Exercise 1.1 - Chapter 1 - Sets, Relations and Functions - 11th Maths Guide Samacheer Kalvi Solutions

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Sets, Relations and Functions
EX 1.1
Question 1.
Write the following in roster form.
(i) $\left\{\mathrm{x} \in \mathrm{N}: \mathrm{x}^2<121\right.$ and $\mathrm{x}$ is a prime $\}$.
(ii) the set of all positive roots of the equation $(x-1)(x+1)\left(x^2-1\right)=0$.
(iii) $\{\mathrm{x} \in \mathrm{N}: 4 \mathrm{x}+9<52\}$.
(iv) $\left\{\mathrm{x}: \frac{x-4}{x+2}=3, \mathrm{x} \in \mathrm{R}-\{-2\}\right\}$
Solution:
(i) $\mathrm{A}=\{2,3,5,7\}$
(ii) $\mathrm{B}=\{1\}$
(iii) $4 \mathrm{x}+9<52$
$4 \mathrm{x}+9-9<52-9$
$4 \mathrm{x}<43$
$x<\frac{43}{4}$ (i.e.) $x<10.754$
But $\mathrm{x} \in \mathrm{N}$
$
\therefore \mathrm{A}=\{1,2,3,4,5,6,7,8,9,10\}
$
(iv) $\left\{x: \frac{x-4}{x+2}=3, x \in \mathrm{R}-\{-2\}\right\}$.
$
\begin{aligned}
& \frac{x-4}{x+2}=3 \\
& \frac{(x-4)(x+2)}{x+2}=3(x+2)\{\therefore x \neq-2\} \\
&
\end{aligned}
$
Question 2.
Write the set $\{-1,1\}$ in set builder form.
Solution:
$
A=\left\{x: x^2=1\right\}
$

Question 3.
State whether the following sets are finite or infinite.
(i) $\{\mathrm{x} \in \mathrm{N}: \mathrm{x}$ is an even prime number $\}$
(ii) $\{\mathrm{x} \in \mathrm{N}: \mathrm{x}$ is an odd prime number $\}$
(iii) $\{\mathrm{x} \in \mathrm{Z}: \mathrm{x}$ is even and less than 10$\}$
(iv) $\{\mathrm{x} \in \mathrm{R}: \mathrm{x}$ is a rational number $\}$
(v) $\{\mathrm{x} \in \mathrm{N}: \mathrm{x}$ is a rational number $\}$
Solution:
(i) Finite set
(ii) Infinite set
(iii) Infinite
(iv) and
(v) infinite
Question 4.
By taking suitable sets A, B, C, verify the following results:
(i) $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$.
(if) $A \times(B \cup C)=(A \times B) \cup(A \times C)$.
(iii) $(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})=(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})$.
(iv) $C-(B-A)=(C \cap A) \cup(C \cap B)$.
(v) $(\mathrm{B}-\mathrm{A}) \cap \mathrm{C}=(\mathrm{B} \cap \mathrm{C})-\mathrm{A}=\mathrm{B} \cap(\mathrm{C}-\mathrm{A})$.

(vi) To prove $(\mathrm{B}-\mathrm{A}) \cup \mathrm{C}=\{1,5,8,9,10\}$
Solution:
To prove the following results let us take $\mathrm{U}=\{1,2,5,7,8,9,10\}$
$A=\{1,2,5,7\}$
$\mathrm{B}=\{2,7,8,9\}$
$\mathrm{C}=\{1,5,8,7\}$
(i) To prove: $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$
$\mathrm{B} \cap \mathrm{C}=\{8\} ; \mathrm{A}=\{1,2,5,7\}$
So $\mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=\{1,2,5,7\} \times\{8\}$
$=\{(1,8),(2.8),(5,8),(7,8)\}$
Now A $\times$ B $=\{(1,2),(1,7),(1,8),(1,9),(2,2),(2,7),(2,8),(2,9),(5,2),(5,7),(5,8)$, $(5,9),(7,2),(7,7),(7,8),(7,9)\} \ldots(1)$
$\mathrm{A} \times \mathrm{C}=\{(1,1),(1,5),(1,8),(1,10),(2,1),(2,5),(2,8),(2,10),(5,1),(5,5),(5,8),(5$, 10), $(7,1),(7,5),(7,8),(7,10)\}$
$(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})=\{(1,8),(2,8),(5,8),(7,8)\}$
(1) $=(2)$
$\Rightarrow \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})$

$
\begin{aligned}
& \text { (ii) To prove } \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \times \mathrm{B})(\mathrm{A} \times \mathrm{C}) \\
& \mathrm{B}=\{2,7,8,9\}, \mathrm{C}=\{1,5,8,10) \\
& \mathrm{B} \cup \mathrm{C}=\{1,2,5,7,8,9,10\} \\
& \mathrm{A}=\{1,2,5,7\}
\end{aligned}
$
$\mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=\{(1,1),(1,2),(1,5),(1,7),(1,8),(1,9),(1,10),(2,1),(2,2),(2,5),(2$,
$7),(2,8),(2,9),(2,10),(5,1),(5,2),(5,5),(5,7),(5,8),(5,9),(5,10),(7,1),(7,2)$,
$(7,5),(7,7),(7,8),(7,9),(7,10)) \ldots(1)$
$\mathrm{A} \times \mathrm{B}=\{(1,2),(1,7),(1,8),(1,9),(2,2),(2,7),(2,8),(2,9),(5,2),(5,7),(5,8),(5$,
$9)$,
$(7,2),(7,7),(7,8),(7,9)\}$
$\mathrm{A} \times \mathrm{C}=\{(1,1),(1,5),(1,8),(1,10),(2,1),(2,5),(2,8),(2,10),(5,1),(5,5),(5,8)$,
$(5,10),(7,1),(7,5),(7,8),(7,10)\}$
$(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})=(1,1),(1,2),(1,5),(1,7),(1,8),(1,9),(1,10),(2,1),(2,2),(2,5)$,
$(2,7),(2,8),(2,9),(2,10),(5,1),(5,2),(5,5),(5,7),(5,8),(5,9),(5,10),(7,1),(7$,
$2),(7,5),(7,7),(7,8),(7,9),(7,10)\} \ldots \ldots(2)$
$(1)=(2) \Rightarrow \mathrm{A} \times(\mathrm{B} \cup \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})$
(iii) $\mathrm{A} \times \mathrm{B}=\{(1,2),(1,7),(1,8),(1,9)(2,2),(2,7),(2,8),(2,9)(5,2),(5,7),(5,8)$, $(5,9)(7,2),(7,7),(7,8),(7,9)\}$ $\mathrm{B} \times \mathrm{A}=\{(2,1),(2,2),(2,5),(2,7)(7,1),(7,2),(7,5),(7,7)(8,1),(8,2),(8,5),(8,7)$ $(9,1),(9,2),(9,5),(9,7)\}$
L.H.S. $(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A})=\{(2,2),(2,7),(7,2),(7,7)\} \ldots(1)$
R.H.S. $A \cap B=\{2,7\}$
$B \cap A=\{2,7\}$
$(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{B} \cap \mathrm{A})=\{2,7\} \times\{2,7\}$
$=\{(2,2),(2,7),(7,2),(7,7)\}$
(1) $=(2) \Rightarrow$ LHS $=$ RHS

(iv) To prove $\mathrm{C}-(\mathrm{B}-\mathrm{A})=(\mathrm{C} \cap \mathrm{A}) \cup(\mathrm{C} \cap \mathrm{B}$
$\mathrm{B}-\mathrm{A}=\{8,9\}$
$\mathrm{C}=\{1,5,8,10\}$
$\therefore \mathrm{LHS}=\mathrm{C}-(\mathrm{B}-\mathrm{A})=\{1,5,10\} \ldots \ldots(1)$
$\mathrm{C} \cap \mathrm{A}=\{1\}$
$\mathrm{U}=\{1,2,5,7,8,9,10\}$
$\mathrm{B}=\{2,7,8,9\} \therefore \mathrm{B}^{\prime}=\{1,5,10\}$
$\mathrm{C} \cap \mathrm{B}=\{1,5,10\}$
R.H.S. $(\mathrm{C} \cap \mathrm{A}) \cup(\mathrm{C} \cap \mathrm{B})=\{1\} \cup\{1,5,10\}$
$=\{1,5,10\} \ldots \ldots(2)$
$(1)=(2) \Rightarrow \mathrm{LHS}=\mathrm{RHS}$
(v) To prove $(\mathrm{B}-\mathrm{A}) \cap \mathrm{C}=(\mathrm{B} \cap \mathrm{C})-\mathrm{A}=\mathrm{B} \cap(\mathrm{C}-\mathrm{A})$
$\mathrm{A}=\{1,2,5,7\}, \mathrm{B}=\{2,7,8,9\}, \mathrm{C}=\{1,5,8,10\}$
Now $\mathrm{B}-\mathrm{A}=\{8,9\}$
$(\mathrm{B}-\mathrm{A}) \cap \mathrm{C}=\{8\}$
$\mathrm{B} \cap \mathrm{C}=\{8\}$
$\mathrm{A}=\{1,2,5,7\}$
So $(\mathrm{B} \cap \mathrm{C})-\mathrm{A}=\{8\}$
$\mathrm{C}-\mathrm{A}=\{8,10\}$
$\mathrm{B}=\{2,7,8,9\}$
$\mathrm{B} \cap(\mathrm{C}-\mathrm{A})=\{8\} \ldots(3)$
$(1)=(2)=(3)$

(vi) To prove $(\mathrm{B}-\mathrm{A}) \cup \mathrm{C}=\{1,5,8,9,10\}$
$
\begin{aligned}
& \mathrm{B}-\mathrm{A}=\{8,9\} \\
& \mathrm{C}=\{1,5,8,10\} \\
& (\mathrm{B}-\mathrm{A}) \cup \mathrm{C}=\{1,5,8,9,10\} \ldots \ldots(1) \\
& \mathrm{B} \cup \mathrm{C}=\{1,2,5,7,8,9,10\} \\
& \mathrm{A}-\mathrm{C}=\{2,7\} \\
& (\mathrm{B} \cup \mathrm{C})-(\mathrm{A}-\mathrm{C})=\{1,5,8,9,10\} \ldots \ldots \ldots(2) \\
& (1)=(2) \\
& \Rightarrow(\mathrm{B}-\mathrm{A}) \cup \mathrm{C}=(\mathrm{B} \cup \mathrm{C})-(\mathrm{A}-\mathrm{C})
\end{aligned}
$
Question 5.
Justify the trueness of the statement.
"An element of a set can never be a subset of itself."
Solution:
A set itself can be a subset of itself (i.e.) $\mathrm{A} \subseteq \mathrm{A}$. But it cannot be a proper subset.
Question 6.
If $\mathrm{n}(\mathrm{P}(\mathrm{A}))=1024, \mathrm{n}(\mathrm{A} \cup \mathrm{B})=15$ and $\mathrm{n}(\mathrm{P}(\mathrm{B}))=32$, then find $\mathrm{n}(\mathrm{A} \cap \mathrm{B})$.
Solution:
$
\begin{aligned}
& \mathrm{n}(\mathrm{P}(\mathrm{A}))=1024=2^{10} \Rightarrow \mathrm{n}(\mathrm{A})=10 \\
& \mathrm{n}(\mathrm{A} \cup \mathrm{B})=15 \\
& \mathrm{n}(\mathrm{P}(\mathrm{B}))=32=2^5 \Rightarrow \mathrm{n}(\mathrm{B})=5
\end{aligned}
$
We know $\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}\{\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
(i.e.) $15=10+5-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
$
\Rightarrow \mathrm{n}(\mathrm{A} \cap \mathrm{B})=15-15=0
$
Question 7.
If $n(A \cap B)=3$ and $n(A \cup B)=10$, then find $n(P(A(A \Delta B))$.
Solution:
$
\begin{aligned}
& \mathrm{n}(\mathrm{A} \cup \mathrm{B})=10 ; \mathrm{n}(\mathrm{A} \cap \mathrm{B})=3 \\
& \mathrm{n}(\mathrm{A} \Delta \mathrm{B})=10-3=7 \\
& \text { and } \mathrm{n}(\mathrm{P}(\mathrm{A} \Delta \mathrm{B}))=27=128
\end{aligned}
$
Question 8.
For a set $A, A \times A$ contains 16 elements and two of its elements are $(1,3)$ and $(0,2)$.

Find the elements of $\mathrm{A}$.
Solution:
$
\begin{aligned}
& A \times A=16 \text { elements }=4 \times 4 \\
& \Rightarrow A \text { has } 4 \text { elements } \\
& \therefore A=\{0,1,2,3\}
\end{aligned}
$
Question 9.
Let $A$ and $B$ be two sets such that $n(A)=3$ and $n(B)=2$. If $(x, 1),(y, 2),(\mathrm{z}, 1)$ are in $A$ $\times \mathrm{B}$, find $\mathrm{A}$ and $\mathrm{B}$, where $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are distinct elements.
Solution:
$\mathrm{n}(\mathrm{A})=3 \Rightarrow$ set $\mathrm{A}$ contains 3 elements
$\mathrm{n}(\mathrm{B})=2 \Rightarrow$ set $\mathrm{B}$ contains 2 elements -
we are given $(\mathrm{x}, 1),(\mathrm{y}, 2),(\mathrm{z}, 1)$ are elements in $\mathrm{A} \times \mathrm{B} \Rightarrow \mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}$ and $\mathrm{B}=\{1,2\}$
Question 10.
If $\mathrm{A} \times \mathrm{A}$ has 16 elements, $\mathrm{S}=\{(\mathrm{a}, \mathrm{b}) \in \mathrm{A} \times \mathrm{A}: \mathrm{a}<\mathrm{b}\} ;(-1,2)$ and $(0,1)$ are two elements of $\mathrm{S}$, then find the remaining elements of $\mathrm{S}$.
Solution:
$\mathrm{n}(\mathrm{A} \times \mathrm{A})=16 \Rightarrow \mathrm{n}(\mathrm{A})=4$
$\mathrm{S}=\{(-1,0),(-1,1),(0,2),(1,2)\}$