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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.2

**10th Maths Exercise 1.2 Samacheer Kalvi Question 1.**

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

(i) R_{1} = {(2, 1), (7, 1)}

(ii) R_{2} = {(-1, 1)}

(iii) R_{3} = {(2, -1), (7, 7), (1, 3)}

(iv) R_{4} = {(7,-1), (0, 3), (3, 3), (0, 7)}

(i) A = {1, 2, 3, 7}, B = {3, 0,-1, 7}

Solution:

R_{1} = {(2,1), (7,1)}

It is not a relation there is no element as 1 in B.

(ii) R_{2} = {(-1, 1)}

It is not [∵ -1 ∉ A, 1 ∉ B]

(iii) R_{3} = {(2, -1), (7, 7), (1, 3)}

It is a relation.

R_{4} = {(7,-1), (0, 3), (3, 3), (0, 7)}

It is also not a relation. [∵ 0 ∉ A]

**Ex 1.2 Class 10 Samacheer Question 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.

Answer:

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}

**Exercise 1.2 Class 10 Maths Samacheer Question 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.

Solution:

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}

**10th Maths Exercise 1.2 Question 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}

Solution:

(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)}

**10th Maths Exercise 1.2 Answers Question 5.**

A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A_{1}, A_{2}, A_{3}, A_{4} and As were Assistants; C_{1}, C_{2}, C_{3}, C_{4} were Clerks; M_{1}, M_{2}, M_{3} were managers and E_{1},E_{2} 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.

Solution:

A – Assistants → A_{1}, A_{2}, A_{3}, A_{4}, A_{5}

C – Clerks → C_{1}, C_{2}, C_{3}, C_{4}

D – Managers → M_{1}, M_{2}, M_{3}

E – Executive officer → E_{1}, E_{2}

(a) R = {(10,000, A_{1}), (10,000, A_{2}), (10,000, A_{3}),

(10,000, A_{4}), (10,000, A_{5}), (25,000, C_{1}),

(25,000, C_{2}), (25,000, C_{3}), (25,000, C_{4}),

(50,000, M_{1}), (50,000, M_{2}), (50,000, M_{3}),

(1,00,000, E_{1}), (1,00,000, E_{2})}

(b)