Exercise 1.1 - Chapter 1 Relations & Functions 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex 1.1 : Chapter 1 - Relations & Functions - 10th Maths Guide Samacheer Kalvi Solutions
Question $1 .$
Find $\mathrm{A} \times \mathrm{B}, \mathrm{A} \times \mathrm{A}$ and $\mathrm{B} \times \mathrm{A}$
(i) $\mathrm{A}=\{2,-2,3\}$ and $\mathrm{B}=\{1,-4\}$
(ii) $\mathrm{A}=\mathrm{B}=\{\mathrm{p}, \mathrm{q}]$
(iii) $\mathrm{A}=\{\mathrm{m}, \mathrm{n}\} ; \mathrm{B}=(\Phi)$
Solution:
(i) $\mathrm{A}=\{2,-2,3\}, \mathrm{B}=\{1,-4\}$
$A \times B=\{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\}$
$\mathrm{A} \times \mathrm{A}=\{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\}$
$\mathrm{B} \times \mathrm{A}=\{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\}$
(ii) $\mathrm{A}=\mathrm{B}=\{(\mathrm{p}, \mathrm{q})]$
$A \times B=\{(p, p),\{p, q),(q, p),(q, q)\}$
$A \times A=\{(p, p),(p, q),(q, p),(q, q)\}$
$B \times A=\{(p, p),\{p, q),(q, p),(q, q)\}$
(iii) $\mathrm{A}=\{\mathrm{m}, \mathrm{n}\} \times \Phi$
$\mathrm{A} \times \mathrm{B}=\{\}$
$A \times A=\{(m, m),(m, n),(n, m),(n, n)\}$
$\mathrm{B} \times \mathrm{A}=\{\}$

Question 2.
Let $\mathrm{A}=\{1,2,3\}$ and $\mathrm{B}=\{\times \mid \mathrm{x}$ is a prime number less than 10$\}$. Find $\mathrm{A} \times \mathrm{B}$ and $\mathrm{B} \times \mathrm{A}$.
Answer:
$\mathrm{A}=\{1,2,3\}, \mathrm{B}=\{2,3,5,7\}$
$\mathrm{A} \times \mathrm{B}=\{1,2,3\} \times\{2,3,5,7\}$
$=\{(1,2)(1,3)(1,5)(1,7)(2,2)$
$(2,3)(2,5)(2,7)(3,2)(3,3)(3,5)(3,7)\}$
$\mathrm{B} \times \mathrm{A}=\{2,3,5,7\} \times\{1,2,3\}$
$=\{(2,1)(2,2)(2,3)(3,1)(3,2)(3,3)(5,1)(5,2)(5,3)(7,1)(7,2)(7,3)\}$
 

Question 3.
If $\mathbf{B} \times \mathbf{A}=\{(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)\}$ find $A$ and $B$.

Solution:
$\begin{aligned}
&\mathrm{B} \times \mathrm{A}=\{(-2,3),(-2,4),(0,3),(0,4),(3,3),(3,4)\} \\
&\mathrm{A}=\{3,4), \mathrm{B}=\{-2,0,3\}
\end{aligned}$
 

Question $4 .$
If $\mathrm{A}=\{5,6\}, \mathrm{B}=\{4,5,6\}, \mathrm{C}=\{5,6,7\}$, Show that $\mathrm{A} \times \mathrm{A}=(\mathrm{B} \times \mathrm{B}) \cap(\mathrm{C} \times \mathrm{C})$
Answer:
$\mathrm{A}=\{5,6\}, \mathrm{B}=\{4,5,6\}, \mathrm{C}=\{5,6,7\}$ $\mathrm{A} \times \mathrm{A}=\{5,6\} \times\{5,6\}$ $=\{(5,5)(5,6)(6,5)(6,6)\} \ldots(1)$ $\mathrm{B} \times \mathrm{B}=\{4,5,6\} \times\{4,5,6\}$ $=\{(4,4)(4,5)(4,6)(5,4)(5,5)(5,6)(6,4)(6,5)(6,6)\}$ $\mathrm{C} \times \mathrm{C}=\{5,6,7\} \times\{5,6,7\}$ $=\{(5,5)(5,6)(5,7)(6,5)(6,6)(6,7)(7,5)(7,6)(7,7)\}$ $(\mathrm{B} \times \mathrm{B}) \cap(\mathrm{C} \times \mathrm{C})=\{(5,5)(5,6)(6,5)(6,6)\} \ldots .(2)$ From $(1)$ and $(2)$ we get $\mathrm{A} \times \mathrm{A}=(\mathrm{B} \times \mathrm{B}) \cap(\mathrm{C} \times \mathrm{C})$
 

Question $5 .$
Given $A=\{1,2,3\}, B=\{2,3,5\}, C=\{3,4\}$ and $D=\{1,3,5\}$, check if $(A \cap C) x(B \cap D)=(A \times$
B) $\cap(\mathrm{C} \times \mathrm{D})$ is true?
Solution:
LHS $=\{(\mathrm{A} \cap \mathrm{C}) \times(\mathrm{B} \cap \mathrm{D})$
$A \cap C=\{3\}$
$\mathrm{B} \cap \mathrm{D}=\{3,5\}$
$(A \cap C) \times(B \cap D)=\{(3,3)(3,5)\}$
RHS $=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{C} \times \mathrm{D})$
$\mathrm{A} \times \mathrm{B}=\{(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)\}$
$\mathrm{C} \times \mathrm{D}=\{(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)\}$
$(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{C} \times \mathrm{D})=\{(3,3),(3,5)\} \ldots(2)$
$\therefore(1)=(2) \therefore$ It is true.

Question 6.
Let $A=\{x \in W \mid x<2\}$, $B=\{x \in N \mid 1<1 (i) $\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})$
(ii) $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$
(iii) $(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=(\mathrm{A} \times \mathrm{C}) \cup(\mathrm{B} \times \mathrm{C})$
Answer:
(i) $\mathrm{A}=\{0,1\}$
$\mathrm{B}=\{2,3,4\}$
$\mathrm{C}=\{3,5\}$

$\begin{aligned}
& \text { (i) } \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{c}) \\
& \mathrm{B} \cup \mathrm{C}=\{2,3,4\} \cup\{3,5\} \\
& =\{2,3,4,5\} \\
& \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=\{0,1\} \times\{2,3,4,5\} \\
& =\{(0,2)(0,3)(0,4)(0,5)(1,2)(1,3)(1,4)(1,5)\} \ldots(1) \\
& \mathrm{A} \times \mathrm{B}=\{0,1\} \times\{2,3,4\} \\
& =\{(0,2)(0,3)(0,4)(1,2)(1,3)(1,4)\} \\
& \mathrm{A} \times \mathrm{C}=\{0,1\} \times\{3,5\} \\
& \{(0,3)(0,5)(1,3)(1,5)\} \\
& (\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})=\{(0,2)(0,3)(0,4)(0,5)(1,2)(1,3)(1,4)(1,5)\} \\
& \text { From }(1) \text { and }(2) \text { we get } \\
& \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})
\end{aligned}$

$\begin{aligned}
& \text { (ii) } \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \mathrm{n}(\mathrm{A} \times \mathrm{C}) \\
& \mathrm{B} \cap \mathrm{C}=\{2,3,4\} \cap\{3,5\} \\
& =\{3\} \\
& \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=\{0,1\} \times\{3\} \\
& =\{(0,3)(1,3)\} \ldots(1) \\
& \mathrm{A} \times \mathrm{B}=\{0,1\} \times\{2,3,4\} \\
& =\{(0,2)(0,3)(0,4)(1,2)(1,3)(1,4)\} \\
& \mathrm{A} \times \mathrm{C}=\{0,1\} \times\{3,5\} \\
& \{(0,3)(0,5)(1,3)(1,5)\} \\
& (\mathrm{A} \times \mathrm{B}) \mathrm{n}(\mathrm{A} \times \mathrm{C})=\{(0,3)(1,3)\} \ldots .(2 \\
& \text { From }(1) \text { and }(2) \text { we get } \\
& \mathrm{A} \times(\mathrm{B} \mathrm{n})=(\mathrm{A} \times \mathrm{B}) \mathrm{n}(\mathrm{A} \times \mathrm{C})
\end{aligned}$

$\begin{aligned}
& \text { (iii) }(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=(\mathrm{A} \times \mathrm{C}) \cup(\mathrm{B} \times \mathrm{C}) \\
& \mathrm{A} \cup \mathrm{B}=\{0,1\} \cup\{2,3,4\} \\
& =\{0,1,2,3,4\} \\
& (\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=\{0,1,2,3,4\} \times\{3,5\} \\
& =\{(0,3)(0,5)(1,3)(1,5)(2,3)(2,5)(3,3)(3,5)(4,3)(4,5)\} \ldots(1) \\
& \mathrm{A} \times \mathrm{C}=\{0,1\} \times\{3,5\} \\
& =\{(0,3)(0,5)(1,3)(1,5)\} \\
& \mathrm{B} \times \mathrm{C}=\{2,3,4\} \times\{3,5\} \\
& =\{(2,3)(2,5)(3,3)(3,5)(4,3)(4,5)\} \\
& (\mathrm{A} \times \mathrm{C}) \cup(\mathrm{B} \times \mathrm{C})=\{(0,3)(0,5)(1,3)(1,5)(2,3)(2,5)(3,3)(3,5)(4,3)(4,5)\} \\
& \text { From }(1) \text { and }(2) \text { we get } \\
& (\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}=(\mathrm{A} \times \mathrm{C}) \cup(\mathrm{B} \times \mathrm{C})
\end{aligned}$

Question 7.
Let $\mathrm{A}=$ The set of all natural numbers less than $8, \mathrm{~B}=$ The set of all prime numbers less than $8, \mathrm{C}$ $=$ The set of even prime number. Verify that

(i) $(\mathrm{A} \cap \mathrm{B}) \times \mathrm{c}=(\mathrm{A} \times \mathrm{C}) \cap(\mathrm{B} \times \mathrm{C})$
(ii) $\mathrm{A} \times(\mathrm{B}-\mathrm{C})=(\hat{\mathrm{A}} \times \mathrm{B})-(\mathrm{A} \times \mathrm{C})$
$A=\{1,2,3,4,5,6,7\}$
$\mathrm{B}=\{2,3,5,7\}$
$\mathrm{C}=\{2\}$
Solution:
(i) $(\mathrm{A} \cap \mathrm{B}) \times \mathrm{C}=(\mathrm{A} \times \mathrm{c}) \cap(\mathrm{B} \times \mathrm{C})$
LHS $=(A \cap B) \times C$
$A \cap B=\{2,3,5,7\}$
$(A \cap B) \times C=\{(2,2),(3,2),(5,2),(7,2)\}$
RHS $=(\mathrm{A} \times \mathrm{C}) \cap(\mathrm{B} \times \mathrm{C})$
$(A \times C)=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)\}$
$(B \times C)=\{2,2),(3,2),(5,2),(7,2)\}$
$(A \times C) \cap(B \times C)=\{(2,2),(3,2),(5,2),(7,2)\}$
$(1)=(2)$
$\therefore$ LHS $=$ RHS. Hence it is verified.
(ii) $A \times(B-C)=(A \times B)-(A \times C)$
LHS $=A \times(B-C)$
$(B-C)=\{3,5,7\}$
$\mathrm{A} \times(\mathrm{B}-\mathrm{C})=\{(1,3),(1,5),(1,7),(2,3),(2,5),(2,7),(3,3),(3,5),(3,7),(4,3),(4,5),(4,7),(5$,
3), $(5,5),(5,7),(6,3),(6,5),(6,7),(7,3),(7,5),(7,7)\} \ldots \ldots \ldots \ldots(1)$
RHS $=(A \times B)-(A \times C)$
$(A \times B)=\{(1,2),(1,3),(1,5),(1,7)$,
$(2,2),(2,3),(2,5),(2,7)$,
$(3,2),(3,3),(3,5),(3,7)$,
$(4,2),(4,3),(4,5),(4,7)$,
$(5,2),(5,3),(5,5),(5,7)$, $(6,2),(6,3),(6,5),(6,7)$,
$(6,2),(6,3),(6,5),(6,7)$
$(A \times C)=\{(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)\}$
$(\mathrm{A} \times \mathrm{B})-(\mathrm{A} \times \mathrm{C})=\{(1,3),(1,5),(1,7),(2,3),(2,5),(2,7),(3,3),(3,5),(3,7),(4,3),(4,5),(4$,
$7),(5,3),(5,5),(5,7),(6,3),(6,5),(6,7),(7,3),(7,5),(7,7)\} \ldots \ldots \ldots \ldots .(2)$
$(1)=(2) \Rightarrow$ LHS $=$ RHS.
Hence it is verified.