10th Maths Exercise 1.2

10th Maths Chapter 1

Relations and Functions

Exercise 1.2

10th Maths Chapter 1 Relations and Functions Exercise 1.2

TN Samacheer Kalvi 10th Maths / Solutions Chapter 1 Relations and Functions Ex 1.2

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10th Maths Chapter 1 video classes

• Class 10 Chapter 1 All Exercise Sums 
• Class 10 Chapter 1 Example 1.1 to 1.5
• Class 10 Chapter 1 Example 1.6 to 1.10
• Class 10 Chapter 1 Example 1.11 to 1.15
• Class 10 Chapter 1 Example 1.16 to 1.20
  

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10th Maths Exercise 1.2 | 10th Maths Ex 1.2 | 10th Maths Chapter 1 Relations and Functions Exercise 1.2 |  10th Maths Chapter 1 Relations and Functions Exercise 1.1 |  10th Maths Chapter 1 Relations and Functions Exercise 1.3 |  10th Maths Chapter 1 Relations and Functions Exercise 1.4 |  10th Maths Chapter 1 Relations and Functions Exercise 1.5 |  10th Maths Chapter 1 Relations and Functions Exercise 1.6 |  Class 10 Chapter 1 Relations and Functions Example 1.11 to 1.15   | Class 10 Chapter 1 Relations and Functions Example 1.6 to 1.10   | 10th Maths Example 1.6 | 10th Maths Example 1.7 | 10th Maths Example 1.8 | 10th Maths Example 1.9 | 10th Maths Example 1.10 | Class 10 Chapter 1  Relations and Functions Example 1.1 to 1.5  TN New Syllabus 10th Maths Chapter 1 Relations And Functions Example 1.1 1 Relations and Functions | 1.1 Introduction |1.2 Ordered Pair | 1.3 Cartesian Product | 1.4 Relations | 1.5 Functions | 1.6 Representation of Functions | 1.7 Types of functions | 1.8 Special cases of Functions | 1.9 Composition of Functions 1.10 Identifying the graphs of Linear, Quadratic, | Cubic and Reciprocal functions | TN (Samacheer) 10 Maths Relations and Functions New syllabus Ex 1.1 TN (Samacheer) 10 Maths New Syllabus Relations and Functions. 

 10th Maths Chapter 1 Relations and Functions Exercise 1.2

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)}

Answer:

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

A × B = {1,2,3} × {3, 0,-1, 7}

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

(2, -1) (2, 7) (3, 3) (3,0) (3,-1)

(3, 7) (7, 3) (7, 0) (7,-1) (7, 7)}

(i) R1 = {(2, 1)} (7, 1)

It is not a relation, there is no element of (2, 1) and (7, 1) in A × B

(ii) R2 = {(-1),1)}

It is not a relation, there is no element of

(-1, 1) in A × B

(iii) R3 = {(2,-1) (7, 7) (1,3)}

Yes, It is a relation

(iv) R4 = {(7,-1) (0,3) (3, 3) (0,7)}

It is not a relation, there is no element of (0, 3) and (0, 7) in A × B

Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.2

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.

Solution:

A = {1, 2, 3, 4, . . . 45}, A × A = {(1, 1), (2, 2) ….. (45, 45)}

R – is square of’

R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}

R ⊂ (A × A)

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.

Answer:

x = {0, 1, 2, 3, 4, 5}

y = x + 3

when x = 0 ⇒ y = 0 + 3 = 3

when x = 1 ⇒ y = 1 + 3 = 4

when x = 2 ⇒ y = 2 + 3 = 5

when x = 3 ⇒ y = 3 + 3 = 6

when x = 4 ⇒ y = 4 + 3 = 7

when x = 5 y = 5 + 3 = 8

R = {(0, 3) (1,4) (2, 5) (3, 6) (4, 7) (5, 8)}

Domain = {0, 1, 2, 3, 4, 5}

Range = {3, 4, 5, 6, 7, 8}

4. Represent each of the given relations 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}

Answer:

(i) x = {2, 3, 4, 5} y = {1, 2, 3, 4}

x = 2y

wheny y = 1 ⇒ x = 2 × 1 = 2

when y = 2 ⇒ x = 2 × 2 = 4

when y = 3 ⇒ r = 2 × 3 = 6

when y = 4 ⇒ x = 2 × 4 = 8

(a) Arrow diagram

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(b) Graph

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(c) Roster form R = {(2, 1) (4, 3)}

(ii) x = {1, 2, 3, 4, 5, 6, 7, 8, 9}

y = {1,2, 3, 4, 5, 6, 7, 8,9}

y = x + 3

when x = 1 ⇒ y = 1 + 3 = 4

when x = 2 ⇒ y = 2 + 3 = 5

when x = 3 ⇒ y = 3 + 3 = 6

when x = 4 ⇒ y = 4 + 3 = 7

when x = 5 ⇒ y = 5 + 3 = 8

when x = 6 ⇒ y = 6 + 3 = 9

when x = 7 ⇒ y = 7 + 3 = 10

when x = 8 ⇒ y = 8 + 3 = 11

when x = 9 ⇒ y = 9 + 3 = 12

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

(a) Arrow diagram

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(b) Graph

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

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 A1, A2, A3, A4 and A5 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.

Answer:

Assistants → A1, A2, A3, A4, A5

Clerks → C1, C2, C3, C4

Managers → M1, M2, M3

Executive officers → E1, E2

R = {00000, A1) (10000, A2) (10000, A3) (10000, A4) (10000, A5)

(25000, C1) (25000, C2) (25000, C3) (25000, C4)

(50000, M1) (50000, M2) (50000, M3) (100000, E1) (100000, E2)}

(a) Arrow diagram

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Functions Definition

A relation f between two non – empty sets X and Y is called a function from X to Y if for each x ∈ X there exists only one Y ∈ Y such that (x, y) ∈ f

f = {(x, y) / for all x ∈ X, y ∈ f}

Note: The range of a function is a subset of its co-domain.