Samacheer Kalvi Class 10 Maths Solutions Chapter 1 Relations and Functions Exercise 1.2

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) 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: First, let's find the Cartesian product \( A \times B \).
\( A \times B = \{(1,3), (1,0), (1,-1), (1,7), (2,3), (2,0), (2,-1), (2,7), (3,3), (3,0), (3,-1), (3,7), (7,3), (7,0), (7,-1), (7,7)\} \)
A relation from A to B means all its ordered pairs must be present in \( A \times B \).
(i) \( R_1 = \{(2,1), (7,1)\} \)
This is not a relation from A to B because the pairs (2,1) and (7,1) are not found in \( A \times B \).
(ii) \( R_2 = \{(-1,1)\} \)
This is not a relation from A to B because the pair (-1,1) is not present in \( A \times B \).
(iii) \( R_3 = \{(2,-1), (7, 7), (1,3)\} \)
Yes, this is a relation from A to B. All the pairs (2,-1), (7,7), and (1,3) are found within \( A \times B \). This means every element of \( R_3 \) is also an element of \( A \times B \).
(iv) \( R_4 = \{(7,-1), (0,3), (3, 3), (0,7)\} \)
This is not a relation from A to B because the pairs (0,3) and (0,7) are not present in \( A \times B \). The elements of A must be the first parts of the pairs.
In simple words: For a set of pairs to be a relation from A to B, every single pair in that set must also be found in the larger set of all possible pairs from A to B. We check each given relation against the full \( A \times B \) list.

🎯 Exam Tip: To check if a set of ordered pairs is a relation from A to B, always first list out all elements of the Cartesian product \( A \times B \) and then compare each ordered pair in the given relation to see if it exists in \( A \times B \).

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 \( \times \) A. Also, find the domain and range of R.
Answer: Given set \( A = \{1, 2, 3, 4, \dots, 45\} \).
The relation R is defined as "is square of" on A. This means for any pair \( (x, y) \in R \), \( x \) is the square of \( y \). Or, \( y = \sqrt{x} \). Since it's "on A", both \( x \) and \( y \) must be elements of A.
Let's list the pairs where the first number is the square of the second number, and both numbers are in A.
If \( y = 1 \), then \( x = 1^2 = 1 \). So, \( (1, 1) \in R \).
If \( y = 2 \), then \( x = 2^2 = 4 \). So, \( (4, 2) \in R \).
If \( y = 3 \), then \( x = 3^2 = 9 \). So, \( (9, 3) \in R \).
If \( y = 4 \), then \( x = 4^2 = 16 \). So, \( (16, 4) \in R \).
If \( y = 5 \), then \( x = 5^2 = 25 \). So, \( (25, 5) \in R \).
If \( y = 6 \), then \( x = 6^2 = 36 \). So, \( (36, 6) \in R \).
If \( y = 7 \), then \( x = 7^2 = 49 \). Since 49 is not in A (A goes up to 45), we stop here.
So, the relation R as a subset of \( A \times A \) is:
\( R = \{(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), (36, 6)\} \)
The domain of R is the set of all first elements in the ordered pairs of R.
Domain of \( R = \{1, 4, 9, 16, 25, 36\} \)
The range of R is the set of all second elements in the ordered pairs of R.
Range of \( R = \{1, 2, 3, 4, 5, 6\} \)
In simple words: We are looking for pairs of numbers where the first number is the square of the second number, and both numbers must be from the set A (numbers from 1 to 45). The domain is all the first numbers we found, and the range is all the second numbers.

🎯 Exam Tip: When defining a relation like "is square of" on a set, always ensure both elements of each ordered pair belong to the given set, and check the limits carefully to avoid including numbers outside the set.

Question 3. A Relation R is given by the set \(\{ (x, y)/y = x + 3, x \in \{0, 1, 2, 3, 4, 5\}\} \). Determine its domain and range.
Answer: The relation R is defined by \( y = x + 3 \), where \( x \) belongs to the set \( \{0, 1, 2, 3, 4, 5\} \).
We need to find the corresponding \( y \) values for each \( x \) value.
When \( x = 0 \),
\( \implies y = 0 + 3 = 3 \)
When \( x = 1 \),
\( \implies y = 1 + 3 = 4 \)
When \( x = 2 \),
\( \implies y = 2 + 3 = 5 \)
When \( x = 3 \),
\( \implies y = 3 + 3 = 6 \)
When \( x = 4 \),
\( \implies y = 4 + 3 = 7 \)
When \( x = 5 \),
\( \implies y = 5 + 3 = 8 \)
So, the relation R in roster form (as a set of ordered pairs) is:
\( R = \{(0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8)\} \)
The domain of R is the set of all the first elements (all the \( x \) values) from the ordered pairs.
Domain of \( R = \{0, 1, 2, 3, 4, 5\} \)
The range of R is the set of all the second elements (all the \( y \) values) from the ordered pairs.
Range of \( R = \{3, 4, 5, 6, 7, 8\} \)
In simple words: We find the 'y' number for each 'x' number by adding 3 to 'x'. The 'x' numbers given are the domain, and the 'y' numbers we get are the range.

🎯 Exam Tip: Always clearly state the given set of \( x \) values and then systematically calculate each corresponding \( y \) value using the given rule to form the ordered pairs. The domain comes from the \( x \) values, and the range comes from the calculated \( y \) values.

Question 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) \( \{(x,y) | x = 2y, x \in \{2, 3, 4, 5\}, y \in \{1, 2, 3, 4\}\} \)
(ii) \( \{(x, y) | y = x + 3, x, y \text{ are natural numbers } < 10\} \)
Answer:
(i) Given relation \( \{(x,y) | x = 2y, x \in \{2, 3, 4, 5\}, y \in \{1, 2, 3, 4\}\} \).
We need to find pairs \( (x, y) \) such that \( x = 2y \), with \( x \) from \( \{2, 3, 4, 5\} \) and \( y \) from \( \{1, 2, 3, 4\} \).
Let's find the pairs by testing \( y \) values:
When \( y = 1 \),
\( \implies x = 2 \times 1 = 2 \). This pair \( (2, 1) \) is valid as \( 2 \in \{2, 3, 4, 5\} \) and \( 1 \in \{1, 2, 3, 4\} \).
When \( y = 2 \),
\( \implies x = 2 \times 2 = 4 \). This pair \( (4, 2) \) is valid as \( 4 \in \{2, 3, 4, 5\} \) and \( 2 \in \{1, 2, 3, 4\} \).
When \( y = 3 \),
\( \implies x = 2 \times 3 = 6 \). This pair \( (6, 3) \) is not valid as \( 6 \notin \{2, 3, 4, 5\} \).
When \( y = 4 \),
\( \implies x = 2 \times 4 = 8 \). This pair \( (8, 4) \) is not valid as \( 8 \notin \{2, 3, 4, 5\} \).

(c) **Roster form:** Based on our calculations, the relation R is:
\( R = \{(2, 1), (4, 2)\} \)

(a) **Arrow diagram:**


(b) **Graph:**


(ii) Given relation \( \{(x, y) | y = x + 3, x, y \text{ are natural numbers } < 10\} \)
Natural numbers less than 10 are \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \).
So, \( x \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \) and \( y \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \).
We need to find pairs \( (x, y) \) such that \( y = x + 3 \).
When \( x = 1 \),
\( \implies y = 1 + 3 = 4 \). This pair \( (1, 4) \) is valid.
When \( x = 2 \),
\( \implies y = 2 + 3 = 5 \). This pair \( (2, 5) \) is valid.
When \( x = 3 \),
\( \implies y = 3 + 3 = 6 \). This pair \( (3, 6) \) is valid.
When \( x = 4 \),
\( \implies y = 4 + 3 = 7 \). This pair \( (4, 7) \) is valid.
When \( x = 5 \),
\( \implies y = 5 + 3 = 8 \). This pair \( (5, 8) \) is valid.
When \( x = 6 \),
\( \implies y = 6 + 3 = 9 \). This pair \( (6, 9) \) is valid.
When \( x = 7 \),
\( \implies y = 7 + 3 = 10 \). This pair \( (7, 10) \) is not valid as \( 10 \notin \{1, \dots, 9\} \).
We stop here.

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

(a) **Arrow diagram:**


(b) **Graph:**

In simple words: For the first part, we find pairs where the 'x' value is double the 'y' value, picking numbers from the given sets. For the second part, we find pairs where 'y' is 3 more than 'x', using natural numbers less than 10. We then show these pairs as a list, with arrows connecting them, and by plotting them on a graph.

🎯 Exam Tip: When constructing graphs or arrow diagrams, ensure all valid pairs are represented accurately. Double-check the boundaries for \( x \) and \( y \) values to avoid including invalid pairs.

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 Rs.10,000, Rs.25,000, Rs.50,000 and Rs.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: Let's define the set of salaries and the set of employees first.
Salaries (S): \( \{10000, 25000, 50000, 100000\} \)
Employees (P): \( \{A1, A2, A3, A4, A5, C1, C2, C3, C4, M1, M2, M3, E1, E2\} \)
The relation R is defined as xRy, meaning x is the salary given to person y.
**Relations based on salary categories:**
Assistants (A) get Rs.10,000. There are 5 assistants: A1, A2, A3, A4, A5.
Clerks (C) get Rs.25,000. There are 4 clerks: C1, C2, C3, C4.
Managers (M) get Rs.50,000. There are 3 managers: M1, M2, M3.
Executive Officers (E) get Rs.1,00,000. There are 2 executive officers: E1, E2.

**R as a set of ordered pairs (salary, person):**
\( R = \{ \)
\( (10000, A1), (10000, A2), (10000, A3), (10000, A4), (10000, A5), \)
\( (25000, C1), (25000, C2), (25000, C3), (25000, C4), \)
\( (50000, M1), (50000, M2), (50000, M3), \)
\( (100000, E1), (100000, E2) \)
\( \} \)

(a) **Arrow diagram:**

In simple words: The company pays different salaries to different types of workers. We list all the pairs showing which salary goes to which specific worker. Then, we draw an arrow diagram to visually link each salary amount to all the employees who receive that amount.

🎯 Exam Tip: For relations involving categories and individuals, ensure every individual is linked to their correct category/value. Carefully list all ordered pairs to avoid missing any relationships and maintain clarity in the arrow diagram.