If A = { x : x ∈ W , x < 2 } , B = { x : x ∈ N , 1 < x < 5 } and C = { 3 , 5 } , find A × ( B ∩ C ) .

Given A = { x : x ∈ W , x < 2 } , B = { x : x ∈ N , 1 < x < 5 } and C = { 3 , 5 } Here, W and N represents set of whole and natural numbers. ∴ A = 0 , 1 , B = 2 , 3 , 4 and C = { 3 , 5 } Now, ( B ∩ C ) = 3 ∴ A × ( B ∩ C ) = { 0 , 1 } × { 3 } ⇒ A × ( B ∩ C ) = 0 , 3 , 1 , 3 .

If A = { x : x ∈ W , x < 2 } , B = { x : x ∈ N , 1 < x < 5 } and C = { 3 , 5 } , find A × ( B ∪ C )

Given, A = { x : x ∈ W , x < 2 } , B = { x : x ∈ N , 1 < x < 5 } and C = { 3 , 5 } Here, W and N represents set of whole and natural numbers. ∴ A = 0 , 1 , B = 2 , 3 , 4 and C = { 3 , 5 } Now, ( B ∪ C ) = 2 , 3 , 4 , 5 ∴ A × ( B ∪ C ) = { 0 , 1 } × { 2 , 3 , 4 , 5 } By the definition of cartesian product, the collection of ordered pairs is A × ( B ∪ C ) = { ( 0 , 2 ) , ( 0 , 3 ) , ( 0 , 4 ) , ( 0 , 5 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) } .

If A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } , verify that A × ( B ∪ C ) = ( A × B ) ∪ ( A × C ) .

Given set-builder form A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } . Here, N and I are set of natural numbers and integers. ∴ A = 1 , 2 , 3 , B = - 1 , 0 , 1 and C = 1 , 2 (Roster form) We need to prove that, A × ( B ∪ C ) = ( A × B ) ∪ ( A × C ) From LHS, we have A × ( B ∪ C ) Now, B ∪ C = - 1 , 0 , 1 , 2 ∴ A × ( B ∪ C ) = { 1 , 2 , 3 } × { - 1 , 0 , 1 , 2 } ∴ A × ( B ∪ C ) = { ( 1 , - 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , - 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , - 1 ) , ( 3 , 0 ) , ( 3 , 1 ) , ( 3 , 2 ) } → 1 From RHS we have , ( A × B ) ∪ ( A × C ) A × B = { ( 1 , - 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , - 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 3 , - 1 ) , ( 3 , 0 ) , ( 3 , 1 ) } And A × C = 1 , 1 , 1 , 2 , 2 , 1 , 2 , 2 , 3 , 1 , 3 , 2 ∴ ( A × B ) ∪ ( A × C ) = { ( 1 , - 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , - 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , - 1 ) , ( 3 , 0 ) , ( 3 , 1 ) , ( 3 , 2 ) } → 2 Hence, from 1 and 2 A × ( B ∪ C ) = ( A × B ) ∪ ( A × C ) Hence proved.

If A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } , verify that A × ( B ∩ C ) = ( A × B ) ∩ ( A × C ) .

Given set-builder form A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } . Here, N and I are set of natural numbers and integers. ∴ A = 1 , 2 , 3 , B = - 1 , 0 , 1 and C = 1 , 2 (Roster form) We need to prove that, A × ( B ∩ C ) = ( A × B ) ∩ ( A × C ) From LHS, we have A × ( B ∩ C ) Now, B ∩ C = 1 ∴ A × ( B ∩ C ) = { 1 , 2 , 3 } × { 1 } ∴ A × ( B ∩ C ) = { ( 1 , 1 ) , ( 2 , 1 ) , ( 3 , 1 ) } → 1 From RHS we have , ( A × B ) ∩ ( A × C ) A × B = { ( 1 , - 1 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 2 , - 1 ) , ( 2 , 0 ) , ( 2 , 1 ) , ( 3 , - 1 ) , ( 3 , 0 ) , ( 3 , 1 ) } And A × C = 1 , 1 , 1 , 2 , 2 , 1 , 2 , 2 , 3 , 1 , 3 , 2 ∴ ( A × B ) ∩ ( A × C ) = { ( 1 , 1 ) , ( 2 , 1 ) , ( 3 , 1 ) } → 2 Hence, from 1 and 2 A × ( B ∩ C ) = ( A × B ) ∩ ( A × C ) Hence proved.

If A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } , verify that ( A - B ) × C = A × C - B × C .

Given set-builder form A = { x : x ∈ N and x ≤ 3 } , B = { x : x ∈ I , - 1 ≤ x ≤ 1 } and C = { 1 , 2 } . Here, N and I are set of natural numbers and integers. ∴ A = 1 , 2 , 3 , B = - 1 , 0 , 1 and C = 1 , 2 (Roster form) We need to prove that, ( A - B ) × C = A × C - B × C From LHS, we have ( A - B ) × C Now, A - B = 2 , 3 ∴ ( A - B ) × C = { 2 , 3 } × { 1 , 2 } ∴ ( A - B ) × C = 2 , 1 , 2 , 2 , 3 , 1 , 3 , 2 From RHS we have , A × C - B × C B × C = { ( - 1 , 1 ) , ( - 1 , 2 ) , ( 0 , 1 ) , ( 0 , 2 ) , ( 1 , 1 ) , ( 1 , 2 ) } And A × C = 1 , 1 , 1 , 2 , 2 , 1 , 2 , 2 , 3 , 1 , 3 , 2 ∴ A × C - B × C = { ( 2 , 1 ) , ( 2 , 2 ) , ( 3 , 1 ) , ( 3 , 2 ) } → 2 Hence, from 1 and 2 ( A - B ) × C = A × C - B × C Hence proved.

If P = { 1 , 2 } , form the set P × P × P .

Given set, P = { 1 , 2 } We know that, A × B = { ( x , y ) : x ∈ A , y ∈ B } Now, P × P = { 1 , 2 } × { 1 , 2 } ⇒ P × P = 1 , 1 , 1 , 2 , 2 , 1 , 2 , 2 Hence, P × P × P = 1 , 1 , 1 , 2 , 2 , 1 , 2 , 2 × { 1 , 2 } ∴ P × P × P = { ( 1 , 1 , 1 ) , ( 1 , 1 , 2 ) , ( 1 , 2 , 1 ) , ( 1 , 2 , 2 ) , ( 2 , 1 , 1 ) , ( 2 , 1 , 2 ) , ( 2 , 2 , 1 ) , ( 2 , 2 , 2 ) }

If A , B are two sets such that n ( A × B ) = 6 , and some elements of A × B are ( - 1 , 2 ) ( 2 , 3 ) , ( 4 , 3 ) then find A × B and B × A .

Given, n ( A × B ) = 6 And some elements of A × B are ( - 1 , 2 ) ( 2 , 3 ) , ( 4 , 3 ) ( - 1 , 2 ) ∈ A × B ⇒ - 1 ∈ A and 2 ∈ B , ( 2 , 3 ) ∈ A × B ⇒ 2 ∈ A and 3 ∈ B And ( 4 , 3 ) ∈ A × B ⇒ 4 ∈ A and 3 ∈ B Thus, - 1 , 2 , 4 ∈ A and 2 , 3 ∈ B Therefore, A = - 1 , 2 , 4 , B = 2 , 3 . Now, A × B = - 1 , 2 , 4 × 2 , 3 = { ( - 1 , 2 ) , ( - 1 , 3 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 4 , 2 ) , ( 4 , 3 ) } And B × A = 2 , 3 × - 1 , 2 , 4 = { ( 2 , - 1 ) , ( 3 , - 1 ) , ( 2 , 2 ) , ( 3 , 2 ) , ( 2 , 4 ) , ( 3 , 4 ) } .

Given B = { 2 , 3 , 5 } and some elements of A × B are ( a , 2 ) , ( b , 3 ) , ( c , 5 ) . Find the set A the remaining ordered pairs of A × B such that A × B is least.

Given, B = { 2 , 3 , 5 } ⇒ n B = 3 And some elements of A × B are ( a , 2 ) , ( b , 3 ) , ( c , 5 ) ( a , 2 ) ∈ A × B ⇒ a ∈ A and 2 ∈ B , ( b , 3 ) ∈ A × B ⇒ b ∈ A and 3 ∈ B And ( c , 5 ) ∈ A × B ⇒ c ∈ A and 5 ∈ B Thus, a , b , c ∈ A and 2 , 3 , 5 ∈ B Therefore, A = a , b , c , B = 2 , 3 , 5 . Now, A × B = a , b , c × 2 , 3 , 5 = { ( a , 2 ) , ( b , 3 ) , ( c , 5 ) , ( a , 3 ) , ( a , 5 ) , ( b , 2 ) , ( b , 5 ) , ( c , 2 ) , ( c , 3 ) } Hence, the remaining ordered pairs are ( a , 3 ) , ( a , 5 ) , ( b , 2 ) , ( b , 5 ) , ( c , 2 ) , ( c , 3 ) .

E M B I B E

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1

M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 1: EXERCISE 2.1

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 1: EXERCISE 2.1

Attempt the practice questions on Chapter 2: Relations and Functions, Exercise 1: EXERCISE 2.1 with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 11 Volume 1 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Relations and Functions, Exercise 1: EXERCISE 2.1 with Hints & Solutions

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 21

If $A = {x : x \in W, x < 2}, B = {x : x \in N, 1 < x < 5}$ and $C = {3, 5}$ , find $A \times (B \cap C)$ .

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 21

If $A = {x : x \in W, x < 2}, B = {x : x \in N, 1 < x < 5}$ and $C = {3, 5}$ , find $A \times (B \cup C)$

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 22

If $A = {x : x \in N and x \leq 3}, B = {x : x \in I, - 1 \leq x \leq 1}$ and $C = {1, 2}$ , verify that $A \times (B \cup C) = (A \times B) \cup (A \times C)$ .

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 22

If $A = {x : x \in N and x \leq 3}, B = {x : x \in I, - 1 \leq x \leq 1}$ and $C = {1, 2}$ , verify that $A \times (B \cap C) = (A \times B) \cap (A \times C)$ .

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 22

If $A = {x : x \in N and x \leq 3}, B = {x : x \in I, - 1 \leq x \leq 1}$ and $C = {1, 2}$ , verify that $(A - B) \times C = (A \times C) - (B \times C)$ .

EASY

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 23

If $P = {1, 2}$ , form the set $P \times P \times P$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 24

If $A, B$ are two sets such that $n (A \times B) = 6$ , and some elements of $A \times B$ are $(- 1, 2)$ $(2, 3), (4, 3)$ then find $A \times B$ and $B \times A$ .

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 1>Chapter 2 - Relations and Functions>EXERCISE 2.1>Q 25

Given $B = {2, 3, 5}$ and some elements of $A \times B$ are $(a, 2), (b, 3), (c, 5)$ . Find the set $A$ the remaining ordered pairs of $A \times B$ such that $A \times B$ is least.

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