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To determine if the relation $R$ is symmetric, we need to check if for every $(x, y)$ in $R$, the pair $(y, x)$ is also in $R$. $R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$. We have $(1, 3)$ in $R$, and $(3, 1)$ is also in $R$. We have $(4, 2)$ in $R$, but $(2, 4)$ is also in $R$. However, we have $(2, 3)$ in $R$, but $(3, 2)$ is not in $R$. Since there exists a pair $(2, 3)$ in $R$ such that $(3, 2)$ is not in $R$, the relation $R$ is not symmetric. Therefore, the correct answer is Option C: $R$ is not symmetric.

To determine the properties of the relation $R$, we need to check for reflexivity, symmetry, and transitivity. Reflexivity: A relation is reflexive if $(a, a) \in R$ for all $a \in A$. Since $(3, 3), (6, 6), (9, 9),$ and $(12, 12)$ are all in $R$, the relation is reflexive. Symmetry: A relation is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$. We have $(6, 12) \in R$, but $(12, 6) \notin R$. Thus, the relation is not symmetric. Transitivity: A relation is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. We have $(3, 6) \in R$ and $(6, 12) \in R$, and $(3, 12) \in R$. Also, $(3, 6) \in R$ and $(6, 6) \in R$, and $(3, 6) \in R$. The relation appears to be transitive. Therefore, the relation $R$ is reflexive and transitive, but not symmetric. So, the correct answer is Option D: $R$ is reflexive and transitive only.

To determine the properties of the relation $R$, we analyze reflexivity, symmetry, and transitivity. Reflexivity: A word always shares at least one letter with itself. So, $(x, x) ∈ R$ for all $x ∈ W$. Therefore, $R$ is reflexive. Symmetry: If a word $x$ has a letter in common with word $y$, then $y$ also has a letter in common with $x$. So, if $(x, y) ∈ R$, then $(y, x) ∈ R$. Therefore, $R$ is symmetric. Transitivity: Consider the words 'cat', 'bat', and 'bee'. 'cat' and 'bat' share the letter 'a', so (cat, bat) ∈ R. 'bat' and 'bee' share the letter 'b', so (bat, bee) ∈ R. However, 'cat' and 'bee' do not share any common letters, so (cat, bee) ∉ R. Therefore, $R$ is not transitive. Thus, the relation $R$ is reflexive, symmetric, and not transitive.

Given $S = {(x, y) : y = x + 1 ext{ and } 0 < x < 2}$. Since $x \ne x+1$ for any $x \in (0,2)$, we have $(x, x) \notin S$. Therefore, $S$ is not reflexive, and hence not an equivalence relation. Also $T = {(x, y) : x - y ext{ is an integer }}$. Since $x - x = 0$ is an integer for all $x \in R$, $T$ is reflexive. If $x - y$ is an integer, then $y - x$ is also an integer, so $T$ is symmetric. If $x - y$ is an integer and $y - z$ is an integer, then $(x - y) + (y - z) = x - z$ is also an integer. Therefore, $T$ is transitive. Hence $T$ is an equivalence relation.

Given $A \cap B = A \cap C$ and $A \cup B = A \cup C$. From the principle of inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$ and $|A \cup C| = |A| + |C| - |A \cap C|$. Since $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $|A \cup B| = |A \cup C|$ and $|A \cap B| = |A \cap C|$. Thus, $|A| + |B| - |A \cap B| = |A| + |C| - |A \cap C|$. This simplifies to $|B| = |C|$. Since $A \cap B = A \cap C$ and $A \cup B = A \cup C$, it means that $B$ and $C$ contain the same elements and therefore are equal to each other ($B=C$).

Relation $R$: $R = \{(x, y) | x, y$ are real numbers and $x = wy$ for some rational number $w\}$. Reflexivity: For all $x$ in $R$, $x = 1 \cdot x$. Since $1$ is a rational number, every element is related to itself. Symmetry: For all $x, y$ in $R$, if $x = wy$ for some rational $w$, then $y = \frac{1}{w}x$. However, if $w = 0$, then $\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry. Transitivity: If $x = wy$ and $y = vz$ for some rational numbers $w$ and $v$, then $x = (wv)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$. Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property. Relation $S$: $S = \{(\frac{m}{n}, \frac{p}{q}) | m, n, p$ and $q$ are integers such that $n, q \neq 0$ and $qm = pn\}$. Reflexivity: For all $\frac{m}{n}$, $\frac{m}{n} = \frac{m}{n}$. Since $n \neq 0$ and $m = m$, every element is related to itself. Symmetry: For all $\frac{m}{n}, \frac{p}{q}$, if $qm = pn$, then $np = mq$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$. Transitivity: If $\frac{m}{n}R\frac{p}{q}$ and $\frac{p}{q}R\frac{r}{s}$ $\Rightarrow mq = np$ and $ps = rq$ $\Rightarrow mq \cdot ps = np \cdot rq$ $\Rightarrow ms = nr$ $\Rightarrow \frac{m}{n} = \frac{r}{s}$ $\Rightarrow \frac{m}{n}R\frac{r}{s}$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$ and $\frac{p}{q}$ is related to $\frac{r}{s}$, then $\frac{m}{n}$ is related to $\frac{r}{s}$. The relation $S$ is transitive. Therefore, $S$ is an equivalence relation on the set of all fractions where the denominator is not zero. In conclusion, $S$ is an equivalence relation but $R$ is not an equivalence relation.

Statement I: $A = \{(x, y) \in R \times R: y - x \text{ is an integer }\}$. Reflexivity: For all $x \in R$, $x - x = 0$ which is an integer. Symmetry: For all $x, y \in R$, if $y - x$ is an integer, then $x - y = -(y - x)$ is also an integer. Transitivity: For all $x, y, z \in R$, if $y - x$ and $z - y$ are integers, then $(z - y) + (y - x) = z - x$ is also an integer. Therefore, $A$ is an equivalence relation on $R$. Statement II: $B = \{(x,y) \in R \times R: x = \alpha y \text{ for some rational number } \alpha\}$. Reflexivity: For all $x \in R$, $x = 1 \cdot x$. Since $1$ is a rational number, every element is related to itself. Symmetry: For all $x, y \in R$, if $x = \alpha y$ for some rational $\alpha$, then $y = \frac{1}{\alpha} x$. However, if $\alpha = 0$, then $\frac{1}{\alpha}$ is undefined, and therefore, $B$ doesn't satisfy symmetry. Transitivity: If $x = \alpha y$ and $y = \beta z$ for some rational numbers $\alpha$ and $\beta$, then $x = (\alpha \beta) z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$. Therefore, $B$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property. In conclusion, Statement I is true, and Statement II is false.

For any element $x_i$ present in $X$, 4 cases arise while making subsets $Y$ and $Z$. Case 1: $x_i \in Y, x_i \in Z \implies Y \cap Z \neq \emptyset$ Case 2: $x_i \in Y, x_i \notin Z \implies Y \cap Z = \emptyset$ Case 3: $x_i \notin Y, x_i \in Z \implies Y \cap Z = \emptyset$ Case 4: $x_i \notin Y, x_i \notin Z \implies Y \cap Z = \emptyset$ Therefore, for every element, the number of ways is $3$ for which $Y \cap Z = \emptyset$. Thus, the total number of ways is $3 \times 3 \times 3 \times 3 \times 3 = 3^5$ since the number of elements in set $X$ is $5$.

Given, $n(A) = 4$, $n(B) = 2$ $\Rightarrow n(A \times B) = 8$ Total number of subsets of set $(A \times B) = 2^8$ Number of subsets of set $A \times B$ having no element (i.e. $\phi$) = $1$ Number of subsets of set $A \times B$ having one element = $^8C_1$ Number of subsets of set $A \times B$ having two elements = $^8C_2$ $\therefore$ Number of subsets having atleast three elements = $2^8 - (1 + ^8C_1 + ^8C_2)$ = $2^8 - 1 - 8 - 28$ = $2^8 - 37$ = $256 - 37 = 219$

Given, for set P: $sin\theta - cos\theta = \sqrt{2}cos\theta$ $sin\theta = (\sqrt{2} + 1)cos\theta$ $cos\theta = (\sqrt{2} - 1)sin\theta$ ... (1) Given, for set Q: $sin\theta + cos\theta = \sqrt{2}sin\theta$ $cos\theta = (\sqrt{2} - 1)sin\theta$ ... (2) From (1) and (2), P = Q

Here, both $R_1$ and $R_2$ are symmetric, as for any $(x, y) \in R_1$, we have $(y, x) \in R_1$ and similarly for any $(x, y) \in R_2$, we have $(y, x) \in R_2$. In $R_1$, $(b, c) \in R_1$, $(c, a) \in R_1$ but $(b, a) \notin R_1$. Similarly, in $R_2$, $(b, a) \in R_2$, $(a, c) \in R_2$ but $(b, c) \notin R_2$. Therefore, $R_1$ and $R_2$ are not transitive. Thus, $R_2$ is symmetric but not transitive.

Given, $4(a-6)^2 + 9(b-5)^2 ≤ 36$. Let $a-6 = x$ and $b-5 = y$. Therefore, $4x^2 + 9y^2 ≤ 36$, which implies $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. This is the equation of an ellipse. According to set A, $|a-5| < 1$, which implies, since $a-6=x$, then $a-5 = x+1$. Thus $|x+1| < 1$, implying $-1 < x+1 < 1$, which leads to $-2 < x < 0$. Similarly, $|b-5| < 1$, implying $|y| < 1$, which leads to $-1 < y < 1$. To check if the entire set A is inside of B, we can test a point of set A in the inequality $\frac{x^2}{9} + \frac{y^2}{4} ≤ 1$. Consider the point $(-2, 1)$. Substituting into the inequality gives LHS = $\frac{(-2)^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{25}{36} < 1$. Since the inequality holds, $(-2, 1)$ is inside the ellipse. Similarly, $(-2, -1)$ is also inside the ellipse. Hence, we can say that entire set A is inside of set B, therefore $A ⊂ B$.

For $R_1$, $2x + y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are: $x = 1, y = 8 \implies (1, 8)$ $x = 2, y = 6 \implies (2, 6)$ $x = 3, y = 4 \implies (3, 4)$ $x = 4, y = 2 \implies (4, 2)$ Therefore, $R_1 = \{(1, 8), (2, 6), (3, 4), (4, 2)\}$. Hence, the range of $R_1$ is $\{2, 4, 6, 8\}$. $R_1$ is not symmetric. $R_1$ is not transitive as $(3, 4), (4, 2) \in R_1$, but $(3, 2) \notin R_1$. For $R_2$, $x + 2y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are: $x = 8, y = 1 \implies (8, 1)$ $x = 6, y = 2 \implies (6, 2)$ $x = 4, y = 3 \implies (4, 3)$ $x = 2, y = 4 \implies (2, 4)$ Therefore, $R_2 = \{(8, 1), (6, 2), (4, 3), (2, 4)\}$. Hence, the range of $R_2 = \{1, 2, 3, 4\}$. $R_2$ is not symmetric and transitive.

Let $A$ be the set of students who opted for Mathematics, $B$ be the set of students who opted for Physics, and $C$ be the set of students who opted for Chemistry. We are given: Total number of students $= 140$ $n(A) = \lfloor\frac{140}{2}\rfloor = 70$ $n(B) = \lfloor\frac{140}{3}\rfloor = 46$ $n(C) = \lfloor\frac{140}{5}\rfloor = 28$ $n(A \cap B) = \lfloor\frac{140}{6}\rfloor = 23$ $n(B \cap C) = \lfloor\frac{140}{15}\rfloor = 9$ $n(C \cap A) = \lfloor\frac{140}{10}\rfloor = 14$ $n(A \cap B \cap C) = \lfloor\frac{140}{30}\rfloor = 4$ Using the principle of inclusion-exclusion: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$ $n(A \cup B \cup C) = 70 + 46 + 28 - 23 - 9 - 14 + 4 = 102$ The number of students who did not opt for any of the three courses is: $140 - n(A \cup B \cup C) = 140 - 102 = 38$

Let $S = {1, 2, 3, ..., 100}$. The total number of non-empty subsets is $2^{100} - 1$. The number of subsets with an odd product consists only of odd numbers. There are 50 odd numbers in the set $S$. So, the number of such subsets is $2^{50} - 1$. Therefore, the number of subsets with an even product is $(2^{100} - 1) - (2^{50} - 1) = 2^{100} - 2^{50} = 2^{50}(2^{50} - 1)$.

Given $A = {x \in Z : 2(x+2)(x^2 - 5x + 6) = 1}$. Since $2(x+2)(x^2 - 5x + 6) = 1$, we can rewrite it as $2(x+2)(x^2 - 5x + 6) = 2^0$. This implies that $x = -2, 2, 3$, so $A = {-2, 2, 3}$. Also, $B = {x \in Z : -3 < 2x - 1 < 9}$. Adding 1 to all sides, we get $-2 < 2x < 10$. Dividing by 2, we get $-1 < x < 5$. Thus, $B = {0, 1, 2, 3, 4}$. Now, $A \times B$ has $3 \times 5 = 15$ elements. The number of subsets of $A \times B$ is $2^{15}$.

Let the total population be $100$. Then: People who read A only: $25 - 8 = 17$ People who read B only: $20 - 8 = 12$ People who read A and look into advertisements: $0.30 \times 17 = 5.1$ People who read B and look into advertisements: $0.40 \times 12 = 4.8$ People who read both A and B and look into advertisements: $0.50 \times 8 = 4$ Total percentage of people who look into advertisements: $5.1 + 4.8 + 4 = 13.9$ Therefore, the percentage of the population who look into advertisement is $13.9$%.

The question provides that $A \cap B \subseteq C$ and $A \cap B \neq \phi$. Analyzing each statement: Statement A: If $(A - B) \subseteq C$, then $A \subseteq C$. This statement can be verified using Venn diagrams. If the part of A that is not in B is contained in C, then A is a subset of C. Statement B: $B \cap C \neq \phi$. This means that the intersection of B and C is not empty. Given that $A\cap B \subseteq C$, this is true. Statement C: $(C \cup A) \cap (C \cup B) = C$. Using the distributive property of sets, $(C \cup A) \cap (C \cup B) = C \cup (A \cap B)$. Since $A \cap B \subseteq C$, then $C \cup (A \cap B) = C$. So, this statement is true. Statement D: If $(A - C) \subseteq B$, then $A \subseteq B$. This statement is NOT necessarily true. It is possible to have $A - C \subseteq B$, $A \cap B \subseteq C$ and $A \cap B \neq \phi$, but it does NOT require that $A \subseteq B$. This is because A can contain elements outside of B that are also outside of C. Therefore, the statement that is not true is D.