## Samacheer Kalvi 10th Maths Book – Unit 1 Exercise 1.2: Book Back Question and Answers

TN Samacheer Kalvi 10th Maths Book uploaded as PDF and its book back question & answers solutions Guide PDF given below. Class 10th Standard New Syllabus Maths Books consist of 8 units.  Each chapter is well designed and organized by the TN textbook corporation. Complete solutions for all 8 units book back questions uploaded and available on our site. For Samacheer kalvi 10th Maths – Relations and Functions Ex 1.2 book back question and answers solutions given below:

TN Samacheer Kalvi 10th Maths Chapter 1 Book Back Exercise given below.

### Exercise 1.2 Relations and Functions

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)