Basic of 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 $x$ such that its square is negative. Thus,this set is empty.
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 $\{ -1, 2 \}$.
Therefore,the correct option is $B$.

(A) Given the conditions for set $A$:
$1$. $x^2 = 16 \implies x = 4$ or $x = -4$.
$2$. $2x = 6 \implies x = 3$.
For an element $x$ to be in set $A$,it must satisfy both conditions simultaneously.
Since there is no value of $x$ that satisfies both $x \in \{4, -4\}$ and $x = 3$,the set $A$ contains no elements.
Therefore,$A = \phi$.

(C) The number of subsets of a set with $n$ elements is given by the sum of binomial coefficients:
$C(n, 0) + C(n, 1) + C(n, 2) + \dots + C(n, n) = \sum_{k=0}^{n} \binom{n}{k} = 2^n$.
Therefore,the total number of subsets is $2^n$.

(C) The number of elements in the set $A = \{1, 2, 3\}$ is $n = 3$.
The total number of subsets of a set with $n$ elements is given by $2^n$.
Here,the total number of subsets is $2^3 = 8$.
$A$ proper subset is any subset of a set except the set itself.
Therefore,the number of proper subsets is $2^n - 1$.
Number of proper subsets $= 2^3 - 1 = 8 - 1 = 7$.

(D) The null set (or empty set) is a set that contains no elements.
In set-builder notation,we define a set by a property that its elements must satisfy.
For the set $\{x : x \neq x\}$,there is no element $x$ that satisfies the condition of not being equal to itself.
Therefore,this set contains no elements and represents the null set.

(B) By the law of identity,for any element $x$,it must be true that $x = x$.
There is no element $x$ such that $x \ne x$.
Therefore,the set $A$ contains no elements.
Thus,$A = \emptyset$ or $\left\{ \right\}$.

(B) The set $Q$ is defined as $Q = \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\}$ where $y \in N = \{1, 2, 3, \dots\}$.
For $y = 1$,$x = \frac{1}{1} = 1$. Thus,$1 \in Q$.
For $y = 2$,$x = \frac{1}{2}$.
Since $0$ is not of the form $\frac{1}{y}$ for any $y \in N$,$0 \notin Q$.
Since $2$ is not of the form $\frac{1}{y}$ for any $y \in N$,$2 \notin Q$.
Since $\frac{2}{3}$ is not of the form $\frac{1}{y}$ for any $y \in N$,$\frac{2}{3} \notin Q$.
Therefore,the correct option is $B$.

(D) The empty set,denoted by $\{\}$ or $\phi$,is a subset of every set.
Therefore,the correct option is $(d)$.

(B) The set $S$ is given by $S = \{ 0, 1, 5, 4, 7 \}$.
The number of elements in the set $S$ is $n = 5$.
The total number of subsets of a set with $n$ elements is given by the formula $2^n$.
Substituting $n = 5$,we get $2^5 = 32$.
Therefore,the total number of subsets of $S$ is $32$.

(A) The total number of subsets of a set with $n$ elements is given by the formula $2^n$.
For the set ${1, 2, 3, 4}$,the number of elements $n = 4$.
Total number of subsets = $2^4 = 16$.
The number of non-empty subsets is obtained by subtracting the empty set (null set) from the total number of subsets.
Number of non-empty subsets = $2^n - 1 = 2^4 - 1 = 16 - 1 = 15$.

(B) Given that $A \cap B = B$.
By the definition of intersection,$A \cap B$ is a subset of both $A$ and $B$.
Specifically,$A \cap B \subseteq A$ and $A \cap B \subseteq B$.
Since $A \cap B = B$,substituting $B$ for $A \cap B$ in the first relation gives $B \subseteq A$.
Therefore,the correct option is $B$.

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

(B) We know that for any two sets $A$ and $B$,the intersection $A \cap B$ is a subset of $A$,and $A$ is a subset of the union $A \cup B$.
Thus,$A \cap B \subseteq A \subseteq A \cup B$.
Therefore,$A \cap B \subseteq A \cup B$.

(C) The set ${N_a}$ consists of all multiples of $a$,i.e.,${N_a} = \{a, 2a, 3a, \dots\}$.
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,the least common multiple of $5$ and $7$ is $5 \times 7 = 35$.
Therefore,the common multiples are multiples of $35$,which is represented as ${N_{35}}$.

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

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

(C) Given the equation $2a^2 + 3b^2 = 35$ for integers $a, b \in Z$.
We test possible values for $b^2$ such that $3b^2 < 35$:
If $b^2 = 0$,$2a^2 = 35$ (no integer solution for $a$).
If $b^2 = 1$,$2a^2 = 35 - 3 = 32 \implies a^2 = 16 \implies a = \pm 4$. This gives pairs $(4, 1), (4, -1), (-4, 1), (-4, -1)$.
If $b^2 = 4$,$2a^2 = 35 - 12 = 23$ (no integer solution for $a$).
If $b^2 = 9$,$2a^2 = 35 - 27 = 8 \implies a^2 = 4 \implies a = \pm 2$. This gives pairs $(2, 3), (2, -3), (-2, 3), (-2, -3)$.
If $b^2 = 16$,$3b^2 = 48 > 35$ (no further solutions).
Thus,the set contains $8$ elements: $(4, 1), (4, -1), (-4, 1), (-4, -1), (2, 3), (2, -3), (-2, 3), (-2, -3)$.

(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 $\phi$,$\{x\}$,$\{y\}$,and $\{x, y\}$.
Therefore,the power set $P(A) = \{\phi, \{x\}, \{y\}, \{x, y\}\}$.

(A) Let the set be $A = \{a, b, c\}$.
$1$. For option $A$: The set $\{a\}$ contains the element $a$,which is also an element of $A$. Therefore,$\{a\} \subseteq A$ is a true statement.
$2$. For option $B$: The element $a$ belongs to $A$ $(a \in A)$,but the set $\{a\}$ is a subset,not an element of $A$. Thus,$\{a\} \in A$ is false.
$3$. For option $C$: The empty set $\phi$ is a subset of every set,but it is not necessarily an element of $A$ unless explicitly stated. Thus,$\phi \in A$ is false.
Therefore,the correct statement is $\{a\} \subseteq \{a, b, c\}$.

(C) The given congruence relation is $x \equiv 3 \pmod{7}$.
By the definition of modular arithmetic,$x \equiv a \pmod{n}$ implies that $x - a$ is a multiple of $n$.
Therefore,$x - 3 = 7p$ for some integer $p \in \mathbb{Z}$.
Rearranging the equation to solve for $x$,we get $x = 7p + 3$.
Thus,the solution set is $\{7p + 3 : p \in \mathbb{Z}\}$.

(B) ${N_3} \cap {N_4} = \{3, 6, 9, 12, 15, \dots\} \cap \{4, 8, 12, 16, 20, \dots\}$
$= \{12, 24, 36, \dots\} = {N_{12}}$.
$\text{Trick: } {N_a} \cap {N_b} = {N_{\text{lcm}(a, b)}}$.
$\because \text{lcm}(3, 4) = 12$,therefore ${N_3} \cap {N_4} = {N_{12}}$.

(D) Let $A, B, C$ be subsets of a universal set $U$.
$1$. Reflexivity: For every set $A$,$A \subseteq A$ is true. Thus,the relation '$\subseteq$' is reflexive.
$2$. Symmetry: If $A \subseteq B$,it is not necessarily true that $B \subseteq A$ (for example,if $A = \{1\}$ and $B = \{1, 2\}$,then $A \subseteq B$ but $B \not\subseteq A$). Thus,the relation '$\subseteq$' is not symmetric.
$3$. Transitivity: If $A \subseteq B$ and $B \subseteq C$,then $A \subseteq C$. Thus,the relation '$\subseteq$' is transitive.
Therefore,the relation '$\subseteq$' is reflexive and transitive,but not symmetric. Since it is not symmetric,it is not an equivalence relation. The correct option is $(D)$.

(C) According to the Remainder Theorem,when a polynomial $P(x)$ is divided by $(x - a)$,the remainder is $P(a)$.
Here,$P(x) = 1 + x + x^3 + x^9 + x^{27} + x^{81} + x^{243}$ and the divisor is $(x - 1)$,so $a = 1$.
The remainder is $P(1) = 1 + 1 + 1^3 + 1^9 + 1^{27} + 1^{81} + 1^{243}$.
$P(1) = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7$.

(C) There are $32$ distinct positions for teeth in a human mouth.
For each position,there are $2$ possibilities: either a tooth is present or it is absent.
Since there are $32$ such positions,the total number of possible combinations of teeth is $2^{32}$.
However,the problem states that there is no person without a tooth,which means we must exclude the case where all $32$ positions are empty.
Therefore,the maximum population of the city is $2^{32} - 1$.

(A) The set $bN$ represents the set of all positive integral multiples of $b$,and $cN$ represents the set of all positive integral multiples of $c$.
The intersection $bN \cap cN$ consists of numbers that are multiples of both $b$ and $c$.
Since $b$ and $c$ are relatively prime,the least common multiple of $b$ and $c$ is $bc$.
Therefore,$bN \cap cN$ is the set of all positive integral multiples of $bc$,which is denoted as $(bc)N$.
Given $bN \cap cN = dN$,we conclude that $d = bc$.

(C) The number of elements in set $A$ is $n = 5$.
The total number of subsets of a set with $n$ elements is given by $2^n$.
Therefore,the total number of subsets of $A$ is $2^5 = 32$.
$A$ proper subset is any subset of $A$ except the set $A$ itself.
Thus,the number of proper subsets is $2^n - 1$.
Substituting $n = 5$,we get $2^5 - 1 = 32 - 1 = 31$.

(D) Since the intelligence of a student in a class is not defined by a specific criterion,it is subjective. Therefore,it is not a well-defined collection and does not form a set.

(B) For option $A$: $x^2 - 1 = 0 \implies x = \pm 1$,which are real numbers. So,the set is $\{-1, 1\}$.
For option $B$: $x^2 + 1 = 0 \implies x^2 = -1$. Since the square of any real number cannot be negative,there is no real value of $x$ that satisfies this equation. Thus,this is an empty set.
For option $C$: $x^2 - 9 = 0 \implies x = \pm 3$,which are real numbers. So,the set is $\{-3, 3\}$.
For option $D$: $x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = 2, -1$,which are real numbers. So,the set is $\{-1, 2\}$.
Therefore,the correct option is $B$.

(A) Given the conditions for set $A$:
$1$) $x^2 = 16 \implies x = 4$ or $x = -4$.
$2$) $2x = 6 \implies x = 3$.
Since there is no value of $x$ that satisfies both equations simultaneously,the intersection of these conditions is empty.
Therefore,$A = \phi$.

(A) Since $A \cap B$ is a subset of $A$,i.e.,$(A \cap B) \subseteq A$.
Therefore,the union of $A$ and $(A \cap B)$ is simply $A$.
$A \cup (A \cap B) = A$.

(A) Given $bN = \{b, 2b, 3b, \dots\}$ and $cN = \{c, 2c, 3c, \dots\}$.
The intersection $bN \cap cN$ consists of all multiples of both $b$ and $c$.
Since $b$ and $c$ are coprime,the least common multiple of $b$ and $c$ is $LCM(b, c) = b \times c$.
Therefore,$bN \cap cN = (bc)N$.
Comparing this with $dN$,we get $d = bc$.

(A) According to the distributive property of the Cartesian product over the union of sets,for any three sets $A, B$,and $C$,we have:
$A \times (B \cup C) = (A \times B) \cup (A \times C)$.

(C) By the distributive property of the Cartesian product over the union of sets,we have:
$R \times (P \cup Q) = (R \times P) \cup (R \times Q)$

(D) set that contains no elements is called an empty set or null set.
In set-builder form,it is represented by defining a condition that can never be satisfied by any element.
For example,$\{x : x \neq x\}$ is an empty set because no element can be unequal to itself.
Therefore,the correct representation in set-builder form is $\{x : x \neq x\}$.

(B) By the law of identity,for any element $x$,it must be true that $x = x$.
Since the condition given is $x \ne x$,there is no element $x$ that can satisfy this condition.
Therefore,the set $A$ contains no elements.
$A$ set with no elements is called an empty set or a null set,denoted by $\{\}$ or $\phi$.

(A) Given set $A = \{ \phi, \{ \phi \} \}$.
The number of elements in $A$ is $n(A) = 2$.
The power set $\mathcal{P}(A)$ contains $2^{n(A)} = 2^2 = 4$ elements.
The elements of the power set are the subsets of $A$.
The subsets of $A$ are $\phi$,$\{ \phi \}$,$\{ \{ \phi \} \}$,and $\{ \phi, \{ \phi \} \} = A$.
Therefore,$\mathcal{P}(A) = \{ \phi, \{ \phi \}, \{ \{ \phi \} \}, A \}$.

(B) The set $Q$ is defined as $Q = \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\}$.
Since $y \in N$ (where $N = \{1, 2, 3, \dots\}$),the smallest value for $y$ is $1$.
For $y = 1$,$x = \frac{1}{1} = 1$.
Therefore,$1 \in Q$ is true.
For $0 \in Q$,we would need $\frac{1}{y} = 0$,which is impossible for any finite $y \in N$.
For $2 \in Q$,we would need $\frac{1}{y} = 2$,so $y = \frac{1}{2}$,which is not a natural number.
For $\frac{2}{3} \in Q$,we would need $\frac{1}{y} = \frac{2}{3}$,so $y = \frac{3}{2}$,which is not a natural number.
Thus,the correct option is $B$.

(D) By definition,the empty set,denoted by $\{\}$ or $\phi$,is a subset of every set.
Therefore,$\{\}$ is a subset of all other sets provided in the options.

(B) The set $S$ is given by $S = \{0, 1, 5, 4, 7\}$.
The number of elements in set $S$ is $n = 5$.
The formula for the number of subsets of a set with $n$ elements is $2^n$.
Therefore,the number of subsets of $S$ is $2^5 = 32$.
Thus,the correct option is $B$.

(C) The number of elements in set $A$ is $n = 5$.
The total number of subsets of a set with $n$ elements is given by the formula $2^n$.
Here,the total number of subsets is $2^5 = 32$.
$A$ proper subset is any subset of $A$ except the set $A$ itself.
Therefore,the number of proper subsets is $2^n - 1$.
Substituting the value of $n$,we get $2^5 - 1 = 32 - 1 = 31$.

(A) The union of two sets $A$ and $B$ is defined as the set of all elements that are in $A$,in $B$,or in both.
If $A \cup B = A$,it implies that every element of $B$ must already be an element of $A$.
By the definition of a subset,if every element of $B$ is in $A$,then $B$ is a subset of $A$,denoted as $B \subseteq A$.
Therefore,the condition $A \cup B = A$ holds if and only if $B \subseteq A$.

(C) By definition,two sets $A$ and $B$ are said to be disjoint if they have no elements in common.
This means that their intersection is the empty set,denoted as $A \cap B = \phi$.

(A) Given that $A \subseteq B$,it means that every element of set $A$ is also an element of set $B$.
By the definition of intersection,$A \cap B$ consists of all elements that are common to both set $A$ and set $B$.
Since all elements of $A$ are already in $B$,the common elements between $A$ and $B$ are simply the elements of $A$.
Therefore,$A \cap B = A$.

(A) Given that $aN = \{ ax : x \in N \}$.
This represents the set of all multiples of $a$.
Therefore,$3N = \{ 3, 6, 9, 12, 15, 18, 21, 24, \dots \}$ and $7N = \{ 7, 14, 21, 28, \dots \}$.
The intersection $3N \cap 7N$ consists of all numbers that are multiples of both $3$ and $7$.
The smallest common multiple of $3$ and $7$ is their least common multiple,$\text{LCM}(3, 7) = 21$.
Thus,$3N \cap 7N = \{ 21, 42, 63, \dots \} = 21N$.
Comparing this with $kN$,we get $k = 21$.

(A) Given that $A \subseteq B$,this means that every element of set $A$ is also an element of set $B$.
Therefore,the intersection of $A$ and $B$ is simply set $A$,i.e.,$A \cap B = A$.
Since $n(A) = 3$,it follows that $n(A \cap B) = n(A) = 3$.

(B) According to the distributive law of sets,the intersection of a set with the union of two other sets is given by the distributive property:
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
This is a fundamental property of set operations.

(C) According to the Remainder Theorem,when a polynomial $P(x)$ is divided by $(x - a)$,the remainder is $P(a)$.
Here,we are dividing by $(x - 1)$,so $a = 1$.
We need to calculate $P(1)$:
$P(1) = 1 + (1) + (1)^3 + (1)^9 + (1)^{27} + (1)^{81} + (1)^{243}$
Since $1^n = 1$ for any positive integer $n$,we have:
$P(1) = 1 + 1 + 1 + 1 + 1 + 1 + 1$
$P(1) = 7$
Therefore,the remainder is $7$.

(C) Given the equation $y = 3[x] + 1 = 4[x - 1] - 10$.
Using the property of the $Greatest$ $Integer$ $Function$ $[x - n] = [x] - n$ for any integer $n$,we have $[x - 1] = [x] - 1$.
Substituting this into the equation:
$3[x] + 1 = 4([x] - 1) - 10$
$3[x] + 1 = 4[x] - 4 - 10$
$3[x] + 1 = 4[x] - 14$
Rearranging the terms to solve for $[x]$:
$4[x] - 3[x] = 1 + 14$
$[x] = 15$.
Now,substitute $[x] = 15$ into the expression for $y$:
$y = 3(15) + 1 = 45 + 1 = 46$.
We need to find the value of $[x + 2y]$:
$[x + 2y] = [x + 2(46)] = [x + 92]$.
Since $92$ is an integer,we use the property $[x + n] = [x] + n$:
$[x + 92] = [x] + 92 = 15 + 92 = 107$.

(B) For set $A$,we have $\sin(\theta) = \tan(\theta)$.
This implies $\sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$,which gives $\sin(\theta)(1 - \frac{1}{\cos(\theta)}) = 0$.
Thus,$\sin(\theta) = 0$ or $\cos(\theta) = 1$.
If $\sin(\theta) = 0$,then $\theta = n\pi$ for $n \in \mathbb{Z}$.
If $\cos(\theta) = 1$,then $\theta = 2n\pi$ for $n \in \mathbb{Z}$.
Combining these,$A = \{ n\pi : n \in \mathbb{Z} \} = \{ 0, \pm\pi, \pm 2\pi, \dots \}$.
For set $B$,we have $\cos(\theta) = 1$,which implies $\theta = 2n\pi$ for $n \in \mathbb{Z}$.
Thus,$B = \{ 2n\pi : n \in \mathbb{Z} \} = \{ 0, \pm 2\pi, \pm 4\pi, \dots \}$.
Comparing the two sets,every element of $B$ is in $A$,so $B \subset A$.
However,$\pi \in A$ but $\pi \notin B$,so $A \not\subset B$.