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$S = \{1, 2, 3, ..., 10\}$ $P(S) =$ power set of $S$ $AR_1B \implies (A \cap B^c) \cup (A^c \cap B) = \emptyset$ $R_1$ is reflexive, symmetric For transitive: $(A \cap B^c) \cup (A^c \cap B) = \emptyset$; $\emptyset = \emptyset \implies A = B$ $(B \cap C^c) \cup (B^c \cap C) = \emptyset \implies B = C$ $\therefore A = C \implies R_1$ is an equivalence relation. $R_2 \equiv A \cup B^c = A^c \cup B$ $R_2 \implies$ Reflexive, symmetric For transitive: $A \cup B^c = A^c \cup B \implies A = B$ $B \cup C^c = B^c \cup C \implies B = C$ $\therefore A = C \implies A \cup C^c = A^c \cup C \implies R_2$ is an equivalence relation.

Given sets are $A = {2, 3, 4}$ and $B = {8, 9, 12}$. We want to find the number of elements of the form $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$. For the first condition, $a_1$ divides $b_2$, with $a_1 \in A$ and $b_2 \in B$, we can list the pairs: $(a_1, b_2) \in {(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)}$. This gives $6$ pairs. For the second condition, $a_2$ divides $b_1$, we can similarly list the pairs. This again has $6$ valid pairs. Now, for every pair from the first condition, we can have any pair from the second condition. This leads to $6 \times 6 = 36$ relations.

Let's analyze the relation $R$. $R$ is a set of ordered pairs of ordered pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1 \in A$, $b_1 \in B$, $a_2 \in A$, $b_2 \in B$, and the conditions $a_1 \le b_2$ and $b_1 \le a_2$ are satisfied. We need to find the number of such ordered pairs. However, the question is formulated wrongly, since we are looking at a relation of pairs. A proper relation is given by $R = {(a_1, b_1) : a_1 \in A, b_1 \in B}$. The number of elements in the relation is not a relation between ordered pairs, it is simply asking to find pairs $(a, b)$ such that $a \le b$. So let $S = {(a,b): a \in A, b \in B, a \le b}$ and find the number of elements in $S$. Now we enumerate: - If $a=1$, then $b$ can be 2, 4, 5, 8, 10. (5 options) - If $a=3$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=4$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=6$, then $b$ can be 8, 10. (2 options) - If $a=9$, then $b$ can be 10. (1 option) So the total number of elements in $S$ is $5+4+4+2+1 = 16$. However, we are looking for elements that obey BOTH rules. That is: Number of pairs satisfying BOTH $a_1 \le b_2$ and $b_1 \le a_2$. Let's re-read the provided solution: - For each $a_1$ in $A = {1, 3, 4, 6, 9}$, we find the possible $b_2$ values in $B = {2, 4, 5, 8, 10}$ such that $a_1 \le b_2$. - $a_1 = 1$: $b_2$ can be 2, 4, 5, 8, 10 (5 choices) - $a_1 = 3$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 4$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 6$: $b_2$ can be 8, 10 (2 choices) - $a_1 = 9$: $b_2$ can be 10 (1 choice) This gives $5+4+4+2+1 = 16$. - For each $b_1$ in $B = {2, 4, 5, 8, 10}$, we find the possible $a_2$ values in $A = {1, 3, 4, 6, 9}$ such that $b_1 \le a_2$. - $b_1 = 2$: $a_2$ can be 3, 4, 6, 9 (4 choices) - $b_1 = 4$: $a_2$ can be 4, 6, 9 (3 choices) - $b_1 = 5$: $a_2$ can be 6, 9 (2 choices) - $b_1 = 8$: $a_2$ can be 9 (1 choice) - $b_1 = 10$: no $a_2$ works (0 choices) This gives $4+3+2+1+0 = 10$. So, $16 \times 10 = 160$. Thus the final number of elements is 160.

Let $S = {1, 2, 3, …, 10}$. $R = {(A, B): A∩B ≠ 𝜙; A, B∈ M}$. For Reflexive, $M$ is subset of 'S'. So $𝜙 ∈ M$ for $𝜙 ∩ 𝜙 = 𝜙 ⇒$ but relation is $A ∩ B ≠ 𝜙$. So it is not reflexive. For symmetric, $A R B ⇒ A ∩ B ≠ 𝜙, ⇒ B R A ⇒ B ∩ A ≠ 𝜙$, So it is symmetric. For transitive, If $A = {(1, 2), (2, 3)}, B = {(2, 3), (3, 4)}, C = {(3, 4), (5, 6)}$. $A R B$ & $B R C$ but $A$ does not relate to $C$. So it not transitive.

Given that the number of subsets of set $A$ is 56 more than that of set $B$, we have: $2^m - 2^n = 56$ $2^n(2^{m-n} - 1) = 56$ $2^n(2^{m-n} - 1) = 2^3 \times 7$ Comparing both sides, we have: $2^n = 2^3$ and $2^{m-n} - 1 = 7$ $n = 3$ and $2^{m-n} = 8$ $n = 3$ and $2^{m-n} = 2^3$ $n = 3$ and $m-n = 3$ $n = 3$ and $m = 6$ So the coordinates of point P are $(6,3)$. The coordinates of point Q are $(-2,-3)$. The distance between $P$ and $Q$ is: $PQ = \sqrt{(6 - (-2))^2 + (3 - (-3))^2}$ $PQ = \sqrt{(6+2)^2 + (3+3)^2}$ $PQ = \sqrt{8^2 + 6^2}$ $PQ = \sqrt{64 + 36}$ $PQ = \sqrt{100}$ $PQ = 10$ Therefore, the distance between $P(6,3)$ and $Q(-2,-3)$ is 10.

Reflexive: Since $a*b - a*b = 0$ is divisible by $5$, $(a, b)R(a, b)$ holds. Symmetric: If $(a, b)R(c, d)$, then $ad - bc$ is divisible by $5$. Thus, $bc - ad$ is divisible by $5$, so $(c, d)R(a, b)$ holds. Not Transitive: Consider $(3, 1)R(10, 5)$ because $3*5 - 1*10 = 5$ is divisible by $5$. Consider $(10, 5)R(1, 1)$ because $10*1 - 5*1 = 5$ is divisible by $5$. But $(3, 1)$ is not related to $(1, 1)$ because $3*1 - 1*1 = 2$ is not divisible by $5$. Hence, $R$ is reflexive and symmetric but not transitive.

Given the set ${1, 2, 3, 4}$, and that $R$ is an equivalence relation. Since ${((1, 2), (1, 3))} \subset R$, $R$ must contain the pairs $(1, 2)$ and $(1, 3)$. An equivalence relation must be reflexive, symmetric, and transitive. Reflexive property: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4) \in R$. Symmetric property: Since $(1, 2) \in R$, $(2, 1) \in R$. Since $(1, 3) \in R$, $(3, 1) \in R$. Transitive property: Since $(1, 2) \in R$ and $(1, 3) \in R$, it follows that $(2, 3) \in R$. By symmetric property, $(3, 2) \in R$. Therefore, the elements in $R$ are: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(1, 2)$, $(2, 1)$, $(1, 3)$, $(3, 1)$, $(2, 3)$, $(3, 2)$. Thus, the number of elements in $R$ is $10$.

To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity. For $R_1$: Reflexivity: For $R_1$ to be reflexive, $aR_1a$ must hold for all $a \in R$. This means $a^2 + a^2 = 1$, or $2a^2 = 1$, which implies $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive. Symmetry: If $aR_1b$, then $a^2 + b^2 = 1$. This also means $b^2 + a^2 = 1$, so $bR_1a$. Thus, $R_1$ is symmetric. Transitivity: If $aR_1b$ and $bR_1c$, then $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. Adding these, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive. $R_1$ is not an equivalence relation. For $R_2$: Reflexivity: For any $(a, b) \in N \times N$, $a + b = b + a$. Therefore, $(a, b)R_2(a, b)$, and $R_2$ is reflexive. Symmetry: If $(a, b)R_2(c, d)$, then $a + d = b + c$. This can be reordered to $c + b = d + a$, so $(c, d)R_2(a, b)$. Thus, $R_2$ is symmetric. Transitivity: If $(a, b)R_2(c, d)$ and $(c, d)R_2(e, f)$, then $a + d = b + c$ and $c + f = d + e$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. Simplifying, we get $a + f = b + e$, so $(a, b)R_2(e, f)$. $R_2$ is reflexive, symmetric, and transitive; therefore, it is an equivalence relation. Therefore, only $R_2$ is an equivalence relation.

Given the relation $(x_1, y_1) R (x_2, y_2)$ if $x_1 \le x_2$ or $y_1 \le y_2$. For reflexivity, we need to check if $(x_1, y_1) R (x_1, y_1)$. Since $x_1 \le x_1$ and $y_1 \le y_1$, the condition $x_1 \le x_1$ or $y_1 \le y_1$ is true. Therefore, $R$ is reflexive. For symmetry, if $(x_1, y_1) R (x_2, y_2)$, then $x_1 \le x_2$ or $y_1 \le y_2$. For $R$ to be symmetric, $(x_2, y_2) R (x_1, y_1)$ must also be true, which means $x_2 \le x_1$ or $y_2 \le y_1$. This is not necessarily true. For example, $(1, 2) R (3, 4)$ since $1 \le 3$ or $2 \le 4$, but $(3, 4) R (1, 2)$ is false because neither $3 \le 1$ nor $4 \le 2$ is true. Thus, $R$ is not symmetric. For transitivity, consider pairs $(3, 9)$, $(4, 6)$, and $(2, 7)$. $(3, 9) R (4, 6)$ because $3 \le 4$. $(4, 6) R (2, 7)$ because $6 \le 7$. However, $(3, 9) R (2, 7)$ is false because neither $3 \le 2$ nor $9 \le 7$ is true. Therefore, $R$ is not transitive. Since $R$ is reflexive but not symmetric, statement (I) is correct. Since $R$ is not transitive, statement (II) is incorrect. Thus, only statement (I) is correct.

Let's find the total number of integers in the range $[100, 700]$. The total number of integers is $700 - 100 + 1 = 601$. Thus, the total number of integers is $601$. Let $n(3)$ be the number of multiples of 3 in the given range. The first multiple of 3 is $102 = 3 \times 34$ and the last multiple of 3 is $699 = 3 \times 233$. So, $n(3) = 233 - 34 + 1 = 200$. Let $n(4)$ be the number of multiples of 4 in the given range. The first multiple of 4 is $100 = 4 \times 25$ and the last multiple of 4 is $700 = 4 \times 175$. So, $n(4) = 175 - 25 + 1 = 151$. Let $n(12)$ be the number of multiples of 12 in the given range. The first multiple of 12 is $108 = 12 \times 9$ and the last multiple of 12 is $696 = 12 \times 58$. So, $n(12) = 58 - 9 + 1 = 50$. The number of elements in A is given by the total number of elements minus (multiples of 3 + multiples of 4) + (multiples of 12) $n(A) = 601 - (200 + 151) + 50 = 601 - 351 + 50 = 250 + 50 = 300$. Therefore, the number of elements in A is $300$.

Given that $(a, b)R(c, d) \Rightarrow 3ad - 7bc \in \text{even}$. For reflexive, we need to check if $(a, b)R(a, b)$. $(a, b)R(a, b) \Rightarrow 3ab - 7ba = -4ab \in \text{even}$. This is true since the product $ab$ will be an integer, and $-4$ times an integer is even. Thus, $R$ is reflexive. For symmetric, if $(a, b)R(c, d)$, then $(c, d)R(a, b)$ must also be true. $(a, b)R(c, d) \Rightarrow 3ad - 7bc = 2m$ for some integer $m$. $(c, d)R(a, b) \Rightarrow 3cb - 7ad$. We want to show this is even. $3cb - 7ad = -7ad + 3cb = -(7ad - 3cb) = -(7ad - 3cb + 3ad - 3ad - 7bc + 7bc) = -(10ad - 10bc - (3ad - 7bc)) = -10(ad - bc) + (3ad-7bc)$

An equivalence relation on a finite set is uniquely determined by its partition into equivalence classes. Counting the number of ways to partition the set ${1, 2, 3}$: 1. Three blocks: Each element in its own block. There is only one way: ${{1}, {2}, {3}}$. 2. Two blocks: We can have ${{1, 2}, {3}}$, ${{1, 3}, {2}}$, or ${{2, 3}, {1}}$. There are 3 ways. 3. One block: All elements together. There is only one way: ${{1, 2, 3}}$. In total, there are $1 + 3 + 1 = 5$ distinct partitions, which means there are 5 equivalence relations on the set ${1, 2, 3}$.

To find the number of elements in set $B$, we consider pairs $(\frac{m}{n})$ where $m, n \in A$ with $m < n$ and $gcd(m, n) = 1$. Here's the breakdown for each possible $m$: For $m = 1$: Possible values for $n$ are $2, 3, 4, 5, 6, 7, 8, 9, 10$. Total pairs: $9$. For $m = 2$: Possible values for $n$ are $3, 5, 7, 9$ (since these have $gcd(2, n) = 1$). Total pairs: $4$. For $m = 3$: Possible values for $n$ are $4, 5, 7, 8, 10$. Total pairs: $5$. For $m = 4$: Possible values for $n$ are $5, 7, 9$. Total pairs: $3$. For $m = 5$: Possible values for $n$ are $6, 7, 8, 9$. Total pairs: $4$. For $m = 6$: Possible value for $n$ is $7$. Total pairs: $1$. For $m = 7$: Possible values for $n$ are $8, 9, 10$. Total pairs: $3$. For $m = 8$: Possible value for $n$ is $9$. Total pairs: $1$. For $m = 9$: Possible value for $n$ is $10$. Total pairs: $1$. Adding all these up, the total number of elements in set $B$ is: $9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31$

For $R$ to be an equivalence relation on $A = \{1, 2, 3, 4\}$, it must be reflexive, symmetric and transitive. 1. **Reflexive:** $R$ must contain $(1, 1), (2, 2), (3, 3), (4, 4)$. Since $(3,3)$ is already in $R$, we need to add $(1, 1), (2, 2), (4, 4)$. 2. **Symmetric:** $R$ must contain $(2, 1)$ and $(3, 2)$ because it contains $(1, 2)$ and $(2, 3)$. 3. **Transitive:** Since $(1, 2)$ and $(2, 3)$ are in $R$, $(1, 3)$ must also be in $R$. And since we added $(3,2)$ now we must add $(1,2)$. Which already exists. So, the minimum elements to be added are: $(1, 1), (2, 2), (4, 4), (2, 1), (3, 2), (1, 3)$. Therefore, the minimum number of elements to be added is $7$.

From the image, we can determine that the points in set C are $(3,0)$, $(-3,0)$, $(0,3)$ and $(0,-3)$. Thus, $C = {(3, 0), (-3, 0), (0, 3), (0, -3)}$. $\sum |x + y| = |3 + 0| + |-3 + 0| + |0 + 3| + |0 + (-3)| = 3 + 3 + 3 + 3 = 12$.

Statement I: Reflexive: $(a_1, b_1) R (a_1, b_1) \Rightarrow b_1 = b_1$ (True). Symmetric: $(a_1, b_1) R (a_2, b_2) \Rightarrow b_1 = b_2$ and $(a_2, b_2) R (a_1, b_1) \Rightarrow b_2 = b_1$ (True). Transitive: $(a_1, b_1) R (a_2, b_2) \Rightarrow b_1 = b_2$ and $(a_2, b_2) R (a_3, b_3) \Rightarrow b_2 = b_3$. Thus, $b_1 = b_3$, so $(a_1, b_1) R (a_3, b_3)$ (True). Hence, relation $R$ is an equivalence relation, and Statement I is true. For Statement II: $(x, y) R (a, b) \Rightarrow y = b$. This represents a line $y=b$, which is parallel to the x-axis. It is not parallel to the line $y=x$. Therefore, Statement II is false.

For set A: $\log_{\frac{2}{\pi}} |\sin x| + \log_{\frac{2}{\pi}} |\cos x| = 2$ $\Rightarrow \log_{\frac{2}{\pi}} |\sin x \cdot \cos x| = 2$ $\Rightarrow |\sin x \cos x| = (\frac{2}{\pi})^2$ $\Rightarrow |\frac{1}{2} \sin 2x| = \frac{4}{\pi^2}$ $\Rightarrow |\sin 2x| = \frac{8}{\pi^2}$ Since $0 < \frac{8}{\pi^2} < 1$, there are 4 solutions for $x \in (0, \pi)$. For set B: Case 1: $x < 2$, $x(x-4) + 3(x-2) + 6 = 0$ $\Rightarrow x^2 - 4x + 3x - 6 + 6 = 0$ $\Rightarrow x^2 - x = 0$ $\Rightarrow x(x-1) = 0$, so $x = 0, 1$. Case 2: $x > 2$, $x(x-4) - 3(x-2) + 6 = 0$ $\Rightarrow x^2 - 4x - 3x + 6 + 6 = 0$ $\Rightarrow x^2 - 7x + 12 = 0$ $\Rightarrow (x-3)(x-4) = 0$, so $x = 3, 4$. Thus, $B = {0, 1, 3, 4}$, so $n(B) = 4$. Since $A$ and $B$ are disjoint, $n(A \cup B) = n(A) + n(B) = 4 + 4 = 8$.

The given relation is $R = {(x, y) : x, y ∈ Z$ and $x + y$ is even $}$. Reflexive: For any $x ∈ Z$, $x + x = 2x$, which is even. So, $(x, x) ∈ R$. Thus, $R$ is reflexive. Symmetric: If $(x, y) ∈ R$, then $x + y$ is even. Since $x + y = y + x$, $y + x$ is also even. So, $(y, x) ∈ R$. Thus, $R$ is symmetric. Transitive: If $(x, y) ∈ R$ and $(y, z) ∈ R$, then $x + y$ is even and $y + z$ is even. Then $(x + y) + (y + z) = x + 2y + z$ is even. Since $2y$ is even, it follows that $x + z$ is even. So, $(x, z) ∈ R$. Thus, $R$ is transitive. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

To check if the relation R is reflexive, symmetric, and transitive: 1. **Reflexive:** Replace $y$ with $x$ in the given relation: $\sec^2x - \tan^2x = 1$. This is a trigonometric identity, so the relation is reflexive. 2. **Symmetric:** Given $\sec^2x - \tan^2y = 1$, we can rewrite this as $1 + \tan^2x - \tan^2y = 1$, which implies $\tan^2x = \tan^2y$. Also, $\sec^2x - \tan^2y = 1$ can be rewritten as $1 + \tan^2x - \sec^2y + 1 = 1$, leading to $\sec^2y - \tan^2x = 1$. Thus, if $xRy$, then $yRx$, so the relation is symmetric. 3. **Transitive:** Suppose $\sec^2x - \tan^2y = 1$ and $\sec^2y - \tan^2z = 1$. Adding these two equations, we get: $\sec^2x - \tan^2y + \sec^2y - \tan^2z = 1 + 1$ $\sec^2x + (\sec^2y - \tan^2y) - \tan^2z = 2$ Since $\sec^2y - \tan^2y = 1$, we have: $\sec^2x + 1 - \tan^2z = 2$ $\sec^2x - \tan^2z = 1$ Thus, if $xRy$ and $yRz$, then $xRz$, so the relation is transitive. Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.

Given $S = \{0, 1, 2, 3, ...\}$. Also, $\log_e y = x \log_e (\frac{2}{5})$. This implies $y = (\frac{2}{5})^x$. Since $x \in S$, $x$ can take values $0, 1, 2, 3, ...$. The required sum is $1 + (\frac{2}{5})^1 + (\frac{2}{5})^2 + (\frac{2}{5})^3 + ... = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.