Exercise 1.2 - Chapter 1 Relations & Functions 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.2$
Question 1.
Let $\mathrm{A}=\{1,2,3,7\}$ and $\mathrm{B}=\{3,0,-1,7\}$, which of the following are relation from $\mathrm{A}$ to $\mathrm{B}$ ?
(i) $\mathrm{R}_{1}=\{(2,1),(7,1)\}$
(ii) $\mathrm{R}_{2}=\{(-1,1)\}$
(iii) $R_{3}=\{(2,-1),(7,7),(1,3)\}$
(iv) $R_{4}=\{(7,-1),(0,3),(3,3),(0,7)\}$
(i) $\mathrm{A}=\{1,2,3,7\}, \mathrm{B}=\{3,0,-1,7\}$
Solution:
It is not a relation there is no element as 1 in $B$.
(ii) $R_{2}=\{(-1,1)\}$
It is not $[\because-1 \notin \mathrm{A}, 1 \notin \mathrm{B}]$
(iii) $\mathrm{R}_{3}=\{(2,-1),(7,7),(1,3)\}$
It is a relation.
$\mathrm{R}_{4}=\{(7,-1),(0,3),(3,3),(0,7)\}$
It is also not a relation. $[\because 0 \notin \mathrm{A}]$

Question 2 .
Let $\mathrm{A}=\{1,2,3,4, \ldots \ldots, 45\}$ and $\mathrm{R}$ be the relation defined as "is square of" on A. Write $\mathrm{R}$ as a subset of $A \times A$. Also, find the domain and range of $R$.
Answer:
$A=\{1,2,3,4 \ldots 45\}$
The relation is defined as "is square of"
$\mathrm{R}=\{(1,1)(2,4)(3,9)$
$(4,16)(5,25)(6,36)\}$
Domain of $R=\{1,2,3,4,5,6\}$
Range of $R=\{1,4,9,16,25,36\}$
 

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.
Solution:
$\begin{aligned}
&x=\{0,1,2,3,4,5\} \\
&y=x+3
\end{aligned}$
i.e. $y=\left\{\begin{array}{l}(0+3)=3 \\ (1+3)=4 \\ (2+3)=5 \\ (3+3)=6 \\ (4+3)=7 \\ (5+3)=8\end{array}\right\}$ $\Rightarrow y=\{3,4,5,6,7,8\}$ $R=\{(x, y)\}$ $-\{(0,3),(1,4),(2,5),(3,6),(4,7),(5,8)\}$ Domain of $R=\{0,1,2,3,4,5\}$ Range of $R=\{3,4,5,6,7,8\}$
Range of $R=\{3,4,5,6,7,8\}$

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) $\{(x, y) \mid x=2 y, x \in\{2,3,4,5\}, y \in\{1,2,3,4)$
(ii) $\{(x, y) \mid y=x+3, x, y$ are natural numbers $<10\}$
Solution:
(i) $\{(x, y) \mid x=2 y, x \in\{2,3,4,5\}, y \in\{1,2,3,4\}\} R=(x=2 y)$
$2-2 \times 1-2$

4 = 2 × 2 = 4

(c) $\{(2,1),(4,2)\}$
(ii) $\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{y}=\mathrm{x}+3, \mathrm{x},+$ are natural numbers $<10\}$
$\mathrm{x}=\{1,2,3,4,5,6,7,8,9\} \mathrm{R}=(\mathrm{y}=\mathrm{x}+3)$
$\mathrm{y}=\{1,2,3,4,5,6,7,8,9\}$
$R=\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$

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 $\square 10,000, \square 25,000, \square 50,000$ and $\square 1,00,000$ as salaries to the people who work in the categories $\mathrm{A}, \mathrm{C}, \mathrm{M}$ and $\mathrm{E}$ respectively. If $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}$,
$\mathrm{A}_{4}$ and $\mathrm{As}$ were Assistants; $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}$ were Clerks; $\mathrm{M}_{1}, \mathrm{M}_{2}, \mathrm{M}_{3}$ were managers and $E_{1}, E_{2}$ were Executive officers and if the relation $R$ is defined by $x R y$, where $x$ is the salary given to person $\mathrm{y}$, express the relation $\mathrm{R}$ through an ordered pair and an arrow diagram.
Solution:
$\mathrm{A}$ - Assistants $\rightarrow \mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \mathrm{~A}_{4}, \mathrm{~A}_{5}$
$\mathrm{C}$ - Clerks $\rightarrow \mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}$
$\mathrm{D}$ - Managers $\rightarrow \mathrm{M}_{1}, \mathrm{M}_{2}, \mathrm{M}_{3}$
$\mathrm{E}$ - Executive officer $\rightarrow \mathrm{E}_{1}, \mathrm{E}_{2}$
(a) $\mathrm{R}=\left\{\left(10,000, \mathrm{~A}_{1}\right),\left(10,000, \mathrm{~A}_{2}\right),\left(10,000, \mathrm{~A}_{3}\right)\right.$,
$\left(10,000, \mathrm{~A}_{4}\right),\left(10,000, \mathrm{~A}_{5}\right),\left(25,000, \mathrm{C}_{1}\right)$,
$\left(25,000, \mathrm{C}_{2}\right),\left(25,000, \mathrm{C}_{3}\right),\left(25,000, \mathrm{C}_{4}\right)$,
$\left(50,000, \mathrm{M}_{1}\right),\left(50,000, \mathrm{M}_{2}\right),\left(50,000, \mathrm{M}_{3}\right)$,
$\left.\left(1,00,000, \mathrm{E}_{1}\right),\left(1,00,000, \mathrm{E}_{2}\right)\right\}$