Venn Diagram and Operation on Sets Questions in English

(B) The symmetric difference of two sets $A$ and $B$ is defined as $A \Delta B = (A - B) \cup (B - A)$.
Since $A$ is a proper subset of $B$ ($A \subset B$ and $A \neq B$),we have $A - B = \emptyset$,so $n(A - B) = 0$.
Because $A$ is a proper subset of $B$,there must be at least one element in $B$ that is not in $A$,which means $n(B - A) \geq 1$.
Therefore,$n(A \Delta B) = n(A - B) + n(B - A) = 0 + n(B - A) \geq 1$.
The minimum possible value of $n(A \Delta B)$ is $1$.

(B) We know that the set difference $A - B$ is defined as $A \cap B^c$.
Substituting this into the expression,we get:
$A - (A - B) = A - (A \cap B^c)$
Using the property $X - Y = X \cap Y^c$,we have:
$A - (A \cap B^c) = A \cap (A \cap B^c)^c$
By De Morgan's Law,$(A \cap B^c)^c = A^c \cup (B^c)^c = A^c \cup B$.
So,$A \cap (A^c \cup B) = (A \cap A^c) \cup (A \cap B)$.
Since $A \cap A^c = \emptyset$,we get $\emptyset \cup (A \cap B) = A \cap B$.
Therefore,$A - (A - B) = A \cap B$.

(D) Given $2\cos^2 \theta + \sin \theta \le 2$.
Using $\cos^2 \theta = 1 - \sin^2 \theta$,we get $2(1 - \sin^2 \theta) + \sin \theta \le 2$.
$2 - 2\sin^2 \theta + \sin \theta \le 2$.
$-2\sin^2 \theta + \sin \theta \le 0$.
$2\sin^2 \theta - \sin \theta \ge 0$.
$\sin \theta (2\sin \theta - 1) \ge 0$.
This inequality holds when $\sin \theta \le 0$ or $\sin \theta \ge \frac{1}{2}$.
For $\sin \theta \le 0$,$\theta \in [\pi, 2\pi]$.
For $\sin \theta \ge \frac{1}{2}$,$\theta \in [\frac{\pi}{6}, \frac{5\pi}{6}]$.
Thus,$A = [\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]$.
Given $B = [\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A \cap B = ([\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]) \cap [\frac{\pi}{2}, \frac{3\pi}{2}]$.
$A \cap B = [\frac{\pi}{2}, \frac{5\pi}{6}] \cup [\pi, \frac{3\pi}{2}]$.

(D) Let $M$,$P$,and $C$ be the sets of students who opted for Mathematics,Physics,and Chemistry,respectively.
$n(M) = \lfloor \frac{140}{2} \rfloor = 70$
$n(P) = \lfloor \frac{140}{3} \rfloor = 46$
$n(C) = \lfloor \frac{140}{5} \rfloor = 28$
Now,find the intersections:
$n(M \cap P) = \lfloor \frac{140}{\text{lcm}(2,3)} \rfloor = \lfloor \frac{140}{6} \rfloor = 23$
$n(M \cap C) = \lfloor \frac{140}{\text{lcm}(2,5)} \rfloor = \lfloor \frac{140}{10} \rfloor = 14$
$n(P \cap C) = \lfloor \frac{140}{\text{lcm}(3,5)} \rfloor = \lfloor \frac{140}{15} \rfloor = 9$
$n(M \cap P \cap C) = \lfloor \frac{140}{\text{lcm}(2,3,5)} \rfloor = \lfloor \frac{140}{30} \rfloor = 4$
Using the Principle of Inclusion-Exclusion:
$n(M \cup P \cup C) = n(M) + n(P) + n(C) - (n(M \cap P) + n(M \cap C) + n(P \cap C)) + n(M \cap P \cap C)$
$n(M \cup P \cup C) = 70 + 46 + 28 - (23 + 14 + 9) + 4 = 144 - 46 + 4 = 102$
The number of students who did not opt for any course is $140 - 102 = 38$.

(A) Given $\phi \ne A \cap B \subseteq C$.
Check option $(A)$: If $(A - C) \subseteq B$ then $A \subseteq B$.
Let $A = \{1, 2\}$,$B = \{2, 3\}$,$C = \{2\}$.
Here $A \cap B = \{2\} \subseteq C$ and $A \cap B \ne \phi$.
$A - C = \{1\}$. Since ${1} \not\subseteq \{2, 3\}$,the condition $(A - C) \subseteq B$ is not satisfied for this example.
However,consider $A = \{1, 2\}$,$B = \{1, 3\}$,$C = \{2\}$.
$A \cap B = \{1\} \subseteq C$ is false.
Let $A = \{1, 2\}$,$B = \{2, 3\}$,$C = \{1, 2\}$.
$A \cap B = \{2\} \subseteq C$.
$A - C = \phi \subseteq B$ is true.
But $A = \{1, 2\} \not\subseteq B = \{2, 3\}$.
Thus,the statement in option $(A)$ is not true.

(C) Given $A = \{x \in R : |x| < 2\} = (-2, 2)$.
Given $B = \{x \in R : |x - 2| \geq 3\}$.
This implies $x - 2 \geq 3$ or $x - 2 \leq -3$.
So,$x \geq 5$ or $x \leq -1$.
Thus,$B = (-\infty, -1] \cup [5, \infty)$.
Now,$B - A$ is the set of elements in $B$ that are not in $A$.
$B - A = ((-\infty, -1] \cup [5, \infty)) - (-2, 2)$.
Since $(-2, 2)$ overlaps with $(-\infty, -1]$ only on the interval $(-2, -1]$,we remove this part.
$B - A = (-\infty, -2] \cup [5, \infty)$.
This can be written as $R - (-2, 5)$.
Therefore,the correct option is $C$.

(D) The sample space of rolling a die is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ (getting a prime number) is $A = \{2, 3, 5\}$.
Event $B$ (getting an odd number) is $B = \{1, 3, 5\}$.
The event '$A$ or $B$' is represented by the union of the two sets,$A \cup B$.
$A \cup B = \{1, 2, 3, 5\}$.

(B) The sample space of rolling a die is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ (getting a prime number) is $A = \{2, 3, 5\}$.
Event $B$ (getting an odd number) is $B = \{1, 3, 5\}$.
The event '$A$ but not $B$' is represented by the set difference $A - B$.
$A - B = \{2, 3, 5\} - \{1, 3, 5\} = \{2\}$.
Thus,the correct set is $\{2\}$.

(A) When a die is thrown,the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
The event $A$ is defined as a number less than $7$,so $A = \{1, 2, 3, 4, 5, 6\}$.
The event $B$ is defined as getting a number $3$,so $B = \{3\}$.
Therefore,$A \cup B = \{1, 2, 3, 4, 5, 6\} \cup \{3\} = \{1, 2, 3, 4, 5, 6\}$.

(A) When a die is thrown,the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ is defined as a number less than $7$,so $A = \{1, 2, 3, 4, 5, 6\}$.
Event $B$ is defined as a number greater than $7$,so $B = \phi$ (since no such number exists on a standard die).
The intersection $A \cap B$ represents the set of elements common to both $A$ and $B$.
Since $B$ is an empty set,$A \cap B = \phi$.

(A) The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
Given $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$.
Combining all elements and listing common elements only once,we get:
$A \cup B = \{2, 4, 6, 8, 10, 12\}$.

(N/A) We have $A = \{a, e, i, o, u\}$ and $B = \{a, i, u\}$.
$A \cup B = \{a, e, i, o, u\} \cup \{a, i, u\} = \{a, e, i, o, u\}$.
Since the resulting set is exactly $A$,we have $A \cup B = A$.
This example illustrates that the union of a set $A$ and its subset $B$ is the set $A$ itself,i.e.,if $B \subset A$,then $A \cup B = A$.

(A) The union of two sets $X$ and $Y$ is the set of all elements that are in $X$,in $Y$,or in both.
$X \cup Y = \{ \text{Ram, Geeta, Akbar} \} \cup \{ \text{Geeta, David, Ashok} \} = \{ \text{Ram, Geeta, Akbar, David, Ashok} \}$.
This set represents all students of Class $XI$ who are in the hockey team,the football team,or both.

(A) The intersection of two sets $A$ and $B$,denoted by $A \cap B$,is the set of all elements that are common to both $A$ and $B$.
Given $A = \{2, 4, 6, 8\}$ and $B = \{6, 8, 10, 12\}$.
The elements common to both sets are $6$ and $8$.
Therefore,$A \cap B = \{6, 8\}$.

(B) The intersection of two sets $X$ and $Y$,denoted by $X \cap Y$,consists of all elements that are common to both sets $X$ and $Y$.
Given $X = \{ \text{Ram, Geeta, Akbar} \}$ and $Y = \{ \text{Geeta, David, Ashok} \}$.
Comparing the elements of both sets,we observe that $\text{Geeta}$ is the only element present in both $X$ and $Y$.
Therefore,$X \cap Y = \{ \text{Geeta} \}$.

(N/A) The intersection of two sets $A$ and $B$,denoted by $A \cap B$,is the set of all elements which are common to both $A$ and $B$.
Given $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $B = \{2, 3, 5, 7\}$.
The common elements in $A$ and $B$ are $2, 3, 5,$ and $7$.
Therefore,$A \cap B = \{2, 3, 5, 7\}$.
Since the set $\{2, 3, 5, 7\}$ is exactly the set $B$,we have $A \cap B = B$.

(A) The difference of two sets $A$ and $B$,denoted by $A - B$,is the set of elements which belong to $A$ but not to $B$.
Given $A = \{1, 2, 3, 4, 5, 6\}$ and $B = \{2, 4, 6, 8\}$.
$A - B = \{x : x \in A \text{ and } x \notin B\} = \{1, 3, 5\}$.
Similarly,$B - A = \{x : x \in B \text{ and } x \notin A\} = \{8\}$.
Thus,$A - B = \{1, 3, 5\}$ and $B - A = \{8\}$.

(N/A) We have,$V - B = \{e, o\}$,since the elements $e, o$ belong to $V$ but not to $B$.
$B - V = \{k\}$,since the element $k$ belongs to $B$ but not to $V$.
We note that $V - B \neq B - V$. Using the set-builder notation,we can rewrite the definition of difference as:
$A - B = \{x : x \in A \text{ and } x \notin B\}$.
The difference of two sets $A$ and $B$ can be represented by a Venn diagram as shown in the figure below.
The shaded portion represents the difference of the two sets $A$ and $B$.

(A) The union of two sets $X$ and $Y$,denoted by $X \cup Y$,is the set of all elements which are in $X$,in $Y$,or in both.
Given $X = \{1, 3, 5\}$ and $Y = \{1, 2, 3\}$.
$X \cup Y = \{1, 3, 5\} \cup \{1, 2, 3\} = \{1, 2, 3, 5\}$.

(A) The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
Given $A = \{a, e, i, o, u\}$ and $B = \{a, b, c\}$.
$A \cup B = \{a, e, i, o, u\} \cup \{a, b, c\}$.
Combining all unique elements,we get $A \cup B = \{a, b, c, e, i, o, u\}$.

First,represent the sets in roster form:
$A = \{ x : x \in \mathbb{N}, 1 < x \le 6 \} = \{ 2, 3, 4, 5, 6 \}$
$B = \{ x : x \in \mathbb{N}, 6 < x < 10 \} = \{ 7, 8, 9 \}$
The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$ or in $B$.
$A \cup B = \{ 2, 3, 4, 5, 6, 7, 8, 9 \}$
In set-builder form,this can be written as:
$A \cup B = \{ x : x \in \mathbb{N} \text{ and } 1 < x < 10 \}$

(A) The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$ or in $B$ or in both.
Given $A = \{1, 2, 3\}$ and $B = \varnothing$ (the empty set).
Since the empty set contains no elements,the union $A \cup B$ consists only of the elements present in set $A$.
Therefore,$A \cup B = \{1, 2, 3\}$.

(A) Given sets are $A = \{a, b\}$ and $B = \{a, b, c\}$.
Since every element of $A$ is also an element of $B$,$A$ is a subset of $B$,i.e.,$A \subset B$.
The union of two sets $A$ and $B$ is the set of all elements which are in $A$ or in $B$.
Therefore,$A \cup B = \{a, b\} \cup \{a, b, c\} = \{a, b, c\} = B$.

(B) Given that $A \subset B$,every element of set $A$ is also an element of set $B$.
Therefore,the union of $A$ and $B$,denoted by $A \cup B$,consists of all elements that are in $A$ or in $B$.
Since all elements of $A$ are already in $B$,the union $A \cup B$ is equal to $B$.

(A) Given sets are $A = \{1, 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$.
The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
$A \cup B = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}$.

(A) Given sets are $A = \{1, 2, 3, 4\}$ and $C = \{5, 6, 7, 8\}$.
The union of two sets $A$ and $C$,denoted by $A \cup C$,is the set of all elements that are in $A$,or in $C$,or in both.
$A \cup C = \{1, 2, 3, 4\} \cup \{5, 6, 7, 8\} = \{1, 2, 3, 4, 5, 6, 7, 8\}$.

(A) Given sets are $B = \{3, 4, 5, 6\}$ and $C = \{5, 6, 7, 8\}$.
The union of two sets $B$ and $C$,denoted by $B \cup C$,is the set of all elements that are in $B$,or in $C$,or in both.
$B \cup C = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\} = \{3, 4, 5, 6, 7, 8\}$.

(A) Given sets are $B = \{3, 4, 5, 6\}$ and $D = \{7, 8, 9, 10\}$.
The union of two sets $B$ and $D$,denoted by $B \cup D$,is the set of all elements that are in $B$,or in $D$,or in both.
$B \cup D = \{3, 4, 5, 6\} \cup \{7, 8, 9, 10\} = \{3, 4, 5, 6, 7, 8, 9, 10\}$.

(A) Given sets are $A = \{1, 2, 3, 4\}$,$B = \{3, 4, 5, 6\}$,and $C = \{5, 6, 7, 8\}$.
The union of sets $A$,$B$,and $C$ is the set of all elements that are in $A$,$B$,or $C$.
$A \cup B \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\}$
Combining all unique elements,we get:
$A \cup B \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$

(A) Given sets are $A = \{1, 2, 3, 4\}$,$B = \{3, 4, 5, 6\}$,and $D = \{7, 8, 9, 10\}$.
The union of sets $A$,$B$,and $D$ is the set of all elements that are in at least one of the sets $A$,$B$,or $D$.
$A \cup B \cup D = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} \cup \{7, 8, 9, 10\}$
Combining all unique elements,we get:
$A \cup B \cup D = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

(A) Given sets are $B = \{3, 4, 5, 6\}$,$C = \{5, 6, 7, 8\}$,and $D = \{7, 8, 9, 10\}$.
The union of sets $B$,$C$,and $D$ is the set of all elements that are in at least one of the sets $B$,$C$,or $D$.
$B \cup C \cup D = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\} \cup \{7, 8, 9, 10\}$
Combining all unique elements,we get:
$B \cup C \cup D = \{3, 4, 5, 6, 7, 8, 9, 10\}$.

(A) The intersection of two sets $X$ and $Y$,denoted by $X \cap Y$,is the set of all elements which are common to both $X$ and $Y$.
Given $X = \{1, 3, 5\}$ and $Y = \{1, 2, 3\}$.
The common elements in both sets are $1$ and $3$.
Therefore,$X \cap Y = \{1, 3\}$.

(A) The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
Given $A = \{a, e, i, o, u\}$ and $B = \{a, b, c\}$.
$A \cup B = \{a, e, i, o, u, b, c\}$ or $\{a, b, c, e, i, o, u\}$.

(A) Given sets are:
$A = \{ 3, 6, 9, 12, \ldots \}$
$B = \{ 1, 2, 3, 4, 5 \}$
The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$ or in $B$ (or in both).
$A \cup B = \{ x : x \in A \text{ or } x \in B \}$
$A \cup B = \{ 1, 2, 3, 4, 5, 6, 9, 12, \ldots \}$

(A) First,we list the elements of set $A$:
$A = \{ x : x \in \mathbb{N}, 1 < x \le 6 \} = \{ 2, 3, 4, 5, 6 \}$
Next,we list the elements of set $B$:
$B = \{ x : x \in \mathbb{N}, 6 < x < 10 \} = \{ 7, 8, 9 \}$
The union of two sets $A$ and $B$,denoted by $A \cup B$,is the set of all elements which are in $A$,in $B$,or in both.
$A \cup B = \{ 2, 3, 4, 5, 6, 7, 8, 9 \}$

(A) The intersection of two sets $A$ and $B$,denoted by $A \cap B$,consists of all elements that are common to both sets.
Given $A = \{3, 5, 7, 9, 11\}$ and $B = \{7, 9, 11, 13\}$.
The common elements in $A$ and $B$ are $7, 9,$ and $11$.
Therefore,$A \cap B = \{7, 9, 11\}$.

(A) The intersection of two sets $B$ and $C$,denoted by $B \cap C$,consists of all elements that are common to both sets.
Given $B = \{7, 9, 11, 13\}$ and $C = \{11, 13, 15\}$.
The common elements in $B$ and $C$ are $11$ and $13$.
Therefore,$B \cap C = \{11, 13\}$.

(A) Given sets are $A = \{3, 5, 7, 9, 11\}$,$C = \{11, 13, 15\}$,and $D = \{15, 17\}$.
First,find the intersection of $A$ and $C$:
$A \cap C = \{3, 5, 7, 9, 11\} \cap \{11, 13, 15\} = \{11\}$.
Next,find the intersection of the result with $D$:
$(A \cap C) \cap D = \{11\} \cap \{15, 17\}$.
Since there are no common elements between $\{11\}$ and $\{15, 17\}$,the result is the empty set $\varnothing$.

(C) The intersection of two sets $A$ and $C$,denoted by $A \cap C$,consists of all elements that are common to both set $A$ and set $C$.
Given $A = \{3, 5, 7, 9, 11\}$ and $C = \{11, 13, 15\}$.
The only element common to both sets is $11$.
Therefore,$A \cap C = \{11\}$.

(NONE) The intersection of two sets $B$ and $D$ consists of all elements that are common to both sets.
Given $B = \{7, 9, 11, 13\}$ and $D = \{15, 17\}$.
Comparing the elements of $B$ and $D$,we see that there are no common elements.
Therefore,$B \cap D = \varnothing$ or $\{\}$.

(A) First,find the union of sets $B$ and $C$:
$B \cup C = \{7, 9, 11, 13\} \cup \{11, 13, 15\} = \{7, 9, 11, 13, 15\}$.
Now,find the intersection of set $A$ with $(B \cup C)$:
$A \cap (B \cup C) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15\}$.
The common elements are $\{7, 9, 11\}$.

(A) The set $A = \{3, 5, 7, 9, 11\}$ and the set $D = \{15, 17\}$.
To find the intersection $A \cap D$,we look for elements that are common to both sets.
Since there are no common elements between $A$ and $D$,the intersection is the empty set,denoted by $\varnothing$.

(N/A) First,find the union of sets $B$ and $D$:
$B \cup D = \{7, 9, 11, 13\} \cup \{15, 17\} = \{7, 9, 11, 13, 15, 17\}$.
Next,find the intersection of set $A$ with the result:
$A \cap (B \cup D) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15, 17\}$.
The common elements are $\{7, 9, 11\}$.
Therefore,$A \cap (B \cup D) = \{7, 9, 11\}$.

First,find the intersection of sets $A$ and $B$:
$(A \cap B) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13\} = \{7, 9, 11\}$.
Next,find the union of sets $B$ and $C$:
$(B \cup C) = \{7, 9, 11, 13\} \cup \{11, 13, 15\} = \{7, 9, 11, 13, 15\}$.
Finally,find the intersection of the two results:
$(A \cap B) \cap (B \cup C) = \{7, 9, 11\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11\}$.

(A) First,find the union of sets $A$ and $D$:
$A \cup D = \{3, 5, 7, 9, 11, 15, 17\}$
Next,find the union of sets $B$ and $C$:
$B \cup C = \{7, 9, 11, 13, 15\}$
Finally,find the intersection of the two resulting sets:
$(A \cup D) \cap (B \cup C) = \{3, 5, 7, 9, 11, 15, 17\} \cap \{7, 9, 11, 13, 15\} = \{7, 9, 11, 15\}$

(A) Given sets are:
$A = \{ 1, 2, 3, 4, 5, \ldots \}$
$B = \{ 2, 4, 6, 8, \ldots \}$
$C = \{ 1, 3, 5, 7, 9, \ldots \}$
$D = \{ 2, 3, 5, 7, \ldots \}$
The intersection $A \cap B$ consists of elements that are common to both set $A$ and set $B$.
Since $B$ is a subset of $A$ $(B \subset A)$,the intersection of $A$ and $B$ is $B$ itself.
$A \cap B = \{ x : x \in A \text{ and } x \in B \} = \{ x : x \text{ is an even natural number} \} = B$.

(A) Given sets are:
$A = \{1, 2, 3, 4, 5, \ldots \}$
$B = \{2, 4, 6, 8, \ldots \}$
$C = \{1, 3, 5, 7, 9, \ldots \}$
$D = \{2, 3, 5, 7, \ldots \}$
The intersection $B \cap C$ represents the set of elements that are common to both $B$ and $C$.
Since $B$ contains only even natural numbers and $C$ contains only odd natural numbers,there are no common elements between them.
Therefore,$B \cap C = \varnothing$.

(A) $A = \{ 1, 2, 3, 4, 5, \ldots \}$
$B = \{ 2, 4, 6, 8, \ldots \}$
$C = \{ 1, 3, 5, 7, 9, \ldots \}$
$D = \{ 2, 3, 5, 7, \ldots \}$
$B \cap D$ represents the set of elements common to both $B$ and $D$.
Since $2$ is the only even prime number,$B \cap D = \{ 2 \}$.

(A) Given sets are:
$A = \{1, 2, 3, 4, 5, \ldots \}$
$B = \{2, 4, 6, 8, \ldots \}$
$C = \{1, 3, 5, 7, 9, \ldots \}$
$D = \{2, 3, 5, 7, 11, \ldots \}$
The intersection $C \cap D$ consists of elements that are common to both set $C$ (odd natural numbers) and set $D$ (prime numbers).
Since $2$ is the only even prime number,all other prime numbers are odd.
Therefore,$C \cap D = \{ x : x \text{ is an odd prime number} \} = \{3, 5, 7, 11, \ldots \}$.