Set Theory Questions in English

(B) The empty set is a set that contains no elements.
For option $A$: $x^2 - 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$. These are real numbers,so the set is $\{1, -1\}$.
For option $B$: $x^2 + 1 = 0 \Rightarrow x^2 = -1$. There is no real number whose square is $-1$. Thus,this set contains no elements and is the empty set.
For option $C$: $x^2 - 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$. These are real numbers,so the set is $\{3, -3\}$.
For option $D$: $x^2 - x - 2 = 0 \Rightarrow (x - 2)(x + 1) = 0 \Rightarrow x = 2, -1$. These are real numbers,so the set is $\{2, -1\}$.
Therefore,the correct option is $B$.

(A) The set $A$ is defined by the intersection of two conditions: $x^2 = 16$ and $2x = 6$.
First,solving $x^2 = 16$ gives $x = 4$ or $x = -4$.
Second,solving $2x = 6$ gives $x = 3$.
Since there is no value of $x$ that satisfies both conditions simultaneously,the set $A$ contains no elements.
Therefore,$A = \phi$ (the empty set).

(B) Step $1$: Find the intersection of sets $B$ and $C$,denoted as $B \cap C$.
The intersection contains elements common to both $B$ and $C$.
$B = \{3, 4\}$,$C = \{4, 5, 6\}$.
$B \cap C = \{4\}$.
Step $2$: Find the union of set $A$ with the result from Step $1$,denoted as $A \cup (B \cap C)$.
The union contains all elements present in either $A$ or the set $\{4\}$.
$A = \{1, 2, 3\}$.
$A \cup \{4\} = \{1, 2, 3, 4\}$.
Therefore,the correct option is $B$.

(D) Using De Morgan's Law,$(A \cap B)^c = A^c \cup B^c$.
Therefore,$A \cap (A \cap B)^c = A \cap (A^c \cup B^c)$.
Applying the Distributive Law,we get $(A \cap A^c) \cup (A \cap B^c)$.
Since $A \cap A^c = \phi$ (the empty set),the expression becomes $\phi \cup (A \cap B^c)$.
Thus,the final result is $A \cap B^c$.

(C) To find the intersection $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations simultaneously.
Given equations are $y = \frac{1}{x}$ and $y = -x$.
Substituting the value of $y$ from the second equation into the first,we get $-x = \frac{1}{x}$.
This simplifies to $-x^2 = 1$,or $x^2 = -1$.
Since $x$ must be a real number $(x \in R)$,the equation $x^2 = -1$ has no real solutions because the square of any real number is non-negative.
Therefore,there are no points $(x, y)$ that belong to both sets $A$ and $B$.
Hence,$A \cap B = \phi$.

(B) Given $A = \{x : x \in R, |x| < 1\}$. This implies $-1 < x < 1$,so $A = (-1, 1)$.
Given $B = \{x : x \in R, |x - 1| \ge 1\}$. This implies $x - 1 \le -1$ or $x - 1 \ge 1$,which simplifies to $x \le 0$ or $x \ge 2$. So $B = (-\infty, 0] \cup [2, \infty)$.
Now,$A \cup B = (-1, 1) \cup (-\infty, 0] \cup [2, \infty)$.
Combining these intervals,we get $A \cup B = (-\infty, 1) \cup [2, \infty)$.
We are given $A \cup B = R - D$. This means $D$ is the complement of $A \cup B$ in $R$.
The complement of $(-\infty, 1) \cup [2, \infty)$ is the set $[1, 2)$.
Therefore,$D = \{x : x \in R, 1 \le x < 2\}$.

(C) The set $A$ represents the graph of the exponential function $y = e^x$. The set $B$ represents the graph of the linear function $y = x$.
For any real number $x$,it is a known property that $e^x > x$. Specifically,the function $f(x) = e^x - x$ has a minimum value of $1$ at $x = 0$,meaning $e^x - x \ge 1$ for all $x \in R$.
Since $e^x$ is never equal to $x$ for any $x \in R$,there are no common points between the two sets.
Therefore,$A \cap B = \phi$.

(C) According to De Morgan's Law,$A^c \cap B^c = (A \cup B)^c$.
Therefore,$n(A^c \cap B^c) = n(U) - n(A \cup B)$.
Using the formula for the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Substituting the given values: $n(A \cup B) = 200 + 300 - 100 = 400$.
Now,$n(A^c \cap B^c) = 700 - 400 = 300$.

(B) Given total families $= 10,000$.
$n(A) = 40\% \text{ of } 10,000 = 4,000$
$n(B) = 20\% \text{ of } 10,000 = 2,000$
$n(C) = 10\% \text{ of } 10,000 = 1,000$
$n(A \cap B) = 5\% \text{ of } 10,000 = 500$
$n(B \cap C) = 3\% \text{ of } 10,000 = 300$
$n(A \cap C) = 4\% \text{ of } 10,000 = 400$
$n(A \cap B \cap C) = 2\% \text{ of } 10,000 = 200$
To find the number of families that buy newspaper $A$ only,we use the formula:
$n(A \text{ only}) = n(A) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C)]$
$n(A \text{ only}) = 4000 - [500 + 400 - 200]$
$n(A \text{ only}) = 4000 - [700] = 3300.$

(C) Let $C$ be the set of people travelling by car and $B$ be the set of people travelling by bus.
Given: $n(C) = 20\%$,$n(B) = 50\%$,and $n(C \cap B) = 10\%$.
We need to find the percentage of persons travelling by car or bus,which is given by $n(C \cup B)$.
Using the formula for the union of two sets:
$n(C \cup B) = n(C) + n(B) - n(C \cap B)$
Substituting the values:
$n(C \cup B) = 20 + 50 - 10 = 60$.
Therefore,the percentage of persons travelling by car or bus is $60\%$.

(D) Let $M, P,$ and $C$ represent the sets of students studying Mathematics,Physics,and Chemistry,respectively.
Given: $n(M) = 23, n(P) = 24, n(C) = 19, n(M \cap P) = 12, n(M \cap C) = 9, n(P \cap C) = 7, n(M \cap P \cap C) = 4$.
To find the number of students who have taken exactly one subject,we calculate:
$1$. Exactly Mathematics: $n(M) - [n(M \cap P) + n(M \cap C) - n(M \cap P \cap C)] = 23 - [12 + 9 - 4] = 23 - 17 = 6$.
$2$. Exactly Physics: $n(P) - [n(P \cap M) + n(P \cap C) - n(M \cap P \cap C)] = 24 - [12 + 7 - 4] = 24 - 15 = 9$.
$3$. Exactly Chemistry: $n(C) - [n(C \cap M) + n(C \cap P) - n(M \cap P \cap C)] = 19 - [9 + 7 - 4] = 19 - 12 = 7$.
The total number of students who have taken exactly one subject is $6 + 9 + 7 = 22$.

(B) The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.
Given $A = \{ 2, 4, 5 \}$,the number of elements in $A$ is $n(A) = 3$.
Given $B = \{ 7, 8, 9 \}$,the number of elements in $B$ is $n(B) = 3$.
The number of elements in the Cartesian product is given by the formula $n(A \times B) = n(A) \times n(B)$.
Therefore,$n(A \times B) = 3 \times 3 = 9$.

(C) Given sets are $A = \{a, b\}$,$B = \{c, d\}$,and $C = \{d, e\}$.
First,find the union of sets $B$ and $C$:
$B \cup C = \{c, d\} \cup \{d, e\} = \{c, d, e\}$.
Now,calculate the Cartesian product of set $A$ and the resulting set $(B \cup C)$:
$A \times (B \cup C) = \{a, b\} \times \{c, d, e\}$.
This results in the set of ordered pairs:
$\{(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)\}$.
Thus,the given set is equal to $A \times (B \cup C)$.

(C) Given $A \cup B = A \cap B$.
Let $x \in A$. Since $A \subseteq A \cup B$,it follows that $x \in A \cup B$.
Because $A \cup B = A \cap B$,we have $x \in A \cap B$.
By the definition of intersection,$x \in A \cap B \implies x \in A$ and $x \in B$. Thus,$x \in B$.
Since $x \in A \implies x \in B$,we conclude $A \subseteq B$.
Similarly,let $y \in B$. Since $B \subseteq A \cup B$,it follows that $y \in A \cup B$.
Because $A \cup B = A \cap B$,we have $y \in A \cap B$.
By the definition of intersection,$y \in A \cap B \implies y \in A$ and $y \in B$. Thus,$y \in A$.
Since $y \in B \implies y \in A$,we conclude $B \subseteq A$.
Since $A \subseteq B$ and $B \subseteq A$,it must be that $A = B$.

(B) To find the intersection $A \cap B$,we set the $y$-values equal: $e^x = e^{-x}$.
Multiplying both sides by $e^x$,we get $e^{2x} = 1$.
Taking the natural logarithm on both sides,$2x = 0$,which implies $x = 0$.
Substituting $x = 0$ into either equation,we get $y = e^0 = 1$.
Thus,the point $(0, 1)$ belongs to both sets $A$ and $B$.
Since there exists at least one common point,$A \cap B \neq \phi$.

(A) Given sets are $A = \{2, 3, 4, 8, 10\}$,$B = \{3, 4, 5, 10, 12\}$,and $C = \{4, 5, 6, 12, 14\}$.
First,find the intersection $A \cap B$:
$A \cap B = \{2, 3, 4, 8, 10\} \cap \{3, 4, 5, 10, 12\} = \{3, 4, 10\}$.
Next,find the intersection $A \cap C$:
$A \cap C = \{2, 3, 4, 8, 10\} \cap \{4, 5, 6, 12, 14\} = \{4\}$.
Finally,find the union $(A \cap B) \cup (A \cap C)$:
$(A \cap B) \cup (A \cap C) = \{3, 4, 10\} \cup \{4\} = \{3, 4, 10\}$.

(A) First,find the union of sets $B$ and $C$:
$B \cup C = \{b, c, d\} \cup \{a, b, d, e\} = \{a, b, c, d, e\}$.
Next,find the intersection of set $A$ with the result $(B \cup C)$:
$A \cap (B \cup C) = \{a, b, c\} \cap \{a, b, c, d, e\}$.
The common elements in both sets are $a, b,$ and $c$.
Therefore,$A \cap (B \cup C) = \{a, b, c\}$.
The correct option is $A$.

(A) By the definition of set difference,$B - A$ is the set of all elements that are in $B$ but not in $A$.
Therefore,for any element $x \in (B - A)$,it is guaranteed that $x \notin A$.
Since $A \cap (B - A)$ represents the set of elements common to both $A$ and $(B - A)$,and there are no elements common to $A$ and $(B - A)$,the intersection must be the empty set.
Thus,$A \cap (B - A) = \phi$.

(C) Using De Morgan's Law,we know that $(A \cup B)' = A' \cap B'$.
Substituting this into the expression,we get $A \cap (A' \cap B')$.
By the associative law of sets,this can be rewritten as $(A \cap A') \cap B'$.
Since $A \cap A' = \phi$ (the empty set),the expression becomes $\phi \cap B'$.
Any set intersected with the empty set results in the empty set,so $\phi \cap B' = \phi$.

(B) Given the universal set $U = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$.
First,find the complement of set $B$,denoted as $B'$.
$B' = U \setminus B = \{ 1, 2, 3, 4, 5, 8, 9, 10 \}$.
Now,find the intersection of set $A$ and $B'$.
$A \cap B' = \{ 1, 2, 5 \} \cap \{ 1, 2, 3, 4, 5, 8, 9, 10 \}$.
The common elements are $\{ 1, 2, 5 \}$.
Since $\{ 1, 2, 5 \} = A$,the result is $A$.

(C) Given that ${N_a} = \{an : n \in N\}$,which represents the set of all multiples of $a$.
Thus,${N_5} = \{5, 10, 15, 20, 25, 30, 35, 40, \dots\}$ and ${N_7} = \{7, 14, 21, 28, 35, 42, \dots\}$.
The intersection ${N_5} \cap {N_7}$ consists of numbers that are multiples of both $5$ and $7$.
Since $5$ and $7$ are relatively prime,their least common multiple $(LCM)$ is $5 \times 7 = 35$.
Therefore,the common multiples are multiples of $35$,which is denoted as ${N_{35}}$.

(A) Given that $aN = \{ ax : x \in N \}$,which represents the set of all multiples of $a$.
$3N = \{ x \in N : x \text{ is a multiple of } 3 \}$.
$7N = \{ x \in N : x \text{ is a multiple of } 7 \}$.
The intersection $3N \cap 7N$ consists of elements that are multiples of both $3$ and $7$.
Since $3$ and $7$ are coprime,their least common multiple is $3 \times 7 = 21$.
Therefore,$3N \cap 7N = \{ x \in N : x \text{ is a multiple of } 21 \} = 21N$.

(A) Using De Morgan's Law,$(A \cup B)' = A' \cap B'$.
So,the expression becomes $(A' \cap B') \cup (A' \cap B)$.
By the distributive law,we can factor out $A'$:
$(A' \cap B') \cup (A' \cap B) = A' \cap (B' \cup B)$.
Since $B' \cup B = U$ (the universal set),the expression simplifies to:
$A' \cap U = A'$.
Thus,$(A \cup B)' \cup (A' \cap B) = A'$.

(C) From the Venn diagram,the set $S = (A - B) \cup (B - C) \cup (C - A)$ represents the region containing elements that belong to exactly one of the sets $A, B,$ or $C$,or elements that belong to exactly two of the sets,but excludes the intersection of all three sets $A \cap B \cap C$.
Specifically,the region $(A - B) \cup (B - C) \cup (C - A)$ covers all parts of the union $A \cup B \cup C$ except for the central region where all three sets overlap.
Therefore,the complement of this set within the universal set $U$ (where $U = A \cup B \cup C$) is the region that remains,which is exactly the intersection of all three sets:
$\{ (A - B) \cup (B - C) \cup (C - A)\} ' = A \cap B \cap C$.

(C) Given that $A \subseteq B$,which means every element of set $A$ is also an element of set $B$.
Therefore,the union of sets $A$ and $B$ is equal to set $B$,i.e.,$A \cup B = B$.
Consequently,the number of elements in $A \cup B$ is equal to the number of elements in $B$.
$n(A \cup B) = n(B) = 6$.

(A) Let the total number of combatants be $100$.
Let $A, B, C, D$ be the sets of combatants who lost an eye,an ear,an arm,and a leg,respectively.
Given: $n(A) = 70, n(B) = 80, n(C) = 75, n(D) = 85$.
The number of people who did $NOT$ lose these parts are:
$n(A^c) = 100 - 70 = 30$
$n(B^c) = 100 - 80 = 20$
$n(C^c) = 100 - 75 = 25$
$n(D^c) = 100 - 85 = 15$
The number of people who lost at least one part is $100 - n(A^c \cap B^c \cap C^c \cap D^c)$.
To minimize $x$ (the intersection of all four),we maximize the union of the complements.
The maximum number of people who lost at least one part is $100$.
Using the principle of inclusion-exclusion for the complements:
$n(A^c \cup B^c \cup C^c \cup D^c) \leq n(A^c) + n(B^c) + n(C^c) + n(D^c) = 30 + 20 + 25 + 15 = 90$.
Thus,the number of people who lost all four parts is at least $100 - 90 = 10$.
Therefore,the minimum value of $x$ is $10$.

(D) Let $C$,$H$,and $B$ represent the sets of boys who play cricket,hockey,and basketball,respectively.
Given:
$n(C) = 224, n(H) = 240, n(B) = 336$
$n(H \cap B) = 64, n(C \cap B) = 80, n(C \cap H) = 40$
$n(C \cap H \cap B) = 24$
Total boys $n(U) = 800$
Using the Principle of Inclusion-Exclusion:
$n(C \cup H \cup B) = n(C) + n(H) + n(B) - n(C \cap H) - n(H \cap B) - n(C \cap B) + n(C \cap H \cap B)$
$n(C \cup H \cup B) = 224 + 240 + 336 - 40 - 64 - 80 + 24$
$n(C \cup H \cup B) = 800 - 184 + 24 = 640$
The number of boys who did not play any game is $n(U) - n(C \cup H \cup B) = 800 - 640 = 160$.

(C) Let $A$ be the set of Americans who like cheese and $B$ be the set of Americans who like apples.
Let the total population of Americans be $100$.
Given: $n(A) = 63$ and $n(B) = 76$.
Using the formula for the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
$n(A \cup B) = 63 + 76 - n(A \cap B) = 139 - n(A \cap B)$.
Since $n(A \cup B) \le 100$,we have $139 - n(A \cap B) \le 100$,which implies $n(A \cap B) \ge 39$.
Also,the intersection of two sets is a subset of each set,so $n(A \cap B) \le n(A)$ and $n(A \cap B) \le n(B)$.
Therefore,$n(A \cap B) \le 63$ and $n(A \cap B) \le 76$. The tighter bound is $n(A \cap B) \le 63$.
Combining these,we get $39 \le n(A \cap B) \le 63$.
Since $x = n(A \cap B)$,we have $39 \le x \le 63$.

(A) Let $M$ be the set of teachers who teach mathematics and $P$ be the set of teachers who teach physics.
Given:
Total teachers $n(M \cup P) = 20$
Teachers teaching mathematics $n(M) = 12$
Teachers teaching both subjects $n(M \cap P) = 4$
Using the formula for the union of two sets:
$n(M \cup P) = n(M) + n(P) - n(M \cap P)$
Substituting the given values:
$20 = 12 + n(P) - 4$
$20 = 8 + n(P)$
$n(P) = 20 - 8 = 12$
Therefore,the number of teachers teaching physics is $12$.

(A) Let $C, H, F$ denote the sets of members who are on the cricket team,hockey team,and football team,respectively.
We are given:
$n(C) = 21, n(H) = 26, n(F) = 29$
$n(H \cap C) = 14, n(H \cap F) = 15, n(F \cap C) = 12$
$n(C \cap H \cap F) = 8$
We need to find the total number of members,which is $n(C \cup H \cup F)$.
Using the Principle of Inclusion-Exclusion:
$n(C \cup H \cup F) = n(C) + n(H) + n(F) - [n(C \cap H) + n(H \cap F) + n(F \cap C)] + n(C \cap H \cap F)$
Substituting the values:
$n(C \cup H \cup F) = (21 + 26 + 29) - (14 + 15 + 12) + 8$
$n(C \cup H \cup F) = 76 - 41 + 8$
$n(C \cup H \cup F) = 43$
Thus,there are $43$ members in total.

(D) Let $n(M)$ be the number of students who passed in Mathematics and $n(P)$ be the number of students who passed in Physics.
Given: $n(M) = 55$,$n(P) = 67$,and the total number of students $n(M \cup P) = 100$.
Using the formula for the union of two sets: $n(M \cup P) = n(M) + n(P) - n(M \cap P)$.
Substituting the values: $100 = 55 + 67 - n(M \cap P)$.
$100 = 122 - n(M \cap P)$.
Therefore,$n(M \cap P) = 122 - 100 = 22$.
This represents the number of students who passed in both Mathematics and Physics.
The number of students who passed in Physics only is given by $n(P \text{ only}) = n(P) - n(M \cap P)$.
$n(P \text{ only}) = 67 - 22 = 45$.

(C) In general,the Cartesian product of two sets is not commutative,i.e.,$A \times B \neq B \times A$.
For the condition $A \times B = B \times A$ to hold,the sets must be equal,i.e.,$A = B$.
If $A = B$,then $A \times B = A \times A$ and $B \times A = A \times A$,which satisfies the equality.

(B) Given sets are $A = \{1, 2, 4\}$,$B = \{2, 4, 5\}$,and $C = \{2, 5\}$.
First,find the set difference $(A - B)$:
$A - B$ represents elements that are in $A$ but not in $B$.
$A - B = \{1, 2, 4\} - \{2, 4, 5\} = \{1\}$.
Next,find the set difference $(B - C)$:
$B - C$ represents elements that are in $B$ but not in $C$.
$B - C = \{2, 4, 5\} - \{2, 5\} = \{4\}$.
Finally,calculate the Cartesian product $(A - B) \times (B - C)$:
$(A - B) \times (B - C) = \{1\} \times \{4\} = \{(1, 4)\}$.

(A) Given that $(1, 3), (2, 5), (3, 3) \in A \times B$.
From the definition of the Cartesian product $A \times B$,the set $A$ consists of all first coordinates and the set $B$ consists of all second coordinates.
Thus,$A = \{1, 2, 3\}$ and $B = \{3, 5\}$.
The total number of elements in $A \times B$ is $n(A) \times n(B) = 3 \times 2 = 6$.
The set $A \times B$ is given by the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
$A \times B = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5)\}$.
We are given three elements: $(1, 3), (2, 5), (3, 3)$.
The remaining elements are $(1, 5), (2, 3), (3, 5)$.

(B) Given sets are $A = \{1, 2, 3\}$ and $B = \{3, 8\}$.
First,find the union of sets $A$ and $B$: $A \cup B = \{1, 2, 3, 8\}$.
Next,find the intersection of sets $A$ and $B$: $A \cap B = \{3\}$.
Now,find the Cartesian product $(A \cup B) \times (A \cap B)$:
$(A \cup B) \times (A \cap B) = \{1, 2, 3, 8\} \times \{3\} = \{(1, 3), (2, 3), (3, 3), (8, 3)\}$.

(C) Given sets are $A = \{2, 3, 5\}$ and $B = \{2, 5, 6\}$.
First,find the set difference $(A - B)$,which contains elements present in $A$ but not in $B$: $A - B = \{3\}$.
Next,find the intersection $(A \cap B)$,which contains elements common to both $A$ and $B$: $A \cap B = \{2, 5\}$.
Finally,calculate the Cartesian product $(A - B) \times (A \cap B) = \{3\} \times \{2, 5\} = \{(3, 2), (3, 5)\}$.

(A) Let $N$,$P$,and $H$ represent the sets of students taking needle work,physics,and history,respectively.
Given: $n(N) = 12$,$n(P) = 16$,$n(H) = 18$,and $n(N \cup P \cup H) = 30$.
Since no one takes all three subjects,$n(N \cap P \cap H) = 0$.
Using the principle of inclusion-exclusion for three sets:
$n(N \cup P \cup H) = n(N) + n(P) + n(H) - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + n(N \cap P \cap H)$.
Substituting the values:
$30 = 12 + 16 + 18 - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + 0$.
$30 = 46 - [n(N \cap P) + n(P \cap H) + n(N \cap H)]$.
Therefore,$n(N \cap P) + n(P \cap H) + n(N \cap H) = 46 - 30 = 16$.
The number of students taking exactly two subjects is given by $[n(N \cap P) + n(P \cap H) + n(N \cap H)] - 3n(N \cap P \cap H)$.
Since $n(N \cap P \cap H) = 0$,the number of students taking exactly two subjects is $16 - 3(0) = 16$.

(D) Given that $n(A) = 4$,$n(B) = 3$,and $n(A \times B \times C) = 24$.
We know that for the Cartesian product of three sets,the number of elements is given by $n(A \times B \times C) = n(A) \times n(B) \times n(C)$.
Substituting the given values into the formula:
$4 \times 3 \times n(C) = 24$
$12 \times n(C) = 24$
$n(C) = \frac{24}{12}$
$n(C) = 2$.

(C) The given set is defined by the equation $2a^2 + 3b^2 = 35$,where $a, b \in \mathbb{Z}$.
We test integer values for $a$ and $b$:
If $a = \pm 2$,then $2(4) + 3b^2 = 35 \implies 8 + 3b^2 = 35 \implies 3b^2 = 27 \implies b^2 = 9 \implies b = \pm 3$.
This gives the pairs: $(2, 3), (2, -3), (-2, 3), (-2, -3)$.
If $a = \pm 4$,then $2(16) + 3b^2 = 35 \implies 32 + 3b^2 = 35 \implies 3b^2 = 3 \implies b^2 = 1 \implies b = \pm 1$.
This gives the pairs: $(4, 1), (4, -1), (-4, 1), (-4, -1)$.
Checking other values of $a$: if $a = 0, 3b^2 = 35$ (no integer solution); if $a = \pm 1, 2 + 3b^2 = 35 \implies 3b^2 = 33 \implies b^2 = 11$ (no integer solution); if $a = \pm 3, 18 + 3b^2 = 35 \implies 3b^2 = 17$ (no integer solution).
Thus,the total number of elements is $4 + 4 = 8$.

(C) The Cartesian product of two sets $A$ and $B$,denoted by $A \times B$,is defined as the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.
Given $A = \{ 1, 2, 3, 4 \}$ and $B = \{ a, b \}$.
Therefore,$A \times B = \{ (1, a), (1, b), (2, a), (2, b), (3, a), (3, b), (4, a), (4, b) \}$.
Thus,option $C$ is the correct answer.

(A) First,find the union of sets $A$ and $B$:
$A \cup B = \{1, 2, 3, 4, 5\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 5, 6\}$.
Next,find the intersection of the resulting set $(A \cup B)$ with set $C$:
$(A \cup B) \cap C = \{1, 2, 3, 4, 5, 6\} \cap \{3, 4, 6\} = \{3, 4, 6\}$.
Therefore,the correct option is $A$.

(D) The power set of a set $A$,denoted by $P(A)$,is the set of all possible subsets of $A$.
Given $A = \{x, y\}$.
The subsets of $A$ are:
$1$. The empty set: $\phi$
$2$. Single-element sets: $\{x\}$ and $\{y\}$
$3$. The set itself: $\{x, y\}$
Therefore,the power set $P(A) = \{\phi, \{x\}, \{y\}, \{x, y\}\}$.

(D) Let the set $S$ contain $N = 2n + 1$ elements.
We need to find the number of subsets containing more than $n$ elements,which means subsets with $(n + 1), (n + 2), \dots, (2n + 1)$ elements.
The number of such subsets is given by the sum: $S = \binom{2n+1}{n+1} + \binom{2n+1}{n+2} + \dots + \binom{2n+1}{2n+1}$.
Using the property of binomial coefficients $\binom{n}{r} = \binom{n}{n-r}$,we can rewrite the terms:
$\binom{2n+1}{n+1} = \binom{2n+1}{n}$,$\binom{2n+1}{n+2} = \binom{2n+1}{n-1}$,and so on.
Thus,$S = \binom{2n+1}{n} + \binom{2n+1}{n-1} + \dots + \binom{2n+1}{0}$.
We know that the sum of all binomial coefficients for a set of size $m$ is $2^m$,i.e.,$\sum_{k=0}^{m} \binom{m}{k} = 2^m$.
For $m = 2n + 1$,the total sum is $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$.
Since $\binom{2n+1}{k} = \binom{2n+1}{2n+1-k}$,the sum of the first half of the coefficients is equal to the sum of the second half.
Therefore,$S = \frac{1}{2} \times 2^{2n+1} = 2^{2n}$.

(A) In set theory, the symbol $\subseteq$ denotes a subset relationship, while $\in$ denotes an element relationship.
For the set $A = \{a, b, c\}$:
$1$. The set ${a}$ is a subset of $A$ because every element of ${a}$ is also an element of $A$. Thus, ${a} \subseteq {a, b, c}$ is a true statement.
$2$. The set ${a}$ is not an element of $A$; only $a$ is an element of $A$. Thus, ${a} \in {a, b, c}$ is false.
$3$. The empty set $\phi$ is a subset of every set, but it is not an element of $A$ unless explicitly stated. Thus, $\phi \in {a, b, c}$ is false.
Therefore, the correct statement is ${a} \subseteq {a, b, c}$.

(B) Given sets are $A = \{4, 8, 12, 16, 20, 24, \dots\}$ and $B = \{6, 12, 18, 24, 30, \dots\}$.
To find $A \cap B$,we look for the common elements in both sets $A$ and $B$.
The common elements are multiples of both $4$ and $6$.
The least common multiple $(LCM)$ of $4$ and $6$ is $12$.
Therefore,the intersection $A \cap B = \{12, 24, 36, \dots\}$,which represents all multiples of $12$.

(C) Let $M$,$P$,and $C$ represent the sets of students taking Mathematics,Physics,and Chemistry,respectively.
Given:
$n(M) = 100$
$n(P) = 70$
$n(C) = 40$
$n(M \cap P) = 30$
$n(M \cap C) = 28$
$n(P \cap C) = 23$
$n(M \cap P \cap C) = 18$
To find the number of students who have offered Mathematics alone,we use the formula:
$n(M \text{ only}) = n(M) - [n(M \cap P) + n(M \cap C) - n(M \cap P \cap C)]$
$n(M \text{ only}) = 100 - [30 + 28 - 18]$
$n(M \text{ only}) = 100 - [40]$
$n(M \text{ only}) = 60$
Thus,$60$ students have offered Mathematics alone.

(D) Let us analyze each relation:
$(1) \, A - B = A - (A \cap B)$: This is a standard set identity. The set $A - B$ consists of elements in $A$ that are not in $B$. Since $A \cap B$ is the set of elements common to both $A$ and $B$,removing these from $A$ leaves exactly $A - B$. Thus,$(1)$ is correct.
$(2) \, A = (A \cap B) \cup (A - B)$: This represents the partition of set $A$ into two disjoint sets: the part common to $B$ $(A \cap B)$ and the part not in $B$ $(A - B)$. Their union is indeed $A$. Thus,$(2)$ is correct.
$(3) \, A - (B \cup C) = (A - B) \cup (A - C)$: According to De Morgan's laws for set difference,$A - (B \cup C) = (A - B) \cap (A - C)$. Therefore,the given relation is incorrect.
Conclusion: Relations $(1)$ and $(2)$ are correct.

(B) The number of elements common to the Cartesian products $A \times B$ and $B \times A$ is given by the intersection of the two sets:
$n((A \times B) \cap (B \times A))$
Using the property of Cartesian products,$(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$.
Since $n(A \cap B) = 99$,we have $n(B \cap A) = 99$.
Therefore,the number of elements is $n(A \cap B) \times n(B \cap A) = 99 \times 99 = 99^2$.

(D) First,we find the number of elements in the union of sets $A$ and $B$ using the formula:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Substituting the given values:
$n(A \cup B) = 12 + 9 - 4 = 17$
Now,to find the number of elements in the complement of $(A \cup B)$,we use the formula:
$n((A \cup B)^C) = n(U) - n(A \cup B)$
Substituting the values:
$n((A \cup B)^C) = 20 - 17 = 3$
Thus,the correct option is $D$.

(A) relation over a set $A$ is defined as any subset of the Cartesian product $A \times A$.
Given $A = \{1, 2, 3\}$,the number of elements in $A$ is $n(A) = 3$.
The number of elements in the Cartesian product $A \times A$ is $n(A \times A) = n(A) \times n(A) = 3 \times 3 = 9$.
The total number of subsets of a set with $m$ elements is $2^m$.
Therefore,the total number of distinct relations that can be defined over $A$ is the number of subsets of $A \times A$,which is $2^9$.