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

$\sqrt{x^2+x-2}>1-x$

$ \Rightarrow \quad x^2+x-2 \geq 0 $

and $x^2+x-2>1+x^2-2 x$

$\Rightarrow(x+2)(x-1) \geq 0$ and $3 x>3$

$\Rightarrow x \leq-2$ or $x \geq 1 \quad$ and $x>1$

On combining the inequality, we get $x>1$ i.e., $(1, \infty)$

Set $A$ has 5 elements.

Set $B$ has 7 elements.

In a function from $A$ to $B$, every element of $A$ must be mapped to an element of $B$. Therefore, total number of functions will be

$ =7 \times 7 \times 7 \times 7 \times 7=7^5 $

Now, number of many one functions $=$ Total functions-number of one-one functions

$ \begin{aligned} & =7^5-{ }^7 C_5 \times 5! \\ & =7^5-\frac{7!}{(7-5)!5!} \times 5! \\ & =7^5-\frac{7!}{(7-5)!}=7^5-{ }^7 P_5 \\ & \qquad \quad\left(\because{ }^n P_r=\frac{n!}{(n-r)!}\right) \\ & =7^5-{ }^7 P_5 \end{aligned} $

In a survey of 1003 families in Calcutta, we have the following information:

63 families have neither a radio nor a TV.

794 families have a radio.

187 families have a TV.

We need to calculate how many families have both a radio and a TV.

Step-by-step Solution:

Calculate the total number of families with either a radio or a TV. Using the formula for the complement set:

$ n(R \cup T) + n(R \cup T)' = \text{Total families} $

$ n(R \cup T) + 63 = 1003 $

Solving for $ n(R \cup T) $:

$ n(R \cup T) = 1003 - 63 = 940 $

Apply the principle of inclusion-exclusion from set theory:

$ n(R \cup T) = n(R) + n(T) - n(R \cap T) $

Substitute the known values:

$ 940 = 794 + 187 - n(R \cap T) $

Simplify to find $ n(R \cap T) $:

$ 940 = 981 - n(R \cap T) $

$ n(R \cap T) = 981 - 940 = 41 $

Therefore, 41 families have both a radio and a TV.

1. We are given the number of medals for events A, B, and C which are 48, 25, and 18 respectively. We are also given that the total number of unique medal recipients across all events is 60 and that 5 people received a medal in all three events.




2. Using the Principle of Inclusion and Exclusion (PIE), we know that the total number of unique medal recipients can be calculated by adding the number of medal recipients in each event, subtracting the number of people who received a medal in any two events (to correct for double counting), and then adding back the number of people who received a medal in all three events (since we subtracted these people too much).

   
   

   Mathematically, this can be represented as :

   
   

   |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C|

   
   

   However, we want to find the total number of people who received a medal in any two events (which is represented by |A ∩ B| + |B ∩ C| + |C ∩ A| in the equation).

   
   



3. To find this, we rearrange the PIE formula to solve for |A ∩ B| + |B ∩ C| + |C ∩ A| :

   
   

   |A ∩ B| + |B ∩ C| + |C ∩ A| = |A| + |B| + |C| + |A ∩ B ∩ C| - |A ∪ B ∪ C|

   
   

   Substituting the given values, we find that the total number of people who received a medal in any two events is 48 + 25 + 18 + 5 - 60 = 36.




4. However, this includes people who received a medal in all three events, and we want to find the number of people who received a medal in exactly two events. Therefore, we need to subtract the people who received a medal in all three events from our calculated value.

   
   

   Since each person who received a medal in all three events is counted three times in |A ∩ B| + |B ∩ C| + |C ∩ A| (once for each pair of events), we subtract three times the number of people who received a medal in all three events from our calculated value:

   
   

   Number of people who received a medal in exactly two events = |A ∩ B| + |B ∩ C| + |C ∩ A| - 3 $ \times $ |A ∩ B ∩ C|

   
   

   Substituting the values we know, we find that the number of people who received a medal in exactly two events is 36 - 3 $ \times $ 5 = 21.

Therefore, 21 people received a medal in exactly two of the three events.

Given sets :
$ A = {2,3,4} $
$ 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 :

1. $ a_1 $ divides $ b_2 $
2. $ a_2 $ divides $ b_1 $

For the first condition :
$ a_1 $ divides $ b_2 $
Given $ 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, the pairs are the same, because it's just the reversed relation. So :
$ a_2 $ divides $ b_1 $
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.

For symmetric $(b,a),(c,b)\in R$

For transitive $(a,c)\in R$

$\Rightarrow (c,a)\in R$

$\therefore (a,b),(b,a)\in R$

$\Rightarrow (a,a)\in R$

$(b,c),(c,b)\in R$

$\Rightarrow (b,b)\in R,(c,c)\in R$

7 elements must be added

a R b if 2a + 3b = 5m, m $\in$ $l$

(1) $(a,a) \in R$ as $2a + 3a = 5a,a \in N$

Hence, R is reflexive

(2) If $(a,b) \in R$ then $2a + 3 = 5m$

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 equivalence relation.

Option (1) is correct.