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Class 12 Science (PCM)

Mathematics Part 1

Relations and Functions

Relations and Functions

Action PlanNotesQuestionsMock-Test
Study Action Plan (Sorted by Difficulty)
Identity Function and Its Role in Inverses
Accuracy: 38%
Total Attempted8
Total Correct3
Total Incorrect5
Difficulty Level:
High
Types of Functions
Accuracy: 38%
Total Attempted55
Total Correct21
Total Incorrect34
Difficulty Level:
High
Invertible Functions
Accuracy: 41%
Total Attempted46
Total Correct19
Total Incorrect27
Difficulty Level:
High
Domain, Codomain and Range in Function Classification
Accuracy: 43%
Total Attempted28
Total Correct12
Total Incorrect16
Difficulty Level:
High
Standard Functions and Their Injectivity-Surjectivity
Accuracy: 43%
Total Attempted14
Total Correct6
Total Incorrect8
Difficulty Level:
High
Onto Function Testing
Accuracy: 44%
Total Attempted80
Total Correct35
Total Incorrect45
Difficulty Level:
High
Properties of Functions in Finite and Infinite Sets
Accuracy: 46%
Total Attempted37
Total Correct17
Total Incorrect20
Difficulty Level:
High
Equivalence Classes in Modular Relations
Accuracy: 47%
Total Attempted17
Total Correct8
Total Incorrect9
Difficulty Level:
High
Composition of Functions
Accuracy: 47%
Total Attempted53
Total Correct25
Total Incorrect28
Difficulty Level:
High
Bijective Functions
Accuracy: 48%
Total Attempted71
Total Correct34
Total Incorrect37
Difficulty Level:
High
One-One Function Testing
Accuracy: 50%
Total Attempted42
Total Correct21
Total Incorrect21
Difficulty Level:
Medium
Types of Relations
Accuracy: 51%
Total Attempted51
Total Correct26
Total Incorrect25
Difficulty Level:
Medium
Finding Inverse of a Function
Accuracy: 53%
Total Attempted40
Total Correct21
Total Incorrect19
Difficulty Level:
Medium
Composition Order and Non-Commutativity
Accuracy: 55%
Total Attempted55
Total Correct30
Total Incorrect25
Difficulty Level:
Medium
Counterexamples in Relation Properties
Accuracy: 57%
Total Attempted51
Total Correct29
Total Incorrect22
Difficulty Level:
Medium
Equivalence Relations
Accuracy: 63%
Total Attempted95
Total Correct60
Total Incorrect35
Difficulty Level:
Medium
Testing Reflexive, Symmetric and Transitive Relations
Accuracy: 66%
Total Attempted32
Total Correct21
Total Incorrect11
Difficulty Level:
Medium
Equivalence Classes and Partitions
Accuracy: 75%
Total Attempted32
Total Correct24
Total Incorrect8
Difficulty Level:
Low
Equivalence Relation Examples from Geometry and Numbers
Accuracy: 100%
Total Attempted12
Total Correct12
Total Incorrect0
Difficulty Level:
Low
Relation as Subset of Cartesian Product
Accuracy: 100%
Total Attempted8
Total Correct8
Total Incorrect0
Difficulty Level:
Low

Learning Objectives

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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.

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Revision Notes & Summary

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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

  1. Determine the properties of various relations defined in different sets.
  2. Show that specific functions are one-one, onto, or neither.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

The function is injective but not surjective.

B.

The function is surjective but not injective.

C.

The function is neither injective nor surjective.

D.

The function is both injective and surjective.
Correct Answer: C

Solution:

The function f(x)=x2f(x) = x^2 is neither injective nor surjective because it maps both xx and −x-x to the same value (not injective) and does not cover negative numbers in the codomain (not surjective).

Chapter Concept:

Domain, Codomain and Range in Function Classification

A.

The identity function IXI_X maps every element of XX to itself.

B.

For inverse verification, gof=IXgof = I_X and fog=IYfog = I_Y must hold.

C.

The identity function IXI_X maps every element of XX to a different element.

D.

The identity function IXI_X is not involved in inverse verification.
Correct Answer: A

Solution:

The identity function IXI_X maps every element of XX to itself, which is crucial for verifying inverse functions where gof=IXgof = I_X and fog=IYfog = I_Y must hold.

Chapter Concept:

Identity Function and Its Role in Inverses

A.

f−1(x)=log⁡(x)f^{-1}(x) = \log(x)

B.

f−1(x)=e−xf^{-1}(x) = e^{-x}

C.

f−1(x)=ln⁡(x)f^{-1}(x) = \ln(x)

D.

f−1(x)=xef^{-1}(x) = x^e
Correct Answer: C

Solution:

The inverse of the exponential function f(x)=exf(x) = e^x is the natural logarithm function, f−1(x)=ln⁡(x)f^{-1}(x) = \ln(x).

Chapter Concept:

Finding Inverse of a Function

A.

f∘g(x)=xf \circ g(x) = x and g∘f(x)=xg \circ f(x) = x for all x∈Rx \in \mathbb{R}

B.

f∘g(x)=xf \circ g(x) = x for x≥0x \geq 0 and g∘f(x)=∣x∣g \circ f(x) = |x|

C.

f∘g(x)=∣x∣f \circ g(x) = |x| and g∘f(x)=xg \circ f(x) = x for x≥0x \geq 0

D.

f∘g(x)=x2f \circ g(x) = x^2 and g∘f(x)=x2g \circ f(x) = \sqrt{x^2}
Correct Answer: B

Solution:

For f∘g(x)=f(g(x))=f(x)=(x)2=xf \circ g(x) = f(g(x)) = f(\sqrt{x}) = (\sqrt{x})^2 = x for x≥0x \geq 0. For g∘f(x)=g(f(x))=g(x2)=x2=∣x∣g \circ f(x) = g(f(x)) = g(x^2) = \sqrt{x^2} = |x|.

Chapter Concept:

Composition of Functions

A.

Only reflexive

B.

Only symmetric

C.

Only transitive

D.

Reflexive, symmetric, and transitive
Correct Answer: D

Solution:

The relation RR is reflexive because every triangle is similar to itself. It is symmetric because if T1T_1 is similar to T2T_2, then T2T_2 is similar to T1T_1. It is transitive because if T1T_1 is similar to T2T_2 and T2T_2 is similar to T3T_3, then T1T_1 is similar to T3T_3. Thus, RR is an equivalence relation.

Chapter Concept:

Equivalence Relations

A.

The function is one-one but not onto.

B.

The function is onto but not one-one.

C.

The function is neither one-one nor onto.

D.

The function is both one-one and onto.
Correct Answer: D

Solution:

The function f(x)=4−xf(x) = 4 - x maps each element of the set {1,2,3}\{1, 2, 3\} to a unique element in the set {1,2,3}\{1, 2, 3\}, covering all elements. Therefore, it is both one-one and onto.

Chapter Concept:

Types of Functions

A.

{(a, b) \in A \times B : a \text{ is brother of } b}

B.

{(a, b) \in A \times A : a \text{ is sister of } b}

C.

{(a, b) \in B \times B : a \text{ is greater than } b}

D.

{(a, b) \in A \times B : a \text{ lives in the same locality as } b}
Correct Answer: A

Solution:

A relation from set AA to set BB is any subset of A×BA \times B. Option A is a subset of A×BA \times B.

Chapter Concept:

Relation as Subset of Cartesian Product

A.

Reflexive only

B.

Symmetric only

C.

Transitive only

D.

None of the above
Correct Answer: D

Solution:

The relation RR is not reflexive because not every natural number is the square of itself. It is not symmetric because if a=b2a = b^2, it does not imply b=a2b = a^2. It is not transitive because if a=b2a = b^2 and b=c2b = c^2, it does not imply a=c2a = c^2. Therefore, it satisfies none of these properties.

Chapter Concept:

Testing Reflexive, Symmetric and Transitive Relations

A.

R is reflexive only.

B.

R is symmetric only.

C.

R is transitive only.

D.

R is an equivalence relation.
Correct Answer: D

Solution:

The relation RR is reflexive because ∣a−a∣=0|a - a| = 0 is even. It is symmetric because if ∣a−b∣|a - b| is even, then ∣b−a∣|b - a| is also even. It is transitive because if ∣a−b∣|a - b| and ∣b−c∣|b - c| are even, then ∣a−c∣|a - c| is even. Thus, RR is an equivalence relation.

Chapter Concept:

Types of Relations

A.

It must be reflexive and symmetric.

B.

It must be symmetric and transitive.

C.

It must be reflexive, symmetric, and transitive.

D.

It must be reflexive and transitive.
Correct Answer: C

Solution:

A relation RR in a set AA is an equivalence relation if it is reflexive, symmetric, and transitive.

Chapter Concept:

Testing Reflexive, Symmetric and Transitive Relations

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True or False

Correct Answer: True

Solution:

In finite sets, if a function is onto, it must also be one-one. This is a characteristic difference between finite and infinite sets.

Chapter Concept :

Properties of Functions in Finite and Infinite Sets

Correct Answer: True

Solution:

The relation of congruence of triangles is reflexive, symmetric, and transitive, hence it is an equivalence relation.

Chapter Concept :

Equivalence Relation Examples from Geometry and Numbers

Correct Answer: True

Solution:

For a function to be invertible, there must exist a function g:Y→Xg: Y \rightarrow X such that g∘f=IXg \circ f = I_X and f∘g=IYf \circ g = I_Y. This is only possible if the function is both one-one (injective) and onto (surjective).

Chapter Concept :

Invertible Functions

Correct Answer: True

Solution:

A relation from set AA to set BB is defined as any subset of the Cartesian product A×BA \times B, as per the definition provided in the excerpt.

Chapter Concept :

Relation as Subset of Cartesian Product

Correct Answer: True

Solution:

An equivalence relation is defined by its reflexive, symmetric, and transitive properties. This is the standard definition of an equivalence relation.

Chapter Concept :

Equivalence Relations

Correct Answer: True

Solution:

By definition, a relation from set AA to set BB is any subset of the Cartesian product A×BA \times B. This means that any collection of ordered pairs from A×BA \times B can be considered a relation.

Chapter Concept :

Relation as Subset of Cartesian Product

Correct Answer: False

Solution:

The composition g∘fg \circ f is defined as (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x)). In general, g∘fg \circ f is not equal to f∘gf \circ g as the compositions are not commutative.

Chapter Concept :

Composition of Functions

Correct Answer: True

Solution:

A function is called bijective if it is both injective (one-one) and surjective (onto), meaning every element of the codomain is mapped to by exactly one element of the domain.

Chapter Concept :

Types of Functions

Correct Answer: True

Solution:

A bijective function pairs every element of the domain with exactly one element of the codomain, making it both injective (one-one) and surjective (onto). This property allows it to have an inverse function.

Chapter Concept :

Bijective Functions

Correct Answer: True

Solution:

Reflexivity requires that every element in the set is related to itself. Therefore, if even one diagonal pair (a,a)(a, a) is missing, the relation cannot be reflexive.

Chapter Concept :

Counterexamples in Relation Properties

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