Sets and Relations

let $A=\{1,2,3,4\}$

$R$ is defined on $A \times A$ such that

$ \begin{aligned} & R=\{((a, b),(c, d)): 2 a+3 b=3 c+4 d\} \\ & \text { since } a, b, c, d \in A \end{aligned} $

$ \text { Let possible values of } 2 a+3 b=L 1 \text { and possible values of } 3 c+4 d=L 2 $

$ \begin{array}{c|c|c|c} a & b & 2 a+3 b & 3 c+4 d \\ \hline 1 & 1 & 5 & 7 \\ 1 & 2 & 8 & 11 \\ 1 & 3 & 11 & 15 \\ 1 & 4 & 14 & 19 \\ 2 & 1 & 7 & 10 \\ 2 & 2 & 10 & 14 \\ 2 & 3 & 13 & 18 \\ 2 & 4 & 16 & 22 \\ 3 & 1 & 9 & 13 \\ 3 & 2 & 12 & 17 \\ 3 & 3 & 15 & 21 \\ 3 & 4 & 18 & 25 \\ 4 & 1 & 11 & 16 \\ 4 & 2 & 14 & 20 \\ 4 & 3 & 17 & 24 \\ 4 & 4 & 20 & 28 \end{array} $

$ \begin{aligned} &\text { Common values of } 2 a+3 b=3 c+4 d\\ &7,10,11,13,14,15,16,17,18,20 \end{aligned} $

value $7,10,13,15,16,17,18,20$

occurs 1 time in L1 and 1 time in L2

e.g value 7 occurs 1 time in L1 $\&$ 1 time in L2 total $=1 \times 1=1$

so, total number of these types of elements

in $\mathrm{R}=8$--- (1)

value 11 and 14 occurs 2 times in L1

and 1 time in L2

e.g value 11 occurs 2 time in L1 $\&$ 1 time in L2 total

elements in $R$ is $2 \times 1=2$

so total number these type of elements $=2+2=4$--- (2)

total number of elements in R = 8+4 = 12

$\begin{aligned} & \text { Set } A=\{x \in \mathbb{Z}:|(|x-3|-3)| \leq 1\} \\\\ & |(|x-3|-3)| \leq 1 \quad\{|x| \leq a \Rightarrow x \in[-a, a]\}\end{aligned}$

$\Rightarrow $ $(|x-3|-3) \in[-1,1]$

$\Rightarrow $ $ |x-3| \in[2,4] $

$ \text { Using }|x| \in[\alpha, \beta] \Rightarrow x \in[-\beta,-\alpha] \cup[\alpha, \beta] $

$ x-3 \in[-4,-2] \cup[2,4]$

$\Rightarrow $ $ x \in[-1,1] \cup[5,7] $

Since $x \in \mathbb{Z}$ (integers), $x=-1,0,1,5,6,7$

So set $A=\{-1,0,1,5,6,7\}$

Set $B=\left\{x \in \mathbb{R}-\{1,2\}, \frac{(x-2)(x-4)}{(x-1)} \log _e(|x-2|)=0\right\}$

$\frac{(x-2)(x-4)}{(x-1)} \log _e(|x-2|)=0$

$\Rightarrow $ $x-2=0 \Rightarrow x=2$ but $x \neq 2$

$\Rightarrow $ $x-4=0 \Rightarrow x=4$

$\Rightarrow $ $\log _e|x-2|=0 \Rightarrow|x-2|=1$

$ \begin{aligned} & x-2= \pm 1 \Rightarrow x=3,1 \text { but } x \neq 1 \\ & \text { So set } B=\{3,4\} \\ & \text { Let } n(A) \& n(B) \text { be number of elements in set } A \text { and } B \\ & \text { respectively } \\ & n(A)=6, n(B)=2 \end{aligned} $

Number of onto function $f: A \rightarrow B$ is

= Total function - Into function

Total function $=(n(B))^{n(A)}=2^6$

number of Into function can be find using inclusion-exclusion principle

Number of into function $={ }^2 C_1(1)^6$

${ }^2 C_1(1)^6=$ only one image or range left from set $B$.

Number of onto function $=2^6-{ }^2 C_1(1)^6$

$=64-2=62$

$ A=\{0,1,2, \ldots 9\} $

A relation $R$ on $A$

$R=|x-y|$ is multiple of 3

$0,3,6,9$ are $3 n$ type numbers

$1,4,7$ are $3 n+1$ type numbers

$2,5,8$ are $3 n+2$ type numbers

Case 1 : Relation $R$ is valid if $x \& y$ both are $3 n$ type.

$ =4 \times 4=16 \text { pairs } $

Case 2 : Relation $R$ is valid if $x$ is $3 n+1$ type $\& y$ is $3 n+2$ type. $=3 \times 3=9$ pairs

Case 3 : similarly relation $R$ is valid if $x$ is $3 n+2$ type $\& y$ is $3 n+1$ type. $=3 \times 3=9$ pair

$ n(R)=16+9+9=34 \neq 36 $

statement I is false.

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Reflexive: For any $x \in A,|x-x|=0$. Since 0 is a multiple of $3,(x, x) \in R$.

Symmetric: If $(x, y) \in R$, then $|x-y|=3 k$. Since $|y-x|=|x-y|$, then $|y-x|= 3 k$, so $(y, x) \in R$.

Transitive: If $|x-y|$ and $|y-z|$ are multiples of 3 , then $x-y=3 k$ and $y-z=3 m$.

Adding these gives $x-z=3(k+m)$, so $|x-z|$ is a multiple of 3.

Statement II is correct.

$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \frac{x}{y} & (-2) & (-1) & 0 & 1 & 2 & 3 & 4 \\ \hline-2 & P & P & P & P & P & A & A \\ \hline-1 & P & P & P & P & A & A & A \\ \hline 0 & P & P & P & P & A & & \\ \hline 1 & P & P & P & A & & & \\ \hline 2 & P & P & P & & A & & \\ \hline 3 & P & P & & & & A & \\ \hline 4 & P & P & & & & & A \\ \hline \end{array} $

Here $P$ implies, $A$ implies absent

$\Rightarrow \quad 10$ elements to be added as $A=10, P=23$

$\Rightarrow A+P=33$

$ \begin{aligned} &\begin{aligned} &\underset{ \,\,\,\,\,\downarrow}{4} x^2+y_{\downarrow}^2<52, x, y \in Z \\ & 0 \quad 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7 \rightarrow 1 \times 15=15 \\ & \pm 1 \quad 0, \pm 1, \pm 2, \pm 3 \ldots \ldots \ldots . \pm 6 \rightarrow 2 \times 13=26 \\ & +2 \quad 0, \pm 1 \pm 2 \pm 3 \ldots \ldots \ldots . \pm 5 \rightarrow 2 \times 11=22 \\ & \pm 3 \quad 0, \pm 1, \pm 2, \pm 3 \rightarrow 2 \times 7=14 \end{aligned}\\ &\text { Number of elements }=77 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & M=\{1,2,3 \ldots .16\} \\ & R=\{(x, y) / 4 y=5 x-3, x, y \in N\} \\ & 4 y=5 x-3 \\ & y=\frac{5 x-3}{4} \end{aligned}\\ &\text { No of order pairs }=4 \end{aligned} $

For set $A$, we have the inequality:

$|x^2 - 10| \leq 6$

This means that the expression inside the modulus lies between $-6$ and $6$:

$-6 \leq x^2 - 10 \leq 6$

By adding $10$ to all parts of the inequality, we get:

$4 \leq x^2 \leq 16$

Taking the square root gives us the range for $|x|$:

$2 \leq |x| \leq 4$

This tells us that the value of $x$ can be positive or negative, giving us the intervals:

$A = [-4, -2] \cup [2, 4]$

Now, let's find the elements of set $B$ from its inequality:

$|x - 2| > 1$

This breaks down into two conditions:

$x - 2 < -1 \quad \text{or} \quad x - 2 > 1$

Solving these, we get:

$x < 1 \quad \text{or} \quad x > 3$

Writing this in interval notation, we have:

$B = (-\infty, 1) \cup (3, \infty)$

Now we check the options by performing the set operations. Let's find $B - A$, which represents the elements present in $B$ but not in $A$.

We take the set $B$ and remove any overlapping parts from $A$.

Looking at the first part of $B$, which is $(-\infty, 1)$, we need to remove the interval $[-4, -2]$ that belongs to $A$. Since the endpoints $-4$ and $-2$ are in $A$, they will be removed, leaving open brackets. This breaks the remaining interval into:

$(-\infty, -4) \cup (-2, 1)$

Next, looking at the second part of $B$, which is $(3, \infty)$, we need to remove the elements from $A$ that overlap here. The set $A$ contains $[2, 4]$. Removing this portion from $(3, \infty)$ cuts off everything up to $4$. Since $4$ is in $A$, it is removed, so we use an open bracket starting from $4$:

$(4, \infty)$

Combining these remaining parts together, we get:

$B - A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$

This exactly matches Option B.

Option B is the correct choice.

For $x=2$

$ \begin{aligned} & 4 \leq 3 y \quad y \in\{2,3,5,7,9\} \\ & (2,2)(2,3)(2,5)(2,7)(2,5) \end{aligned} $

For $x=3$

$ \begin{aligned} & 6 \leq 3 y y \in\{2,3,5,7,9\} \\ & (3,2)(3,3)(3,5)(3,7)(3,5) \end{aligned} $

For $x=5$

$ 10 \leq 3 y y \in\{5,7,9\} $

$ \begin{aligned} & (5,5)(5,7)(5,5) \\ & \text { For } x=7 \\ & 14 \leq 3 y y \in\{5,7,9\} \\ & (7,5)(7,7)(7,9) \\ & \text { For } x=9 \\ & 18 \leq 3 y y \in\{7,9\} \\ & (9,7)(9,9) \\ & R=5+5+3+3+2=18 \\ & (2,3) \in R \text { but }(3,2) \in R(\operatorname{symmetric}) \\ & (2,5) \in R \text { but }(5,2) \notin R \operatorname{Add}(5,2) \\ & (2,7) \in R \text { but }(7,2) \notin R \operatorname{Add}(7,2) \\ & (2,9) \in R \text { but }(9,2) \notin R \operatorname{Add}(9,2) \\ & (3,5) \in R \text { but }(5,3) \notin R \operatorname{Add}(5,3) \\ & (3,7) \in R \text { but }(7,3) \notin R \operatorname{Add}(7,3) \\ & (3,9) \in R \text { but }(9,3) \notin R \operatorname{Add}(9,3) \\ & (5,7) \in R \text { but }(7,5) \in R \\ & (5,9) \in R \text { but }(9,5) \notin R \operatorname{Add}(9,5) \\ & (7,9) \in R \text { but }(9,7) \in R \\ & 18+7=25 \end{aligned} $

Number of such relations:

From $A \rightarrow A$ such that

$ \begin{aligned} & n(A)=N \text { is } \\ & \Rightarrow 2^{\left(\frac{N^2-N}{2}\right)} \end{aligned} $

for $N=4 \Rightarrow 2^6=64$

Alter :

$ \begin{aligned} & {\left[\begin{array}{cccc} 0 & - & - & - \\ \Theta & 0 & - & - \\ \Theta & \Theta & 0 & - \\ \Theta & \Theta & \Theta & 0 \end{array}\right] \Rightarrow \text { each } \Theta \text { has two choices }} \\ & \Rightarrow 2^6=64 \end{aligned} $

We first observe the relation $\boldsymbol{R}$ defined on

$\boldsymbol{S} \times \boldsymbol{S} (\boldsymbol{a}, \boldsymbol{b})$ is an ordered pair $\in \boldsymbol{R}$ iff $(1+\boldsymbol{a} \boldsymbol{b})>0=\boldsymbol{a} \boldsymbol{b}>-1 \ldots \ldots(\boldsymbol{i})$

Refer the chart below :

We also observe the mathematical relationship from $(\boldsymbol{i})$ :

$a b>-1 \Rightarrow b a>-1 \equiv(a, b) \in R \Rightarrow(b, a) \in R \ldots \ldots(i i) \leftarrow$ symmetric

We take 2 ordered pair entries from $\boldsymbol{R}$ :

$ \begin{array}{|r|l|l|l|l|l|l|l|} \hline \text { Ordered pair }(\boldsymbol{a}, \boldsymbol{b}) & (2,0) & \boldsymbol{a} \boldsymbol{b}>-1 & (0,-2) & \boldsymbol{a} \boldsymbol{b}>-1 & (2,-2) & \boldsymbol{a} \boldsymbol{b} \ngtr-1 & \text { Transitive } \\ \hline \boldsymbol{a} \boldsymbol{b} & 0 & 0>-1 & 0 & 0>-1 & -4 & -4<-1 & \\ \hline \in \boldsymbol{R} & \text { Yes } & & \text { Yes } & & \text { No } & & \text { No } \\ \hline \end{array} $

$\therefore $ $ \begin{array}{|l|l|l|l|} \hline \text { Relation } & \text { Reflexive } & \text { Symmetric } & \text { Transitive } \\ \hline \boldsymbol{R} & \text { Yes } & \text { Yes } & \text { No } \\ \hline \end{array} $

Now let us focus on the cardinality of

$\boldsymbol{R} \in \boldsymbol{S} \times \boldsymbol{S} ;(\boldsymbol{a} \boldsymbol{b}>-1)$

$ \begin{array}{|l|l|l|l|} \hline \mathrm{I} & \text { The number of elements in } \mathrm{R} \text { is } 17 & \text { Valid } & \text { Option } \\ \hline \mathrm{II} & \mathrm{R} \text { is an equivalence relation } & \text { Invalid } & \mathrm{A} \\ \hline \end{array} $

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, $(x, y) \in R$ if and only if $\max\{x, y\} \in \{3, 4\}$.

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 $ S_1 $.

Next, we analyze the properties of the relation $ R $:

Reflexivity: A relation is reflexive if $(x, x) \in R$ for all $ x \in 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 $(a, b) \in R$, then $(b, a) \in R$ 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 $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. 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 $ S_2 $ is correct as $ R $ is symmetric but neither reflexive nor transitive.

$\begin{aligned} & \text { A: }|x-1| \leq 4 \text { and }|y-5| \leq 6 \\ & \Rightarrow-4 \leq x-1 \leq 4 \Rightarrow-6 \leq y-5 \leq 6 \\ & \Rightarrow-3 \leq x \leq 5 \quad \Rightarrow-1 \leq y \leq 11 \\ & \text { B : } 16(x-2)^2+9(y-6)^2 \leq 144 \\ & \text { B : } \frac{(x-2)^2}{9}+\frac{(y-6)^2}{16} \leq 1 \end{aligned}$

From Diagram $\mathrm{B} \subset \mathrm{A}$

$\begin{aligned} &\begin{aligned} & x R y \Leftrightarrow 2 x-y \in\{0,1\} \\ & \Rightarrow \quad y=2 x \text { or } y=2 x-1 \\ & A=\{-3,-2,-1,0,1,2,3\} \\ & \mathrm{R}=\{(-1,-2),(0,0),(1,2),(-1,-3),(0,-1),(1,1), \\ & (2,3)\} \\ & \Rightarrow \quad I=7 \end{aligned}\\ &\text { For } R \text { to be reflexive }(0,0),(1,1) \in R \end{aligned}$

But other $(a, a)$ such that $2 a-a \in\{0,1\}$

$\Rightarrow \quad a \in\{0,1\}$

5 other pairs needs to be added $\Rightarrow m=5$

$x R y \Rightarrow y R x$ 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, $n=5$

$\Rightarrow \quad l+m+n=17$

$\begin{aligned} & A=\left\{(x, y) \in R \times R: x^2+y^2=25\right\}, B=\{(x, y) \in \mathbb{R} \times \\ & \left.\mathbb{R}: x^2+9 y^2=144\right\} \\ & x^2+9 y^2-\left(x^2+y^2\right)=144-25 \\ & \text { Plug in } y^2=\frac{119}{8} \text { into either equation to find } x . \\ & x^2=25-\frac{119}{8} \\ & x^2=\frac{200-119}{8} \\ & x^2=\frac{81}{8} \\ & x= \pm \sqrt{\frac{81}{8}}, y= \pm \sqrt{\frac{119}{8}} \end{aligned}$

Now, $C=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x^2+y^2 \leq 4\right\}$

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 $C=13$

$\Rightarrow \quad$ There are 4 distinct real points in set $D$

$\therefore \quad$ The number of one-one functions from $D$ to $C$

$\Rightarrow \quad 13 P_4 \Rightarrow \frac{13!}{(13-4)!}=\frac{13!}{9!}=17160$

To solve the problem, we start by defining the set $A = \{-2, -1, 0, 1, 2, 3\}$ and the relation $R$ on set $A$, where an element $x$ is related to $y$ (written as $x \, R \, y$) if and only if $y = \max\{x, 1\}$.

This leads us to the following pairs in the relation $R$:

For $x = -2$, $y = \max\{-2, 1\} = 1$, so $(-2, 1)$ is in $R$.

For $x = -1$, $y = \max\{-1, 1\} = 1$, so $(-1, 1)$ is in $R$.

For $x = 0$, $y = \max\{0, 1\} = 1$, so $(0, 1)$ is in $R$.

For $x = 1$, $y = \max\{1, 1\} = 1$, so $(1, 1)$ is in $R$.

For $x = 2$, $y = \max\{2, 1\} = 2$, so $(2, 2)$ is in $R$.

For $x = 3$, $y = \max\{3, 1\} = 3$, so $(3, 3)$ is in $R$.

Thus, the relation $R$ consists of the pairs: $\{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and there are $l = 6$ elements in $R$.

Making the Relation Reflexive

A relation is reflexive if every element in the set $A$ relates to itself. Therefore, the missing reflexive pairs are:

$(-2, -2)$

$(-1, -1)$

$(0, 0)$

Adding these three pairs will make the relation reflexive, so $m = 3$.

Making the Relation Symmetric

A relation is symmetric if whenever $(x, y)$ is in $R$, $(y, x)$ must also be in $R$. Therefore, the missing symmetric pairs are:

$(1, -2)$

$(1, -1)$

$(1, 0)$

Thus, we need to add these three pairs for symmetry, so $n = 3$.

Finally, we calculate the sum $l + m + n = 6 + 3 + 3 = 12$.

The relation $ R $ is defined for the set $ \mathrm{A} = \{-3, -2, -1, 0, 1, 2, 3\} $ with the condition $ 0 \leq x^2 + 2y \leq 4 $. Let's determine the pairs $(x, y)$ that satisfy this condition.

For $ y = -3 $:

Solving $ x^2 + 2(-3) \leq 4 $, we find $ x = \{3, -3\} $.

For $ y = -2 $:

Solving $ x^2 + 2(-2) \leq 4 $, we find $ x = \{-2, 2\} $.

For $ y = -1 $:

Solving $ x^2 + 2(-1) \leq 4 $, we find $ x = \{-2, 2\} $.

For $ y = 0 $:

Solving $ x^2 + 0 \leq 4 $, we find $ x = \{-2, -1, 0, 1, 2\} $.

For $ y = 1 $:

Solving $ x^2 + 2(1) \leq 4 $, we find $ x = \{-1, 0, 1\} $.

For $ y = 2 $:

Solving $ x^2 + 2(2) \leq 4 $, we find $ x = \{0\} $.

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)\} $

Currently, $ R $ has $ l = 15 $ elements. To make $ R $ reflexive, it must include all pairs $(x, x)$ for every $ x \in \mathrm{A} $. We identify the missing reflexive pairs $(-1, -1)$, $(2, 2)$, and $(3, 3)$, which are required to satisfy reflexivity.

Thus, $ m = 3 $ more elements are needed. Therefore, the total $ l + m = 15 + 3 = 18 $.

The relation $ R $ is defined on the set $ A = \{1, 2, 3, \ldots, 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), \ldots, (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, \ldots, 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) $

For $R$ to be reflexive, $(f, f)$ must be in $R$.

The means $f(0)=f(1)$ and $f(1)=f(0)$ must be true for all $f$.

But $f(0) \neq f(1)$ always

Therefore, $R$ is not reflexive

If $(f, g) \in R$, then $f(0)=g(1)$ and $f(1)=g(0)$

$\begin{aligned} & \because \quad f(0)=g(1) \Rightarrow g(1)=f(0) \\ & \text { and } f(1)=g(0) \Rightarrow g(0)=f(1) \end{aligned}$

$R$ is symmetric

If $(f, g) \in R$ and $(g, h) \in \mathrm{R}$, then $f(0)=g(1)$, $f(1)=g(0), g(0)=n(1) \& g(1)=h(0)$

Since, $f(0)=g(1)$ and $g(1)=h(0)$, then $f(0)$ is not necessarily equal to $h(0)$.

Therefore, $R$ is not transitive.

$\therefore \quad$ The relation $R$ is symmetric but not reflexive or transitive.

$\begin{aligned} & S=\{0,1,2,3 \ldots . .\} \\ & \log _{\mathrm{e}} \mathrm{y}=\log _{\mathrm{e}}\left(\frac{2}{5}\right) \\ & \Rightarrow \mathrm{y}=\left(\frac{2}{5}\right)^{\mathrm{x}} \end{aligned}$

Required

$\text { Sum }=1+\left(\frac{2}{5}\right)^1+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}$

$\begin{aligned} & \sec ^2 x-\tan ^2 x=1 \quad(\text { on replacing } y \text { with } x) \\ & \Rightarrow \text { Reflexive } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \\ & \Rightarrow \sec ^2 y-\tan ^2 x=1 \\ & \Rightarrow \operatorname{symmetric} \\ & \sec ^2 x-\tan ^2 y=1 \\ & \sec ^2 y-\tan ^2 z=1 \end{aligned}$

Adding both

$\begin{aligned} & \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \\ & \sec ^2 x+1-\tan ^2 z=2 \\ & \sec ^2 x-\tan ^2 z=1 \\ & \Rightarrow \text { Transitive } \end{aligned}$

hence equivalence relation

$R=\{(x, y): x, y \in z$ and $x+y$ is even $\}$

reflexive $x+x=2 x$ even

symmetric of $x+y$ is even, then $(y+x)$ is also even

transitive of $\mathrm{x}+\mathrm{y}$ is even $\& \mathrm{y}+\mathrm{z}$ is even then $x+z$ is also even

So, relation is an equivalence relation.

$\begin{aligned} & \mathrm{A}: \log _{2 /\pi}|\sin \mathrm{x}|+\log _{2 /\pi}|\cos \mathrm{x}|=2 \\ & \Rightarrow \log _{2 /\pi}(|\sin \mathrm{x} \cdot \cos \mathrm{x}|)=2 \\ & \Rightarrow|\sin 2 \mathrm{x}|=\frac{8}{\pi^2} \end{aligned}$

Number of solution 4

B : let $\sqrt{\mathrm{x}}=\mathrm{t}<2$

Then $\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}-4)+3(\sqrt{\mathrm{x}}-2)+6=0$

$\begin{aligned} & \Rightarrow \mathrm{t}^2-4 \mathrm{t}+3 \mathrm{t}-6+6=0 \\ & \Rightarrow \mathrm{t}^2-\mathrm{t}=0, \mathrm{t}=0, \mathrm{t}=1 \\ & \mathrm{x}=0, \mathrm{x}=1 \end{aligned}$

again let $\sqrt{x}=t>2$

then $\mathrm{t}^2-4 \mathrm{t}-3 \mathrm{t}+6+6=0$

$\begin{aligned} & \Rightarrow t^2-7 \mathrm{t}+12=0 \\ & \Rightarrow \mathrm{t}=3,4 \\ & \mathrm{x}=9,16 \end{aligned}$

Total number of solutions

$\mathrm{n}(\mathrm{~A} \cup \mathrm{~B})=4+4=8$

Statement - I :

Reflexive : $\left(a_1, b_1\right) R\left(a_1, b_1\right) \Rightarrow b_1=b_1 \quad$ True

$\left.\begin{array}{rl}\text { Symmetric : } & \left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow \mathrm{b}_1=\mathrm{b}_2 \\ & \left(\mathrm{a}_2, \mathrm{~b}_2\right) \mathrm{R}\left(\mathrm{a}_1, \mathrm{~b}_1\right) \Rightarrow \mathrm{b}_2=\mathrm{b}_1\end{array}\right\}$ True

$\left.\begin{array}{ll}\text { Transitive: } & \left(a_1, b_1\right) R\left(a_2, b_2\right) \Rightarrow b_1=b_2 \\ & \&\left(a_2, b_2\right) R\left(a_3, b_3\right) b_2=b_3\end{array}\right\} b_1=b_3$

$\Rightarrow\left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_3 \cdot \mathrm{~b}_3\right) \Rightarrow \text { True }$

Hence Relation $R$ is an equivence relation Statement-I is true.

For statement - II $\Rightarrow \mathrm{y}=\mathrm{b}$ so False

$\begin{aligned} & \mathrm{C}=\{(3,0),(-3,0),(0,3),(0,-3)\} \\ & \Sigma|\mathrm{x}+\mathrm{y}|=12 \end{aligned}$

$\mathrm{A}=\{1,2,3,4\}$

For relation to be reflexive

$\mathrm{R}=\{(1,2),(2,3),(3,3)\}$

Minimum elements added will be

$\begin{aligned} & (1,1),(2,2),(4,4)(2,1)(3,2)(3,2)(3,1)(1,3) \\ & \therefore \text { Minimum number of elements }=7 \end{aligned}$

To find the number of elements in set $ B$, we consider pairs $\left(\frac{m}{n}\right)$ where $ m, n \in A $ with $ m < n $ and $\text{gcd}(m, n) = 1$.

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 $\text{gcd}(2, n) = 1$).

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 $

An equivalence relation on a finite set is uniquely determined by its partition into equivalence classes. Hence, counting the number of equivalence relations on a set is equivalent to counting the number of ways to partition that set.

Step: Counting partitions of $\{1,2,3\}$

We want all possible ways to split the set $\{1,2,3\}$ into nonempty subsets (its “blocks”).

3 blocks (each element in its own block)

$ \{\{1\}, \{2\}, \{3\}\}. $

2 blocks

$\{\{1,2\}, \{3\}\}$

$\{\{1,3\}, \{2\}\}$

$\{\{2,3\}, \{1\}\}$

1 block (all elements together)

$ \{\{1,2,3\}\}. $

Counting these, there are a total of 5 distinct partitions, and thus 5 equivalence relations on the set $\{1,2,3\}$.

All equivalence relations are automatically nonempty (they include at least $(1,1), (2,2), (3,3)$ because they are reflexive), so the answer to “the number of nonempty equivalence relations” is also 5.

Answer: Option C (5)

$(a, b) R(c, d) \Rightarrow 3 a d-7 b c \in$ even

For reflexive

$(a, b) R(a, b) \Rightarrow 3 a b-7 b a=-4 a b \in$ even

For symmetric

$(a, b) R(c, d)$

then

$(c, d) R(a, b)=3 b c-7 a d$

$\in$ even

Now check for transitive

$\begin{aligned} & \begin{array}{ll} (a, b) R(c, d) \text { and }(c, d) R(e, f) \text { then }(a, b) R(e, f) \\ 3 a d-7 b c=2 m \quad \Rightarrow (2,5) R(6,8) \text { and } \\ 3 c f-7 e d=2 n \quad (6,8) R(9,4) \\ \text { then } 3 a f-7 e b \neq \text { even } \notin(2,5) R(9,4) \\ \Rightarrow \text { Not transitive option }(3) \end{array} \end{aligned}$

$\begin{aligned} & A=\{1,2,3,4,5\} \\ & x R y \Leftrightarrow 4 x \leq 5 y \\ & 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25 \end{aligned}$

$\begin{aligned} & R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \\ & (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \\ & \therefore \quad n(R)=m=16 \end{aligned}$

Elements to be added to $R$ to make it symmetric

$(1,2) \in R \quad \Rightarrow \quad(2,1)$ should be added

Similarly, $(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3), (5,3)$

$\therefore 9$ elements should be added

$\begin{array}{ll} \therefore & n=9 \\ \therefore & m+n=25 \end{array}$

$n \in[100,700]$

$n(A)=$ Total $-$ (multiple of $3$ + multiple of 4) + (multiple of 12)

Total $=601$

Multiple of $3=102,105, \ldots, 699$

$\begin{aligned} & n=699=102+(n-1) 3 \\ & \Rightarrow n=200 \end{aligned}$

Multiple of $4=100,104 \ldots ., 700$

$\begin{aligned} & n=700=100+(n-1) 4 \\ & \frac{600}{4}+1=n \\ & \Rightarrow n=151 \end{aligned}$

Multiple of $12=108,120 \ldots .696$

$\begin{aligned} & n=696=108+(n-1) 12 \\ & n=50 \\ & \therefore n(A)=601-(200+151)+50 \\ & =300 \end{aligned}$

$\begin{aligned} & R_1=\{(x, y): 2 x-3 y=2\} \\ & R_2=\{(x, y):-5 x+4 y=0\} \\ & 2 x-3 y=2 \end{aligned}$

So $2 x$ and $3 y$ both has to be even or odd simultaneously and $2 x$ can't be odd so $2 x$ and $3 y$ both will be even

$R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}$

For symmetric we need to add 6 elements as

$\begin{aligned} & (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \\ & M=6 \end{aligned}$

For $R_2-5 x+4 y=0$

$5 x$ and $4 y$ has to be equal $4 y$ is always even so $5 x$ will also be even

$R_2=\{(4,5),(8,10),(12,15),(16,20)\}$

For symmetric we need to add 4 element as

$\begin{aligned} & (5,4)(10,8)(15,12)(20,16) \\ & N=4 \\ & M+N=6+4=10 \end{aligned}$

$\begin{aligned} & \left(x_1, y_1\right) R\left(x_2, y_2\right) \\ & \text { If } x_1 \leq x_2 \text { or } y_1 \leq y_2 \end{aligned}$

For reflexive;

$\begin{aligned} & \left(x_1, y_1\right) R\left(x_1, y_1\right) \\ & \Rightarrow x_1 \leq x_1 \text { or } y_1 \leq y_1 \end{aligned}$

So, $R$ is reflexive

For symmetric

When $\left(x_1, y_1\right) R\left(x_2, y_2\right)$

$\Rightarrow x_1 \leq x_2 \text { or } y_1 \leq y_2$

For $\left(x_2, y_2\right) R\left(x_1, y_1\right)$

$\Rightarrow x_2 \leq x_1 \text { or } y_2 \leq y_1$

Not true for $(1,2)$ and $(3,4)$

For transitive

Take pairs as $(3,9),(4,6),(2,7)$

$(3,9) R(4,6)$

as $4 \geq 3$

$(4,6) R(2,7)$

As $7 \geq 6$

But $(3,9) R(2,7)$

As neither $2 \geq 3$ nor $7 \geq 9$

So not transitive

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.

Let's start by analysing $R_1$:

Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \in S$, the pair $(a,a)$ is in $R$. In the case of $R_1$, for an arbitrary $a \in \mathbf{R}$, we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$. This would mean $2a^2 = 1$ or $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive.

Symmetry: A relation $R$ is symmetric if whenever $a R b$, then $b R a$. For $R_1$, if $a^2 + b^2 = 1$, then it is also true that $b^2 + a^2 = 1$. Thus, $R_1$ is symmetric.

Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$, then $a R c$. For $R_1$, suppose $a R_1 b$ and $b R_1 c$, this means $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. However, if we add these two, 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.

Conclusion: Relation $R_1$ is not reflexive or transitive, and hence it is not an equivalence relation.

Now let's consider $R_2$:

Reflexivity: For any $(a, b) \in \mathbf{N} \times \mathbf{N}$, it's clear that $a + b = b + a$, which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_2 (a, b)$, and $R_2$ is reflexive.

Symmetry: If $(a, b) R_2 (c, d)$, meaning $a + d = b + c$, then by reordering the terms we can similarly have $c + b = d + a$, which means $(c, d) R_2 (a, b)$, so $R_2$ is symmetric.

Transitivity: Suppose $(a, b) R_2 (c, d)$ and $(c, d) R_2 (e, f)$, i.e., $a + d = b + c$ and $c + f = d + e$, we want to show that $(a, b) R_2 (e, f)$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. By rearranging and simplifying, we get $a + f = b + e$, thus, $(a, b) R_2 (e, f)$.

So $R_2$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.

Conclusion: According to the above examination, only $R_2$ is an equivalence relation. Thus, the correct answer is:

Option C: Only $R_2$ is an equivalence relation.

Given set $\{1,2,3,4\}$

$(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2)$

Thus no. of elements $=10$

$(a, b) R(a, b)$ as $a b-a b=0$

Therefore reflexive

Let $(a, b) R(c, d) \Rightarrow a d-b c$ is divisible by 5

$\Rightarrow \mathrm{bc}-\mathrm{ad}$ is divisible by $5 \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})$

Therefore symmetric

Relation not transitive as $(3,1) \mathrm{R}(10,5)$ and $(10,5) \mathrm{R}(1,1)$ but $(3,1)$ is not related to $(1,1)$

$\begin{aligned} & 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\ & 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\ & 2^{\mathrm{n}}=2^3 \text { and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\ & \Rightarrow \mathrm{n}=3 \text { and } 2^{\mathrm{m}-\mathrm{n}}=8 \\ & \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}-\mathrm{n}=3 \\ & \Rightarrow \mathrm{n}=3 \text { and } \mathrm{m}=6 \\ & \mathrm{P}(6,3) \text { and } \mathrm{Q}(-2,-3) \\ & \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10 \end{aligned}$

Hence option (1) is correct

Let $S=\{1,2,3, \ldots, 10\}$

$R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}$

For Reflexive,

$M$ is subset of '$S$'

So $\phi \in \mathrm{M}$

for $\phi \cap \phi=\phi$

$\Rightarrow$ but relation is $\mathrm{A} \cap \mathrm{B} \neq \phi$

So it is not reflexive.

For symmetric,

$\begin{array}{ll} \text { ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \\ \Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi, \end{array}$

So it is symmetric.

For transitive,

$\begin{aligned} \text { If } A & =\{(1,2),(2,3)\} \\ B & =\{(2,3),(3,4)\} \\ C & =\{(3,4),(5,6)\} \end{aligned}$

$\mathrm{ARB}$ & $\mathrm{BRC}$ but $\mathrm{A}$ does not relate to $\mathrm{C}$ So it not transitive