Exercise 1.5 - Chapter 1 Set Language 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.5$
Question 1.
Using the adjacent venn diagram, find the following sets:
(i) $\mathrm{A}-\mathrm{B}$
(ii) $\mathrm{B}-\mathrm{C}$
(iii) $A^{\prime} \cup B^{\prime}$
(vi) $\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$
(v) $(B \cup C)^{\prime}$
(vi) $A-(B \cup C)$
(vii) $\mathrm{A}-(\mathrm{B} \cap \mathrm{C})$

Solution:
(i) $\mathrm{A}-\mathrm{B}=\{3,4,6\}$
(ii) $\mathrm{B}-\mathrm{C}=\{-1,5,7\}$
(iii) $A^{\prime} \cup B^{\prime}$
$\begin{aligned}
&\mathrm{A}^{\prime}=\{1,2,0,-3,5,7,8\} \\
&\mathrm{B}^{\prime}=\{-3,0,1,2,3,4,6) \\
&\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{-3,0,1,2,3,4,5,6,7,8)
\end{aligned}$
(iv) A' $\cap$ B'
$A^{\prime} \cap B^{\prime}=\{-3,0,1,2\}$
(v) $\mathrm{B} \cup \mathrm{C}=\{-3,-2,-1,0,3,5,7,8\}$
$(B \cup C)^{\prime}=U-(B \cup C)$
$=\{-3,-2,-1,0,1,2,3,4,5,6,7,8\}-\{-3,-2,-1,0,3,5,7,8\}$
$(\mathrm{B} \cup \mathrm{C})^{\prime}=\{1,2,4,6\}$
(vi) $\mathrm{A}-(\mathrm{B} \cup \mathrm{C})=\{-2,-1,3,4,6\}-\{-3,-2,-1,0,3,5,7,8\}=\{4,6\}$
$\mathrm{A}-(\mathrm{B} \cap \mathrm{C})$
$\begin{aligned}
&B \cap C=\{-2,8\} \\
&A-(B \cap C)=\{-2,-1,3,4,6\}-\{-2,8\}=\{-1,3,4,6\}
\end{aligned}$

Question $2 .$
If $K=\{a, b, d, e, f\}, L=\{b, c, d, g\}$ and $M\{a, b, c, d, h\}$ then find the following:
(i) $\mathrm{K} \cup(\mathrm{L} \cap \mathrm{M})$
(ii) $\mathrm{K} \cap$ ( $\mathrm{L} \cup \mathrm{M})$
(iii) $(\mathrm{K} \cup \mathrm{L}) \cap(\mathrm{K} \cup \mathrm{M})$
(iv) $(\mathrm{K} \cap \mathrm{L}) \cup(\mathrm{K} \cap \mathrm{M})$ and verify distributive laws.
Solution:
$K=\{a, b, d, e, f\}, L=\{b, c, d, g\}$ and $M\{a, b, c, d, h\}$
(i) $\mathrm{K} \cup(\mathrm{L} \cap \mathrm{M})$
$L \cap M=\{b, c, d, g\} \cap\{a, b, c, d, h\}=\{b, c, d\}$
$K \cup(L \cap M)=\{a, b, d, e, f\} \cup\{b, c, d)=\{a, b, c, d, e, f\}$
(ii) $\mathrm{K} \cap(\mathrm{L} \cup \mathrm{M})$
$\mathrm{L} \cup \mathrm{M}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{g}, \mathrm{h}\}$
$\mathrm{K} \cap(\mathrm{L} \cup \mathrm{M})=\{\mathrm{a}, \mathrm{b}, \mathrm{d}, \mathrm{e}, \mathrm{f}\} \cap\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{g}, \mathrm{h}\}=\{\mathrm{a}, \mathrm{b}, \mathrm{d}\}$
(iii) $(\mathrm{K} \cup \mathrm{L}) \cap(\mathrm{K} \cup \mathrm{M})$
$\mathrm{K} \cup \mathrm{L}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}, \mathrm{g}\}$
$\mathrm{K} \cup \mathrm{M}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}, \mathrm{h}\}$
$(K \cup L) \cap(K \cup M)=\{a, b, c, d, e, f\}$
(iv) $(\mathrm{K} \cap \mathrm{L}) \cup(\mathrm{K} \cap \mathrm{M})$
$(\mathrm{K} \cap \mathrm{L})=\{\mathrm{b}, \mathrm{d})$
$(K \cap M)=\{a, b, d\}$
$(K \cap L) \cup(K \cap M)=\{b, d\} \cup\{a, b, d\}=\{a, b, d\}$
Distributive laws
$\mathrm{K} \cup(\mathrm{L} \cap \mathrm{M})=(\mathrm{K} \cup \mathrm{L}) \cap(\mathrm{K} \cup \mathrm{M})$
$\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f})=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}, \mathrm{g}\} \cap\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}, \mathrm{h}\}$
$=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}, \mathrm{f}\}$
Thus Verified.
$\mathrm{K} \cap(\mathrm{L} \cup \mathrm{M})=(\mathrm{K} \cap \mathrm{L}) \cup(\mathrm{K} \cap \mathrm{M})$
$\{a, b, d\}=\{a, b, c, d, e, f, g\} \cup\{a, b, c, d, e, f, h\}$
$=\{a, b, d\}$
Thus Verified.

Question $3 .$
If $\mathrm{A}=\{\mathrm{x}: \mathrm{x} \in Z,-2<\mathrm{x} \leq 4\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{W}, \mathrm{x} \leq 5\}, \mathrm{C}=\{-4,-1,0,2,3,4\}$, then verify $\mathrm{A} \cup$ $(B \cap C)=(A \cup B) \cap(A \cup C)$.
Solution:
$\begin{aligned}
&\mathrm{A}=\{\mathrm{x}: \mathrm{x} \in \mathrm{Z},-2<\mathrm{x} \leq 4\}=\{-1,0,1,2,3,4\} \\
&\mathrm{B}=\{\mathrm{x}: \mathrm{x} \in \mathrm{W}, \mathrm{x} \leq 5\}=\{0,1,2,3,4,5\} \\
&\mathrm{C}=\{-4,-1,0,2,3,4\}
\end{aligned}$
$A \cup(B \cap C)$
$\mathrm{B} \cap \mathrm{C}=\{0,1,2,3,4,5\} \cap\{-4,-1,0,2,3,4\}=\{0,2,3,4\}$
$\mathrm{A} \cup(\mathrm{B} \cap \mathrm{C})=\{-1,0,1,2,3,4\} \cup(0,2,3,4\}=\{-1,0,1,2,3,4\} \ldots \ldots \ldots \ldots$ (1)
$(A \cup B) \cap(A \cup C)$
$A \cap B=\{0,1,2,3,4\}$
$A \cap C=\{-1,0,2,3,4\}$
$(A \cap B) \cup(A \cap C)=\{0,1,2,3,4\} \cup\{-1,0,2,3,4\}=\{-1,0,1,2,3,4\}$
From (1) and (2), it is verified that
$A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$
 

Question $4 .$
Verify $A \cup(B \cap C)=(A \cup B) \cap(A \cup C)$ using Venn diagrams.
Solution:
L.H.S A U $(B \cap C)$

Question $5 .$
If $A=\{b, c, e, g, h\}, B=\{a, c, d, g, i\}$ and $C=\{a, d, e, g, h\}$, then show that $A-(B \cap C)=$ $(A-B) \cup(A-C)$.
Solution:
$\mathrm{A}=\{\mathrm{b}, \mathrm{c}, \mathrm{e}, \mathrm{g}, \mathrm{h}\}$
$B=\{a, c, d, g, i\}$
$\mathrm{C}=\{\mathrm{a}, \mathrm{d}, \mathrm{e}, \mathrm{g}, \mathrm{h}\}$
$\mathrm{B} \cap \mathrm{C}=\{\mathrm{a}, \mathrm{d}, \mathrm{g}\}$
$A-(B \cap C)=\{b, c, e, g, h\}-\{a, d, g\}=\{b, c, e, h\} \ldots \ldots . \ldots . .(1)$
$A-B=\{b, c, e, g, h\}-\{a, c, d, g, i\}=\{b, e, h\}$
$\mathrm{A}-\mathrm{C}=\{\mathrm{b}, \mathrm{c}, \mathrm{e}, \mathrm{g}, \mathrm{h}\}-\{\mathrm{a}, \mathrm{d}, \mathrm{e}, \mathrm{g}, \mathrm{h}\}=\{\mathrm{b}, \mathrm{c}\}$
$(A-B) \cup(A-C)=\{b, c, e, h\} \ldots \ldots . \ldots . . .(2)$
From (1) and (2) it is verified that
$A-(B \cap C)=(A-B) \cup(A-C)$

Question $6 .$
If $\mathrm{A}=\{\mathrm{x}: \mathrm{x}=6 \mathrm{n} \in \mathrm{W}$ and $\mathrm{n}<6\}, \mathrm{B}=\{\mathrm{x}: \mathrm{x}=2 \mathrm{n}, \mathrm{n} \in \mathrm{N}$ and $2<\mathrm{n} \leq 9\}$ and $\mathrm{C}=\{\mathrm{x}: \mathrm{x}=$ $3 \mathrm{n}, \mathrm{n} \in \mathrm{N}$ and $4 \leq \mathrm{n}<10\}$, then show that $\mathrm{A}-(\mathrm{B} \cap \mathrm{C})=(\mathrm{A}-\mathrm{B}) \cup(\mathrm{A}-\mathrm{C})$
Solution:
$A=\{x: x=6 n, n \in W, n<6\}$
$x=6 n$
$\mathrm{n}=\{0,1,2,3,4,5\}$
$\Rightarrow \mathrm{x}=6 \times 0=0$
$x=6 \times 1=6$
$x=6 \times 2=12$
$x=6 \times 3=18$
$x=6 \times 4=24$
$x=6 \times 5=30$
$\therefore \mathrm{A}=\{0,6,12,18,24,30\}$
$\mathrm{B}=\{\mathrm{x}: \mathrm{x}=2 \mathrm{n}, \mathrm{n} \in \mathrm{N}, 2<\mathrm{n} \leq 9\}$
$\mathrm{n}=\{3,4,5,6,7,8,9\}$
$\mathrm{x}=2 \mathrm{n}$
$\Rightarrow x=2 \times 3=6$
$2 \times 4=8$
$2 \times 5=10$
$2 \times 6=12$
$2 \times 7=14$
$2 \times 8=16$
$2 \times 9=18$
$\therefore \mathrm{B}\{6,8,10,12,14,16,18\}$
$C=\{x: x=3 n, n \in N, 4 \leq n<10\}$
$N=\{4,5,6,7,8,9\}$

Question 7

If $\mathrm{A}=\{-2,0,1,3,5\}, \mathrm{B}=\{-1,0,2,5,6\}$ and $\mathrm{C}=\{-1,2,5,6,7\}$, then show that $\mathrm{A}-(\mathrm{B} \cup$
$C)=(A-B) \cap(A-C)$.
Solution:
$\mathrm{A}=\{-2,0,1,3,5\}$
$\mathrm{B}=\{-1,0,2,5,6\}$
$\mathrm{C}=\{-1,2,5,6,7\}$
$\mathrm{B} \cup \mathrm{C}=\{-1,0,2,5,6,7\}$
$A-(B \cup C)=\{-2,1,3\} \ldots \ldots . . \ldots .(1)$
$(A-B)=\{-2,1,3\}$
$(A-C)=\{-2,0,1,3\}$
$(A-B) \cap(A-C)=\{-2,1,3\} \ldots . \ldots . \ldots . .(2)$
From $(1)$ and $(2)$, it is verified that . $A-(B \cup C)=(A-B) \cap(A-C)$

Question $8 .$
if $\mathrm{A}=\{\mathrm{y}: \mathrm{y}=\backslash(\operatorname{frac}\{\mathrm{a}+1\}\{2\} \backslash), \mathrm{a} \in \mathrm{W}$ and $\mathrm{a} \leq 5\}, \mathrm{B}=\{\mathrm{y}: \mathrm{y}=\backslash($ $<5\}$ and $C=\{-1,(-\mid f r a c\{1\}\{2\} \backslash), 1$, V ( frac $\{3\}\{2\}), 2\}$ then show that $A-(B \cup C)=(A-B)$ $\cap(\mathrm{A}-\mathrm{C})$.
Solution:

$\begin{aligned}
&\mathrm{A}=\left\{y: y=\frac{a+1}{2}, a \in \mathbf{W}, a \leq 5\right\}\\
&a=\{0,1,2,3,4,5\} \Rightarrow y=\frac{a+1}{2}=\frac{1}{2}\\
&y=\frac{1+1}{2}=\frac{2}{2}=1\\
&y=\frac{2+1}{2}=\frac{3}{2}\\
&y=\frac{3+1}{2}=\frac{A^{2}}{2}=2\\
&y=\frac{4+1}{2}=\frac{5}{2}\\
&y=\frac{5+1}{2}=\frac{\frac{3}{6}}{2}=3 \quad \therefore \mathrm{A}=\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3\right\}\\
&\mathrm{B}=\left\{y: y=\frac{2 n-1}{2}, n \in \mathbf{W} \text { and } n<5\right\}\\
&n=\{0,1,2,3,4\} \Rightarrow y=\frac{2 \times 0-1}{2}=\frac{-1}{2}
\end{aligned}$

$y=\frac{2 \times 1-1}{2}=\frac{1}{2}$
$y=\frac{2 \times 2-1}{2}=\frac{3}{2}$
$y=\frac{2 \times 3-1}{2}=\frac{5}{2}$
$y=\frac{2 \times 4-1}{2}=\frac{7}{2} \quad \therefore \mathrm{B}=\left\{-\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{7}{2}\right\}$
$\mathrm{C}=\left\{-1,-\frac{1}{2}, 1, \frac{3}{2}, 2\right\}$
$\mathrm{B} \cup \mathrm{C}=\left\{-1,-\frac{1}{2}, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, \frac{7}{2}\right\}$
$A-(B \cup C) .=\{3\}$
$\mathrm{A}-\mathrm{B}=\{1,2,3\}$
$\mathrm{A}-\mathrm{C}=\left\{\frac{1}{2}, \frac{5}{2}, 3\right\}$
$(A-B) \cap(A-C)=\{3\} \ldots \ldots . . . . .(2)$
From $(1)$ and (2), it is verified that $A-(B \cup C)=(A-B) \cap(A-C)$.

Question 9.
Verify $A-(B \cap C)=(A-B) \cup(A-C)$ using Venn diagrams.
Solution:

$\therefore A-(B \cap C)=(A-B) \cup(A-C)$
Hence it is proved.
 

Question 10.
If $\mathrm{U}=\{4,7,8,10,11,12,15,16\}, \mathrm{A}=\{7,8,11,12\}$ and $\mathrm{B}=\{4,8,12,15\}$, then verify De Morgan's Laws for complementation.
$\mathrm{U}=\{4,7,8,10,11,12,15,16\}$
$A=\{7,8,11,12\}, B=\{4,8,12,15\}$
De Morgan's Laws for complementation.
$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
$A \cup B=\{4,7,8,11,12,15\}$
$(\mathrm{A} \cup \mathrm{B})^{\prime}=\{4,7,8,10,11,12,15,16\}-\{4,7,8,11,12,15\}$
$=\{10,16\} \ldots \ldots . \ldots \ldots . .(1)$
$A^{\prime}=\{4,10,15,16\}$
$B^{\prime}=\{7,10,11,16\}$
$A^{\prime} \cap B^{\prime}=\{10,16\} \ldots . . \ldots . . . . . . .(2)$
From (1) and (2) it is verified that $(\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}$.
 

Question $11 .$
Verify $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$ using Venn diagrams.
Solution:
$(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$

$\begin{aligned}
&(2)=(5) \\
&\therefore(A \cap B)^{\prime}=A \prime \cup B
\end{aligned}$