Basic of Set theory Questions in English

(C) Given $D_{30} = \{1, 2, 3, 5, 6, 10, 15, 30\}$.
We need to calculate $f(2, 5, 15) = (2+5) \cdot (5^{\prime} + 15)$.
First,calculate $2+5 = \operatorname{LCM}(2, 5) = 10$.
Next,calculate $5^{\prime} = \frac{30}{5} = 6$.
Then,$5^{\prime} + 15 = 6 + 15 = \operatorname{LCM}(6, 15) = 30$.
Finally,$f(2, 5, 15) = 10 \cdot 30 = \operatorname{GCD}(10, 30) = 10$.

(A) To find the dual of a Boolean expression,we replace the $OR$ operator $(+)$ with the $AND$ operator $(\cdot)$ and vice versa,and replace $0$ with $1$ and $1$ with $0$.
Given expression: $(x+y) \cdot (x^{\prime} \cdot 1)$.
Replacing $+$ with $\cdot$ and $\cdot$ with $+$,and replacing $1$ with $0$ (though $0$ is not present,we focus on the operators):
The dual is $(x \cdot y) + (x^{\prime} + 0)$.
Since $x^{\prime} + 0 = x^{\prime}$,the expression simplifies to $(x \cdot y) + x^{\prime}$.
However,looking at the provided options,the transformation of the operator $1$ is often treated as a constant in dual logic. Replacing the operators in $(x+y) \cdot (x^{\prime} \cdot 1)$ gives $(x \cdot y) + (x^{\prime} + 1)$.

(C) The given circuit consists of two parallel branches. The first branch has switches $S_1$ and $S_2$ in series,which can be represented by the logical expression $(S_1 \land S_2)$.
The second branch has switches $S_1$ and $S_3$ in series,which can be represented by the logical expression $(S_1 \land S_3)$.
Since these two branches are in parallel,the total circuit is represented by the expression $(S_1 \land S_2) \lor (S_1 \land S_3)$.
Using the distributive law of Boolean algebra,we can factor out $S_1$:
$(S_1 \land S_2) \lor (S_1 \land S_3) = S_1 \land (S_2 \lor S_3)$.
This expression represents a switch $S_1$ in series with a parallel combination of switches $S_2$ and $S_3$.

(B) Let us evaluate each statement for the set $A = \{2, 3, 4, 5, 6\}$.
$A$: $\exists x \in A$ such that $(x-2) \in \mathbb{N}$. If $x=3$,$3-2=1 \in \mathbb{N}$. This is true.
$B$: $\forall x \in A, x+6$ is divisible by $2$. If $x=3$,$3+6=9$,which is not divisible by $2$. This is false.
$C$: $\exists x \in A$ such that $x+2$ is a prime number. If $x=3$,$3+2=5$,which is prime. This is true.
$D$: $\exists x \in A$ such that $x^{2}+1$ is an even number. If $x=3$,$3^{2}+1=10$,which is even. This is true.
Therefore,the statement in option $B$ is false.

(A) We are given the set $A = \{x \in \mathbb{R} : x^2 + 5|x| + 6 = 0\}$.
Since $x^2 = |x|^2$,the equation can be written as $|x|^2 + 5|x| + 6 = 0$.
Let $|x| = t$,where $t \ge 0$. The equation becomes $t^2 + 5t + 6 = 0$.
Factoring the quadratic,we get $(t + 2)(t + 3) = 0$.
This gives $t = -2$ or $t = -3$.
Since $t = |x|$ must be non-negative $(t \ge 0)$,there are no real values of $x$ that satisfy these conditions.
Therefore,the set $A$ is an empty set,i.e.,$A = \emptyset$.
Thus,the number of elements in $A$ is $n(A) = 0$.

(C) The set $A$ consists of prime numbers between $0$ and $9$.
$A = \{2, 3, 5, 7\}$.
The number of elements in set $A$ is $n(A) = 4$.
The number of elements in the power set of $A$ is given by $2^{n(A)}$.
$2^{4} = 16$.

(D) Given the set $A = \{x \in R : x^2 - 5|x| + 6 = 0\}$.
Since $x^2 = |x|^2$,the equation becomes $|x|^2 - 5|x| + 6 = 0$.
Let $|x| = t$,then $t^2 - 5t + 6 = 0$.
Factoring the quadratic equation: $(t - 2)(t - 3) = 0$.
So,$|x| = 2$ or $|x| = 3$.
If $|x| = 2$,then $x = 2$ or $x = -2$.
If $|x| = 3$,then $x = 3$ or $x = -3$.
Thus,the set $A = \{-3, -2, 2, 3\}$.
The number of elements in set $A$ is $n(A) = 4$.

(B) According to the Handshaking Lemma,the sum of the degrees of all vertices in a graph is equal to twice the number of edges,which is an even number.
Let $V_{odd}$ be the set of vertices with odd degree and $V_{even}$ be the set of vertices with even degree.
The sum of degrees is $\sum_{v \in V_{odd}} \text{deg}(v) + \sum_{v \in V_{even}} \text{deg}(v) = 2|E|$.
Since $\sum_{v \in V_{even}} \text{deg}(v)$ is always even,the sum $\sum_{v \in V_{odd}} \text{deg}(v)$ must also be even for the total sum to be even.
For the sum of $m$ odd numbers to be even,$m$ must be an even number.
Therefore,the number of vertices of odd degree,$m$,is always even.

(D) The group $(Z_{5}, +_{5})$ is a cyclic group of prime order $p = 5$.
By Lagrange's Theorem,the order of any subgroup must divide the order of the group.
Since $5$ is a prime number,its only divisors are $1$ and $5$.
Therefore,the only subgroups are the trivial subgroup ${0}$ of order $1$ and the group itself $Z_{5}$ of order $5$.
Thus,the total number of subgroups is $2$.

(C) Let set $A$ have $m$ elements and set $B$ have $n$ elements.
The number of subsets of set $A$ is $2^m$ and the number of subsets of set $B$ is $2^n$.
According to the problem,$2^m - 2^n = 56$.
We can write this as $2^n(2^{m-n} - 1) = 56$.
Factorizing $56$,we get $56 = 8 \times 7 = 2^3 \times (2^3 - 1)$.
Comparing $2^n(2^{m-n} - 1) = 2^3(2^3 - 1)$,we get $n = 3$ and $m - n = 3$.
Substituting $n = 3$ into $m - n = 3$,we get $m - 3 = 3$,which implies $m = 6$.
Thus,the values of $m$ and $n$ are $6$ and $3$ respectively.

(A) For option $A$: $x^2+1=0 \Rightarrow x^2=-1$. Since the square of any real number $x \in \mathbb{R}$ is always non-negative $(x^2 \ge 0)$,there is no real number $x$ that satisfies $x^2=-1$. Thus,the set is empty.
For option $B$: $x^2-9=0$ $\Rightarrow x^2=9$ $\Rightarrow x = \pm 3$. This set is $\{3, -3\}$,which is not empty.
For option $C$: $x^2-x-2=0$ $\Rightarrow (x-2)(x+1)=0$ $\Rightarrow x=2, -1$. This set is $\{2, -1\}$,which is not empty.
For option $D$: $x^2-1=0$ $\Rightarrow x^2=1$ $\Rightarrow x = \pm 1$. This set is $\{1, -1\}$,which is not empty.
Therefore,the correct option is $A$.

(A) Given,$A = \{1, 2, 3, \ldots, 10\}$.
The set of odd numbers in $A$ is $S = \{1, 3, 5, 7, 9\}$.
The number of elements in $S$ is $n = 5$.
The total number of subsets of $S$ is $2^n = 2^5 = 32$.
Since the question asks for subsets containing only odd numbers,we exclude the empty set (which contains no elements and thus no odd numbers).
Therefore,the number of non-empty subsets is $2^5 - 1 = 32 - 1 = 31$.

(C) Given the set $ A = \{-1, 1\} $.
We analyze the given options:
For option $ C $,the equation is $ x^{2} = 1 $.
Solving $ x^{2} = 1 $,we get $ x = \pm 1 $.
Thus,the set of roots is $ \{-1, 1\} $.
Therefore,the set builder form is $ A = \{x : x \text{ is a root of the equation } x^{2} = 1\} $.

(A) Let $f(n) = n(n+1)(2n+1)$.
We know that $2n+1 = (n-1) + (n+2)$,so $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.
Note that $n(n+1)(n-1)$ is the product of three consecutive integers,which is always divisible by $3! = 6$.
Similarly,$n(n+1)(n+2)$ is also the product of three consecutive integers,which is always divisible by $3! = 6$.
Therefore,$f(n)$ is divisible by $6$ for all $n \in \mathbb{Z}$.
Thus,$\{n(n+1)(2n+1) : n \in \mathbb{Z}\} \subset \{6k : k \in \mathbb{Z}\}$.

(A) set formed with three real numbers $a, b, c$ where the order of elements is fixed or preassigned is defined as an ordered triad,denoted as $(a, b, c)$.

(A) Let $f(n) = n(n+1)(2n+1)$.
We know that $n(n+1)(2n+1) = n(n+1)(n-1+n+2) = n(n+1)(n-1) + n(n+1)(n+2)$.
Each term $n(n+1)(n-1)$ and $n(n+1)(n+2)$ is a product of three consecutive integers.
The product of three consecutive integers is always divisible by $3! = 6$.
Therefore,$n(n+1)(n-1) = 6k_1$ and $n(n+1)(n+2) = 6k_2$ for some integers $k_1, k_2$.
Thus,$f(n) = 6(k_1 + k_2) = 6k$ for some integer $k$.
Hence,the set is a subset of $\{6k : k \in \mathbb{Z}\}$.

(B) Given that $x_{1}, x_{2}, \ldots, x_{15}$ are $15$ distinct numbers chosen from the set $\{1, 2, 3, \ldots, 15\}$.
Since there are exactly $15$ distinct numbers in the set and we are choosing $15$ distinct numbers,the set $\{x_{1}, x_{2}, \ldots, x_{15}\}$ must be exactly the set $\{1, 2, 3, \ldots, 15\}$ in some order.
Therefore,one of the values $x_{i}$ must be equal to $1$.
If $x_{i} = 1$ for some $i \in \{1, 2, \ldots, 15\}$,then the term $(x_{i}-1) = (1-1) = 0$.
Since the product contains a factor of $0$,the entire product $(x_{1}-1)(x_{2}-1) \ldots (x_{15}-1)$ is equal to $0$.

(B) We evaluate the expression $(A \setminus B) \setminus C$:
By definition of set difference,$A \setminus B = A \cap B'$.
Thus,$(A \setminus B) \setminus C = (A \cap B') \cap C'$.
Using the associative property of intersection,we get $A \cap (B' \cap C')$.
By De Morgan's Law,$B' \cap C' = (B \cup C)'$.
Therefore,$(A \setminus B) \setminus C = A \cap (B \cup C)' = A \setminus (B \cup C)$.
Hence,the statement in option $B$ is valid.

(B) The set $P$ represents the arc of the unit circle $x^2 + y^2 = 1$ in the first quadrant $(x > 0, y > 0)$.
The set $T$ represents the region inside the curve $x^8 + y^8 = 1$ in the first quadrant.
For any point $(x, y)$ on the circle $x^2 + y^2 = 1$ in the first quadrant,we have $0 < x < 1$ and $0 < y < 1$.
Since $0 < x < 1$,we have $x^8 < x^2$,and since $0 < y < 1$,we have $y^8 < y^2$.
Therefore,$x^8 + y^8 < x^2 + y^2 = 1$.
This means that every point $(x, y)$ in $P$ also satisfies the condition $x^8 + y^8 < 1$,so $P \subset T$.
Thus,$P \cap T = P$.