Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6 Important

Example 7

Example 8 Important

Example 9

Example 10

Example 11 Important

Example 12 Important

Example 13 Important

Example 14 Important

Example 15

Example 16

Example 17 Important

Example 18

Example 19 Important

Example 20 Important You are here

Example 21

Example 22 Important

Example 23 Important

Example 24 Important

Example 25

Example 26 Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Important Deleted for CBSE Board 2024 Exams

Question 9 Deleted for CBSE Board 2024 Exams

Question 10 Important Deleted for CBSE Board 2024 Exams

Question 11 (a) Deleted for CBSE Board 2024 Exams

Question 11 (b) Deleted for CBSE Board 2024 Exams

Question 11 (c) Deleted for CBSE Board 2024 Exams

Question 12 Deleted for CBSE Board 2024 Exams

Question 13 Deleted for CBSE Board 2024 Exams

Question 14 Important Deleted for CBSE Board 2024 Exams

Question 15 Deleted for CBSE Board 2024 Exams

Question 16 Deleted for CBSE Board 2024 Exams

Question 17 Deleted for CBSE Board 2024 Exams

Question 18 Deleted for CBSE Board 2024 Exams

Question 19 Deleted for CBSE Board 2024 Exams

Question 20 Important Deleted for CBSE Board 2024 Exams

Question 21 Deleted for CBSE Board 2024 Exams

Question 22 Deleted for CBSE Board 2024 Exams

Question 23 Deleted for CBSE Board 2024 Exams

Question 24 (a) Deleted for CBSE Board 2024 Exams

Question 24 (b) Deleted for CBSE Board 2024 Exams

Question 25 Deleted for CBSE Board 2024 Exams

Chapter 1 Class 12 Relation and Functions

Serial order wise

Last updated at June 6, 2023 by Teachoo

Example 20 (Method 1) Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R1 = R2. R1 = {(x, y) : x – y is divisible by 3} R2 = { (x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} } We will prove R1 = R2 by proving R1 ⊂ R2 and R2 ⊂ R1 i.e, all elements of R1 are in the set R2 and all elements of R2 are in the set R1 Proving R1 ⊂ R2 Let (x, y) ∈ R1 ⇒ x – y is a divisible 3 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ (x, y) ∈ R2. Hence, R1⊂ R2. Rough In {1, 4, 7} x – y = 1 – 4 = – 3 = 4 – 7 = –3 = 7 – 1 = 6 So, x – y is divisible by 3 Similarly, in {2, 5, 8} & {3, 6, 9} x – y is divisible by 3 Proving R2 ⊂ R1 Let (x, y) ∈ R2 ⇒ {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} ⇒ x – y is divisible 3 ⇒ (x, y) ∈ R1. Hence, R2 ⊂ R1. Hence, R1 ⊂ R2 & R2 ⊂ R1. ∴ R1 = R2 Hence shown Rough In {1, 4, 7} x – y = 1 – 4 = – 3 = 4 – 7 = –3 = 7 – 1 = 6 So, x – y is divisible by 3 Similarly, in {2, 5, 8} & {3, 6, 9} x – y is divisible by 3 Example 20 (Method 2) Let X = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let R1 be a relation in X given by R1 = {(x, y) : x – y is divisible by 3} and R2 be another relation on X given by R2 = {(x, y): {x, y} ⊂ {1, 4, 7}} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9}}. Show that R1 = R2. X = {1, 2, 3, 4, 5, 6, 7, 8, 9} R1 = {(x, y) : x – y is divisible by 3} Finding R1 R1 = { (1, 4), (1, 7) , (2, 5) , (2, 8), (3, 6) , (3, 9) , (4, 1), (4, 7), (5, 2), (5, 8) , (6, 3) , (6, 9), (7, 1) , (7, 4) , (8, 2), (8, 5), (9, 3), (9, 6) } Now, finding R2 R2 = { (x, y): {x, y} ⊂ {1, 4, 7} or {x, y} ⊂ {2, 5, 8} or {x, y} ⊂ {3, 6, 9} } R2 = { (1, 4), (4, 1) , (1, 7) ,(7, 1), (4, 7) , (7, 4) , (2, 5), (5, 2) , (2, 8) ,(8, 2), (5, 8) , (8, 5) , (3, 6), (6, 3), (3, 9) , (9, 3), (6, 9) , (9, 6) } ∴ R1 = R2 Hence proved