Question 15 Marks
If A, B, C are three sets such that $\text{A}\subset\text{B},$ then prove that $\text{C} - \text{B}\subset\text{C} - \text{A}.$
AnswerWe have, ACB
To show: $\text{C} - \text{B}\subset\text{C} - \text{A}$
Let, $\text{x}\in\text{C} - \text{B}$
$\Rightarrow\text{x} \in \text{C and x}\not\in\text{B}$
$\Rightarrow\text{x} \in \text{C and x}\not\in\text{A}$ $[\because \text{A} \subset \text{B}]$
Thus, $\text{x} \in \text{C} - \text{B}\Rightarrow\text{x}\in \text{C} - \text{A}$
This is true for all $\text{x}\in\text{C} - \text{B}$
$\therefore \text{C} - \text{B} \subset \text{C} - \text{A}.$
View full question & answer→Question 25 Marks
Show that for any sets A and B,$\text{A}\cup(\text{B}-\text{A})=(\text{A}\cup\text{B})$
AnswerLet $\text{x}\in\text{A}\cup\text{(B - A)}$ $\Rightarrow\text{x}\in\text{A or x}\in(\text{B - A})$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B}$ $\Rightarrow\text{x}\in\text{(A}\cup\text{B})$ $\therefore\text{A}\cup\text{(B - A)}\subset\text{(A}\cup\text{B}).....\text{(i)}$ Let and $\text{x}\in\text{(A}\cup\text{B})$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{B and x}\not\in\text{A}$ $\Rightarrow\text{x}\in\text{A or x}\in\text{(B - A)}$ $\Rightarrow\text{x}\in\text{A}\cup\text{(B - A)}$ $\therefore(\text{A}\cup\text{B})\subset\text{A}\cup\text{(B - A)}.....\text{(ii)}$ From (i) and (ii), we get $\text{A}\cup\text{(B - A)}=\text{A}\cup\text{B.}$
View full question & answer→Question 35 Marks
For any two sets of A and B, prove that:$\text{B}'\subset\text{A}'\Rightarrow\text{A}\subset\text{B.}$
AnswerWe have $\text{B}'\subset\text{A}'$
To show: $\text{A}\subset\text{B}$
Let, $\text{x}\in\text{A}$
$\Rightarrow\text{x}\not\in\text{A}'$ $[\because\text{A}\cap\text{A}'=\phi]$
$\Rightarrow\text{x}\not\in\text{B}'$ $[\because\text{B}'\subset\text{A}']$
$\Rightarrow\text{x}\in\text{B}$ $[\because\text{B}\cap\text{B}'=\phi]$
Thus, $\text{x}\in\text{A}\Rightarrow\text{x}\in\text{B}$
This is true foe all $\text{x}\in\text{A}$
$\therefore\text{ A}\subset\text{B}.$
View full question & answer→Question 45 Marks
For any two sets A and B, prove that
$\text{A}-(\text{A}-\text{B})=\text{A}\cap\text{B}$
AnswerLet $\text{x}\in \text{A}- \text{(A} - \text{B)}\Leftrightarrow\text{x}\in \text{A}$ and
$\Leftrightarrow\text{x}\in\text{A and x}\in \text{(A} ∩ \text{B)}$
$\Leftrightarrow\text{x}\in\text{A}\cap\text{(A} \cap \text{B})$
$\Leftrightarrow \text{x}\in\text{(A}\cap\text{B})$
$\therefore\text{A} - \text{(A – B) = (A}\cap\text{B})$
View full question & answer→Question 55 Marks
In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea. Find:
How may drink tea and coffee both.
AnswerLet,
n(P) denote the total number of person,
n(T) denote the total number of person who drink tea and
n(C) denote the total number of person who drink coffee.
Then, $\text{n(P)} = 50, \text{ n(T} - \text{C)} = 14,\text{ n(T)}= 30$
To find: $\text{n(T}\cap\text{C)}$
Clearly T is the disjoint union T - C and $\text{T}\cup\text{C}$
$\therefore\text{T = (T}- \text{C)}\cup\text{(T}\cap\text{C)}$
$\therefore\text{n(T) = n(T} - \text{C)}+\text{n(T}\cap\text{C)}$
$\Rightarrow30=14+\text{n(T}\cap\text{C)}$
$\Rightarrow\text{n(T}\cap\text{C)}=30-14$
$= 16$
Hence, 16 persons drink tea and coffee both.
View full question & answer→Question 65 Marks
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
The number of people who read exactly one the newspapers.
AnswerThe venn diagram representing people reading newspaper H, T and I is shown above.
The shaded region showns the number of people who read newspaper H only, newspaper T only and newspaper I only respectively.
The number of people who read newspaper H only equels
= 25 - (8 + 3 + 6)
= 25 - 17
= 8
The number of people who read newspaper T only
= 26 - (8 + 3 + 5)
= 26 - 16
= 10
And, the number of people who read newspaper I only
= 26 - (6 + 3+ 5)
= 26 - 14
= 12
Hence, the number of people, who read exactly one newspaper = 8 + 10 + 12 = 30.
View full question & answer→Question 75 Marks
Using properties of sets, show that for any two sets A and $\text{B},\text{ (A}\cup\text{B})\cap(\text{A}\cup\text{B}')=\text{A}.$
AnswerWe need to show $(\text{A}\cup\text{B})\cap(\text{A}\cap\text{B}')=\text{A}$
Now,
$(\text{A}\cup\text{B})\cap(\text{A}\cap\text{B}')=((\text{A}\cup\text{B})\cap\text{A})\cap\text{B}'$
$=((\text{A}\cap\text{A})\cup(\text{B}\cap\text{A}))\cap\text{B}'$
$=\text{A}\cap\text{B}'$
$=\text{A}.$
View full question & answer→Question 85 Marks
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
The number of people who read at least one of the newspapers.
AnswerLet n(P) denote total number of people n(H) denote number of people who read newspaper H n(T) denote number of people who read newspaper T n(I) denote number of people who read newspaper I.
Then, $\text{n(P)} = 60, \text{ n(H)} = 25, \text{ n(T)} = 26, \text{n(I)} = 26,$ $\text{n(H}\cap\text{I})=9,\ \text{n(H}\cap\text{T})=11,\text{n(T}\cap\text{I})=8,\ \text{n(H}\cap\text{T}\cap\text{I})=3$
We need to find the number of people who read at least one of the newspaper, i.e., n(H or T or I), i.e., $\text{n(H}\cap\text{T}\cap\text{I)}$ we know that if A, B, C are 3 sets.
Then,
$\text{n(A}\cup\text{B}\cup\text{C) = n(A) + n(B) + n(C) - n(A}\cap\text{B)}-\\\text{n(B}\cap\text{C)}-\text{n(A}\cap\text{C) + n(A}\cap\text{B}\cap\text{C)}$
$\therefore\text{ n(H}\cup\text{T}\cup\text{I) = n(H) + n(T) + n(I) - n(H}\cap\text{T)}\\-\text{n(T}\cap\text{I) - n(H}\cap\text{I) + n(H}\cap\text{T}\cap\text{I)}$
$= 25 + 26 + 26 - 9 - 11 - 8 + 3$
$= 25 + 52 - 28 + 3$
$= 25 + 52 - 25$
$= 52$
Hence, 52 people read at least one of the newspaper.
View full question & answer→Question 95 Marks
For any two sets of A and B, prove that:$\text{A}'\cup\text{B}=\text{U}\Rightarrow\text{A}\subset\text{B.}$
AnswerWe have $\text{A}'\cup\text{B}=\cap,$ the universal set
To show: $\text{A}\subset\text{B}$
Let, $\text{x}\in\text{A}$
$\Rightarrow\text{x}\not\in\text{A}'$ $[\because\text{A}\cap\text{A}'=\phi]$
$\because\text{x}\in\text{A and A}\subset\cup$
$\Rightarrow\text{x}\in\cup$
$\Rightarrow\text{x}\in(\text{A}'\cup\text{B})$ $[\because\cup=\text{A}'\cup\text{B}]$
But, $\text{x}\not\in\text{A}',$
$\therefore\text{ x}\in\text{B}$
Thus, $\text{x}\in\text{A}\Rightarrow\text{x}\in\text{B}$
This is true for all $\text{x}\in\text{A}$
$\therefore\text{ A}\subset\text{B.}$
View full question & answer→Question 105 Marks
Let$ A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}$. Verify the following identities:
$\text{A}-(\text{B}\cap\text{C})=(\text{A}-\text{B})\cup(\text{A}-\text{C})$
Answer$\text{A} = \{1, 2, 4, 5\},$
$\text{B}= \{2, 3, 5, 6\},$
and $\text{C} = \{4, 5, 6, 7\}$
$\text{B}\cap\text{C}= \{5, 6\}$
$\text{A}-(\text{B}\cap\text{C}) = \{1, 2, 4\}\ ....(1)$
$\text{(A}- \text{B)} = \{1, 4\}$
$\text{(A} - \text{C)} = \{1, 2\}$
$(\text{A}-\text{B})\cup(\text{A}-\text{C})= \{1, 2, 4\}\ ....(2)$
From eq ${ }^n$ (1) and eqn (2), we get
$\text{A}-(\text{B}\cap\text{C})=(\text{A}-\text{B})\cup(\text{A}-\text{C}).$
View full question & answer→Question 115 Marks
Each set X, contains 5 elements and each set Y, contains 2 elements and $\bigcup^\limits{20}_{\text{r=1}}\text{X}_\text{r}=\text{S =}\bigcup\limits^\text{n}_\text{r=1}\text{Y}_\text{r}.$ If each element of S belongs to exactly 10 of the $\text{X}'^\text{s}_\text{r}$ and to exactly 4 of $\text{Y}'^\text{s}_\text{r}$, then find the value of n.
AnswerSince each $X_r$ has 5 elements and each element of S belong to exactly 10 of $\text{X}'^\text{s}_\text{r}.$
$\therefore\text{ S}=\bigcup\limits^{20}_\text{r=1}\text{X}_\text{r}\Rightarrow\frac{1}{10}\sum\limits^{20}_\text{r=1}\text{n(X}_\text{r})=\frac{1}{10}(5\times20)=10.....\text{(i)}$
Since each $Y_r$ has 2 elements and each element of S belong to exactly 4 of $\text{Y}'^\text{s}_\text{r}.$
$\therefore\text{ S}=\bigcup\limits^\text{n}_\text{r=1}\text{Y}_\text{r}\Rightarrow\frac{1}{4}\sum\limits^\text{n}_\text{r=1}\text{n(Y}_\text{r})=\frac{1}{4}\text{(2n)}=\frac{\text{n}}{2}.....\text{(ii)}$
From (i) and (ii), we get
$10=\frac{\text{n}}{2}\Rightarrow\text{n}=20.$
View full question & answer→Question 125 Marks
For three sets A, B and C, show that.
$\text{A}\cap\text{B}=\text{A}\cap\text{C}$ need not imply B = C.
AnswerLet A = {1, 2, 3}, B = {2, 4, 6} and C = {2, 5, 7}
Then,
$\text{A}\cap\text{B}=\{2\}$
and $\text{A}\cap\text{C}=\{2\}$
Hence, $\text{A}\cap\text{B}=\text{A}\cap\text{C},$ but clearly $\text{B}\not=\text{C}.$
View full question & answer→Question 135 Marks
For three sets A, B and C, show that.
$\text{A}\subset\text{B}\Rightarrow\text{C}-\text{B}\subset\text{C}-\text{A}.$
AnswerGiven $\text{A}\subset\text{B}$
To show: $\text{C} - \text{B }\subset\text{ C} - \text{A}$
Let $\text{x}\in\text{C}-\text{B}$
$\Rightarrow \text{x}\in\text{C and x}\not\in\text{B}$ [by defination of C - B]
$\Rightarrow \text{x}\in\text{C and x}\not\in\text{A}$ $[\because\text{A}\subset\text{B}]$
This can be seen by the venn diagram above
$\Rightarrow \text{x}\in\text{C - A}$ [by defination of C - A]
Thus $\text{x}\in\text{C - B}\Rightarrow\text{x}\in\text{C - A.}$ Thus is true for all $\text{x}\in\text{C}-\text{B}$
$\therefore\text{C - B}\subset\text{C - A}.$
View full question & answer→Question 145 Marks
Let A and B be two stes such that: $\text{n(P)}= 20,$$\text{n(A}\cup\text{B)=42 and n(A}\cap\text{B})=4.$ Find:
$\text{n(B} - \text{A)}.$
AnswerTo find: $\text{B}- \text{A}$
On a similar lines we have B is the disjoint union of $\text{B} - \text{A}$ and $\text{A}\cap\text{B}$
i.e., $\text{B = (B} - \text{A)}\cup\text{(A}\cap\text{B})$
$\therefore\text{ n(B)=n(B}- \text{A)}+\text{n}\text{(A}\cap\text{B})$
$\Rightarrow26 = \text{n(B} - \text{A)} + 4$
$\Rightarrow\text{n(B} - \text{A)} = 26 - 4$
$= 22$
$\therefore\text{n(B} - \text{A)} = 22.$
View full question & answer→Question 155 Marks
For any two sets A and B, prove that
$\text{A}-(\text{A}\cap\text{B})=\text{A} - \text{B}$
Answer$\text{A – (A}\cap \text{B) = (A} -\text{A) }\cap\text{(A} - \text{B)}$
$= \phi \cap\text{(A – B)}$
$= \text{A} – \text{B}.$
View full question & answer→Question 165 Marks
For any two sets A and B, prove that: $\text{A}\cap\text{B}=\phi\Rightarrow\text{A}\subseteq\text{B}'.$
AnswerGiven $\text{A}\cap\text{B}=\phi,$ i.e., A and B are disjoint set this can represented by venn diagram as follows.
To show: $\text{A}\subseteq\text{B}'$
This is clear from the venn diagram it self
$\because$ A is lying in the complement of B, but we give a proof of it.
So let $\text{x}\in\text{A}$
$\because\text{A}\cap\text{B}=\phi,$
$\therefore\text{x}\not\in\text{B}$
and so $\text{x}\in\text{B}'$ $[\because\text{x}\not\in\text{B}\Rightarrow\text{x}\in\text{B}']$
Thuse $\text{x}\in\text{A}\Rightarrow\text{x}\in\text{B}'.$ This is true for all $\text{x}\in\text{A}$
Hence, $\text{A}\subseteq\text{B}'.$
View full question & answer→Question 175 Marks
Let $A = {1, 2, 4, 5}, B = {2, 3, 5, 6}, C = {4, 5, 6, 7}$. Verify the following identities:
$\text{A}\cap(\text{B}\Delta\text{C})=(\text{A}\cap\text{B})\Delta(\text{A}\cap\text{C}).$
Answer$\text{A} = \{1, 2, 4, 5\},$
$\text{B} = \{2, 3, 5, 6\},$
and $\text{C} = \{4, 5, 6, 7\}$
$\text{B}\Delta\text{C}=(\text{B}-\text{C})\cup(\text{C}-\text{B})= \{2, 3\}\cup\{4, 7\} = \{2, 3, 4, 7\}$
$\text{A}\cap(\text{B}\Delta\text{C})= \{2, 4\} .........(1)$
$(\text{A}\cap\text{B})= \{2, 5\}$
$(\text{A}\cap\text{B})= \{4, 5\}$
$(\text{A}\cap\text{B})\Delta(\text{A}\cap\text{C})=[(\text{A}\cap\text{B})-(\text{A}\cap\text{C})]\cup[(\text{A}\cap\text{C})-(\text{A}\cap\text{B})]$
$(\text{A}\cap\text{B})\Delta(\text{A}\cap\text{C})=\{2\}\cup\{4\} = \{2, 4\} ..............(2)$
From $eq^n (1)$ and $eq^n (2)$, we get
$\text{A}\cap(\text{B}\Delta\text{C})=(\text{A}\cap\text{B})\Delta(\text{A}\cap\text{C}).$
View full question & answer→Question 185 Marks
Of the members of three athletic teams in a certain school, 21 are in the basketball team, 26 in hockey team and 29 in the football team. 14 play hockey and basketball, 15 play hockey and football, 12 play football and 8 play all the three game. How many member are there in all?
AnswerLet,
n(P) denote total number of members,
n(B) denote number of member in the basketball team
n(H) denote number of member in the hockey team and
n(F) denote number of member in the football team
Then, $\text{n(B)} = 21, \text{ n(H)} = 26, \text{ n(F)} = 29$
Also $\text{n(H}\cap\text{B})=14,\ \text{n(H}\cap\text{F)}=15,\\ \text{n(F}\cap\text{B)}=12,\ \text{n(B}\cap\text{H}\cap\text{F)}=8$
Now,
$\text{P}=\text{B}\cup\text{H}\cup\text{F}$
$\therefore\text{ n(P)}=\text{n(B}\cup\text{H}\cup\text{F)}$
$=\text{n(B) + n(H) + n(F)} - \text{n(B}\cap\text{H)}- \text{n(H}\cap\text{F)}-\\\text{n(B}\cap\text{F) + n(B}\cap\text{H}\cap\text{F)}$
$\Rightarrow\text{n(P)} = 21 + 26 + 29 - 14 - 15 - 12 + 8$
$= 76 - 41 + 8$
$= 43$
Hence, there are 43 membersb in all.
View full question & answer→Question 195 Marks
In a group of 950 person, 750 can speak Hindi and 460 can speak English. Find:
how many can speak Hindi only.
AnswerClearly H is the disjoint union of $\text{H} - \text{E}\ \&\ \text{H}\cap\text{E}$
i.e., $\text{H = (H} - \text{E)}\cup\text{(H}\cap\text{E})$
$\therefore\text{ n(H) = n(H} - \text{E)}+\text{(H}\cap\text{E})$ $\begin{bmatrix}\because\text{if A & B arer disjoint then}\\\text{n(A }\cup\text{B)}=\text{n(A) + n(B)}\end{bmatrix}$
$\Rightarrow750 = \text{n(H} - \text{E)} + 260$
$\Rightarrow\text{n(H} - \text{E)} = 750 - 260$
$= 490$
Hence, 490 persons can speak Hindi only.
View full question & answer→Question 205 Marks
Find sets A, B and C such that $\text{A}\cap\text{B},\text{ A}\cap\text{C and B}\cap\text{C}$ are non-empty sets and $\text{A}\cap\text{B}\cap\text{C}=\phi.$
AnswerTo find sets A, B and C such that $\text{A}\cap\text{B}\not=\phi,\ \text{A}\cap\text{C}=\phi$ and $\text{B}\cap\text{C}=\phi,\text{ and }\text{A}\cap\text{B}\cap\text{C}=\phi$
Take $\text{A} = \{1, 2, 3\}$
$\text{B} = \{2, 4, 6\}$
and $\text{C} = \{3, 4, 7\}$
Then,
$\text{A}\cap\text{B}=\{2\}$
$\therefore\text{A}\cap\text{B}\not=\phi$
$\text{A}\cap\text{C}=\{3\}$
$\therefore\text{A}\cap\text{C}\not=\phi$
$\text{B}\cap\text{C}=\{4\}$
$\therefore\text{B}\cap\text{C}\not=\phi$
However A, B and C have no elements in common,
$\therefore\text{A}\cap\text{B}\not=\phi.$
View full question & answer→Question 215 Marks
In a survey it was found that 21 persons liked product $P_1, 26$ liked product $P_2$ and 29 liked product $P_3$. If 14 persons liked products $P_1$ and $P_2 ; 12$ persons liked product $P_3$ and $P_1 ; 14$ persons liked products $P_2$ and $P_3$ and 8 liked all the three products. Find how many liked product $P_3$ only.
AnswerLet,
$n\left(P_1\right)$ be the number of persons liking products $P_1$
$n\left(P_2\right)$ be the number of persons liking products $P_2$ and $n\left(P_3\right)$ be the number of persons liking products $P_2$.
Then,$\text{n(P}_1) = 100,\text{n(P}_2) = 26,\text{n(P}_3) = 29,\text{n(P}_1\cap\text{P}_2) =14,\\\text{n(P}_1\cap\text{P}_3) =12,\text{n(P}_2\cap\text{P}_3) =14,\text{n(P}_1\cap\text{P}_2\cap\text{P}_3) =8$
$\therefore$ Number of people liking product $P_3$ only
$= 29 - (4 + 8 + 6)$
$= 29 - 18$
$= 11$
Hence, 11 persons liked product $P_3$ only.
View full question & answer→Question 225 Marks
Let A and B be two stes such that: $\text{n(P)} = 20, \text{n(A}\cup\text{B)=42 and n(A}\cap\text{B})=4.$ Find:
$\text{n(A} - \text{B)}.$
AnswerTo find: $\text{n(A} - \text{B)}$
We know that if A and B are disjoint sets, then $\text{A}\cap\text{B}=\phi$
$\therefore\text{n(A}\cup\text{B) = n(A)+n(B)}-\text{n(A}\cap\text{B)}$
$\text{n(A}\cup\text{B) = n(A)+n(B)}-\text{n}(\phi)$
$\Rightarrow\text{n(A}\cup\text{B)=n(A)+n(B)}$
Now,
$\text{A}=\text{(A}-\text{B)}\cup\text{(A}\cap\text{B)}$
i.e., A is the disjoint union of A - B and $\text{A}\cap\text{B}$
$\therefore\text{ n(A)}=\text{n(A}-\text{B)}\cup\text{(A}\cap\text{B)}$
$\text{ n(A)}=\text{n(A}-\text{B)}+\text{n(A}\cap\text{B)}$
$\Rightarrow 20 = \text{n(A} - \text{B)} + 4$
$\Rightarrow\text{n(A} - \text{B)} = 20 - 4$
$= 16$
$\therefore\text{n(A} -\text{B)} = 16.$
View full question & answer→Question 235 Marks
For any two sets A and B, prove that: A' - B' = B - A.
AnswerTo show $\text{A}' - \text{B}' = \text{B} - \text{A}$
We show that $\text{A}' – \text{B}' = \subseteq\text{B} - \text{A} $ and vice versa
Let, $\text{x}\in\text{A}'-\text{B}'$
$\Rightarrow\text{x}\in\text{A}'\text{and x}\not\in\text{B}'$
$\Rightarrow\text{x}\not\in\text{A }\text{and x}\in\text{B}$ $[\because\text{A}\cap\text{A}'=\oint\text{ and B}\cap\text{B}'=\oint]$
$\Rightarrow\text{x}\in\text{B}\text{ and x}\not\in\text{A}$
$\text{x}\in\text{B}- \text{A}$
This is true for all $\text{x}\in\text{A}'-\text{B}'$
Hence $\text{A}'-\text{B}'\subseteq\text{B}-\text{A}$
Conversely,
Let, $\text{x}\in\text{B} - \text{A}$
$\Rightarrow\text{x}\in\text{B and x}\not\in\text{A}$
$\Rightarrow\text{x}\not\in\text{B}'\text{ and x}\in\text{A}'$
$\Rightarrow\text{x}\in\text{A}'\text{ and x}\not\in\text{B}'$ $[\because\text{B}\cap\text{B}'=\oint\text{ and A}\cap\text{A}'=\oint]$
$\Rightarrow\text{x}\in\text{A}'-\text{B}'$
This is true for all $\text{x}\in\text{B} - \text{A}$
Hence $\text{B}- \text{A}\subseteq\text{A}'=\text{B}'$
$\therefore\text{ A}' - \text{B}' = \text{B} - \text{A}$ Proved.
View full question & answer→Question 245 Marks
For any two sets A and B, prove the following:
$\text{A}\cap\text{(A}'\cup\text{B})=\text{A}\cap\text{B}$
Answer$\text{LHS}=\text{A}\cap\text{(A}'\cup\text{B)}$
$=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B)}$ $[\because\cap$ distributes over (i)$]$
$= \oint\cup\text{ (A}\cap\text{B)}$ $[\because\text{A}\cap\text{A}' = \oint]$
$=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x for any set x$]$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.
View full question & answer→Question 255 Marks
For any two sets A and B, prove that
$\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}$
AnswerLet $\text{x}\in \text{A} \cup \text{(B} - \text{A)}\Rightarrow\text{x}\in\text{A or x}\in\text{(B} - \text{A)}$
$\Rightarrow \text{x} \in \text{A or x} \in \text{B and x}\not\in \text{A}$
$\Rightarrow \text{x}\in \text{B}$
$\Rightarrow \text{x}\in \text{A}\cup\text{B}$ $[\because \text{B}\subset\text{(A}\cup\text{B})]$
This is true for all $\text{x}\in\text{A}\cup\text{(B} - \text{A)}$
$\therefore \text{A}\cup\text{(B} - \text{A)}\subset\text{(A}\cup \text{B})....\text{(i)}$
Conversely,
Let, $\text{x}\in \text{(A}\cup\text{B)}$
$\Rightarrow\text{x} \in \text{A or x} \in\text{B}$
$\Rightarrow \text{x}\in \text{A or x} \in \text{(B – A)}$ $[\because\text{B}\subset \text{(B – A)]}$
$\therefore\text{(A}\cup\text{B}) \subset \text{A}\cup \text{(B}- \text{A)}.....\text{(ii)}$
From (i) and (ii), we get
$\text{A}\cup\text{(B} - \text{A) = (A}\cup\text{B}).$
View full question & answer→Question 265 Marks
For any two sets A and B, prove the following:
$\text{A}\cap\text{(A}\cup\text{B})=\phi$
Answer$\text{LHS}=\text{A}\cap\text{(A}\cup\text{B}')$
$=\text{A}\cap\text{(A}'\cap\text{B}')$ [By De-morgan's law]
$=\text{(A}\cap\text{A}')\cap\text{B}'$ [By associative law]
$= \oint \cap \text{ B}'$ $[\because \text{A}\cap{A}' =\oint]$
$= \oint$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.
View full question & answer→Question 275 Marks
For any two sets A and B, prove that
$(\text{A}\cap\text{B})-\text{B = A}- \text{B}$
Answer$\text{(A} \cup\text{B) – B = (A} - \text{B)}\cup\text{(A} - \text{B)}$
$= \oint\cap \text{ (A} - \text{B)}$
$= \text{A} - \text{B}.$
View full question & answer→Question 285 Marks
For any two sets A and B, prove the following:
$\text{A}-\text{(A}-\text{B})=\text{A}\cap\text{B}$
AnswerFor any sets A and B we have by De-morgan's laws
$\text{(A}\cup\text{B)}' = \text{A}'\cap\text{B}',\text{(A}\cap\text{B)}' = \text{A}'\cup\text{B}'$
Also
$\text{LHS}= \text{A} - \text{(A} - \text{B)}$
$=\text{A}\cap\text{(A}- \text{B)}'$
$=\text{A}\cap\text{(A}\cap\text{B}')'$
$=\text{A}\cap\text{(A}\cap\text{(B})')'$ [By De-morgan's law]
$=\text{A}\cap\text{(A}'\cup\text{B})$ $[\because\text{(B}')'=\text{B}]$
$=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B})$
$=\oint\cup\text{(A}\cap\text{B})$ $[\because\text{A}\cap\text{A}'=\oint]$
$=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x, for and set x$]$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.
View full question & answer→Question 295 Marks
For any two sets A and B, prove that
$\text{(A}-\text{B)}\cup(\text{B}\cap\text{A})=\text{A}$
AnswerLet $\text{x}\in\text{A.}$
Then either $\text{x} \in \text{(A}- \text{B) or x} \in \text{(A}\cap \text{B})$
$\Rightarrow \text{x}\in\text{(A – B)}\cup \text{(A} \cap\text{B})$
$\therefore\text{A} \subset \text{(A – B)}\cup \text{(A}\cap\text{B}).....\text{(i)}$
Conversely,
Let $\text{x}\in\text{(A} - \text{B)} \cup \text{(A}\cap\text{B)}$
$\Rightarrow\text{x} \in \text{(A – B) or x (A} \cap \text{B})$
$\Rightarrow\text{x} \in \text{A and x}\not\in\text{B or x}\in \text{A and x} \in\text{B}$
$\Rightarrow \text{x} \in \text{A}$
$\therefore\text{(A} - \text{B)}\cup\text{(A} \cap \text{B}) \subset \text{A}.....\text{(ii)}$
From (i) and (ii), we get
$\text{(A} - \text{B)}\cup\text{(A} \cap \text{B) = A}.$
View full question & answer→Question 305 Marks
For any two sets A and B, prove the following:
$\text{A} - \text{B}=\text{A }\Delta\text{ (A}\cap\text{B})$
Answer$\text{RHS}=\text{A }\Delta\text{ (A} \cap8)$ $=\text{(A} - \text{(A}\cap\text{B}))\cup\text{(A}\cap\text{B} - \text{A})$ $[\because\text{E }\Delta\text{ F} =\text{(E} - \text{F)}\cup\text{(F} - \text{E)]}$ $=\text{(A}\cap\text{(A}\cap\text{B})')\cup\text{(A}\cap\text{B}\cap\text{A}')$ $[\because \text{E - F = E }\cap \text{F}']$ $=\text{(A}\cap \text{(A}'\cup\text{B}'))\cup\text{(A}\cap\text{A}'\cap\text{B})$ [By de-morgan's law & associative law] $=\text{(A}\cap\text{A}')\cup(\text{A}\cap\text{B}')\cup(\oint\cap\text{B})$ $[\because\cap\text{ distributes over}\cup\text{and A }\cap \text{A}' = \oint]$ $= \oint \cup \text{(A}\cap\text{B}')\cup \oint$ $[\because \oint \cap \text{ B} = \oint]$ $= \text{A}\cap\text{B}'$ $[\because\oint\cup\text{ x = x for any set x}]$ $= \text{A} - \text{B }$ $[\because \text{A} \cap\text{B}' = \text{A - B}]$ $=\text{LHS}$ $\therefore$ LHS = RHS Proved.
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