You are currently viewing 10TH MATHS CHAPTER 2 EXERCISE 1.2 RELATIONS AND FUNCTIONS Guide

10TH MATHS CHAPTER 2 EXERCISE 1.2 RELATIONS AND FUNCTIONS Guide

10TH MATHS CHAPTER 2 EXERCISE 1.2 RELATIONS AND FUNCTIONS Guide

10th Standard Maths Chapter 1 Exercise 1.2 Relations and Functions Guide Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here

10TH MATHS CHAPTER 1 RELATIONS AND FUNCTIONS Guide

 

1. Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B ?

(i) R1 = {(2, 1), (7, 1)}
(ii) R2 = {(-1, 1)}
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
(iv) R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
(i) A = {1, 2, 3, 7}, B = {3, 0,-1, 7}
Solu.:
R1 = {(2,1), (7,1)}

 

It is not a relation there is no element as 1 in B.
(ii) R2 = {(-1, 1)}
It is not [∵ -1 ∉ A, 1 ∉ B]
(iii) R3 = {(2, -1), (7, 7), (1, 3)}
It is a relation.
R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
It is also not a relation. [∵ 0 ∉ A]

 

 

2. Let A = {1, 2, 3, 4, ….., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.

Solu.:
A = {1,2, 3, 4 . . . . 45}
The relation is defined as “is square of’
R = {(1,1) (2, 4) (3, 9)
(4, 16) (5,25) (6, 36)}
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1, 4, 9, 16, 25, 36}

 

3. A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.

Solu.:
x = {0, 1, 2, 3, 4, 5}
y = x + 3

 

⇒ y = {3, 4, 5, 6, 7, 8}
R = {(x, y)}
= {(0, 3),(1, 4),(2, 5),(3, 6), (4, 7), (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

 

4. Represent each of the given relation by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.

(i) {(x, y)|x = 2y,x ∈ {2, 3, 4, 5},y ∈ {1, 2, 3, 4)
(ii) {(x, y)|y = x + 3, x, y are natural numbers <10}
Solu.:
(i){(x, y)|x = 2y, x ∈ {2, 3, 4, 5},y ∈ {1, 2, 3, 4}} R = (x = 2y)
2 = 2 × 1 = 2                        4 = 2 × 2 = 4

 

(c) {(2, 1), (4, 2)}
(ii) {(x, y)|y = x + 3, x,+ are natural numbers <10}
x = {1, 2, 3, 4, 5, 6, 7, 8, 9} R = (y = x + 3)
y = {1, 2, 3, 4, 5, 6, 7, 8, 9}
R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

(c) R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

 

5. A company has four categories of employees given Assistants (A), Clerks (C), Managers (M), and an Executive Officer (E). The company provides Rs.10,000, Rs.25,000, Rs.50,000, and Rs.1,00,000 as salaries to the people who work in categories A, C, M, and E respectively. If A1, A2, A3, A4, and As were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers, and E1, E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.

Solu.:
A – Assistants → A1, A2, A3, A4, A5
C – Clerks → C1, C2, C3, C4
D – Managers → M1, M2, M3
E – Executive officer → E1, E2
(a) R = {(10,000, A1), (10,000, A2), (10,000, A3),
(10,000, A4), (10,000, A5), (25,000, C1),
(25,000, C2), (25,000, C3), (25,000, C4),
(50,000, M1), (50,000, M2), (50,000, M3),
(1,00,000, E1), (1,00,000, E2)}

 

(b)

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