Sets and Relations

$R = \{ (x,y):y \in {A_i},\,iff\,x \in {A_i}\,1 \le i \ge k\} $

(1) Reflexive

(a, a) $\Rightarrow$ $a \in {A_i}$ iff $a \in {A_i}$

(2) Symmetric

(a, b) $\Rightarrow$ $a \in {A_i}$ iff $b \in {A_i}$

(b, a) $\in$R as $b \in {A_i}$ iff $a \in {A_i}$

(3) Transitive

(a, b) $\in$R & (b, c) $\in$R.

$\Rightarrow$ $a \in {A_i}$ iff $b \in {A_i}$ & $b \in {A_i}$ iff $c \in {A_i}$

$\Rightarrow$ $a \in {A_i}$ iff $c \in {A_i}$

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

$\Rightarrow$ RElation is equivalnece.

$R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\}$ and

$R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$

In $R_{1}: \because|2-11|=9 \leq 13$

$\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$

$\therefore \quad R_{1}$ is not transitive

Hence $R_{1}$ is not equivalence

In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$ $(\because|13-26|=13)$

$\therefore R_{2}$ is not transitive

Hence $R_{2}$ is not equivalence.

Note that (a, b) and (b, c) satisfy 0 < |x $-$ y| $\le$ 1 but (a, c) does not satisfy it so 0 $\le$ |x $-$ y| $\le$ 1 is symmetric but not transitive.

For example,

x = 0.2, y = 0.9, z = 1.5

0 ≤ |x – y| = 0.7 ≤ 1

0 ≤ |y – z| = 0.6 ≤ 1

But |x – z| = 1.3 > 1

So, (b) is correct.

This solution begins by applying the principle of inclusion and exclusion, which in the context of this problem, is represented by the formula :

n(A ∪ B) ≥ n(A) + n(B) - n(A ∩ B)

Here, n(A ∪ B) represents the total number of patients in the hospital, which is 100%. n(A) represents the proportion of patients with a heart ailment (89%), and n(B) represents the proportion of patients with a lung infection (98%).

By rearranging this formula, the solution establishes an inequality for n(A ∩ B), the proportion of patients suffering from both ailments :

100% ≥ 89% + 98% - n(A ∩ B)

Therefore,

n(A ∩ B) ≥ 87%

Next, the solution notes that n(A ∩ B) cannot be greater than the smaller of n(A) and n(B), since it cannot be larger than the smallest group. Thus, we have another inequality :

n(A ∩ B) ≤ 89%

Combining these two inequalities gives :

87% ≤ n(A ∩ B) ≤ 89%

Hence, the proportion of patients suffering from both ailments must be a value between 87% and 89% inclusive. So, the set of values {79,81,83,85} which are all less than 87% are values that n(A ∩ B) cannot belong to. Therefore, option C is the correct answer.

${x^3} - 3{x^2}y - x{y^2} + 3{y^3} = 0$

$ \Rightarrow x({x^2} - {y^2}) - 3y({x^2} - {y^2}) = 0$

$ \Rightarrow (x - 3y)(x - y)(x + y) = 0$

Now, x = y $\forall$(x, y) $\in$N $\times$ N so reflexive but not symmetric & transitive.

See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, $-$1) satisfies but (3, $-$1) does not.

For reflexive relation,

$\forall(A, A) \in R$ for matrix $P$.

$\Rightarrow A=P A P^{-1}$ is true for $P=1$

So, $R$ is reflexive relation.

For symmetric relation,

Let $(A, B) \in R$ for matrix $P$.

$ \Rightarrow \quad A=P B P^{-1} $

After pre-multiply by $P^{-1}$ and post-multiply by $P$, we get

$ P^{-1} A P=B $

So, $(B, A) \in R$ for matrix $P^{-1}$.

So, $R$ is a symmetric relation.

For transitive relation,

Let $A R B$ and $B R C$

So, $A=P B P^{-1}$ and $B=P C P^{-1}$

Now, $A=P\left(P C P^{-1}\right) P^{-1}$

$\Rightarrow A=(P)^2 C\left(P^{-1}\right)^2 \Rightarrow A=(P)^2 \cdot C \cdot\left(P^2\right)^{-1}$

$\therefore(A, C) \in R$ for matrix $P^2$.

$\therefore R$ is transitive relation.

Hence, $R$ is an equivalence relation.

As none play all three games the intersection of all three circles must be zero.

Hence none of P, Q, R justify the given statement

ad = bc

(a, b) R (4, 3) $ \Rightarrow $ 3a = 4b

a = ${4 \over 3}$b

b must be multiple of 3

b = {3, 6, 9 ..... 30}

(a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)}

$ \Rightarrow $ 7 ordered pair

Case 1 :

x $ \le $ $-$4

($-$x $-$ 3)($-$x $-$ 4) = 6

$ \Rightarrow $ (x + 3)(x + 4) = 6

$ \Rightarrow $ x² + 7x + 6 = 0

$ \Rightarrow $ x = $-$1 or $-$6

but x $ \le $ $-$4

x = $-$6

Case 2 :

x $\in$ ($-$4, 0)

($-$x $-$ 3)(x + 4) = 6

$ \Rightarrow $ $-$x² $-$ 7x $-$ 12 $-$ 6 = 0

$ \Rightarrow $ x² + 7x + 18 = 0

D < 0 No solution

Case 3 :

x $ \ge $ 0

(x $-$ 3)(x + 4) = 6

$ \Rightarrow $ x² + x $-$ 12 $-$ 6 = 0

$ \Rightarrow $ x² + x $-$ 18 = 0

x = ${{ - 1 \pm \sqrt {1 + 72} } \over 2}$

$ \therefore $ x = ${{\sqrt {73} - 1} \over 2}$ only

Given R = {(P, Q) | P and Q are at the same distance from the origin}.

Then equivalence class of (1, $-$1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, $-$1).

i.e., radius of circle = $\sqrt {{1^2} + {1^2}} = \sqrt 2 $

$ \therefore $ Required equivalence class of (S)

$ = \{ (x,y)|{x^2} + {y^2} = 2\} $.

C $ \to $ person like coffee

T $ \to $ person like Tea

n(C) = 73

n(T) = 65

n(C $ \cup $ T) $ \le $ 100

n(C) + n(T) – n (C $ \cap $ T) $ \le $ 100

73 + 65 – x $ \le $ 100

x $ \ge $ 38

73 – x $ \ge $ 0 $ \Rightarrow $ x $ \le $ 73

65 – x $ \ge $ 0 $ \Rightarrow $ x $ \le $ 65

$ \therefore $ 38 $ \le $ x $ \le $ 65

$\mathop \cup \limits_{i = 1}^{50} {X_i} = $ X₁, X₂,....., X₅₀ = 50 sets. Given each sets having 10 elements.

So total elements = 50 $ \times $ 10

$\mathop \cup \limits_{i = 1}^n {Y_i} =$ $ Y₁, Y₂,....., Y_n = n sets. Given each sets having 5 elements.

So total elements = 5 $ \times $ n

Now each element of set T contains exactly 20 of sets X_i.

So number of effective elements in set T = ${{50 \times 10} \over {20}}$

Also each element of set T contains exactly 6 of sets Y_i.

So number of effective elements in set T = ${{50 \times 10} \over {20}}$

$ \therefore $ ${{50 \times 10} \over {20}}$ = ${{5 \times n} \over {20}}$

$ \Rightarrow $ n = 30

A $ \cup $ B = 63 - x + x + 76 - x = 139 - x

As 139 - x $ \le $ 100

$ \Rightarrow $ x $ \ge $ 39

From venn diagram, you can see x should be less than 76 and 63 otherwise only A newspaper or only B newspaper reader will be negative number.

Intersection of x $ \le $ 63 and x $ \le $76 is = x $ \le $ 63.

$ \therefore $ 39 $ \le $ x $ \le $ 63

From options possible value of x = 55.

For R₁ :

Let a = 1 + $\sqrt 2 $, b = 1 $-$ $\sqrt 2 $, c = ${8^{{1 \over 4}}}$

aR₁b : a² + b² = 6 $ \in $ Q

bR₁c : b² + c² = 3 $-$ 2$\sqrt 2 $ + 2$\sqrt 2 $ = 3 $ \in $ Q

aR₁c : a² + c² = 3 + 2$\sqrt 2 $ + 2$\sqrt 2 $ $ \notin $ Q

$ \therefore $ R₁ is not transitive.

For R₂ :

Let a = 1 + $\sqrt 2 $, b = $\sqrt 2 $, c = 1 $-$ $\sqrt 2 $

aR₂b : a² + b² = 5 + 2$\sqrt 2 $ $ \notin $ Q

bR₂c : b² + c² = 5 $-$ 2$\sqrt 2 $ $ \notin $ Q

aR₂c : a² + c² = 3 + 2$\sqrt 2 $ + 3 $-$ 2$\sqrt 2 $ = 6 $ \in $ Q

$ \therefore $ R₂ is not transitive.

As roots are real so, $D \ge 0$

${(m + 1)^2} - 4(m + 4) \ge 0$

$ \Rightarrow {m^2} - 2m - 15 \ge 0$

$ \Rightarrow $ $(m - 5)(m + 3) \ge 0$

$m\, \in \,$($ - $$ \propto $, $ - $3] $ \cup $ [5, $ \propto $)

$A= ( - $$ \propto $, $ - $3] $ \cup $ [5, $ \propto $)

Given B = [$ - $3, 5)

Now, let's examine the options.

Option A : A ∩ B = {–3} The intersection of sets A and B would be the set of elements common to both sets. In this case, the only common element is -3. So, option A is true.

Option B : B – A = (–3, 5) The subtraction (or difference) of sets A from B is the set of elements that are in B but not in A. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5. So, option B is also true.

Option C : A ∪ B = R The union of sets A and B is the set of elements that are in A, or B, or both. Here, A U B would cover all real numbers. So, option C is true.

Option D : A - B = (-∞, -3) ∪ (5, ∞) The subtraction (or difference) of set B from A is the set of elements that are in A but not in B. B is [–3, 5), and A is (-∞, -3] U [5, ∞). Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5. But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞). So, option D is not true.

Given R = {(x, y) : x, y $ \in $ Z, x² + 3y² $ \le $ 8}

So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1)
(-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)}

$ \Rightarrow $ R : { -2, -1, 0, 1, 2} $ \to $ {-1, 0, 1}

$ \therefore $ R^-1 : {-1, 0, 1} $ \to $ { -2, -1, 0, 1, 2}

$ \therefore $ Domain of R^–1 = {-1, 0, 1}

A : x $ \in $ (–2, 2);

B : x $ \in $ (–$\infty $, –1] $ \cup $ [5, $\infty $)

$ \Rightarrow $ B – A = R – (–2, 5)

According to the question, we have the following Venn diagram.

Here, $A \cap B \subseteq C$ and $A \cap B \neq \phi$


Now, from the Venn diagram, it is clear that $B \cap C \neq \phi$, is true

Also, $(C \cup A) \cap(C \cup B)=C \cup(A \cap B)=C$ is true.

If $(A-B) \subseteq C$, for this statement the Venn diagram is


From the Venn diagram, it is clear that if $A-B \subseteq C$, then $A \subseteq C$.

Now, if $(A-C) \subseteq B$, for this statement the Venn diagram.


From the Venn diagram, it is clear that

$A \cap B \neq \phi, A \cap B \subseteq C$ and $A-C=\phi \subseteq B$ but $A \subseteq B$

The total population to be 100 (for simplicity's sake) and the percentages can be treated as actual numbers of people in this context.

The percentage of people who read newspaper A is given as 25. However, among these, there are people who read both newspapers A and B, given as 8. To find the number of people who read only newspaper A, we subtract the number of people who read both from the total number of people who read A. That is,

n(A only) = 25 – 8 = 17

Similarly, the number of people who read only newspaper B is calculated as :

n(B only) = 20 – 8 = 12

Now, we are given the percentage of each of these groups that look into the advertisements:

• 30% of those who read A but not B,
• 40% of those who read B but not A,
• 50% of those who read both A and B.

To find the total percentage of the population that looks into advertisements, we add up the contributions from each of these groups. We calculate each group's contribution by multiplying the size of the group by the percentage of that group that looks at advertisements :

= ${{30} \over {100}} \times 17$ (from A only) + ${{40} \over {100}} \times 12 $(from B only) + ${{50} \over {100}} \times 8 $(from both A and B)

= 5.1 (from A only) + 4.8 (from B only) + 4 (from both A and B)

Adding these up, we get

= 13.9

This means that 13.9% of the total population looks into the advertisements.

So, the correct answer is :

Option D : 13.9.

A ={x $ \in $ z : 2^{(x+2)(x2 $-$ 5x + 6)} = 1}

2^{(x+2)(x2 $-$ 5x + 6)} = 2⁰ $ \Rightarrow $ x = $-$ 2, 2, 3

A = {$-$2, 2, 3}

B = {x $\varepsilon $ Z : $-$ < 2x $-$ 1 < 9}

B = {0, 1, 2, 3, 4}

A $ \times $ B has is 15 elements so number of subsets of A $ \times $ B is 2¹⁵.

S = {1,2,3, . . . .100}

= Total non empty subsets-subsets with product of element is odd

= 2¹⁰⁰ $-$ 1 $-$ 1[(2⁵⁰ $-$ 1)]

= 2¹⁰⁰ $-$ 2⁵⁰

= 2⁵⁰ (2⁵⁰ $-$ 1)

We're given that there are 140 students numbered from 1 to 140.

1. Define the set $A$ to be the set of even numbered students. The cardinality of $A$ (the number of elements in $A$), denoted as $n(A)$, can be computed as the greatest integer less than or equal to $140/2$. Hence, $n(A) = \left[\frac{140}{2}\right] = 70$. ([.] denotes greatest integer function)

2. Similarly, let $B$ be the set of students whose number is divisible by 3. Hence, $n(B) = \left[\frac{140}{3}\right] = 46$. ([.] denotes greatest integer function)

3. Let $C$ be the set of students whose number is divisible by 5. Hence, $n(C) = \left[\frac{140}{5}\right] = 28$.

So far, we've found the number of students who opted for Mathematics ($n(A)$), Physics ($n(B)$), and Chemistry ($n(C)$).

We also need to consider the students who have opted for multiple subjects :

1. $n(A \cap B)$ represents the count of numbers that are divisible by both 2 and 3 (i.e., divisible by 6).

So, $n(A \cap B) = \left[\frac{140}{6}\right] = 23$.

2. $n(B \cap C)$ represents the count of numbers that are divisible by both 3 and 5 (i.e., divisible by 15).

So, $n(B \cap C) = \left[\frac{140}{15}\right] = 9$.

3. $n(C \cap A)$ represents the count of numbers that are divisible by both 2 and 5 (i.e., divisible by 10).

So, $n(C \cap A) = \left[\frac{140}{10}\right] = 14$.

Finally, $n(A \cap B \cap C)$ represents the count of numbers that are divisible by 2, 3, and 5 (i.e., divisible by 30).

So, $n(A \cap B \cap C) = \left[\frac{140}{30}\right] = 4$.

Now we use the principle of inclusion and exclusion to compute the number of students who have opted for at least one subject. The principle states :

$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)$

Substituting the values we calculated above :

$n(A \cup B \cup C) = (70+46+28) - (23+9+14) + 4 = 102$

Hence, the number of students who opted for at least one subject is 102. Therefore, the number of students who did not opt for any of the subjects is

Total $-$ n(A $ \cup $ B $ \cup $ C)

= 140 $-$ 102 = 38

For R₁; 2x + y = 10 and x, y $ \in $ N possible values for x and y are :

x = 1, y = 8    i.e.   (1, 8);

x = 2, y = 6    i.e    (2, 6);

x = 3, y = 4    i.e    (3, 4);

x = 4, y = 2    i.e    (4, 2)

$\therefore\,\,\,$ R₁ = { (1, 8), (2, 6), (3, 4), (4, 2) }

$\therefore\,\,\,$ Range of R₁ is {2, 4, 6, 8}

R₁ is not symmetric.

R₁ is not transitive also as

(3, 4), (4, 2) $ \in $ R , but (3, 2) $ \notin $ R₁

For R₂ : x + 2y = 10 and x, y $ \in $ N

Possible values of x, and y are :

x = 8, y= 1    i.e    (8, 1)

x = 6, y = 2    i.e    (6, 2)

x = 4, y = 3    i.e    (4, 3) and

x = 2, y = 4    i.e    (2, 4)

$\therefore\,\,\,$ R₂ = {(8, 1) (6, 2) (4, 3) (2, 4)}

$\therefore\,\,\,$ Range of R₂ = $\left\{ {1,2,3,4} \right\}$

R₂ is not symmetric and transitive

Given,

$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$

Let $a - 6 = x$ and $b - 5 = y$

$\therefore\,\,\,\,$ $4{x^2} + 9{y^2} \le 36$

$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$

This is a equation of ellipse.

This ellipse will look like this,



According to set A,

$\left| {a - 5} \right| < 1$

as $a - 6 = x$ then $a - 5 = x + 1$

$\therefore\,\,\,$ $\left| {x + 1} \right| < 1$

$ \Rightarrow \,\,\, - 1 < x + 1 < 1$

$ \Rightarrow \,\,\, - 2 < x < 0$

$\left| {b - 5} \right| < 1$

as $b - 5 = y$

$\therefore\,\,\,$ $\left| y \right| < 1$

$ \Rightarrow \,\,\,\, - 1 < y < 1$

This will look like this,



By combining both those graphs it will look like this,



To check entire set A is inside of B or not put any paint of set A on the equation ${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$ and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $-$1) inside the graph. But to check point ($-2$, 1) or ($-$2, $-$1) is inside of the ellipse or not, put on the ${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$

Putting ($-$2, 1) on the inequality,

LHS $ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$

$ = {4 \over 9} + {1 \over 4}$

$ = {{25} \over {36}} < 1$

$\therefore\,\,\,$ Inequality holds.

So, $\left( { - 2,1} \right)$ is inside the ellipse.

Similarly by checking we can see $\left( { - 2, - 1} \right)$ is also inside the ellipse.

Hence, we can say entire set A is inside of the set B.

$\therefore\,\,\,\,$ A $ \subset $ B

Here both R₁ and R₂ are symmetric as for any (x, y) $ \in $ R₁, we have (y, x) $ \in $ R₁ and similarly for any (x, y) $ \in $ R₂, we have (y, x) $ \in $ R₂

In R₁, (b, c) $ \in $ R₁, (c, a) $ \in $ R₁ but (b,a) $ \notin $ R₁

Similarly in R₂, (b, a) $ \in $ R₂, (a, c) $ \in $ R₂ but (b, c) $ \notin $ R₂

$ \therefore $ R₁ and R₂ are not transitive.

Given,

sin$\theta $ $-$ cos$\theta $ = $\sqrt 2 $cos$\theta $

$ \Rightarrow $   sin$\theta $ = $\left( {\sqrt 2 + 1} \right)$cos$\theta $

$ \Rightarrow $   sin$\theta $ = ${{2 - 1} \over {\sqrt 2 - 1}}$cos$\theta $

$ \Rightarrow $   $\left( {\sqrt 2 - 1} \right)$sin$\theta $ = cos$\theta $      . . . (1)

This is for set P.

sin$\theta $ + cos$\theta $ = ${\sqrt 2 }$sin$\theta $

$ \Rightarrow $   cos$\theta $ = $\left( {\sqrt 2 - 1} \right)$sin$\theta $      . . . .(2)

This is from set Q.

So,   P = Q

Given,

$ \begin{aligned} &n(A)=4, n(B) =2 \\\\ &\Rightarrow n(A \times B) =8 \end{aligned} $

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 $={ }^8 C_1$

Number of subsets of set $A \times B$ having two elements $={ }^8 C_2$

$\therefore$ Number of subsets having atleast three elements

$ \begin{aligned} &=2^8-\left(1+{ }^8 C_1+{ }^8 C_2\right) \\\\ &=2^8-1-8-28 \\\\ &=2^8-37 \\\\ &=256-37=219 \end{aligned} $

For any element x_i present in X, 4 cases arises while making subsets Y and Z.

Case- 1 : x_i $ \in $ Y, x_i $ \in $ Z $ \Rightarrow $ Y $ \cap $ Z $ \ne $ $\phi $

Case- 2 : x_i $ \in $ Y, x_i $ \notin $ Z $ \Rightarrow $ Y $ \cap $ Z = $\phi $

Case- 3 : x_i $ \notin $ Y, x_i $ \in $ Z $ \Rightarrow $ Y $ \cap $ Z = $\phi $

Case- 4 : x_i $ \notin $ Y, x_i $ \notin $ Z $ \Rightarrow $ Y $ \cap $ Z = $\phi $

$ \therefore $ For every element, number of ways = 3 for which Y $ \cap $ Z = $\phi $

$ \Rightarrow $ Total ways = 3 × 3 × 3 × 3 × 3 [$ \because $ no. of elements in set X = 5]

= 3⁵

An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).

Statement I : $A={(x, y) \in R \times R: }$ y-x is an integer .

• Reflexivity : For all $x$ in $R$, $x - x = 0$ which is an integer. So, every element is related to itself.


• Symmetry : For all $x, y$ in $R$, if $y - x$ is an integer, then $x - y = - (y - x)$ is also an integer. So, if $x$ is related to $y$, then $y$ is related to $x$.


• 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. So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $A$ is an equivalence relation on $R$.

Statement II : $ B={(x, y) \in R \times R: x=\alpha y}$ for some rational number $\alpha$.

• Reflexivity : For all $x$ in $R$, $x = 1x$. 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, the correct answer is

Option B : Statement I is true, Statement II is false.

Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Relation R : $R=(x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w$.

• Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.


• Symmetry : For all $x, y$ in $R$, if $x = w y$ 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 $x = w y$ and $y = v z$ for some rational numbers $w$ and $v$, then $x = (w v)z$. 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 : $S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p m\}$

• 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 $q m = p n$, then $n p = m n$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$.


• Transitivity :
  
  $ \begin{array}{rlr} \frac{m}{n} R \frac{p}{q} \text { and } \frac{p}{q} R \frac{r}{s} \\\\ \Rightarrow m q=n p \text { and } p s=r q \\\\ \Rightarrow m q \cdot p s=n p \cdot r q \\\\ \Rightarrow \quad m s=n r \\\\ \Rightarrow \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} R \frac{r}{s} \end{array} $
  
  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 denominator is not zero.

In conclusion, the correct answer is

Option C : $S$ is an equivalence relation but $R$ is not an equivalence relation.

From the given conditions, we have :

1. A ∩ B = A ∩ C : The intersection of set A with set B is the same as the intersection of set A with set C. This indicates that all elements common to A and B are also common to A and C, and vice versa.




2. A ∪ B = A ∪ C : The union of set A with set B is the same as the union of set A with set C. This indicates that all elements in A, B, and C are the same.

From these two conditions, we can infer that set B is equal to set C because every element of B is also an element of C and vice versa. Hence, Option B : B = C is the correct answer.

Given $S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,$

and $\,\,\,\left. {0 < x < 2} \right\}$

As $\,\,\,\,x \ne x + 1\,\,\,$

for any $\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S$

$\therefore$ $S$ is not reflexive.

Hence $S$ in not an equivalence relation.

Also $\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y$ is an integer $\left. {} \right\}$

as $x - x = 0$ is an integer $\forall x \in R$

$\therefore$ $T$ is reflexive.

If $x-y$ is an integer then $y-x$ is also an integer

$\therefore$ $T$ 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

Let's evaluate the relation $R$ for the properties of reflexivity, symmetry, and transitivity.

Relation R : $R={(x, y) \in W \times W \mid}$ the words $x$ and $y$ have at least one letter in common}.

• Reflexivity : Each word in English obviously has at least one letter in common with itself, so the relation is reflexive.


• Symmetry : If a word $x$ has at least one letter in common with a word $y$, then $y$ necessarily has that same letter in common with $x$. So, the relation is symmetric.


• Transitivity : This property is not always satisfied. For instance, consider the three words 'cat', 'bat', and 'bee'. 'Cat' and 'bat' share a letter (the 'a'), and 'bat' and 'bee' share a letter (the 'b'), but 'cat' and 'bee' do not share any letters. Therefore, even though 'cat' is related to 'bat' and 'bat' is related to 'bee', 'cat' is not related to 'bee', so the relation is not transitive.

In conclusion, the correct answer is Option A : $R$ is reflexive, symmetric and not transitive.

We have to examine whether the relation $R$ satisfies the properties of reflexivity, symmetry, and transitivity.

Relation R : $R=\{(3,3),(6,6),(9,9),(12,12),(6,12)$, $(3,9),(3,12),(3,6)\}$ on set $A=\{3,6,9,12\}$.

We will evaluate each of the three properties :

• Reflexivity : A relation is reflexive if all elements in the set are related to themselves. Here, (3,3), (6,6), (9,9), and (12,12) are all present in $R$, so $R$ is reflexive.




• Symmetry : A relation is symmetric if for every pair (a,b) in the relation, the pair (b,a) is also in the relation. In $R$, we have pairs such as (6,12) and (3,6), but their corresponding reverse pairs (12,6) and (6,3) are not in $R$, so $R$ is not symmetric.




• Transitivity : A relation is transitive if for every pair of pairs ((a,b), (b,c)) in the relation, the pair (a,c) is also in the relation. For the pairs (3,6) and (6,12) in $R$, we have (3,12) in $R$, and for the pairs (3,6) and (6,6) in $R$, we also have (3,6) in $R$. Therefore, $R$ is 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.

Let's evaluate each of the properties for the relation $R$.

Relation R : $R=\{(1,3),(4,2),(2,4),(2,3),(3,1)\}$

• A relation is a function if each element in the domain is related to exactly one element in the codomain. In this case, for example, 2 is related to both 4 and 3, so $R$ is not a function.




• A relation is transitive if for every pair of elements $(x,y)$ and $(y,z)$ in the relation, $(x,z)$ is also in the relation. In this case, for example, $(1,3)$ and $(3,1)$ are in the relation but $(1,1)$ is not, so $R$ is not transitive.




• A relation is symmetric if for every pair $(x,y)$ in the relation, $(y,x)$ is also in the relation. In this case, for example, $(1,3)$ is in the relation but $(3,1)$ is not, so $R$ is not symmetric.




• A relation is reflexive if every element is related to itself, i.e., if all pairs of the form $(x,x)$ are in the relation for all $x$ in the set. In this case, none of the pairs are of the form $(x,x)$, so $R$ is not reflexive.

Therefore, the correct answer is Option C : $R$ is not symmetric.