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Given that $aRb$ if $2a + 3b = 5m$, $m \in I$. (1) $(a, a) \in R$ as $2a + 3a = 5a$, $a \in N$. Hence, R is reflexive. (2) If $(a, b) \in R$ then $2a + 3b = 5m$. Now, $5(a + b) = 5n$. $3a + 2b + 2a + 3b = 5n$ $\therefore 3a + 2b = 5(n - m)$. $\therefore (b, a) \in R$. $\therefore R$ is symmetric. (3) If $(a, b) \in R$ and $(b, c) \in R$ then $2a + 3b = 5m$, $2b + 3c = 5n$. $\Rightarrow 2a + 5b + 3c = 5(m + n)$. $\Rightarrow 2a + 3c = 5(m + n - b)$. $\therefore (a, c) \in R$. $\therefore R$ is transitive. Hence, R is an equivalence relation.

Let $|A|$ be the number of medals in event A, $|B|$ be the number of medals in event B, and $|C|$ be the number of medals in event C. We are given $|A| = 48$, $|B| = 25$, and $|C| = 18$. We are also given that the total number of people who received medals is $60$, so $|A \cup B \cup C| = 60$, and the number of people who received medals in all three events is $5$, so $|A \cap B \cap C| = 5$. We want to find the number of people who received medals in exactly two of the three events. Let $x$ be the number of people who received medals in exactly two events. Using the Principle of Inclusion-Exclusion, we have: $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$ $60 = 48 + 25 + 18 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $60 = 91 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $|A \cap B| + |A \cap C| + |B \cap C| = 91 + 5 - 60 = 36$ Now, let $N_{2}$ be the number of people who received medals in exactly two events. Then: $|A \cap B| + |A \cap C| + |B \cap C| = N_{2} + 3|A \cap B \cap C|$ $36 = N_{2} + 3(5)$ $N_{2} = 36 - 15 = 21$ Therefore, the number of people who received medals in exactly two of the three events is $21$.

Given the set $A = {1, 2, 3, 4, 5}$ and the relation $xRy$ if and only if $4x \le 5y$. We need to find the number of elements in R (denoted by m) and the minimum number of elements that need to be added to R to make it symmetric (denoted by n). First, let's find the relation R: $4x \le 5y \implies \frac{x}{y} \le \frac{5}{4} = 1.25$ Now, let's find the pairs $(x, y)$ that satisfy this condition: If $x = 1$, then $y$ can be $1, 2, 3, 4, 5$. If $x = 2$, then $y$ can be $2, 3, 4, 5$. If $x = 3$, then $y$ can be $3, 4, 5$. If $x = 4$, then $y$ can be $4, 5$. If $x = 5$, then $y$ can be $4, 5$. So, $R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)}$. The number of elements in $R$ is $m = 16$. Now, we need to find the elements to be added to R to make it symmetric. A relation is symmetric if $(x, y) \in R$ implies $(y, x) \in R$. Currently, the elements $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (5,4) \in R$. The symmetric pairs that are missing are: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$. The number of elements to be added is $n = 9$. Therefore, $m + n = 16 + 9 = 25$.

To determine if the relation $R$ is reflexive, symmetric, and transitive, we analyze each property separately. Reflexive: For $R$ to be reflexive, $(f, f)$ must be in $R$. This means $f(0) = f(1)$ and $f(1) = f(0)$ must be true for all $f$. However, $f(0)$ is not necessarily equal to $f(1)$ for all functions $f$. Therefore, $R$ is not reflexive. Symmetric: If $(f, g) \in R$, then $f(0) = g(1)$ and $f(1) = g(0)$. To check for symmetry, we need to verify if $(g, f) \in R$. If $f(0) = g(1)$, then $g(1) = f(0)$. And if $f(1) = g(0)$, then $g(0) = f(1)$. This shows that if $(f, g) \in R$, then $(g, f) \in R$, so $R$ is symmetric. Transitive: If $(f, g) \in R$ and $(g, h) \in R$, then $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. For $R$ to be transitive, we need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$, so $f(0) = h(0)$. Also, $f(1) = g(0) = h(1)$, so $f(1) = h(1)$. Therefore, $f(0) = g(1)$ and $g(1) = h(0)$ implies $f(0) = h(0)$ and $f(1) = g(0)$ and $g(0) = h(1)$ implies $f(1) = h(1)$. Since $f(0)$ is not necessarily equal to $h(1)$ and $f(1)$ is not necessarily equal to $h(0)$, $R$ is not transitive. Actually $f(0) = h(0)$ and $f(1) = h(1)$ so this doesn't imply that $f(0) = h(1)$ and $f(1) = h(0)$ so the relation is not transitive Therefore, the relation $R$ is symmetric but neither reflexive nor transitive.

The relation $R$ is defined on the set $A = {1, 2, 3, ..., 100}$ such that $R = {(a, b) : a = 2b + 1}$. We need to find the largest integer $k$ for which there exists a sequence of $k$ ordered pairs from $R$ where the second element of each pair is the first element of the next pair. The sequence in terms of $k$ is: $(a_1, a_2), (a_2, a_3), ..., (a_k, a_{k+1})$. Here, each $a_i$ satisfies the equation $a_i = 2a_{i+1} + 1$. Consequently, $a_1 = 2a_2 + 1$, making $a_1$ an odd number. Let's examine the pattern: $a_2 = 2a_3 + 1$, implying $a_1 = 2(2a_3 + 1) + 1 = 4a_3 + 3$. $a_3 = 2a_4 + 1$, leading to $a_1 = 4(2a_4 + 1) + 3 = 8a_4 + 7$. Continuing this pattern, we find: $a_k = 2a_{k+1} + 1 \implies a_1 = 2^k \cdot a_{k+1} + (2^k - 1)$ where $a_{k+1}$ needs to be in set $A$. This implies: $a_{k+1} = \frac{a_1 + 1 - 2^k}{2^k}$ Thus, $2^k \mid (a_1 + 1)$. The task is to find the highest $k$ where $2^k$ divides any $e_i$ in ${2, ..., 101}$. The largest power of 2 that divides an element within this range determines $k$. After computation, we find that $k$ can be a maximum of 6 because $2^6 = 64$ divides $95+1 = 96$, but $2^7 = 128$ does not divide any $e_i$ for $e_i \in A$. Therefore, the maximum $k$ is 6. The sequence corresponding to this maximum $k$ is: $(95, 47), (47, 23), (23, 11), (11, 5), (5, 2)$

The relation $R$ is defined such that $y = \max(x, 1)$. Therefore, we find the following pairs: $x = -2 \implies y = \max(-2, 1) = 1$, so $(-2, 1) \in R$. $x = -1 \implies y = \max(-1, 1) = 1$, so $(-1, 1) \in R$. $x = 0 \implies y = \max(0, 1) = 1$, so $(0, 1) \in R$. $x = 1 \implies y = \max(1, 1) = 1$, so $(1, 1) \in R$. $x = 2 \implies y = \max(2, 1) = 2$, so $(2, 2) \in R$. $x = 3 \implies y = \max(3, 1) = 3$, so $(3, 3) \in R$. Thus, $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and $l = 6$. To make $R$ reflexive, we need to add $(-2, -2), (-1, -1), (0, 0)$. Thus, $m = 3$. To make $R$ symmetric, we need to add $(1, -2), (1, -1), (1, 0)$. Thus, $n = 3$. Therefore, $l + m + n = 6 + 3 + 3 = 12$.

$A = \{(x, y) \in R \times R : x^2 + y^2 = 25\}$, $B = \{(x, y) \in R \times R: x^2 + 9y^2 = 144\}$ $x^2 + 9y^2 - (x^2 + y^2) = 144 - 25$ Plug in $y^2 = \frac{119}{8}$ into either equation to find $x$. $x^2 = 25 - \frac{119}{8}$ $x^2 = \frac{200 - 119}{8}$ $x^2 = \frac{81}{8}$ $x = \pm \sqrt{\frac{81}{8}}$, $y = \pm \sqrt{\frac{119}{8}}$ Now, $C = \{(x, y) \in Z \times Z: x^2 + y^2 \leq 4\}$ Valid points are $(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1)$ $\therefore$ Total valid points in $C = 13$ $\Rightarrow$ There are 4 distinct real points in set $D$ $\therefore$ The number of one-one functions from $D$ to $C$ $\Rightarrow {}^{13}P_4 \Rightarrow \frac{13!}{(13-4)!} = \frac{13!}{9!} = 17160$

To evaluate the relation $R$ on the set $A = {0, 1, 2, 3, 4, 5}$, we first need to understand the conditions for an element $(x, y)$ to be in $R$. Specifically, $(x, y) ∈ R$ if and only if $max{x,y} ∈ {3,4}$. Considering this, let's list the pairs: For $max{x, y} = 3$, the possible pairs are: $(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3)$ For $max{x, y} = 4$, the possible pairs are: $(0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)$ Combining these, the set $R$ consists of the following elements: $R = {(0, 3), (3, 0), (1, 3), (3, 1), (2, 3), (3, 2), (3, 3), (0, 4), (4, 0), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3), (4, 4)}$ This gives us a total of 16 elements in $R$, not 18 as initially claimed in statement $S1$. Next, we analyze the properties of the relation $R$: Reflexivity: A relation is reflexive if $(x, x) ∈ R$ for all $x ∈ A$. For example, $(0, 0), (1, 1), (2, 2)$ are not in $R$, so $R$ is not reflexive. Symmetry: A relation is symmetric if whenever $(a, b) ∈ R$, then $(b, a) ∈ R$ as well. For all pairs $(x, y)$ listed, both $(x, y)$ and $(y, x)$ are present. Thus, $R$ is symmetric. Transitivity: A relation is transitive if whenever $(a, b) ∈ R$ and $(b, c) ∈ R$, then $(a, c) ∈ R$. An example where transitivity fails is $(0, 3)$ and $(3, 1)$ are in $R$ but $(0, 1)$ is not in $R$. Therefore, $R$ is not transitive. In conclusion, statement $S2$ is correct as $R$ is symmetric but neither reflexive nor transitive.

To find $P(X \geq 3)$, we first determine the constant $k$ using the total probability for $X$. The probability $P(X = x)$ is given by: $P(X= x) = k(x+1) \cdot 3^{-x}, x= 0,1,2,3,\dots$ The total probability must equal 1: $s = \sum_{x=0}^{\infty} k(x+1) \cdot 3^{-x}$ Calculating that series: $s = k3^0 + 2\frac{k}{3} + 3\frac{k}{3^2} + \dots$ Therefore, dividing the series by 3: $\frac{s}{3} = \frac{k}{3} + 2\frac{k}{3^2} + \dots$ Subtracting these: $s - \frac{s}{3} = k + \frac{k}{3} + \frac{k}{3^2} + \dots$ The resulting series is a geometric series: $2\frac{s}{3} = k(1 + \frac{1}{3} + \frac{1}{3^2} + \dots )$ The sum of the infinite geometric series is: $2\frac{s}{3} = k \cdot \frac{1}{1 - \frac{1}{3}} = \frac{3k}{2}$ Equating: $s = \frac{9k}{4} = 1$ Thus, solving for $k$: $k = \frac{4}{9}$ Next, compute $P(X \geq 3)$: $P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2))$ Calculating these: $P(X=0) = k = \frac{4}{9}$ $P(X=1) = 2\frac{k}{3} = \frac{8}{27}$ $P(X=2) = 3\frac{k}{9} = \frac{4}{27}$ Adding these probabilities: $P(X=0) + P(X=1) + P(X=2) = \frac{4}{9} + \frac{8}{27} + \frac{4}{27} = \frac{8}{9}$ Finally, calculate $P(X \geq 3)$: $P(X \geq 3) = 1 - \frac{8}{9} = \frac{1}{9}$

$X$\n$P(X)S$\n$XP(X)$\n$\left(X_i-\mu\right)^2$ SP_i X\left(X_i-\mu\right)^2$\n$X=0$\n$\frac{{}^7 C_2}{{}^{10} C_2}$\n0\nS\left(0-\frac{3}{5}\right)^2$\n$\frac{7}{15}\left(\frac{9}

Symmetric but neither reflective nor transitive

The relation R is defined on the set A = {1, 2, 3, ..., 100} such that R = {(a, b): a = 2b + 1}. We need to find the largest integer k for which there exists a sequence of k ordered pairs from R where the second element of each pair is the first element of the next pair.

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