Sets and Relations

Now, x = y $\forall$(x, y)

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.

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 $a^2 = \dfrac{1}{2}$.

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.

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: $a_{k+1} = \dfrac{a_1 + 1 - 2^k}{2^k}$ 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 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 $P$.

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

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.

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.

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, $n(A) = \left[\dfrac{140}{2}\right] = 70$.

([.] denotes greatest integer function) 2.

Similarly, let $B$ be the set of students whose number is divisible by 3.

Hence, $n(B) = \left[\dfrac{140}{3}\right] = 46$.

([.] denotes greatest integer function) 3.

Let $C$ be the set of students whose number is divisible by 5.

Hence, $n(C) = \left[\dfrac{140}{5}\right] = 28$.

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

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

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

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

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

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

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 :

Substituting the values we calculated above :

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

Given, sin$\theta$ $-$ cos$\theta$ =

cos$\theta$ $\Rightarrow$ sin$\theta$ =

cos$\theta$ $\Rightarrow$ sin$\theta$ =

cos$\theta$ $\Rightarrow$

sin$\theta$ = cos$\theta$ . . . (1) This is for set P. sin$\theta$ + cos$\theta$ =

sin$\theta$ $\Rightarrow$ cos$\theta$ =

sin$\theta$ . . . .(2) This is from set Q. So, P = Q

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.

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.

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 $A \subseteq B$