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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 $1 + (\frac{2}{5})^1 + (\frac{2}{5})^2 + (\frac{2}{5})^3 + ... = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$.

First, we determine the pairs $(x, y)$ that satisfy the condition $0 \le x^2 + 2y \le 4$ for the given set $A = \{-3, -2, -1, 0, 1, 2, 3\}$. For $y = -3$, $x^2 + 2(-3) \le 4 \implies x^2 \le 10$, so $x$ can be $-3$ or $3$. For $y = -2$, $x^2 + 2(-2) \le 4 \implies x^2 \le 8$, so $x$ can be $-2$ or $2$. For $y = -1$, $x^2 + 2(-1) \le 4 \implies x^2 \le 6$, so $x$ can be $-2$ or $2$. For $y = 0$, $x^2 + 2(0) \le 4 \implies x^2 \le 4$, so $x$ can be $-2, -1, 0, 1, 2$. For $y = 1$, $x^2 + 2(1) \le 4 \implies x^2 \le 2$, so $x$ can be $-1, 0, 1$. For $y = 2$, $x^2 + 2(2) \le 4 \implies x^2 \le 0$, so $x$ can only be $0$. Thus, the relation R consists of the following pairs: $R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\}$. The number of elements in R is $l = 15$. To make R reflexive, it must include all pairs $(x, x)$ for every $x \in A$. The set A has 7 elements, so the reflexive relation must contain $(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$. Currently, R has $(-3,-3), (-2, -2), (0,0), (1,1)$. So the missing pairs are $(-1, -1), (2, 2), (3, 3)$, meaning $m = 3$. Therefore, $l + m = 15 + 3 = 18$.

The given sets are $A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$ and $B = \{(x, y) \in R \times R : x^2 + 9y^2 = 144\}$. To find the intersection of A and B, we solve the system of equations. Subtracting the equation for A from the equation for B gives: $x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$, which simplifies to $8y^2 = 119$, so $y^2 = \frac{119}{8}$. Substituting this into the equation for A, we get $x^2 = 25 - \frac{119}{8} = \frac{200 - 119}{8} = \frac{81}{8}$. Thus, $x = \pm \frac{9}{\sqrt{8}}$ and $y = \pm \frac{\sqrt{119}}{\sqrt{8}}$. Since $D = A \cap B$, the set D contains 4 points. Now, consider the set $C = \{(x, y) \in Z \times Z : x^2 + y^2 \le 4\}$. The integer pairs (x, y) that satisfy this condition are: $(\pm 2, 0)$, $(0, \pm 2)$, $(\pm 1, 0)$, $(0, \pm 1)$, $(\pm 1, \pm 1)$, and $(0, 0)$. Counting these points, we find that there are 13 such pairs. The number of one-to-one functions from D to C is the number of ways to choose 4 elements from C and arrange them, which is $P(13, 4) = \frac{13!}{(13-4)!} = \frac{13!}{9!} = 13 \times 12 \times 11 \times 10 = 17160$.

For set A, we have $|x - 1| \leq 4$ and $|y - 5| \leq 6$. This implies $-4 \leq x - 1 \leq 4$ and $-6 \leq y - 5 \leq 6$. Therefore, $-3 \leq x \leq 5$ and $-1 \leq y \leq 11$. For set B, we have $16(x - 2)^2 + 9(y - 6)^2 \leq 144$, which simplifies to $\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} \leq 1$. This represents an ellipse centered at $(2, 6)$ with semi-major axis 4 and semi-minor axis 3. From the diagram, it's clear that $B \subset A$.

To determine the elements in the relation $R$, we need to find all pairs $(x, y)$ such that $\max(x, y) \in {3, 4}$. The set $A = {0, 1, 2, 3, 4, 5}$. If $\max(x, y) = 3$, the pairs are $(0, 3), (1, 3), (2, 3), (3, 3), (3, 0), (3, 1), (3, 2)$. If $\max(x, y) = 4$, the pairs are $(0, 4), (1, 4), (2, 4), (3, 4), (4, 4), (4, 0), (4, 1), (4, 2), (4, 3)$. Thus, the total number of elements in $R$ is $7 + 9 = 16$. Therefore, statement S1 is false. For statement S2, we check the properties of $R$: Reflexivity: $(0, 0)$ is not in $R$, so $R$ is not reflexive. Symmetry: If $(x, y) \in R$, then $\max(x, y) \in {3, 4}$, which implies $\max(y, x) \in {3, 4}$, so $(y, x) \in R$. Thus, $R$ is symmetric. Transitivity: $(0, 3) \in R$ and $(3, 1) \in R$, but $(0, 1)$ is not in $R$. Therefore, $R$ is not transitive. Statement S2 is true since $R$ is symmetric but neither reflexive nor transitive.

$xRy \Leftrightarrow 2x - y \in \{0, 1\}$ $\Rightarrow y = 2x$ or $y = 2x - 1$ $A=\{-3,-2,-1, 0, 1, 2, 3\}$ $R = \{(-1,-2), (0,0), (1, 2), (-1, -3), (0, -1), (1, 1), (2,3)\}$ $\Rightarrow I=7$ For $R$ to be reflexive $(0,0), (1,1) \in R$ But other $(a, a)$ such that $2a - a \in \{0, 1\}$ $\Rightarrow a \in \{0, 1\}$ 5 other pairs needs to be added $\Rightarrow m=5$ $xRy \Rightarrow yRx$ to be symmetric $(-1, -2) \Rightarrow (-2, -1)$ $(1, 2) \Rightarrow (2, 1)$ $(-1, -3) \Rightarrow (-3, -1)$ $(0, -1) \Rightarrow (-1, 0)$ $(2, 3) \Rightarrow (3, 2)$ $\Rightarrow 5$ needs to be added, $n=5$ $\Rightarrow l+m+n=17$

$A: |x-1| \leq 4$ and $|y-5| \leq 6$ $\Rightarrow -4 \leq x-1 \leq 4 \Rightarrow -6 \leq y-5 \leq 6$ $\Rightarrow -3 \leq x \leq 5 \Rightarrow -1 \leq y \leq 11$ $B : 16(x-2)^{2} + 9(y-6)^{2} \leq 144$ $B : \frac{(x-2)^{2}}{9} + \frac{(y-6)^{2}}{16} \leq 1$ From Diagram $B \subset A$

Probability distribution $x = 0$, $p = \frac{^7C_2}{^{10}C_2} = \frac{42}{90}$ $x = 1$, $p = \frac{^7C_1 \times ^3C_1}{^{10}C_2} = \frac{42}{90}$ $x = 2$, $p = \frac{^3C_2}{^{10}C_2} = \frac{6}{90}$ Now, $\mu = \sum x_i p_i = \frac{42}{90} + \frac{12}{90} = \frac{54}{90}$ $\sigma^2 = \sum p_i x_i^2 - \mu^2 = \frac{42}{90} + \frac{24}{90} - (\frac{54}{90})^2$ $\Rightarrow \frac{66}{90} - (\frac{54}{90})^2$ $\sigma^2 \Rightarrow \frac{28}{75}$

$i^{k_1} + i^{k_2} \ne 0 \Rightarrow i^{k_1} \rightarrow 4$ option for $i, -1, -i, 1$ Total cases $ \Rightarrow 4 \times 4 = 16$ Unfovourble cases $ \Rightarrow i^{k_1} + i^{k_2} = 0$ $ \{\begin{array}{c}1, -1 \\ -1, 1 \\ i, -i \\ -i, i\end{array}\}$ 4 Cases $\Rightarrow$ Probability $ = \frac{16-4}{16} = \frac{3}{4}$

A, E,G R D N Probabllity $(P) = \frac{\text{favourable case}}{\text{Total case}}$ (when A & E are in order) Total case = $6!$ Favourable case = ${6}C_2 . 4! = (15)4! = (30)4!$ $P = \frac{(15)4!}{(30)4!} = \frac{1}{2}$ Probability when not in order = $1 - \frac{1}{2} = \frac{1}{2}$