Relations and Functions — Class 11 Maths Notes

Relations and Functions · Class 11 Maths · 9 topics.

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Topics covered in Relations and Functions

1.Introduction of Relations & Functions
Let's start with an interesting real-life example to understand the concept of relations and functions.
Imagine You Are a Delivery Manager:
You manage a team of delivery drivers who deliver packages to various locations. Each driver is assigned specific areas, and you need to keep track of which driver delivers to which area.
- Relation: This is like a list or a set of pairs showing which driver (input) delivers to which area (output).
- Function: If each driver delivers to exactly one area, this relation is called a function.
Step-by-Step Explanation
1. Relation:
  - A relation is a connection between sets of information.
  - For example, let’s consider two sets:
    - Set A (Drivers): {D1, D2, D3}
    - Set B (Areas): {A1, A2, A3, A4}
  - A relation can be represented as pairs showing which driver delivers to which area. For example, {(D1, A1), (D2, A2), (D3, A3), (D1, A4)}.
  - Here, D1 delivers to both A1 and A4, showing that one driver can cover multiple areas.
2. Function:
  - A function is a special type of relation where each input (driver) is associated with exactly one output (area).
  - For example, let’s redefine the sets:
    - Set A (Drivers): {D1, D2, D3}
    - Set B (Areas): {A1, A2, A3}
  - A function can be represented as {(D1, A1), (D2, A2), (D3, A3)}.
  - Here, each driver delivers to only one unique area.
Key Points
- Relation: A set of ordered pairs. It can have multiple outputs for one input.
- Function: A set of ordered pairs with a unique output for each input.
- Domain: The set of all possible inputs (e.g., all drivers).
- Range: The set of all possible outputs (e.g., all areas).
Real-Life Applications
- Mapping Cities to Pin Codes: Each city has a unique pin code, making it a function.
- School Timetable: Each period is assigned to one specific subject, demonstrating a function.
- Career Applications:
  - Computer Science: Functions are used in algorithms and programming.
  - Economics: Relations and functions are used to model economic behaviors and trends.
  - Engineering: Functions describe physical phenomena like electrical currents, forces, etc.
Simple Activity to Understand Functions
1. Take a list of your classmates and their favorite subjects.
2. Create pairs like (Classmate, Favorite Subject).
3. Check if each classmate has exactly one favorite subject.
4. If yes, you’ve created a function!
2.Cartesian Products of Sets
Step-by-Step Explanation
1. Definition:
  - The Cartesian product of two sets 𝐴A and 𝐵B, denoted as 𝐴×𝐵A×B, is the set of all ordered pairs (𝑎,𝑏)(a,b) where 𝑎∈𝐴a∈A and 𝑏∈𝐵b∈B.
  - Mathematically:𝐴×𝐵={(𝑎,𝑏)∣𝑎∈𝐴 and 𝑏∈𝐵}A×B={(a,b)∣a∈A and b∈B}
2. Example:
  - Let 𝐴={1,2}A={1,2} and 𝐵={𝑥,𝑦}B={x,y}.
  - The Cartesian product 𝐴×𝐵A×B is:𝐴×𝐵={(1,𝑥),(1,𝑦),(2,𝑥),(2,𝑦)}A×B={(1,x),(1,y),(2,x),(2,y)}
3. Visual Representation:
  - Think of set 𝐴A as a list of elements written horizontally and set 𝐵B as a list written vertically. Each pair (𝑎,𝑏)(a,b) is a point on a grid where the row corresponds to 𝑎a and the column corresponds to 𝑏b.
Detailed Problem-Solving Examples
Example 1:
Sets: 𝐴={1,3},𝐵={2,4}A={1,3},B={2,4}
Solution:
𝐴×𝐵={(1,2),(1,4),(3,2),(3,4)}A×B={(1,2),(1,4),(3,2),(3,4)}
Example 2:
Sets: 𝐴={𝑎,𝑏,𝑐},𝐵={1,2}A={a,b,c},B={1,2}
Solution:
𝐴×𝐵={(𝑎,1),(𝑎,2),(𝑏,1),(𝑏,2),(𝑐,1),(𝑐,2)}A×B={(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}
Example 3:
Sets: 𝐴={𝑥,𝑦},𝐵={𝑚,𝑛,𝑝}A={x,y},B={m,n,p}
Solution:
𝐴×𝐵={(𝑥,𝑚),(𝑥,𝑛),(𝑥,𝑝),(𝑦,𝑚),(𝑦,𝑛),(𝑦,𝑝)}A×B={(x,m),(x,n),(x,p),(y,m),(y,n),(y,p)}
Example 4:
Sets: 𝐴={1,2},𝐵={𝑎,𝑏},𝐶={𝑥,𝑦}A={1,2},B={a,b},C={x,y}
Solution: First, find 𝐴×𝐵A×B:
𝐴×𝐵={(1,𝑎),(1,𝑏),(2,𝑎),(2,𝑏)}A×B={(1,a),(1,b),(2,a),(2,b)}
Now, find (𝐴×𝐵)×𝐶(A×B)×C:
(𝐴×𝐵)×𝐶={((1,𝑎),𝑥),((1,𝑎),𝑦),((1,𝑏),𝑥),((1,𝑏),𝑦),((2,𝑎),𝑥),((2,𝑎),𝑦),((2,𝑏),𝑥),((2,𝑏),𝑦)}(A×B)×C={((1,a),x),((1,a),y),((1,b),x),((1,b),y),((2,a),x),((2,a),y),((2,b),x),((2,b),y)}
Real-Life Example
Imagine you are a restaurant owner who offers a set of main courses and a set of desserts. You want to create a menu that lists all possible combinations of main courses and desserts. The Cartesian product is a mathematical way to list all these combinations.
For example:
- Main Courses: {Pizza, Burger}
- Desserts: {Ice Cream, Cake}
The Cartesian product of these two sets will give us all possible pairs of main courses and desserts. Real-Life Applications
- Database Management: Cartesian products are used in SQL queries to combine tables.
- Computer Graphics: Creating pixel grids for images.
- Combinatorics: Generating all possible outcomes in probability and statistics.
3.Exercise Questions
Question 1:
Problem: If (𝑥3+1,𝑦3−23)=(53,13)(3x+1,3y−32)=(35,31), find the values of 𝑥x and 𝑦y.
Solution: To solve this, we equate the corresponding elements of the ordered pairs:
𝑥3+1=533x+1=35𝑦3−23=133y−32=31
Step-by-step solution:
1. Solve for 𝑥x:
𝑥3+1=533x+1=35
Subtract 1 from both sides:
𝑥3=53−13x=35−1
Convert 1 to 3333:
𝑥3=53−33=233x=35−33=32
Multiply both sides by 3:
𝑥=2x=2
1. Solve for 𝑦y:
𝑦3−23=133y−32=31
Add 2332 to both sides:
𝑦3=13+23=33=13y=31+32=33=1
Multiply both sides by 3:
𝑦=3y=3
So, the values of 𝑥x and 𝑦y are 𝑥=2x=2 and 𝑦=3y=3.
Question 2:
Problem: If the set 𝐴A has 3 elements and the set 𝐵={3,4,5}B={3,4,5}, then find the number of elements in 𝐴×𝐵A×B.
Solution: The number of elements in the Cartesian product 𝐴×𝐵A×B is given by the product of the number of elements in 𝐴A and 𝐵B.
Let the number of elements in 𝐴A be 𝑛(𝐴)=3n(A)=3.
The number of elements in 𝐵B is 𝑛(𝐵)=3n(B)=3.
The number of elements in 𝐴×𝐵A×B is:
𝑛(𝐴×𝐵)=𝑛(𝐴)×𝑛(𝐵)=3×3=9n(A×B)=n(A)×n(B)=3×3=9
So, the number of elements in 𝐴×𝐵A×B is 9.
Question 3:
Problem: If 𝐺={7,8}G={7,8} and 𝐻={5,4,2}H={5,4,2}, find 𝐺×𝐻G×H and 𝐻×𝐺H×G.
Solution:
1. Find 𝐺×𝐻G×H:
𝐺×𝐻={(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}G×H={(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
1. Find 𝐻×𝐺H×G:
𝐻×𝐺={(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}H×G={(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}
Question 4:
Problem: State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If 𝑃={𝑚,𝑛}P={m,n} and 𝑄={𝑛,𝑚}Q={n,m}, then 𝑃×𝑄={(𝑚,𝑛),(𝑛,𝑚)}P×Q={(m,n),(n,m)}.
(ii) If 𝐴A and 𝐵B are non-empty sets, then 𝐴×𝐵A×B is a non-empty set of ordered pairs (𝑥,𝑦)(x,y) such that 𝑥∈𝐴x∈A and 𝑦∈𝐵y∈B.
(iii) If 𝐴={1,2}A={1,2}, 𝐵={3,4}B={3,4}, then 𝐴×(𝐵∩∅)=∅A×(B∩∅)=∅.
Solution:
(i) False. The correct statement is:
𝑃×𝑄={(𝑚,𝑛),(𝑚,𝑚),(𝑛,𝑛),(𝑛,𝑚)}P×Q={(m,n),(m,m),(n,n),(n,m)}
(ii) True. The statement is correct as is.
(iii) True. The statement is correct as is because 𝐵∩∅=∅B∩∅=∅ and any set multiplied by the empty set results in the empty set.
Question 5:
Problem: If 𝐴={−1,1}A={−1,1}, find 𝐴×𝐴×𝐴A×A×A.
Solution: To find 𝐴×𝐴×𝐴A×A×A, we list all possible ordered triples:
𝐴×𝐴×𝐴={(−1,−1,−1),(−1,−1,1),(−1,1,−1),(−1,1,1),(1,−1,−1),(1,−1,1),(1,1,−1),(1,1,1)}A×A×A={(−1,−1,−1),(−1,−1,1),(−1,1,−1),(−1,1,1),(1,−1,−1),(1,−1,1),(1,1,−1),(1,1,1)}
So, 𝐴×𝐴×𝐴A×A×A has 8 elements.
Question 6:
Problem: If 𝐴×𝐵={(𝑎,𝑥),(𝑎,𝑦),(𝑏,𝑥),(𝑏,𝑦)}A×B={(a,x),(a,y),(b,x),(b,y)}, find 𝐴A and 𝐵B.
Solution: Given the pairs in 𝐴×𝐵A×B, we can infer the elements of 𝐴A and 𝐵B:
- From (𝑎,𝑥)(a,x) and (𝑎,𝑦)(a,y), we know 𝑎∈𝐴a∈A and 𝑥,𝑦∈𝐵x,y∈B.
- From (𝑏,𝑥)(b,x) and (𝑏,𝑦)(b,y), we know 𝑏∈𝐴b∈A and 𝑥,𝑦∈𝐵x,y∈B.
So, 𝐴={𝑎,𝑏}A={a,b} and 𝐵={𝑥,𝑦}B={x,y}.
4.Exercise Questions
Question 1:
Problem: Let 𝐴={1,2}A={1,2} and 𝐵={3,4}B={3,4}. Write 𝐴×𝐵A×B. How many subsets will 𝐴×𝐵A×B have? List them.
Solution:
1. Find 𝐴×𝐵A×B:
𝐴×𝐵={(1,3),(1,4),(2,3),(2,4)}A×B={(1,3),(1,4),(2,3),(2,4)}
1. Number of subsets of 𝐴×𝐵A×B:
The number of subsets of a set with 𝑛n elements is 2𝑛2n.
Here, 𝑛=4n=4 (since 𝐴×𝐵A×B has 4 elements).
So, the number of subsets is 24=1624=16.
1. List the subsets:
We can list the subsets in the order of increasing size:
- Empty set: ∅∅
- Subsets with 1 element: {(1,3)},{(1,4)},{(2,3)},{(2,4)}{(1,3)},{(1,4)},{(2,3)},{(2,4)}
- Subsets with 2 elements: {(1,3),(1,4)},{(1,3),(2,3)},{(1,3),(2,4)},{(1,4),(2,3)},{(1,4),(2,4)},{(2,3),(2,4)}{(1,3),(1,4)},{(1,3),(2,3)},{(1,3),(2,4)},{(1,4),(2,3)},{(1,4),(2,4)},{(2,3),(2,4)}
- Subsets with 3 elements: {(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)}{(1,3),(1,4),(2,3)},{(1,3),(1,4),(2,4)},{(1,3),(2,3),(2,4)},{(1,4),(2,3),(2,4)}
- Subset with 4 elements: {(1,3),(1,4),(2,3),(2,4)}{(1,3),(1,4),(2,3),(2,4)}
Question 2:
Problem: Let 𝐴A and 𝐵B be two sets such that 𝑛(𝐴)=3n(A)=3 and 𝑛(𝐵)=2n(B)=2. If (𝑥,1),(𝑦,2),(𝑧,1)(x,1),(y,2),(z,1) are in 𝐴×𝐵A×B, find 𝐴A and 𝐵B, where 𝑥,𝑦x,y and 𝑧z are distinct elements.
Solution:
Given that the pairs (𝑥,1),(𝑦,2),(𝑧,1)(x,1),(y,2),(z,1) are in 𝐴×𝐵A×B:
1. The set 𝐵B must be {1,2}{1,2} because those are the second elements in the ordered pairs.
2. The elements 𝑥,𝑦,x,y, and 𝑧z are distinct and belong to set 𝐴A. Since 𝑛(𝐴)=3n(A)=3, the set 𝐴A must be {𝑥,𝑦,𝑧}{x,y,z}.
Therefore, the sets are:
- 𝐴={𝑥,𝑦,𝑧}A={x,y,z}
- 𝐵={1,2}B={1,2}
Question 3:
Problem: The Cartesian product 𝐴×𝐴A×A has 9 elements among which are found (−1,0)(−1,0) and (0,1)(0,1). Find the set 𝐴A and the remaining elements of 𝐴×𝐴A×A.
Solution:
1. Given that 𝐴×𝐴A×A has 9 elements, the set 𝐴A must have 3 elements because 3×3=93×3=9.
2. The elements (−1,0)(−1,0) and (0,1)(0,1) suggest that −1,0,−1,0, and 11 are elements of 𝐴A.
So, 𝐴={−1,0,1}A={−1,0,1}.
1. List the remaining elements of 𝐴×𝐴A×A:
Since 𝐴={−1,0,1}A={−1,0,1}, the Cartesian product 𝐴×𝐴A×A is:
𝐴×𝐴={(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}A×A={(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}
We already know (−1,0)(−1,0) and (0,1)(0,1) are part of the product. The remaining elements are:
{(−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)}{(−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)}
So, the remaining elements are:
{(−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)}{(−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)}
5.Introduction to Relations
Definition of Relations
A relation is a connection or association between elements of two sets. Mathematically, a relation from a set 𝐴A to a set 𝐵B is a subset of the Cartesian product 𝐴×𝐵A×B.
Formal Definition:
Let 𝐴A and 𝐵B be two sets. A relation 𝑅R from 𝐴A to 𝐵B is a subset of the Cartesian product 𝐴×𝐵A×B. That is, 𝑅⊆𝐴×𝐵R⊆A×B
If (𝑎,𝑏)∈𝑅(a,b)∈R, we say that 𝑎a is related to 𝑏b by the relation 𝑅R, often written as 𝑎𝑅𝑏aRb.
Examples of Relations
1. Example 1:
  - Sets: 𝐴={1,2,3}A={1,2,3} and 𝐵={4,5}B={4,5}
  - Relation 𝑅R: {(1,4),(2,5),(3,4)}{(1,4),(2,5),(3,4)}
  - Explanation: Here, the relation 𝑅R relates the elements of set 𝐴A to elements of set 𝐵B.
2. Example 2:
  - Sets: 𝐴={Apple,Banana}A={Apple,Banana} and 𝐵={Red,Yellow}B={Red,Yellow}
  - Relation 𝑅R: {(Apple,Red),(Banana,Yellow)}{(Apple,Red),(Banana,Yellow)}
  - Explanation: This relation shows the colors of the fruits.
Types of Relations
1. Empty Relation:
  - No elements of 𝐴A are related to any elements of 𝐵B.
  - Example: 𝑅=∅R=∅
2. Universal Relation:
  - Every element of 𝐴A is related to every element of 𝐵B.
  - Example: If 𝐴={1,2}A={1,2} and 𝐵={3,4}B={3,4}, then 𝑅={(1,3),(1,4),(2,3),(2,4)}R={(1,3),(1,4),(2,3),(2,4)}.
3. Identity Relation:
  - Every element is related to itself.
  - Example: If 𝐴={1,2}A={1,2}, then the identity relation is 𝑅={(1,1),(2,2)}R={(1,1),(2,2)}.
Real-Life Example
Imagine you are at a library. There are two lists: one of the books and another of the authors. A relation in this context would describe which books are written by which authors.
For example:
- Books: {Book1, Book2, Book3}
- Authors: {AuthorA, AuthorB}
A relation would tell us pairs like (Book1, AuthorA), (Book2, AuthorB), showing which book is written by which author. Real-Life Applications of Relations
1. Database Management: Relations between tables (e.g., customer IDs and their orders).
2. Social Networks: Relationships like friendships (e.g., Alice is friends with Bob).
3. Function Mapping: Defining functions in mathematics where each input is related to exactly one output.
Simple Activity to Understand Relations
1. Take a list of students and their favorite subjects.
2. Create pairs like (Student, Subject).
3. Check if there is any pattern or specific relation (e.g., students preferring science subjects).
6.Definitions of Relations
1. Empty Relation
A relation 𝑅R from set 𝐴A to set 𝐵B is called an empty relation if no element of 𝐴A is related to any element of 𝐵B. Mathematically, 𝑅=∅R=∅.
Example:
- Set 𝐴A: {1, 2}
- Set 𝐵B: {3, 4}
Empty Relation 𝑅R:
𝑅=∅R=∅
There are no pairs (𝑎,𝑏)(a,b) such that 𝑎∈𝐴a∈A and 𝑏∈𝐵b∈B.
Problem 1: If 𝐴={𝑎,𝑏}A={a,b} and 𝐵={𝑥,𝑦}B={x,y}, define the empty relation 𝑅R from 𝐴A to 𝐵B.
Solution:
𝑅=∅R=∅
Problem 2: If 𝐴={1,2,3}A={1,2,3} and 𝐵={4,5}B={4,5}, list the pairs in the empty relation 𝑅R from 𝐴A to 𝐵B.
Solution:
𝑅=∅R=∅
2. Universal Relation
A relation 𝑅R from set 𝐴A to set 𝐵B is called a universal relation if every element of 𝐴A is related to every element of 𝐵B. Mathematically, 𝑅=𝐴×𝐵R=A×B.
Example:
- Set 𝐴A: {1, 2}
- Set 𝐵B: {3, 4}
Universal Relation 𝑅R:
𝑅={(1,3),(1,4),(2,3),(2,4)}R={(1,3),(1,4),(2,3),(2,4)}
Problem 1: If 𝐴={𝑎,𝑏}A={a,b} and 𝐵={𝑥,𝑦}B={x,y}, define the universal relation 𝑅R from 𝐴A to 𝐵B.
Solution:
𝑅=𝐴×𝐵={(𝑎,𝑥),(𝑎,𝑦),(𝑏,𝑥),(𝑏,𝑦)}R=A×B={(a,x),(a,y),(b,x),(b,y)}
Problem 2: If 𝐴={1,2}A={1,2} and 𝐵={3}B={3}, list the pairs in the universal relation 𝑅R from 𝐴A to 𝐵B.
Solution:
𝑅=𝐴×𝐵={(1,3),(2,3)}R=A×B={(1,3),(2,3)}
3. Identity Relation
A relation 𝑅R on set 𝐴A is called an identity relation if every element of 𝐴A is related to itself only. Mathematically, 𝑅={(𝑎,𝑎)∣𝑎∈𝐴}R={(a,a)∣a∈A}.
Example:
- Set 𝐴A: {1, 2, 3}
Identity Relation 𝑅R:
𝑅={(1,1),(2,2),(3,3)}R={(1,1),(2,2),(3,3)}
Problem 1: If 𝐴={𝑎,𝑏,𝑐}A={a,b,c}, define the identity relation 𝑅R on 𝐴A.
Solution:
𝑅={(𝑎,𝑎),(𝑏,𝑏),(𝑐,𝑐)}R={(a,a),(b,b),(c,c)}
Problem 2: If 𝐴={1,2}A={1,2}, list the pairs in the identity relation 𝑅R on 𝐴A.
Solution:
𝑅={(1,1),(2,2)}R={(1,1),(2,2)}
4. Reflexive Relation
A relation 𝑅R on set 𝐴A is called reflexive if every element of 𝐴A is related to itself. That is, for every 𝑎∈𝐴a∈A, (𝑎,𝑎)∈𝑅(a,a)∈R.
Example:
- Set 𝐴A: {1, 2}
Reflexive Relation 𝑅R:
𝑅={(1,1),(2,2),(1,2)}R={(1,1),(2,2),(1,2)}
Problem 1: If 𝐴={𝑎,𝑏}A={a,b}, define a reflexive relation 𝑅R on 𝐴A.
Solution:
𝑅={(𝑎,𝑎),(𝑏,𝑏),(𝑎,𝑏)}R={(a,a),(b,b),(a,b)}
Problem 2: If 𝐴={1,2,3}A={1,2,3}, give an example of a reflexive relation 𝑅R on 𝐴A.
Solution:
𝑅={(1,1),(2,2),(3,3),(1,2),(2,3)}R={(1,1),(2,2),(3,3),(1,2),(2,3)}
5. Symmetric Relation
A relation 𝑅R on set 𝐴A is called symmetric if (𝑎,𝑏)∈𝑅(a,b)∈R implies (𝑏,𝑎)∈𝑅(b,a)∈R for all 𝑎,𝑏∈𝐴a,b∈A.
Example:
- Set 𝐴A: {1, 2}
Symmetric Relation 𝑅R:
𝑅={(1,1),(2,2),(1,2),(2,1)}R={(1,1),(2,2),(1,2),(2,1)}
Problem 1: If 𝐴={𝑎,𝑏}A={a,b}, define a symmetric relation 𝑅R on 𝐴A.
Solution:
𝑅={(𝑎,𝑎),(𝑏,𝑏),(𝑎,𝑏),(𝑏,𝑎)}R={(a,a),(b,b),(a,b),(b,a)}
Problem 2: If 𝐴={1,2,3}A={1,2,3}, give an example of a symmetric relation 𝑅R on 𝐴A.
Solution:
𝑅={(1,1),(2,2),(3,3),(1,2),(2,1)}R={(1,1),(2,2),(3,3),(1,2),(2,1)}
6. Transitive Relation
A relation 𝑅R on set 𝐴A is called transitive if (𝑎,𝑏)∈𝑅(a,b)∈R and (𝑏,𝑐)∈𝑅(b,c)∈R imply (𝑎,𝑐)∈𝑅(a,c)∈R for all 𝑎,𝑏,𝑐∈𝐴a,b,c∈A.
Example:
- Set 𝐴A: {1, 2, 3}
Transitive Relation 𝑅R:
𝑅={(1,2),(2,3),(1,3)}R={(1,2),(2,3),(1,3)}
Problem 1: If 𝐴={𝑎,𝑏,𝑐}A={a,b,c}, define a transitive relation 𝑅R on 𝐴A.
Solution:
𝑅={(𝑎,𝑏),(𝑏,𝑐),(𝑎,𝑐)}R={(a,b),(b,c),(a,c)}
Problem 2: If 𝐴={1,2,3}A={1,2,3}, give an example of a transitive relation 𝑅R on 𝐴A.
Solution:
𝑅={(1,2),(2,3),(1,3),(1,1),(2,2),(3,3)}R={(1,2),(2,3),(1,3),(1,1),(2,2),(3,3)}
7.Exercise Questions
Question 1:
Problem: Let 𝐴={1,2,3,…,14}A={1,2,3,…,14}. Define a relation 𝑅R from 𝐴A to 𝐴A by 𝑅={(𝑥,𝑦)∣3𝑥−𝑦=0},R={(x,y)∣3x−y=0}, where 𝑥,𝑦∈𝐴x,y∈A. Write down its domain, codomain, and range.
Solution:
1. Find the pairs (𝑥,𝑦)(x,y) satisfying the relation 3𝑥−𝑦=03x−y=0:
  𝑦=3𝑥y=3x
  For 𝑥x in set 𝐴A, find corresponding 𝑦y:
  - If 𝑥=1x=1, 𝑦=3×1=3y=3×1=3 (Valid, as 𝑦∈𝐴y∈A)
  - If 𝑥=2x=2, 𝑦=3×2=6y=3×2=6 (Valid, as 𝑦∈𝐴y∈A)
  - If 𝑥=3x=3, 𝑦=3×3=9y=3×3=9 (Valid, as 𝑦∈𝐴y∈A)
  - If 𝑥=4x=4, 𝑦=3×4=12y=3×4=12 (Valid, as 𝑦∈𝐴y∈A)
  - If 𝑥=5x=5, 𝑦=3×5=15y=3×5=15 (Not in 𝐴A)
  - If 𝑥=6x=6 and beyond, 𝑦y exceeds the set 𝐴A
  So, the relation 𝑅R is:
  𝑅={(1,3),(2,6),(3,9),(4,12)}R={(1,3),(2,6),(3,9),(4,12)}
2. Domain: The domain of 𝑅R is the set of all first elements of the pairs in 𝑅R:
  Domain={1,2,3,4}Domain={1,2,3,4}
3. Codomain: The codomain of 𝑅R is the set 𝐴A:
  Codomain={1,2,3,…,14}Codomain={1,2,3,…,14}
4. Range: The range of 𝑅R is the set of all second elements of the pairs in 𝑅R:
  Range={3,6,9,12}Range={3,6,9,12}
Question 2:
Problem: Define a relation 𝑅R on the set 𝑁N of natural numbers by 𝑅={(𝑥,𝑦)∣𝑦=𝑥+5},R={(x,y)∣y=x+5}, where 𝑥x is a natural number less than 4; 𝑥,𝑦∈𝑁x,y∈N. Depict this relationship using roster form. Write down the domain and the range.
Solution:
1. Find the pairs (𝑥,𝑦)(x,y) satisfying the relation 𝑦=𝑥+5y=x+5:
  For 𝑥x being a natural number less than 4:
  - If 𝑥=1x=1, 𝑦=1+5=6y=1+5=6
  - If 𝑥=2x=2, 𝑦=2+5=7y=2+5=7
  - If 𝑥=3x=3, 𝑦=3+5=8y=3+5=8
  So, the relation 𝑅R is:
  𝑅={(1,6),(2,7),(3,8)}R={(1,6),(2,7),(3,8)}
2. Domain: The domain of 𝑅R is the set of all first elements of the pairs in 𝑅R:
  Domain={1,2,3}Domain={1,2,3}
3. Range: The range of 𝑅R is the set of all second elements of the pairs in 𝑅R:
  Range={6,7,8}Range={6,7,8}
Question 3:
Problem: Let 𝐴={1,2,3,5}A={1,2,3,5} and 𝐵={4,6,9}B={4,6,9}. Define a relation 𝑅R from 𝐴A to 𝐵B by 𝑅={(𝑥,𝑦)∣the difference between 𝑥 and 𝑦 is odd;𝑥∈𝐴,𝑦∈𝐵}R={(x,y)∣the difference between x and y is odd;x∈A,y∈B}. Write 𝑅R in roster form.
Solution:
1. Find the pairs (𝑥,𝑦)(x,y) satisfying the relation where the difference between 𝑥x and 𝑦y is odd:
  - If 𝑥=1x=1:
    - ∣1−4∣=3∣1−4∣=3 (Odd)
    - ∣1−6∣=5∣1−6∣=5 (Odd)
    - ∣1−9∣=8∣1−9∣=8 (Even)
  - If 𝑥=2x=2:
    - ∣2−4∣=2∣2−4∣=2 (Even)
    - ∣2−6∣=4∣2−6∣=4 (Even)
    - ∣2−9∣=7∣2−9∣=7 (Odd)
  - If 𝑥=3x=3:
    - ∣3−4∣=1∣3−4∣=1 (Odd)
    - ∣3−6∣=3∣3−6∣=3 (Odd)
    - ∣3−9∣=6∣3−9∣=6 (Even)
  - If 𝑥=5x=5:
    - ∣5−4∣=1∣5−4∣=1 (Odd)
    - ∣5−6∣=1∣5−6∣=1 (Odd)
    - ∣5−9∣=4∣5−9∣=4 (Even)
  So, the relation 𝑅R is:
  𝑅={(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}R={(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
8.Exercise Questions
Question 1:
Problem: Write the relation 𝑅={(𝑥,𝑥)∣𝑥 is a prime number less than 10}R={(x,x)∣x is a prime number less than 10} in roster form.
Solution:
1. Identify Prime Numbers Less Than 10:
  - Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves.
  - Prime numbers less than 10 are: 2, 3, 5, 7.
2. Form Ordered Pairs:
  - For each prime number 𝑥x, create a pair (𝑥,𝑥)(x,x).
  - For 𝑥=2x=2, the pair is (2,2)(2,2).
  - For 𝑥=3x=3, the pair is (3,3)(3,3).
  - For 𝑥=5x=5, the pair is (5,5)(5,5).
  - For 𝑥=7x=7, the pair is (7,7)(7,7).
3. Write the Relation in Roster Form:
  𝑅={(2,2),(3,3),(5,5),(7,7)}R={(2,2),(3,3),(5,5),(7,7)}
So, the relation 𝑅R is:
𝑅={(2,2),(3,3),(5,5),(7,7)}R={(2,2),(3,3),(5,5),(7,7)}
Question 2:
Problem: Let 𝐴={𝑥,𝑦,𝑧}A={x,y,z} and 𝐵={1,2}B={1,2}. Find the number of relations from 𝐴A to 𝐵B.
Solution:
1. Determine the Number of Elements in Each Set:
  - Set 𝐴A has 3 elements: 𝑥,𝑦,𝑧x,y,z.
  - Set 𝐵B has 2 elements: 1,21,2.
2. Find the Number of Elements in the Cartesian Product 𝐴×𝐵A×B:
  - The Cartesian product 𝐴×𝐵A×B consists of all possible ordered pairs where the first element is from 𝐴A and the second element is from 𝐵B.
  - There are 𝑛(𝐴)×𝑛(𝐵)=3×2=6n(A)×n(B)=3×2=6 ordered pairs.
3. Calculate the Number of Possible Relations:
  - A relation from 𝐴A to 𝐵B is any subset of the Cartesian product 𝐴×𝐵A×B.
  - The number of subsets of a set with 𝑛n elements is 2𝑛2n.
  - Here, 𝑛=6n=6, so the number of subsets is 2626.
4. Compute the Number of Relations:
  26=6426=64
So, the number of relations from 𝐴A to 𝐵B is 64.
Question 3:
Problem: Let 𝑅R be the relation on 𝑍Z defined by 𝑅={(𝑎,𝑏)∣𝑎,𝑏∈𝑍,𝑎−𝑏 is an integer}R={(a,b)∣a,b∈Z,a−b is an integer}. Find the domain and range of 𝑅R.
Solution:
1. Understand the Relation 𝑅R:
  - The relation 𝑅R consists of pairs (𝑎,𝑏)(a,b) where both 𝑎a and 𝑏b are integers, and 𝑎−𝑏a−b is an integer.
  - Since 𝑎a and 𝑏b are always integers, 𝑎−𝑏a−b will always be an integer. Therefore, every pair of integers (𝑎,𝑏)(a,b) will satisfy this condition.
2. Determine the Domain of 𝑅R:
  - The domain of 𝑅R is the set of all first elements of the pairs in 𝑅R.
  - Since 𝑎a can be any integer, the domain is all integers:
  Domain=𝑍Domain=Z
3. Determine the Range of 𝑅R:
  - The range of 𝑅R is the set of all second elements of the pairs in 𝑅R.
  - Since 𝑏b can be any integer, the range is all integers:
  Range=𝑍Range=Z
So, the domain and range of 𝑅R are both 𝑍Z (the set of all integers).
9.Functions: Introduction and Definitions
Definitions of Functions
1. Function:
  - A function 𝑓f from a set 𝐴A to a set 𝐵B is a rule that assigns each element 𝑥x in 𝐴A exactly one element 𝑦y in 𝐵B. We write 𝑓:𝐴→𝐵f:A→B.
  - Mathematically, if 𝑓f is a function, then for every 𝑥∈𝐴x∈A, there is a unique 𝑦∈𝐵y∈B such that 𝑦=𝑓(𝑥)y=f(x).
2. Domain:
  - The set of all possible inputs for the function.
  - Example: For 𝑓(𝑥)=𝑥2f(x)=x2, the domain is all real numbers 𝑅R.
3. Range:
  - The set of all possible outputs of the function.
  - Example: For 𝑓(𝑥)=𝑥2f(x)=x2, the range is all non-negative real numbers 𝑅+R+.
4. Codomain:
  - The set 𝐵B in which all outputs of the function are contained.
  - Example: For 𝑓:𝑅→𝑅f:R→R, 𝑅R is the codomain.
5. One-to-One Function:
  - A function 𝑓f is one-to-one (injective) if different inputs map to different outputs.
  - Mathematically, 𝑓(𝑥1)=𝑓(𝑥2)f(x1)=f(x2) implies 𝑥1=𝑥2x1=x2.
6. Onto Function:
  - A function 𝑓f is onto (surjective) if every element in the codomain is the output for some input in the domain.
  - Mathematically, for every 𝑦∈𝐵y∈B, there is some 𝑥∈𝐴x∈A such that 𝑓(𝑥)=𝑦f(x)=y.
7. Bijective Function:
  - A function 𝑓f is bijective if it is both one-to-one and onto.
  - This means every element in the codomain is the output for exactly one input in the domain.
Real-Life Example
Imagine you are at a coffee shop. You can choose a type of coffee (like Espresso, Latte, Cappuccino), and for each type, there is a fixed price. The type of coffee is your input, and the price is the output. This relationship between coffee type and price can be described as a function. Problem Solving Examples
Example 1: Basic Function
Problem: Let 𝑓:𝑅→𝑅f:R→R be defined by 𝑓(𝑥)=2𝑥+3f(x)=2x+3. Find 𝑓(1)f(1), 𝑓(−2)f(−2), and the domain and range of 𝑓f.
Solution:
1. Calculate 𝑓(1)f(1):
  𝑓(1)=2(1)+3=5f(1)=2(1)+3=5
2. Calculate 𝑓(−2)f(−2):
  𝑓(−2)=2(−2)+3=−4+3=−1f(−2)=2(−2)+3=−4+3=−1
3. Domain:
  - The function 𝑓(𝑥)=2𝑥+3f(x)=2x+3 is defined for all real numbers 𝑅R.
4. Range:
  - As 𝑥x takes all real values, 2𝑥+32x+3 also takes all real values. Thus, the range is 𝑅R.
Example 2: One-to-One Function
Problem: Determine if the function 𝑓:𝑅→𝑅f:R→R defined by 𝑓(𝑥)=3𝑥+1f(x)=3x+1 is one-to-one.
Solution:
1. Assume 𝑓(𝑥1)=𝑓(𝑥2)f(x1)=f(x2):
  3𝑥1+1=3𝑥2+13x1+1=3x2+1
2. Subtract 1 from both sides:
  3𝑥1=3𝑥23x1=3x2
3. Divide by 3:
  𝑥1=𝑥2x1=x2
Since 𝑥1=𝑥2x1=x2 implies 𝑓f is one-to-one.
Example 3: Onto Function
Problem: Determine if the function 𝑓:𝑅→𝑅f:R→R defined by 𝑓(𝑥)=𝑥2f(x)=x2 is onto.
Solution:
1. For 𝑓f to be onto, every 𝑦∈𝑅y∈R must be 𝑓(𝑥)f(x) for some 𝑥∈𝑅x∈R.
2. 𝑓(𝑥)=𝑥2f(x)=x2 gives non-negative outputs.
3. Hence, there is no 𝑥x such that 𝑓(𝑥)f(x) is a negative number.
4. Therefore, 𝑓(𝑥)=𝑥2f(x)=x2 is not onto, as it does not cover all 𝑅R.

Relations and Functions — Class 11 Maths Notes

Topics covered in Relations and Functions

1.Introduction of Relations & Functions

Step-by-Step Explanation

Key Points

Real-Life Applications

Simple Activity to Understand Functions

2.Cartesian Products of Sets

Step-by-Step Explanation

Detailed Problem-Solving Examples

Example 1:

Example 2:

Example 3:

Example 4:

Real-Life Example

The Cartesian product of these two sets will give us all possible pairs of main courses and desserts. Real-Life Applications

3.Exercise Questions

Question 1:

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

4.Exercise Questions

Question 1:

Question 2:

Question 3:

5.Introduction to Relations

Definition of Relations

Formal Definition:

Examples of Relations

Types of Relations

Real-Life Example

A relation would tell us pairs like (Book1, AuthorA), (Book2, AuthorB), showing which book is written by which author. Real-Life Applications of Relations

Simple Activity to Understand Relations

6.Definitions of Relations

1. Empty Relation

2. Universal Relation

3. Identity Relation

4. Reflexive Relation

5. Symmetric Relation

6. Transitive Relation

7.Exercise Questions

Question 1:

Question 2:

Question 3:

8.Exercise Questions

Question 1:

Question 2:

Question 3:

9.Functions: Introduction and Definitions

Definitions of Functions

Real-Life Example

Example 1: Basic Function

Example 2: One-to-One Function

Example 3: Onto Function

More Class 11 Maths chapters