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

EXERCISE 2.1

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for EXERCISE 2.1

Simple step-by-step solutions to EXERCISE 2.1 questions of Relations and Functions from Understanding ISC Mathematics Class 11 Volume 1. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from 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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