10 maths ex 1.1

Exercise 1.2 Solutions

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

Note: A relation \(R\) from \(A\) to \(B\) is a subset of \(A \times B\) (i.e., \(R \subseteq A \times B\)).

(i) \(R_1 = \{(2,1), (7,1)\}\)

Here, \((2,1) \notin A \times B\) and \((7,1) \notin A \times B\) because \(1 \notin B\).

So, \(R_1\) is NOT a relation from \(A\) to \(B\).

(ii) \(R_2 = \{(-1,1)\}\)

Here, \(-1 \notin A\), so \((-1,1) \notin A \times B\).

So, \(R_2\) is NOT a relation from \(A\) to \(B\).

(iii) \(R_3 = \{(2,-1), (7,7), (1,3)\}\)

Here, \((2,-1) \in A \times B\), \((7,7) \in A \times B\), and \((1,3) \in A \times B\).

Since \(R_3 \subseteq A \times B\),

So, \(R_3\) IS a relation from \(A\) to \(B\).

(iv) \(R_4 = \{(7,-1), (0,3), (3,3), (0,7)\}\)

Here, \((0,3) \notin A \times B\) and \((0,7) \notin A \times B\) because \(0 \notin A\).

So, \(R_4\) is NOT a relation from \(A\) to \(B\).

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

Solution:

Relation \(R = \{(x, y) \mid y = x^2 \text{ and } x, y \in A\}\)

Possible values of \(x\) whose squares are \(\le 45\):

• \(1^2 = 1 \in A\)
• \(2^2 = 4 \in A\)
• \(3^2 = 9 \in A\)
• \(4^2 = 16 \in A\)
• \(5^2 = 25 \in A\)
• \(6^2 = 36 \in A\)

Roster form:

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

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 \(\{(x,y) \mid y=x+3, x \in \{0,1,2,3,4,5\}\}\). Determine its domain and range.

Solution:

Given \(y = x + 3\):

• If \(x = 0 \implies y = 0 + 3 = 3\)
• If \(x = 1 \implies y = 1 + 3 = 4\)
• If \(x = 2 \implies y = 2 + 3 = 5\)
• If \(x = 3 \implies y = 3 + 3 = 6\)
• If \(x = 4 \implies y = 4 + 3 = 7\)
• If \(x = 5 \implies y = 5 + 3 = 8\)

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

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) \(\{(x,y) \mid x = 2y, x \in \{2,3,4,5\}, y \in \{1,2,3,4\}\}\)

Since \(x = 2y \implies y = \frac{x}{2}\):

• If \(x = 2 \implies y = 1 \in y\)
• If \(x = 4 \implies y = 2 \in y\)

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

(a) Arrow Diagram: Represent sets \(X=\{2,3,4,5\}\) and \(Y=\{1,2,3,4\}\) with arrows from \(2 \to 1\) and \(4 \to 2\).

(b) Graph: Plot points \((2,1)\) and \((4,2)\) on an X-Y axis.

(ii) \(\{(x,y) \mid y = x+3, x, y \text{ are natural numbers } < 10\}\)

Natural numbers \(< 10\) are \(\{1,2,3,4,5,6,7,8,9\}\):

• \(x=1 \implies y=4\)
• \(x=2 \implies y=5\)
• \(x=3 \implies y=6\)
• \(x=4 \implies y=7\)
• \(x=5 \implies y=8\)
• \(x=6 \implies y=9\)

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

(a) Arrow Diagram: Draw arrows from \(x \to x+3\) for the pairs above.

(b) Graph: Plot the points \((1,4), (2,5), (3,6), (4,7), (5,8), (6,9)\).

5. A company has four categories of employees...

Given:

• Assistants (\(A\)): \(A_1, A_2, A_3, A_4, A_5\) → Salary: ₹10,000
• Clerks (\(C\)): \(C_1, C_2, C_3, C_4\) → Salary: ₹25,000
• Managers (\(M\)): \(M_1, M_2, M_3\) → Salary: ₹50,000
• Executive Officers (\(E\)): \(E_1, E_2\) → Salary: ₹1,000,000

Relation \(R\) is defined by \(xRy\), where \(x\) is Salary and \(y\) is Person.

(a) Ordered Pair:

\(R = \{(10000, A_1), (10000, A_2), (10000, A_3), (10000, A_4), (10000, A_5), (25000, C_1), (25000, C_2), (25000, C_3), (25000, C_4), (50000, M_1), (50000, M_2), (50000, M_3), (100000, E_1), (100000, E_2)\}\)

(b) Arrow Diagram:

Set 1 (Salaries): \(\{10000, 25000, 50000, 100000\}\)
Set 2 (Employees): \(\{A_1..A_5, C_1..C_4, M_1..M_3, E_1, E_2\}\)
Map each salary to its respective employees with arrows.