relation R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Use Diagrams: When studying functions, sketch diagrams to visualize mappings between sets. This can help clarify whether a function is one-one or onto.
Check Extremes: For relations, remember the definitions of empty and universal relations. These can often be overlooked but are crucial for understanding the broader concepts.
Review Definitions Regularly: Make flashcards for key definitions and properties of relations and functions to reinforce your memory.
Solve Past Papers: Familiarize yourself with the format and types of questions that appear in exams by solving past papers.

Practice & Assessment

Multiple Choice Questions

Correct Answer: B

Solution:

To be reflexive, all pairs (1, 1), (2, 2), and (3, 3) must be included. For symmetry, if (a, b) is included, then (b, a) must also be included. There are 2 such relations:

\{(1, 1), (2, 2), (3, 3)\}

and

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}

R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}

R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}

Correct Answer: A

Solution:

An equivalence relation must be reflexive, symmetric, and transitive. Option (a) satisfies all these properties: it is reflexive as it includes

(1, 1)

(2, 2)

, and

(3, 3)

; symmetric as it includes both

(1, 2)

and

(2, 1)

; and transitive as there are no pairs that violate transitivity.

The function is one-one but not onto.

The function is onto but not one-one.

The function is both one-one and onto.

The function is neither one-one nor onto.

Correct Answer: C

Solution:

The function

f: A \to B

is one-one because each element of

A

maps to a unique element in

B

. It is also onto because every element in

B

is an image of some element in

A

. Therefore, the function is both one-one and onto.

R

is reflexive, symmetric, and transitive.

R

is reflexive and symmetric, but not transitive.

R

is symmetric and transitive, but not reflexive.

R

is reflexive and transitive, but not symmetric.

Correct Answer: A

Solution:

The relation

R

is reflexive because for any

a \in A

|a - a| = 0

is even. It is symmetric because if

|a - b|

is even, then

|b - a|

is also even. It is transitive because if

|a - b|

and

|b - c|

are even, then

|a - c|

is also even.

R = {(a, a) : a is a student in a class}

R = {(a, b) : a is taller than b}

R = {(a, b) : a is a friend of b}

R = {(a, b) : a is the father of b}

Correct Answer: A

Solution:

A reflexive relation on a set A is one where every element is related to itself. In option A, each student is related to themselves.

f \circ f(x) = e^{2x}

f \circ f(x) = e^{e^x}

f \circ f(x) = e^{x^2}

f \circ f(x) = x

Correct Answer: B

Solution:

The composition

f \circ f(x)

means applying

f

twice. Since

f(x) = e^x

, we have

f(f(x)) = f(e^x) = e^{e^x}

. Therefore,

f \circ f(x) = e^{e^x}

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

\{(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)\}

\{(1, 1), (2, 2), (3, 3)\}

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}

Correct Answer: A

Solution:

The relation

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

is reflexive and symmetric but not transitive because it does not contain the pair (1, 3) which is needed for transitivity.

h

is injective but not surjective

h

is surjective but not injective

h

is both injective and surjective

h

is neither injective nor surjective

Correct Answer: C

Solution:

The function

h

maps each element of

A

to a unique element in

B

, hence it is injective. It also covers all elements of

B

, making it surjective. Therefore,

h

is both injective and surjective.

f

is one-one but not onto.

f

is onto but not one-one.

f

is both one-one and onto.

f

is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f

is defined from a set

A

with 3 elements to a set

B

with 4 elements. Since each element of

A

maps to a unique element in

B

f

is one-one. However, not every element in

B

is mapped to, so

f

is not onto.

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: B

Solution:

The function

f(n) = 2n

is one-one because each natural number maps to a unique even number. However, it is not onto because odd numbers are not in the range.

R is reflexive.

R is symmetric.

R is transitive.

R is an equivalence relation.

Correct Answer: B

Solution:

The relation

R

is symmetric because if

(a, b) \in R

, then

a + b = 6

implies

b + a = 6

, so

(b, a) \in R

. However,

R

is not reflexive because there is no

a \in A

such that

a + a = 6

. It is also not transitive because if

(a, b) \in R

and

(b, c) \in R

a + b = 6

and

b + c = 6

does not imply

a + c = 6

. Thus,

R

is not an equivalence relation.

The function is one-one but not onto.

The function is onto but not one-one.

The function is both one-one and onto.

The function is neither one-one nor onto.

Correct Answer: C

Solution:

The function

f(x) = x^3 - 3x + 1

is a cubic polynomial, which is continuous and differentiable everywhere. The derivative

f'(x) = 3x^2 - 3

is zero at

x = 1

and

x = -1

. However, the function changes sign around these points, indicating that

f(x)

is strictly increasing on

(-\infty, -1)

(-1, 1)

, and

(1, \infty)

. Therefore,

f(x)

is one-one. Since the range of a cubic polynomial is

\mathbb{R}

, the function is onto as well.

The function is one-one and onto.

The function is many-one and onto.

The function is one-one but not onto.

The function is neither one-one nor onto.

Correct Answer: D

Solution:

The modulus function

f(x) = |x|

is neither one-one nor onto because different values of

x

can have the same absolute value, and not all real numbers are positive or zero.

f

is one-one and onto.

f

is one-one but not onto.

f

is onto but not one-one.

f

is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = x^2

is neither one-one (as

f(1) = f(-1)

) nor onto (as negative numbers are not in the range).

Reflexive and symmetric but not transitive.

Symmetric and transitive but not reflexive.

Reflexive and transitive but not symmetric.

Transitive but neither reflexive nor symmetric.

Correct Answer: D

Solution:

The relation

R = \{(1, 2), (2, 3), (1, 3)\}

is transitive because

(1, 2)

and

(2, 3)

imply

(1, 3)

. However, it is not reflexive since it does not contain

(1, 1)

(2, 2)

, or

(3, 3)

. It is also not symmetric because

(1, 2)

does not imply

(2, 1)

. Therefore, the relation is transitive but neither reflexive nor symmetric.

f: \mathbb{N} \to \mathbb{N}

given by

f(x) = 2x

f: \mathbb{R} \to \mathbb{R}

given by

f(x) = x^2

f: \mathbb{R} \to \mathbb{R}

given by

f(x) = 3x + 1

f: \mathbb{Z} \to \mathbb{Z}

given by

f(x) = x + 1

Correct Answer: A

Solution:

The function

f(x) = 2x

is one-one because different inputs give different outputs, but it is not onto because odd numbers in the codomain cannot be expressed as

2x

for any natural number

x

Correct Answer: C

Solution:

An equivalence relation partitions a set into disjoint subsets. For a set with 3 elements, the possible partitions are: all elements in one subset (1), each element in its own subset (1), and two elements in one subset with the third in another (3). Therefore, there are a total of 1 + 1 + 3 = 5 possible partitions, but each partition corresponds to an equivalence relation, so the total number of equivalence relations is 8.

{(1, 4), (2, 5), (3, 6)}

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

{(1, 4), (2, 4), (3, 4)}

{(1, 5), (2, 6), (3, 4)}

Correct Answer: B

Solution:

A reflexive relation on a set

A

must include all pairs

(a, a)

for all

a \in A

. Since

B

is the codomain, option b is not reflexive on

A \times B

, but it is reflexive on

A

. However, the question is misleading as reflexivity is not typically defined for relations from

A

B

unless

A = B

. Thus, the correct option is a trick question.

f^{-1}(x) = \frac{x - 3}{2}

f^{-1}(x) = 2x - 3

f^{-1}(x) = \frac{x + 3}{2}

f^{-1}(x) = \frac{x}{2} - 3

Correct Answer: A

Solution:

To find the inverse of

f(x) = 2x + 3

, set

y = 2x + 3

and solve for

x

x = \frac{y - 3}{2}

. Therefore, the inverse function is

f^{-1}(x) = \frac{x - 3}{2}

f^{-1}(x) = \frac{x - 7}{3}

f^{-1}(x) = \frac{x + 7}{3}

f^{-1}(x) = 3x - 7

f^{-1}(x) = \frac{x}{3} - 7

Correct Answer: A

Solution:

To find the inverse, solve for

x

y = 3x + 7 \Rightarrow x = \frac{y - 7}{3}

. Thus, the inverse function is

f^{-1}(x) = \frac{x - 7}{3}

R is reflexive and symmetric but not transitive.

R is symmetric and transitive but not reflexive.

R is reflexive, symmetric, and transitive.

R is not reflexive, symmetric, or transitive.

Correct Answer: C

Solution:

For reflexivity,

a - a = 0

is a multiple of 4, so

R

is reflexive. For symmetry, if

(a, b) \in R

, then

a - b

is a multiple of 4, and so is

b - a = -(a - b)

, hence

(b, a) \in R

. For transitivity, if

(a, b) \in R

and

(b, c) \in R

, then

a - b

and

b - c

are multiples of 4, so

a - c = (a - b) + (b - c)

is also a multiple of 4, hence

(a, c) \in R

. Therefore,

R

is reflexive, symmetric, and transitive.

{(1, 2), (2, 1)}

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

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

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

Correct Answer: A

Solution:

The relation {(1, 2), (2, 1)} is symmetric because if (1, 2) is in the relation, then (2, 1) is also in the relation. However, it is neither reflexive (since not all elements are related to themselves) nor transitive (since there is no (1, 3) or (3, 1) to complete a transitive closure).

The function is one-one and onto.

The function is one-one but not onto.

The function is onto but not one-one.

The function is neither one-one nor onto.

Correct Answer: D

Solution:

The function

g(x) = \sin x

is periodic and not injective, hence it is not one-one. It is also not onto because its range is limited to

[-1, 1]

, not covering all real numbers.

It is one-one and onto.

It is one-one but not onto.

It is onto but not one-one.

It is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = \sin x

is periodic with period

2\pi

, which means it is not one-one over the entire set of real numbers. Additionally, the range of

f(x) = \sin x

[-1, 1]

, meaning it is not onto when considering the entire set of real numbers as the codomain. Therefore,

f(x) = \sin x

is neither one-one nor onto.

Reflexive only

Symmetric only

Transitive only

Reflexive, symmetric, and transitive

Correct Answer: D

Solution:

The relation

R

is reflexive because for any real number

a

a - a = 0

, which is an integer. It is symmetric because if

a - b

is an integer, then

b - a

is also an integer. It is transitive because if

a - b

and

b - c

are integers, then

a - c = (a - b) + (b - c)

is also an integer. Thus,

R

is an equivalence relation.

f

is one-one and onto

f

is many-one and onto

f

is one-one but not onto

f

is neither one-one nor onto

Correct Answer: D

Solution:

The function

f(x) = x^4

is neither one-one nor onto because it maps multiple

x

values to the same

f(x)

value (e.g.,

f(1) = f(-1)

) and does not cover all real numbers.

The composition

g \circ f

is onto.

The composition

f \circ g

is one-one.

The composition

g \circ f

is one-one.

The composition

f \circ g

is onto.

Correct Answer: A

Solution:

The composition

g \circ f(x) = (x^2) + 1

is onto because for any real number

y

, there exists

x = \sqrt{y-1}

such that

g \circ f(x) = y

. However, it is not one-one because

g \circ f(x) = g \circ f(-x)

\mathbb{R}

[0, \infty)

(-\infty, 0]

(-\infty, \infty)

Correct Answer: B

Solution:

The function

f(x) = x^2

maps all real numbers to non-negative real numbers, so the range is

[0, \infty)

f

is one-one but not onto

f

is onto but not one-one

f

is both one-one and onto

f

is neither one-one nor onto

Correct Answer: C

Solution:

The function

f(x) = 2x + 1

is a linear function with a non-zero slope, hence it is one-one. Since it maps every real number to another real number, covering the entire real line, it is also onto. Therefore,

f

is both one-one and onto.

Reflexive only

Symmetric only

Transitive only

Symmetric and Transitive

Correct Answer: B

Solution:

The relation

R

is symmetric because if

|a - b|

is odd, then

|b - a|

is also odd, hence

(b, a) \in R

. However,

R

is not reflexive because

|a - a| = 0

is not odd, and it is not transitive because if

(a, b) \in R

and

(b, c) \in R

|a - c|

is not necessarily odd.

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = 2x + 1

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = x^3

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = |x|

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = x^2

Correct Answer: A

Solution:

The function

f(x) = 2x + 1

is both one-one and onto, making it a bijection.

f \circ f

is one-one.

f \circ f

is onto.

f \circ f

is neither one-one nor onto.

f \circ f

is both one-one and onto.

Correct Answer: C

Solution:

The function

f(x) = \sin x

is neither one-one nor onto as it maps

\mathbb{R}

[-1, 1]

. The composition

f \circ f(x) = \sin(\sin x)

further restricts the range to

[-\sin 1, \sin 1]

, making it neither one-one nor onto.

Correct Answer: B

Solution:

The smallest equivalence relation containing

(1, 2)

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

. Adding

(2, 3)

and

(3, 2)

makes it the universal relation. Thus, there are 2 such relations.

Injective

Surjective

Bijective

None of the above

Correct Answer: C

Solution:

A function that is both one-one (injective) and onto (surjective) is called bijective.

The function is one-one but not onto.

The function is onto but not one-one.

The function is neither one-one nor onto.

The function is both one-one and onto.

Correct Answer: C

Solution:

The function

f(x) = 3x^2 - 4x + 5

is a quadratic function, which is not one-one because it is not strictly increasing or decreasing over its entire domain. Additionally, it is not onto because a quadratic function with a minimum value (parabola opening upwards) cannot cover the entire range of real numbers. Thus, the function is neither one-one nor onto.

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (3, 1)\}

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (3, 2)\}

Correct Answer: A

Solution:

The relation

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

is reflexive and symmetric because it contains all pairs

(a, a)

for

a \in A

and for every

(a, b)

(b, a)

is also present. It is not transitive because

(1, 2)

and

(2, 1)

are present, but

(1, 1)

is not required for transitivity.

g \circ f

f \circ g

f + g

g - f

Correct Answer: A

Solution:

The composition of two functions

f

and

g

is denoted by

g \circ f

, which means applying

f

first and then

g

\{4, 5, 6\}

\{4, 5\}

\{4, 5, 6, 7\}

\{4, 5, 6, 7, 8\}

Correct Answer: A

Solution:

For a function to be bijective, the sets

A

and

B

must have the same number of elements. Since

A

has 3 elements,

B

must also have 3 elements.

It is an equivalence relation.

It is reflexive and symmetric but not transitive.

It is reflexive and transitive but not symmetric.

It is symmetric and transitive but not reflexive.

Correct Answer: A

Solution:

The relation

R

is reflexive because

a - a = 0

is divisible by 4 for any integer

a

. It is symmetric because if

a - b

is divisible by 4, then

b - a

is also divisible by 4. It is transitive because if

a - b

and

b - c

are divisible by 4, then

a - c

is divisible by 4. Thus,

R

is an equivalence relation.

Correct Answer: B

Solution:

There are two equivalence relations that include the pair (1, 2):

\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}

and the universal relation.

g \circ f = f \circ g

for all

x \in \mathbb{R}

g \circ f \neq f \circ g

for all

x \in \mathbb{R}

g \circ f = f \circ g

for

x = 0

, but not for other values.

g \circ f \neq f \circ g

for

x = 0

, but equal for other values.

Correct Answer: B

Solution:

For

g \circ f(x) = g(f(x)) = \cos(3x^2)

and

f \circ g(x) = f(g(x)) = 3(\cos x)^2

. These expressions are not equal for any

x

, including

x = 0

The function is one-one and onto.

The function is one-one but not onto.

The function is onto but not one-one.

The function is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = x^2

is neither one-one (since

f(x) = f(-x)

) nor onto (since negative numbers are not in the range).

The function

f

is one-one but not onto.

The function

f

is onto but not one-one.

The function

f

is both one-one and onto.

The function

f

is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(x) = e^x

is one-one because it is strictly increasing for all

x \in \mathbb{R}

. However, it is not onto because its range is

(0, \infty)

, not all of

\mathbb{R}

Reflexive and symmetric but not transitive.

Reflexive and transitive but not symmetric.

Symmetric and transitive but not reflexive.

Reflexive, symmetric, and transitive.

Correct Answer: D

Solution:

For reflexivity,

a + a = 2a

is even for all

a \in A

. For symmetry, if

a + b

is even, then

b + a

is also even. For transitivity, if

a + b

and

b + c

are even, then

a + c

is also even. Thus,

R

is reflexive, symmetric, and transitive.

Lines that are parallel to each other.

Lines that intersect at a point.

Lines that are perpendicular to each other.

Lines that have the same slope.

Correct Answer: A

Solution:

The relation of being parallel is reflexive (a line is parallel to itself), symmetric (if line

L_1

is parallel to

L_2

, then

L_2

is parallel to

L_1

), and transitive (if

L_1

is parallel to

L_2

and

L_2

is parallel to

L_3

, then

L_1

is parallel to

L_3

The function is one-one but not onto.

The function is onto but not one-one.

The function is both one-one and onto.

The function is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(n) = 2n

is one-one because

f(n_1) = f(n_2)

implies

n_1 = n_2

. However, it is not onto because there is no natural number

n

such that

f(n) = 1

One-one only

Onto only

Both one-one and onto

Neither one-one nor onto

Correct Answer: D

Solution:

The function

f(x) = 2x^3 - 3x^2 + 1

is neither one-one nor onto. It is not one-one because it is a cubic polynomial which can have turning points, and it is not onto because there is no

x

such that

f(x)

covers all real numbers.

Correct Answer: B

Solution:

A relation is reflexive if it contains all pairs

(a, a)

for

a \in A

. It is symmetric if for every

(a, b)

in the relation,

(b, a)

is also in the relation. To satisfy these conditions and not be transitive, we can have two such relations.

Books with the same number of pages

Books written by the same author

Books published in the same year

Books with the same title

Correct Answer: A

Solution:

The relation 'books with the same number of pages' is reflexive, symmetric, and transitive, making it an equivalence relation.

f(x) = x^2

f(x) = x + 1

f(x) = 2x

f(x) = \frac{1}{x}

Correct Answer: A

Solution:

The function

f(x) = x^2

is neither one-one nor onto. It is not one-one because

f(x) = f(-x)

for any

x

. It is not onto because negative numbers are not in the range of

f(x) = x^2

Yes,

f

is invertible

No,

f

is not invertible

Yes,

f

is invertible but not bijective

No,

f

is not invertible but bijective

Correct Answer: A

Solution:

The function

f(x) = x + 3

maps each element of set

A = \{1, 2, 3\}

to a unique element in set

B = \{4, 5, 6\}

, making it one-one. Since each element of

B

is covered, the function is also onto. Therefore,

f

is bijective and hence invertible.

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: D

Solution:

The function

f(x) = x^2 - 4x + 3

is a quadratic function, which is neither one-one (since it is not injective) nor onto (since it does not cover all real numbers, e.g., there is no

x

such that

f(x) = -1

Correct Answer: B

Solution:

The number of one-one functions from a set to itself is the number of permutations of the set. For a set with 3 elements, this is

3! = 6

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: D

Solution:

The function

f(x) = x^2 - 4x + 3

is a quadratic function, which is neither one-one (as it is not injective) nor onto (as it does not cover all real numbers as outputs).

Reflexive

Symmetric

Transitive

None of the above

Correct Answer: D

Solution:

The relation

R

is not reflexive as it does not include pairs like

(1, 1)

. It is not symmetric because

(1, 2)

is in

R

but

(2, 1)

is not. It is not transitive because

(1, 2)

and

(2, 3)

are in

R

, but

(1, 3)

is not. Hence, none of the properties are satisfied.

{(1, 4), (2, 5), (3, 6)}

{(1, 5), (2, 4), (3, 5)}

{(1, 4), (1, 5), (2, 6)}

{(2, 5), (3, 6), (3, 4)}

Correct Answer: A

Solution:

A relation from

A

B

is a subset of

A \times B

. Option A is a valid subset.

One-one

Onto

Bijective

Neither one-one nor onto

Correct Answer: A

Solution:

The function

f

is one-one because each element of

A

maps to a unique element in

B

. However, it is not onto because not every element in

B

is an image of some element in

A

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = x^2

f: \mathbb{R}^+ \to \mathbb{R}^+

defined by f(x) = \ln(x)$

f: \mathbb{N} \to \mathbb{N}

defined by

f(x) = 2x

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = x^3

Correct Answer: D

Solution:

The function

f(x) = x^3

is a cubic polynomial, which is strictly increasing over the entire set of real numbers. This means it is one-one. Additionally, for every real number

y

, there exists a real number

x

such that

f(x) = y

, meaning it is onto. Therefore,

f(x) = x^3

is both one-one and onto.

f

is one-one and onto.

f

is one-one but not onto.

f

is onto but not one-one.

f

is neither one-one nor onto.

Correct Answer: B

Solution:

The function

f(n) = n^2 + 1

is one-one because if

f(a) = f(b)

, then

a^2 + 1 = b^2 + 1

, which implies

a = b

. However, it is not onto because there is no natural number

n

such that

f(n) = 1

, for instance.

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: D

Solution:

The modulus function

f(x) = |x|

is neither one-one nor onto. It is not one-one because

f(x) = f(-x)

for

x \neq 0

, and it is not onto because it does not map to negative numbers.

The function is one-one but not onto.

The function is onto but not one-one.

The function is both one-one and onto.

The function is neither one-one nor onto.

Correct Answer: C

Solution:

A bijective function is both one-one (injective) and onto (surjective).

The function

g

is one-one and onto.

The function

g

is one-one but not onto.

The function

g

is onto but not one-one.

The function

g

is neither one-one nor onto.

Correct Answer: B

Solution:

The function

g(x) = e^{2x} - 5

is strictly increasing because the derivative

g'(x) = 2e^{2x} > 0

for all

x

. Therefore,

g

is one-one. However,

g(x)

approaches

-5

x \to -\infty

and grows without bound as

x \to \infty

, thus it is not onto since it cannot achieve values less than

-5

R = {(1, 2), (2, 1)}

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

R = {(1, 1), (2, 2)}

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

Correct Answer: A

Solution:

A symmetric relation satisfies

(a, b) \in R \implies (b, a) \in R

. Option A satisfies this condition.

f is one-one but not onto

f is onto but not one-one

f is both one-one and onto

f is neither one-one nor onto

Correct Answer: C

Solution:

A bijective function is both one-one (injective) and onto (surjective).

R = {(1, 2), (2, 1)}

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

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

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

Correct Answer: A

Solution:

The relation R = {(1, 2), (2, 1)} is symmetric because if (1, 2) is in R, then (2, 1) is also in R. However, it is not reflexive because not all elements are related to themselves, and it is not transitive because (1, 2) and (2, 1) do not imply (1, 1).

It is one-one and onto.

It is one-one but not onto.

It is onto but not one-one.

It is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = x^2

is neither one-one nor onto. It is not one-one because

f(x) = f(-x)

for any

x

. It is not onto because negative numbers are not in the range of

f(x) = x^2

R = {(a, a) : a ∈ A}

R = {(a, b) : a < b}

R = {(a, b) : a ≠ b}

R = {(a, b) : a > b}

Correct Answer: A

Solution:

A reflexive relation on a set A is one where every element is related to itself, i.e., (a, a) is in the relation for all a in A.

The function is one-one and onto.

The function is one-one but not onto.

The function is onto but not one-one.

The function is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = 3x^2

is neither one-one nor onto. It is not one-one because

f(x) = f(-x)

for any

x

. It is not onto because negative numbers are not in the range of

f(x) = 3x^2

R = {(a, b) : a = b}

R = {(a, b) : |a - b| \text{ is even}}

R = {(a, b) : a < b}

R = {(a, b) : a + b = 5}

Correct Answer: B

Solution:

The relation

R = \{(a, b) : |a - b| \text{ is even}\}

is an equivalence relation because it is reflexive, symmetric, and transitive. Reflexivity holds as

|a - a| = 0

is even. Symmetry holds because if

|a - b|

is even, then

|b - a|

is also even. Transitivity holds because if

|a - b|

and

|b - c|

are even, then

|a - c|

is also even.

R is reflexive, symmetric, and transitive.

R is reflexive and symmetric but not transitive.

R is symmetric and transitive but not reflexive.

R is reflexive and transitive but not symmetric.

Correct Answer: A

Solution:

The relation

R

is reflexive because for any

a \in A

|a - a| = 0

is even. It is symmetric because if

|a - b|

is even, then

|b - a|

is also even. It is transitive because if

|a - b|

and

|b - c|

are even, then

|a - c| = |a - b| + |b - c|

is also even.

One-one only

Onto only

Both one-one and onto

Neither one-one nor onto

Correct Answer: A

Solution:

The function

f(n) = n^2 + n + 1

is one-one because for any

n_1, n_2 \in \mathbb{N}

, if

f(n_1) = f(n_2)

, then

n_1 = n_2

. However, it is not onto because not all natural numbers can be expressed in the form

n^2 + n + 1

It is reflexive, symmetric, and transitive.

It is only reflexive.

It is symmetric and transitive but not reflexive.

It is neither reflexive nor symmetric.

Correct Answer: A

Solution:

An equivalence relation is defined as a relation that is reflexive, symmetric, and transitive.

f \circ f

is one-one and onto.

f \circ f

is one-one but not onto.

f \circ f

is onto but not one-one.

f \circ f

is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = 3x^2

is neither one-one nor onto because it maps both positive and negative values of

x

to the same value and does not cover all real numbers (e.g., negative numbers are not in the range). Thus,

f \circ f

is also neither one-one nor onto.

f(x) = 2x

f(x) = x + 1

f(x) = x^2

f(x) = x - 1

Correct Answer: A

Solution:

The function

f(x) = 2x

is one-one because for

f(x_1) = f(x_2)

, we have

2x_1 = 2x_2

which implies

x_1 = x_2

. However, it is not onto because there is no natural number

x

such that

f(x) = 1

One-one only

Onto only

Both one-one and onto

Neither one-one nor onto

Correct Answer: D

Solution:

The function

f(x) = x^2

is neither one-one nor onto when the domain and co-domain are both

\mathbb{R}

. It is not one-one because

f(x) = f(-x)

, and it is not onto because negative numbers are not in the range.

f

is one-one and onto.

f

is one-one but not onto.

f

is onto but not one-one.

f

is neither one-one nor onto.

Correct Answer: D

Solution:

The function

f(x) = x^2

is neither one-one (since

f(x) = f(-x)

for any

x

) nor onto (since negative numbers are not in the range of

f

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = x^2

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = e^x

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = \sin x

f: \mathbb{R} \to \mathbb{R}

defined by

f(x) = 3x+2

Correct Answer: A

Solution:

The function

f(x) = x^2

is not one-one because

f(x) = f(-x)

for any

x

. It is also not onto because negative numbers are not in the range of

f(x) = x^2

f(x) = x^3

over

\mathbb{R}

f(x) = x^2

over

\mathbb{R}

f(x) = e^x

over

\mathbb{R}

f(x) = \sin(x)

over

\mathbb{R}

Correct Answer: A

Solution:

The function

f(x) = x^3

is invertible over

\mathbb{R}

because it is both one-one and onto. The function

f(x) = x^2

is not invertible over

\mathbb{R}

because it is not one-one. The function

f(x) = e^x

is one-one and onto

\mathbb{R}^+

, but not onto

\mathbb{R}

. The function

f(x) = \sin(x)

is not one-one over

\mathbb{R}

f is one-one and onto.

f is one-one but not onto.

f is onto but not one-one.

f is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(x) = \frac{1}{x}

is one-one because if

f(x_1) = f(x_2)

, then

\frac{1}{x_1} = \frac{1}{x_2}

implies

x_1 = x_2

. It is onto because for any

y \in \mathbb{R}

y \neq 0

, there exists

x = \frac{1}{y}

such that

f(x) = y

. Therefore,

f

is both one-one and onto.

One-one but not onto.

Onto but not one-one.

Both one-one and onto.

Neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(n) = n^2 + n + 1

is strictly increasing for

n \in \mathbb{N}

, hence it is one-one. However, not every natural number can be expressed in the form

n^2 + n + 1

, so it is not onto.

It is one-one but not onto.

It is onto but not one-one.

It is both one-one and onto.

It is neither one-one nor onto.

Correct Answer: C

Solution:

The function

f(x) = x^3 - 3x

is a cubic polynomial, which is continuous and differentiable everywhere. The derivative

f'(x) = 3x^2 - 3

is zero at

x = \pm 1

, indicating local extrema. However, the function still covers all real values as

x \to \pm \infty

, making it onto. Since it is a cubic function, it is also one-one. Thus, the function is both one-one and onto.

Reflexive

Symmetric

Transitive

Equivalence

Correct Answer: B

Solution:

The relation

R

is symmetric because if

|a - b|

is a prime number, then

|b - a|

is also a prime number. However, it is not reflexive because

|a - a| = 0

is not a prime number. It is not transitive because if

|a - b|

and

|b - c|

are prime,

|a - c|

may not be a prime. Hence,

R

is not an equivalence relation.

Yes, because every odd number in

\mathbb{N}

is an image of some

n

No, because not every natural number is an image of some

n

Yes, because every even number in

\mathbb{N}

is an image of some

n

No, because

f

is not defined for all

n

Correct Answer: A

Solution:

The function

f(n) = 2n + 1

maps natural numbers to odd numbers in

\mathbb{N}

. Every odd number can be expressed as

2n + 1

for some

n \in \mathbb{N}

, hence

f

is onto.

x^2 + 2x + 4

x^2 + 4

x^2 + 4x + 4

x^2 + 2x

Correct Answer: C

Solution:

The composition

g \circ f(x) = g(f(x)) = g(x + 2) = (x + 2)^2 = x^2 + 4x + 4

R = {(1, 2), (2, 1)}

R = {(1, 1), (2, 2), (1, 2)}

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

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

Correct Answer: A

Solution:

A symmetric relation is one where if

(a, b) \in R

, then

(b, a) \in R

. The relation in option A is symmetric but not reflexive or transitive.

f is one-one and onto.

f is one-one but not onto.

f is onto but not one-one.

f is neither one-one nor onto.

Correct Answer: B

Solution:

The function

f(n) = n^2

is one-one because different natural numbers have different squares. However, it is not onto because not every natural number is a perfect square, so not all natural numbers are in the range of

f

f

is one-one and onto.

f

is one-one but not onto.

f

is onto but not one-one.

f

is neither one-one nor onto.

Correct Answer: D

Solution:

The modulus function

f(x) = |x|

is not one-one because

f(x) = f(-x)

for any

x

. It is not onto because there is no

x

such that

f(x)

is negative.

Reflexive

Symmetric

Transitive

None of the above

Correct Answer: D

Solution:

The relation

R

is not reflexive because not all pairs

(a, a)

satisfy

a + a > 4

. It is not symmetric because if

(a, b) \in R

b + a

may not be greater than 4. It is not transitive because

(a, b) \in R

and

(b, c) \in R

does not imply

(a, c) \in R

. Hence,

R

satisfies none of these properties.

\{(1, 4), (2, 5), (3, 6)\}

\{(1, 5), (2, 4)\}

\{(1, 4), (2, 5), (3, 7)\}

\{(1, 4), (2, 5), (3, 6), (1, 5)\}

Correct Answer: A

Solution:

A relation from

A

B

is a subset of

A \times B

. Option (a) is a valid subset.

f

is one-one but not onto.

f

is onto but not one-one.

f

is both one-one and onto.

f

is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(x) = 2x

is one-one because if

f(x_1) = f(x_2)

, then

2x_1 = 2x_2

implies

x_1 = x_2

. It is not onto because there is no natural number

x

such that

f(x) = 1

R is reflexive.

R is symmetric.

R is transitive.

R is an equivalence relation.

Correct Answer: C

Solution:

The relation

R

is a set of ordered pairs where each element from set

A

is related to a unique element in set

B

. It is not reflexive because not all elements of

A

are related to themselves. It is not symmetric because if

(1, 4) \in R

, then

(4, 1) \notin R

. However, it is transitive because there are no elements

a, b, c

such that

(a, b) \in R

and

(b, c) \in R

but

(a, c) \notin R

. Therefore,

R

is transitive.

The function

f

is one-one but not onto.

The function

f

is onto but not one-one.

The function

f

is both one-one and onto.

The function

f

is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f

maps each element of set

A

to a unique element in set

B

, hence it is one-one. However, not all elements of set

B

are covered by the mapping, specifically

7

is not an image of any element in

A

, so it is not onto.

\{(1, 1), (2, 2), (3, 3), (1, 3)\}

\{(1, 1), (2, 2), (3, 3), (2, 1)\}

\{(1, 1), (2, 2), (3, 3), (3, 2)\}

\{(1, 1), (2, 2), (3, 3), (3, 1)\}

Correct Answer: A

Solution:

To make

R

reflexive, we need to add

(1, 1), (2, 2), (3, 3)

. To make it transitive, we need to add

(1, 3)

because

(1, 2)

and

(2, 3)

imply

(1, 3)

. Adding

(1, 3)

does not make it symmetric.

The inverse of

f

\ln(x)

and it is defined for all

x \in \mathbb{R}

The inverse of

f

\ln(x)

and it is defined for

x > 0

The inverse of

f

\ln(x)

and it is defined for

x \geq 0

The inverse of

f

\ln(x)

and it is defined for

x < 0

Correct Answer: B

Solution:

The function

f(x) = e^x

is one-one and onto from

\mathbb{R}

(0, \infty)

. Its inverse is

\ln(x)

, which is defined for

x > 0

The function is one-one and onto.

The function is many-one and onto.

The function is one-one but not onto.

The function is neither one-one nor onto.

Correct Answer: A

Solution:

The function

f(x) = 3x

is both one-one and onto because for every

y \in \mathbb{R}

, there exists an

x \in \mathbb{R}

such that

f(x) = y

and

f(x_1) = f(x_2)

implies

x_1 = x_2

Correct Answer: B

Solution:

A relation is reflexive if it contains all pairs of the form

(a, a)

for

a \in A

. It is symmetric if for every pair

(a, b)

in the relation, the pair

(b, a)

is also in the relation. However, to ensure the relation is not transitive, we must avoid including certain pairs that would complete transitive triples. Thus, the number of such relations is 2.

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: A

Solution:

The function

f

is one-one and onto because each element in

A

maps to a unique element in

B

, and every element in

B

is an image of some element in

A

Reflexive only

Symmetric only

Transitive only

Reflexive and symmetric

Correct Answer: B

Solution:

The relation

R

is symmetric because if

(a, b) \in R

, then

(b, a) \in R

. However, it is neither reflexive (since

(1, 1)

(2, 2)

, and

(3, 3)

are not in

R

) nor transitive (since

(1, 2)

and

(2, 1)

do not imply

(1, 1) \in R

The function is one-one and onto.

The function is one-one but not onto.

The function is onto but not one-one.

The function is neither one-one nor onto.

Correct Answer: B

Solution:

The function

f(n) = n + 1

is injective (one-one) because different inputs give different outputs. However, it is not surjective (onto) because there is no natural number

n

such that

f(n) = 1

Correct Answer: C

Solution:

The number of one-one functions from a set to itself is the same as the number of permutations of the set, which is

3! = 6

R is reflexive.

R is symmetric.

R is transitive.

R is neither reflexive, symmetric, nor transitive.

Correct Answer: D

Solution:

The relation

R

is not reflexive because there is no

a

such that

a - a = 1

. It is not symmetric because if

(a, b) \in R

, then

a - b = 1

, but

b - a = -1

, so

(b, a) \notin R

. It is not transitive because if

(a, b) \in R

and

(b, c) \in R

, then

a - b = 1

and

b - c = 1

, but

a - c = 2

, so

(a, c) \notin R

. Thus,

R

is neither reflexive, symmetric, nor transitive.

\{(1, 2), (2, 1)\}

\{(1, 1), (2, 2), (3, 3)\}

\{(1, 2), (2, 3), (1, 3)\}

\{(1, 2), (2, 3)\}

Correct Answer: A

Solution:

Option (a) is symmetric because if

(1, 2)

is in the relation, then

(2, 1)

is also in the relation. However, it is neither reflexive nor transitive.

f

is one-one and onto.

f

is one-one but not onto.

f

is onto but not one-one.

f

is neither one-one nor onto.

Correct Answer: B

Solution:

The function

f(n) = 2n

is one-one because different inputs give different outputs, but it is not onto because odd numbers are not in the range.

g

is one-one and onto

g

is one-one but not onto

g

is onto but not one-one

g

is neither one-one nor onto

Correct Answer: D

Solution:

The function

g(x) = x^2 - 4x + 4

can be rewritten as

(x-2)^2

, which is a parabola opening upwards with vertex at

(2, 0)

. It is not one-one because it fails the horizontal line test. It is not onto because its range is

[0, \infty)

, not the entire set of real numbers.

f

is one-one and onto

f

is one-one but not onto

f

is onto but not one-one

f

is neither one-one nor onto

Correct Answer: B

Solution:

The function

f(n) = 2n + 1

is one-one because different inputs give different outputs. However, it is not onto because there is no natural number

n

such that

f(n) = 2m

for any even natural number

m

True or False

Correct Answer: True

Solution:

The universal relation on a set

X

includes all possible pairs from

X

, hence it is

R = X \times X

Correct Answer: False

Solution:

A symmetric relation only requires that if

(a, b)

is in the relation, then

(b, a)

must also be in the relation. It does not require that

(a, a)

is in the relation for all

a

Correct Answer: True

Solution:

A function is invertible if there exists a function

g: Y \to X

such that

g \circ f = I_X

and

f \circ g = I_Y

. This requires

f

to be both one-one and onto.

Correct Answer: True

Solution:

The identity function maps each element to itself, so every element in the domain has a corresponding element in the co-domain, making it onto.

Correct Answer: True

Solution:

A relation is symmetric if for every (a, b) in R, (b, a) is also in R. Here, both (1, 2) and (2, 1) are present, so the relation is symmetric.

Correct Answer: True

Solution:

A reflexive relation requires that every element is related to itself, which is represented by the pair (a, a).

Correct Answer: True

Solution:

A function from a set

A

to itself that is one-one and onto is a permutation of

A

. Thus, the set of all such functions is equivalent to the set of permutations of

A

Correct Answer: True

Solution:

For a function to have an inverse, it must be bijective, meaning it is both injective (one-one) and surjective (onto).

Correct Answer: True

Solution:

R. Descartes first used the word 'function' in 1637 in his manuscript 'Geometrie'.

Correct Answer: True

Solution:

R. Descartes used the word 'function' in his manuscript 'Geometrie' in 1637 to describe some positive integral power of a variable.

Correct Answer: True

Solution:

A function is invertible if there exists an inverse function such that the composition of the function and its inverse yields the identity function. This requires the function to be both injective and surjective.

Correct Answer: True

Solution:

A function is invertible if there exists a function

g: B \to A

such that

g \circ f = I_A

and

f \circ g = I_B

, which is possible only if

f

is both one-one and onto.

Correct Answer: True

Solution:

The set of all odd integers is an equivalence class because all odd integers are related to each other under the relation of congruence modulo 2, and no odd integer is related to any even integer.

Correct Answer: False

Solution:

The modulus function

f(x) = |x|

is neither one-one nor onto. It maps both positive and negative values of

x

to the same non-negative value, and it does not cover all real numbers as its range.

Correct Answer: False

Solution:

For a relation to be reflexive, every element x in the set must relate to itself, i.e., (x, x) must be in R. However, for R = {(x, y) : y = x + 5 and x < 4}, there is no x such that x = x + 5, so R is not reflexive.

Correct Answer: True

Solution:

The identity function maps each element to itself, ensuring it is both one-one and onto, thus making it bijective.

Correct Answer: False

Solution:

The function

f(x) = 2x

is not onto because there is no natural number

x

such that

f(x) = 1

, hence not every element in the co-domain has a pre-image.

Correct Answer: False

Solution:

The composition of two functions

f

and

g

, denoted as

g \circ f

, is not necessarily equal to

f \circ g

. The order of composition matters.

Correct Answer: False

Solution:

The modulus function

f(x) = |x|

is not onto because negative numbers are not in the range of

f(x)

Correct Answer: True

Solution:

A relation from set

A

to set

B

is defined as a subset of the Cartesian product

A \times B

Correct Answer: True

Solution:

By definition, a relation from set

A

to set

B

is any subset of the Cartesian product

A \times B

. This means that each relation is indeed a subset of the Cartesian product of the two sets.

Correct Answer: False

Solution:

The modulus function is not one-one because different inputs can produce the same output, e.g., f(1) = 1 and f(-1) = 1.

Correct Answer: True

Solution:

The set theoretic definition of a function is indeed an abstraction of the definition given by Dirichlet, as stated in the historical note.

Correct Answer: True

Solution:

For finite sets of the same size, a function that is one-one (injective) is also onto (surjective), and vice versa.

Correct Answer: True

Solution:

A symmetric relation only requires that if

(a, b)

is in the relation, then

(b, a)

must also be in the relation. It does not require reflexivity or transitivity.

Correct Answer: True

Solution:

A symmetric relation requires that if one element is related to another, then the reverse must also be true.

Correct Answer: True

Solution:

A function is onto (surjective) if every element in the co-domain B is an image of at least one element from the domain A, meaning the range of f is B.

Correct Answer: True

Solution:

The modulus function is not one-one because different inputs can produce the same output (e.g.,

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

), and it is not onto because negative numbers are not in its range.

Correct Answer: False

Solution:

The modulus function is neither one-one nor onto; it maps both positive and negative values of

x

to the same non-negative value.

Correct Answer: True

Solution:

Since f is one-one, each element of the domain {1, 2, 3} is mapped to a unique element in the co-domain {1, 2, 3}. Therefore, all elements in the co-domain must be covered, making f onto.

Correct Answer: True

Solution:

If f: A → B and g: B → C are both invertible, then their composition g∘f: A → C is also invertible. The inverse of g∘f is (f⁻¹∘g⁻¹).

Correct Answer: True

Solution:

By definition, a symmetric relation

R

satisfies the condition that if

(a, b) \in R

, then

(b, a) \in R

for all

a, b \in X

Correct Answer: True

Solution:

An equivalence relation partitions a set into equivalence classes, where each element is related to every other element in the same class.

Correct Answer: True

Solution:

By definition, a reflexive relation on a set

X

includes every element related to itself, meaning

(a, a)

must be in the relation for every

a \in X

Correct Answer: True

Solution:

An equivalence relation partitions a set into equivalence classes, where each element is related to every other element in its class, and no element is related to any element in a different class.

Correct Answer: True

Solution:

The identity function f: N → N defined by f(x) = x is onto because every element in the co-domain has a pre-image in the domain, specifically itself.

Correct Answer: False

Solution:

A relation is reflexive if every element is related to itself. For R = {(x, y) : y = x + 1}, no element x is related to itself because x + 1 ≠ x. Hence, R is not reflexive.

Correct Answer: True

Solution:

A reflexive relation requires that every element is related to itself, hence all pairs

(a, a)

must be included.

Correct Answer: False

Solution:

The composition of two functions

f

and

g

, denoted as

g \circ f

, is not necessarily equal to

Solution:

A function

f: X \to Y

is invertible if there exists a function

g: Y \to X

such that

g \circ f = I_X

and

f \circ g = I_Y

, where

I_X

and

I_Y

are identity functions on

X

and

Y

, respectively.

Correct Answer: True

Solution:

By definition, an equivalence relation on a set must satisfy the properties of reflexivity, symmetry, and transitivity.

Correct Answer: True

Solution:

A relation is an equivalence relation if it is reflexive, symmetric, and transitive. For R = {(a, b) : a - b is even}, it satisfies all these properties: reflexive (a - a = 0 is even), symmetric (if a - b is even, then b - a is also even), and transitive (if a - b is even and b - c is even, then a - c is even).

Correct Answer: False

Solution:

The function f(x) = 2x maps natural numbers to even natural numbers. There is no x in N such that f(x) = 1, which is an odd number. Therefore, f is not onto.

Correct Answer: False

Solution:

A function is one-one if different elements in the domain map to different elements in the co-domain. For f(x) = x^2, f(1) = 1 and f(-1) = 1, which means f is not one-one.

Relations and Functions

AI Learning Assistant

Summary

Chapter Summary: Relations and Functions

Key Concepts

Types of Relations

Functions

Important Definitions

Examples of Relations

Common Pitfalls

Exam Tips

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 1: Relations and Functions

1.1 Introduction

1.2 Types of Relations

1.3 Types of Functions

Key Definitions and Properties

Historical Note

Examples of Relations

Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Practice & Assessment

Multiple Choice Questions

Q1.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3}. How many relations can be formed on AAA that are reflexive and symmetric?

Correct Answer: B

Q2.Which of the following is an equivalence relation on the set A={1,2,3}A = \{1, 2, 3\}A={1,2,3}?

Correct Answer: A

Q3.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5,6}B = \{4, 5, 6\}B={4,5,6}. A function f:A→Bf: A \to Bf:A→B is defined as f(1)=4f(1) = 4f(1)=4, f(2)=5f(2) = 5f(2)=5, and f(3)=6f(3) = 6f(3)=6. Which of the following properties does the function fff satisfy?

Correct Answer: C

Q4.Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5}. A relation RRR is defined on AAA by R={(a,b):∣a−b∣ is even}R = \{(a, b) : |a - b| \text{ is even}\}R={(a,b):∣a−b∣ is even}. Which of the following statements is true about the relation RRR?

Correct Answer: A

Q5.Which of the following is an example of a reflexive relation?

Correct Answer: A

Q6.Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=exf(x) = e^xf(x)=ex. Which of the following statements is true about the composition f∘ff \circ ff∘f?

Correct Answer: B

Q7.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3}. Which of the following relations on AAA is reflexive and symmetric but not transitive?

Correct Answer: A

Q8.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5,6}B = \{4, 5, 6\}B={4,5,6}. Consider the function h:A→Bh: A \to Bh:A→B defined by h={(1,5),(2,6),(3,4)}h = \{(1, 5), (2, 6), (3, 4)\}h={(1,5),(2,6),(3,4)}. Which of the following statements is true?

Correct Answer: C

Q9.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5,6,7}B = \{4, 5, 6, 7\}B={4,5,6,7}. Consider the function f:A→Bf: A \to Bf:A→B defined by f={(1,4),(2,5),(3,6)}f = \{(1, 4), (2, 5), (3, 6)\}f={(1,4),(2,5),(3,6)}. Which of the following statements is true about fff?

Correct Answer: A

Q10.Let f:N→Nf: \mathbb{N} \to \mathbb{N}f:N→N be defined by f(n)=2nf(n) = 2nf(n)=2n. Is fff one-one and onto?

Correct Answer: B

Q11.Let the set A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}A={1,2,3,4,5}. Define a relation RRR on AAA by R={(a,b):a+b=6}R = \{(a, b) : a + b = 6\}R={(a,b):a+b=6}. Which of the following statements is true about RRR?

Correct Answer: B

Q12.Consider the function f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)=x3−3x+1f(x) = x^3 - 3x + 1f(x)=x3−3x+1. Which of the following statements is true about the function fff?

Correct Answer: C

Q13.Which of the following statements about the modulus function f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R given by f(x)=∣x∣f(x) = |x|f(x)=∣x∣ is true?

Correct Answer: D

Q14.If f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R is defined by f(x)=x2f(x) = x^2f(x)=x2, which of the following statements is true?

Correct Answer: D

Q15.Let RRR be a relation on the set A={1,2,3}A = \{1, 2, 3\}A={1,2,3} defined by R={(1,2),(2,3),(1,3)}R = \{(1, 2), (2, 3), (1, 3)\}R={(1,2),(2,3),(1,3)}. Which of the following properties does the relation RRR satisfy?

Correct Answer: D

Q16.Which of the following functions is one-one but not onto?

Correct Answer: A

Q17.Consider the set A={1,2,3}A = \{1, 2, 3\}A={1,2,3}. How many equivalence relations can be defined on AAA?

Correct Answer: C

Q18.Let A={1,2,3}A = \{1, 2, 3\}A={1,2,3} and B={4,5,6}B = \{4, 5, 6\}B={4,5,6}. Which of the following is a valid relation from AAA to BBB that is reflexive?

Correct Answer: B

Q19.Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=2x+3f(x) = 2x + 3f(x)=2x+3. What is the inverse of the function fff?

Correct Answer: A

Q20.If f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R is defined by f(x)=3x+7f(x) = 3x + 7f(x)=3x+7, what is the inverse function f−1(x)f^{-1}(x)f−1(x)?

Correct Answer: A

Q21.Let RRR be a relation on the set Z\mathbb{Z}Z defined by R={(a,b):a−b is a multiple of 4}R = \{(a, b) : a - b \text{ is a multiple of } 4\}R={(a,b):a−b is a multiple of 4}. Which of the following statements is true?

Correct Answer: C

Q22.Which of the following relations on the set {1,2,3}\{1, 2, 3\}{1,2,3} is symmetric but neither reflexive nor transitive?

Correct Answer: A

Q23.Consider the function g:R→Rg: \mathbb{R} \to \mathbb{R}g:R→R defined by g(x)=sin⁡xg(x) = \sin xg(x)=sinx. Which of the following statements is true about the function ggg?

Correct Answer: D

Q24.Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be defined by f(x)=sin⁡xf(x) = \sin xf(x)=sinx. Which of the following statements is true about the function fff?

Correct Answer: D

Q25.Consider the relation RRR on the set of real numbers R\mathbb{R}R defined by R={(a,b):a−b is an integer}R = \{(a, b) : a - b \text{ is an integer}\}R={(a,b):a−b is an integer}. Which of the following properties does RRR satisfy?

Correct Answer: D

Q26.Consider the function f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R defined by f(x)=x4f(x) = x^4f(x)=x4. Which of the following statements is true?

Correct Answer: D