10th Maths Chapter 1 Relations And Functions Example 1.1 to Example 1.5

TN New Syllabus 10th Maths 

Chapter 1 

Relations And Functions Example 1.1 to 1.5 | Alex Maths Official 

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10th maths Guide English Medium 

TN New Syllabus 10th Maths  Chapter 1  Relations And Functions Example 1.1 | Alex Maths Official 

10th maths Exercise 1.1 Solution (Answers)

 10th maths Example 1.1 Solutions

10th Maths Chapter 1 Relations and Functions Example 1.1  | Alex Maths

Example 1.1 If A = {1, 3, 5} and B = {2, 3} then 

(i) find A × B and B × A.

 (ii) Is A × B = B × A? If not why ? 

(iii) Show that n(A × B) = n(B × A) = n(A) × n(B).

Solution : 

Given that A = {1, 3, 5} and B = {2, 3}

(i) A × B = {1, 3, 5} × {2, 3}  

= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}

B × A = {2, 3} × {1, 3, 5}  

= {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)} .   

*************************************** 

  

(ii)  From (1) and (2) we conclude that  

 A × B ≠ B × A as (1, 2) ≠ (2, 1) 

and   (1, 3) ≠) (3, 1), etc.

*************************************** 

(iii)  n(A) = 3; n(B) = 2. 

From (1) and (2) we observe that,  

 n(A × B) = n(B × A) = 6;

we see that, n(A) × n(B) = 3 × 2 = 6 and  

n(B) × n(A) = 2 × 3 = 6Hence, 

n(A × B) = n(B × A) = n(A) × n(B) = 6.

Thus, n(A × B) = n(B × A) = n(A) × n(B).

10th Maths Chapter 1 Relations and Functions Example 1.2 | Alex Maths

10th Maths Example 1.2 solutions

Exampl 1.2  If A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find A and B.

Solution :

 A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} 

We have  

 A = {set of all first coordinates of elements of A × B}. Therefore A = {3, 5} 

B = {set of all second coordinates of ele-ments of A × B}. ThereforeB = {2, 4} 

Thus A = {3, 5} and B = {2, 4}.

10th Maths Chapter 1 Relations and Functions Example 1.3 | Alex Maths

Samacheer kalvi 10th maths Example 1.3 solution

Example 1.3 Let A = {x ∈ N | 1 < x < 4}, B = {x ∈ W |  0 ≤ x < 2} and C = {x ∈ N | x < 3}.Then verify that  (i) A×(B∪C)= (A×B) ∪ (A×C), (ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Solution : 

A = {x ∈ N | 1 < x < 4} = {2, 3}, 

 B = {x ∈ W | 0 ≤ x < 2} = {0, 1}.

 C = {x ∈ N | x < 3} = {1, 2}

(i) A × (B ∪ C) = (A × B) ∪ (A × C) B ∪ C = {0, 1} ∪ {1, 2} = {0, 1, 2} A × (B ∪ C)  

= {2, 3} × {0, 1, 2} 

 = {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}       

A × B = {2, 3} × {0, 1}           

  = {(2, 0), (2, 1), (3, 0), (3, 1)}

A × C  = {2, 3} × {1, 2}            

 = {(2, 1), (2, 2), (3, 1), (3, 2)} (A × B) ∪ (A × C) 

 = {(2, 0), (2, 1), (3, 0),   (3, 1)} ∪ {(2, 1), (2, 2), (3, 1), (3, 2)}  

= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}      ... (2) 

From (1) and (2), A × (B ∪ C)

 = (A × B) ∪ (A × C) is verified.

****************

(ii) A × (B ∩ C) = (A × B) ∩ (A × C) 

B ∩ C  = {0, 1} ∩ {1, 2}= {1}

 A × (B ∩ C)  = {2, 3} × {1}    = {(2, 1), (3, 1)} ... (3)

 A × B  = {2, 3} × {0, 1}         = {(2, 0), (2, 1), (3, 0), (3, 1)} 

A × C  = {2, 3} ×{1, 2}         = {(2, 1), (2, 2), (3, 1), (3, 2)}

 (A × B) ∩ (A × C)   

    = {(2, 0), (2, 1), (3, 0), (3, 1)} ∩     {(2, 1), (2, 2), (3, 1), (3, 2)}  

= {(2, 1), (3, 1)}   ... (4)

 From (3) and (4),  

A×(B ∩ C) = (A ×B) ∩ (A × C) is verified.

10th Maths Chapter 1 Relations and Functions Example 1.4 | Alex Maths

Example 1.4 Question : Let A = {3, 4, 7, 8} and B = {1, 7, 10}. Which of the following sets are relations from A to B ?

(i)  R1 = {(3, 7), (4, 7), (7, 10), (8, 1)} 

(ii)  R2 = {(3, 1), (4, 12)}

(iii)  R3 = {(3, 7), (4, 10), (7, 7), (7, 8), (8, 11),  (8, 7), (8, 10)}

Solution :

A×B = {(3, 1), (3, 7), (3, 10), (4, 1), (4, 7), 

        (4,10), (7,1), (7,7), (7,10), (8, 1), (8, 7), (8, 10)}

(i) We note that R1 ⊆ A × B. Thus, R1 is a relation from A to B. 

(ii) Here, (4, 12) ∈ R2, but (4, 12) ∉ A × B. So, R2 is not a relation from A to B.

 (iii) Here, (7, 8) ∈ R3, but (7, 8) ∉ A × B. So R3 is not a relation from A to B.

 10th Maths Chapter 1 Relations and Functions Example 1.5 | Alex Maths

Example 1.5 : The arrow diagram shows a relationship be-tween the sets P and Q. Write the relation in 

(i) Set builder form 

(ii) Roster form 

(iii) What is the domain and range of R.

Solution :

(i)  Set builder form of  R  = {(x, y) | y = x – 2, x ∈ P, y ∈ Q}

(ii)  Roster form R  = {(5, 3), (6, 4), (7, 5)}

(iii)  Domain of R  = {5, 6, 7} and range of  R = {3, 4, 5}

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