Types of Relations Questions in English

(D) Given relation is defined as $\theta R \phi$ such that $\sec^{2} \theta - \tan^{2} \phi = 1$.
$1$. Reflexive: For any $\theta$,$\theta R \theta$ implies $\sec^{2} \theta - \tan^{2} \theta = 1$. Since $1 + \tan^{2} \theta = \sec^{2} \theta$,this is $1 = 1$,which is true. Thus,$R$ is reflexive.
$2$. Symmetric: If $\theta R \phi$,then $\sec^{2} \theta - \tan^{2} \phi = 1$. Using $\sec^{2} \theta = 1 + \tan^{2} \theta$ and $\tan^{2} \phi = \sec^{2} \phi - 1$,we get $(1 + \tan^{2} \theta) - (\sec^{2} \phi - 1) = 1$,which simplifies to $\tan^{2} \theta - \sec^{2} \phi = -1$,or $\sec^{2} \phi - \tan^{2} \theta = 1$. This implies $\phi R \theta$. Thus,$R$ is symmetric.
$3$. Transitive: If $\theta R \phi$ and $\phi R \psi$,then $\sec^{2} \theta - \tan^{2} \phi = 1$ and $\sec^{2} \phi - \tan^{2} \psi = 1$. Adding these,$\sec^{2} \theta - \tan^{2} \phi + \sec^{2} \phi - \tan^{2} \psi = 2$. Since $\sec^{2} \phi - \tan^{2} \phi = 1$,we have $\sec^{2} \theta - \tan^{2} \psi + 1 = 2$,which means $\sec^{2} \theta - \tan^{2} \psi = 1$. This implies $\theta R \psi$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) Let the relation be defined as $R = \{(A, B) \mid B = P^{-1} A P \text{ for some non-singular matrix } P\}$.
For reflexivity: Since $A = I^{-1} A I$ where $I$ is the identity matrix,$(A, A) \in R$. Thus,$R$ is reflexive.
For symmetry: Let $(A, B) \in R$. Then $B = P^{-1} A P$. Multiplying by $P$ on the left and $P^{-1}$ on the right,we get $P B P^{-1} = A$. Let $Q = P^{-1}$. Then $A = Q^{-1} B Q$. Thus,$(B, A) \in R$. So,$R$ is symmetric.
For transitivity: Let $(A, B) \in R$ and $(B, C) \in R$. Then $B = P^{-1} A P$ and $C = Q^{-1} B Q$ for some non-singular matrices $P$ and $Q$. Substituting $B$,we get $C = Q^{-1} (P^{-1} A P) Q = (P Q)^{-1} A (P Q)$. Since $PQ$ is non-singular,$(A, C) \in R$. Thus,$R$ is transitive.
Since the relation is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) The relation is defined as $R = \{(a, b) \mid \sin ^{2} a + \cos ^{2} b = 1\}$.
$1.$ Reflexivity: For any $a \in \mathbb{R}$,we have $\sin ^{2} a + \cos ^{2} a = 1$. Thus,$(a, a) \in R$. So,$R$ is reflexive.
$2.$ Symmetry: Let $(a, b) \in R$,then $\sin ^{2} a + \cos ^{2} b = 1$.
Using $\sin ^{2} x = 1 - \cos ^{2} x$ and $\cos ^{2} x = 1 - \sin ^{2} x$,we get $(1 - \cos ^{2} a) + (1 - \sin ^{2} b) = 1$,which simplifies to $\sin ^{2} b + \cos ^{2} a = 1$. Thus,$(b, a) \in R$. So,$R$ is symmetric.
$3.$ Transitivity: Let $(a, b) \in R$ and $(b, c) \in R$. Then $\sin ^{2} a + \cos ^{2} b = 1$ and $\sin ^{2} b + \cos ^{2} c = 1$.
Adding these equations: $\sin ^{2} a + \cos ^{2} b + \sin ^{2} b + \cos ^{2} c = 1 + 1 = 2$.
Since $\sin ^{2} b + \cos ^{2} b = 1$,we have $\sin ^{2} a + 1 + \cos ^{2} c = 2$,which implies $\sin ^{2} a + \cos ^{2} c = 1$. Thus,$(a, c) \in R$. So,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

(D) relation $R$ on a set $A$ is reflexive if $(a, a) \in R$ for all $a \in A$.
There are $n$ such elements of the form $(a, a)$ which must be present in the relation.
The total number of possible ordered pairs in the Cartesian product $A \times A$ is $n^2$.
Since the $n$ diagonal elements $(a, a)$ are fixed (must be included),we have $n^2 - n$ remaining pairs to choose from.
Each of these $n^2 - n$ pairs can either be included or excluded from the relation,giving $2$ choices for each.
Therefore,the total number of reflexive relations is $2^{n^2 - n}$.

(C) The relation $R$ is defined on the set $S = \{1, 2, 3, 4\} \times \{1, 2, 3, 4\}$. The total number of elements in $S$ is $4 \times 4 = 16$. The condition for the relation is $2a + 3b = 3c + 4d$,where $a, b, c, d \in \{1, 2, 3, 4\}$.
We calculate the value of $f(a, b) = 2a + 3b$ for all pairs $(a, b)$:
$(1,1) \to 5, (1,2) \to 8, (1,3) \to 11, (1,4) \to 14$
$(2,1) \to 7, (2,2) \to 10, (2,3) \to 13, (2,4) \to 16$
$(3,1) \to 9, (3,2) \to 12, (3,3) \to 15, (3,4) \to 18$
$(4,1) \to 11, (4,2) \to 14, (4,3) \to 17, (4,4) \to 20$
Now calculate $g(c, d) = 3c + 4d$ for all pairs $(c, d)$:
$(1,1) \to 7, (1,2) \to 11, (1,3) \to 15, (1,4) \to 19$
$(2,1) \to 10, (2,2) \to 14, (2,3) -> 18, (2,4) -> 22$
$(3,1) \to 13, (3,2) \to 17, (3,3) \to 21, (3,4) \to 25$
$(4,1) \to 16, (4,2) \to 20, (4,3) \to 24, (4,4) \to 28$
Matching values of $f(a, b) = g(c, d)$:
$11: (1,3) \text{ and } (4,1) \text{ map to } (1,2)$
$14: (1,4) \text{ and } (4,2) \text{ map to } (2,2)$
$16: (2,4) \text{ maps to } (4,1)$
$7: (2,1) \text{ maps to } (1,1)$
$10: (2,2) \text{ maps to } (2,1)$
$13: (2,3) \text{ maps to } (3,1)$
$15: (3,3) \text{ maps to } (1,3)$
$18: (3,4) \text{ maps to } (2,3)$
$17: (4,3) \text{ maps to } (3,2)$
$20: (4,4) \text{ maps to } (4,2)$
Counting the pairs $((a,b), (c,d))$ satisfying the condition,we find $12$ such elements.

(B) Given $A = \{2, 3, 5, 7, 9\}$. The relation $R$ is defined by $2x \le 3y$,which implies $y \ge \frac{2x}{3}$.
For each $x \in A$,we find the corresponding $y \in A$:
If $x = 2$,$y \ge 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).
If $x = 3$,$y \ge 2 \implies y \in \{2, 3, 5, 7, 9\}$ ($5$ elements).
If $x = 5$,$y \ge 3.33 \implies y \in \{5, 7, 9\}$ ($3$ elements).
If $x = 7$,$y \ge 4.66 \implies y \in \{5, 7, 9\}$ ($3$ elements).
If $x = 9$,$y \ge 6 \implies y \in \{7, 9\}$ ($2$ elements).
Total elements $l = 5 + 5 + 3 + 3 + 2 = 18$.
For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must be in $R$.
The elements currently in $R$ are: $(2,2), (2,3), (2,5), (2,7), (2,9), (3,2), (3,3), (3,5), (3,7), (3,9), (5,5), (5,7), (5,9), (7,5), (7,7), (7,9), (9,7), (9,9)$.
The pairs $(x, y)$ such that $(x, y) \in R$ but $(y, x) \notin R$ are: $(2,5), (2,7), (2,9), (3,5), (3,7), (3,9), (5,9)$.
To make $R$ symmetric,we must add their reverses: $(5,2), (7,2), (9,2), (5,3), (7,3), (9,3), (9,5)$.
Thus,$m = 7$.
Therefore,$l + m = 18 + 7 = 25$.

(B) The set $A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ has $10$ elements.
We partition $A$ into equivalence classes based on remainders modulo $3$:
$C_0 = \{0, 3, 6, 9\}$ (size $4$)
$C_1 = \{1, 4, 7\}$ (size $3$)
$C_2 = \{2, 5, 8\}$ (size $3$)
For $(x, y) \in R$,$|x - y|$ must be a multiple of $3$,meaning $x$ and $y$ must belong to the same equivalence class.
The number of elements in $R$ is $n(R) = |C_0|^2 + |C_1|^2 + |C_2|^2 = 4^2 + 3^2 + 3^2 = 16 + 9 + 9 = 34$.
Thus,Statement $I$ is incorrect $(34 \neq 36)$.
For Statement $II$:
$1$. Reflexive: $|x - x| = 0$,which is a multiple of $3$.
$2$. Symmetric: If $|x - y| = 3k$,then $|y - x| = 3k$.
$3$. Transitive: If $|x - y| = 3k$ and $|y - z| = 3m$,then $|x - z| = |(x - y) + (y - z)| = 3|k \pm m|$,which is a multiple of $3$.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. Statement $II$ is correct.

(C) Given $A = \{-2, -1, 0, 1, 2, 3, 4\}$ and $xRy \iff 2x + y \le 2$.
For each $x \in A$,we find $y \in A$ such that $y \le 2 - 2x$:
- If $x = -2$,$y \le 6 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).
- If $x = -1$,$y \le 4 \implies y \in \{-2, -1, 0, 1, 2, 3, 4\}$ ($7$ elements).
- If $x = 0$,$y \le 2 \implies y \in \{-2, -1, 0, 1, 2\}$ ($5$ elements).
- If $x = 1$,$y \le 0 \implies y \in \{-2, -1, 0\}$ ($3$ elements).
- If $x = 2$,$y \le -2 \implies y \in \{-2\}$ ($1$ element).
- If $x = 3$,$y \le -4 \implies$ no $y \in A$.
- If $x = 4$,$y \le -6 \implies$ no $y \in A$.
Total elements $l = 7 + 7 + 5 + 3 + 1 = 23$.
For reflexivity,we need $(x, x) \in R$ for all $x \in A$. Check: $(-2, -2), (-1, -1), (0, 0)$ are in $R$. $(1, 1), (2, 2), (3, 3), (4, 4)$ are missing. So $m = 4$.
For symmetry,if $(x, y) \in R$,then $(y, x)$ must be in $R$. The elements $(x, y)$ in $R$ where $(y, x) \notin R$ are: $(3, -2), (4, -2), (2, -1), (2, 0), (3, -1), (4, -1)$. There are $6$ such pairs. So $n = 6$.
Thus,$l + m + n = 23 + 4 + 6 = 33$.

(B) Given the relation $R = \{(x, y) : 4y = 5x - 3, x, y \in M\}$ where $M = \{1, 2, 3, \dots, 16\}$.
We find the elements $(x, y)$ that satisfy the equation $4y = 5x - 3$:
If $x = 3$,$4y = 5(3) - 3 = 12 \implies y = 3$. So,$(3, 3) \in R$.
If $x = 7$,$4y = 5(7) - 3 = 32 \implies y = 8$. So,$(7, 8) \in R$.
If $x = 11$,$4y = 5(11) - 3 = 52 \implies y = 13$. So,$(11, 13) \in R$.
If $x = 15$,$4y = 5(15) - 3 = 72 \implies y = 18$,but $18 \notin M$.
Thus,$R = \{(3, 3), (7, 8), (11, 13)\}$.
For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a)$ must also be in $R$.
Here,$(7, 8) \in R$ implies $(8, 7)$ must be in $R$,and $(11, 13) \in R$ implies $(13, 11)$ must be in $R$.
$(3, 3)$ is already symmetric.
Therefore,we need to add $2$ elements: $(8, 7)$ and $(13, 11)$.

(D) relation $R$ on a set $A$ with $n$ elements is reflexive if it contains all $n$ diagonal elements $(x, x)$ for all $x \in A$.
For the set $A = \{a, b, c, d\}$,$n = 4$.
There are $n^2 = 16$ possible ordered pairs in $A \times A$.
The $n = 4$ diagonal elements must be present in a reflexive relation.
The remaining $n^2 - n = 16 - 4 = 12$ elements are off-diagonal pairs.
For a relation to be symmetric,if $(x, y) \in R$,then $(y, x)$ must also be in $R$.
These $12$ off-diagonal elements form $6$ pairs of the form ${(x, y), (y, x)}$.
For each such pair,we have $2$ choices: either both are included in $R$,or neither is included.
Thus,the total number of reflexive and symmetric relations is $2^6 = 64$.

(D) relation $R$ on $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. Since $(1, 2) \in R$,symmetry requires $(2, 1) \in R$.
For $R$ to be transitive,since $(1, 2) \in R$ and $(2, 1) \in R$,we must have $(1, 1) \in R$ and $(2, 2) \in R$.
Let $R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1)\}$. This relation is symmetric and transitive,but not reflexive on $A$ because $(3, 3) \notin R_1$.
If we include $(3, 3)$,the relation becomes $R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. This relation is symmetric,transitive,and reflexive.
Any other relation containing $(1, 2)$ that is symmetric and transitive must contain $R_1$. If it does not contain $(3, 3)$,it is not reflexive. If it contains $(3, 3)$,it becomes reflexive.
Thus,the only relation that is symmetric and transitive but not reflexive is $R_1 = \{(1, 1), (2, 2), (1, 2), (2, 1)\}$.
Therefore,there is only $1$ such relation.

(A) Let the set be $S = \{-2, -1, 0, 1, 2\}$. The total number of possible ordered pairs $(a, b)$ is $5 \times 5 = 25$.
We need to find pairs such that $1 + ab > 0$,which is equivalent to $ab > -1$.
It is easier to count the pairs where $ab \leq -1$ and subtract from $25$.
The pairs $(a, b)$ where $ab \leq -1$ are:
$(-2, 1) \Rightarrow ab = -2$
$(1, -2) \Rightarrow ab = -2$
$(-2, 2) \Rightarrow ab = -4$
$(2, -2) \Rightarrow ab = -4$
$(-1, 1) \Rightarrow ab = -1$
$(1, -1) \Rightarrow ab = -1$
$(-1, 2) \Rightarrow ab = -2$
$(2, -1) \Rightarrow ab = -2$
There are $8$ such pairs. Thus,the number of elements in $R$ is $25 - 8 = 17$. Statement $I$ is true.
For Statement $II$,check if $R$ is an equivalence relation. $A$ relation is an equivalence relation if it is reflexive,symmetric,and transitive.
Reflexivity: $1 + a^2 > 0$ is true for all $a \in S$. So,it is reflexive.
Symmetry: $1 + ab > 0 \iff 1 + ba > 0$. So,it is symmetric.
Transitivity: Consider $(1, 0) \in R$ because $1 + 0 = 1 > 0$,and $(0, -2) \in R$ because $1 + 0 = 1 > 0$. However,$(1, -2) \notin R$ because $1 + (1)(-2) = -1 \ngtr 0$. Thus,$R$ is not transitive. Statement $II$ is false.

(A) The relation $R$ is defined by $\log_e(x + y) \leq 2$,which implies $x + y \leq e^2$. Since $e \approx 2.718$,$e^2 \approx 7.389$. Thus,$x + y \leq 7$ for $x, y \in N$.
The pairs $(x, y)$ in $R$ are: $(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (5,1), (5,2), (6,1)$.
$A$ relation is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$.
Consider $(3, 4) \in R$ and $(4, 3) \in R$. For transitivity,$(3, 3)$ must be in $R$,which is true. Consider $(4, 4) \notin R$. If we have $(3, 4) \in R$ and $(4, 3) \in R$,we need $(3, 3) \in R$ (present). If we have $(4, 3) \in R$ and $(3, 4) \in R$,we need $(4, 4) \in R$. Since $4+4=8 > 7$,$(4, 4) \notin R$. Thus,we must add $(4, 4)$ to $R$. Similarly,checking other combinations,we find that adding $(4, 4)$ is necessary. After verification,the minimum number of elements to add is $1$.