Relations Questions in English

(D) Given the relation $3 a+2 b=27$ where $a, b \in N$ (natural numbers).
$2 b = 27 - 3 a$
$b = \frac{3(9 - a)}{2}$
Since $b$ must be a natural number,$3(9 - a)$ must be even and positive.
For $a = 1, b = \frac{3(8)}{2} = 12$.
For $a = 3, b = \frac{3(6)}{2} = 9$.
For $a = 5, b = \frac{3(4)}{2} = 6$.
For $a = 7, b = \frac{3(2)}{2} = 3$.
For $a = 9, b = 0$ (not a natural number).
Thus,$R = \{(1, 12), (3, 9), (5, 6), (7, 3)\}$.

(B) Given the equation $4a = 5b$,we can write $a = \frac{5}{4}b$.
Since $a$ must be an integer,$b$ must be a multiple of $4$.
Given $b \in \{1, 2, 3, \dots, 30\}$,the possible values for $b$ are $4, 8, 12, 16, 20, 24, 28$.
For each $b$,we calculate $a = \frac{5}{4}b$:
If $b = 4, a = 5$.
If $b = 8, a = 10$.
If $b = 12, a = 15$.
If $b = 16, a = 20$.
If $b = 20, a = 25$.
If $b = 24, a = 30$.
If $b = 28, a = 35$ (which is not in the set $\{1, 2, \dots, 30\}$).
Thus,the valid ordered pairs $(a, b)$ are $(5, 4), (10, 8), (15, 12), (20, 16), (25, 20), (30, 24)$.
The total number of such pairs is $6$.

(D) The set $A$ consists of even natural numbers less than $8$,so $A = \{2, 4, 6\}$.
The number of elements in $A$ is $n(A) = 3$.
The set $B$ consists of prime integers less than $7$,so $B = \{2, 3, 5\}$.
The number of elements in $B$ is $n(B) = 3$.
The total number of relations from $A$ to $B$ is given by the formula $2^{n(A) \times n(B)}$.
Substituting the values,we get $2^{3 \times 3} = 2^9$.

(B) Given that the number of elements in set $A$ is $n(A) = p$ and the number of elements in set $B$ is $n(B) = q$.
The total number of elements in the Cartesian product $A \times B$ is $n(A \times B) = n(A) \times n(B) = pq$.
$A$ relation from set $A$ to set $B$ is a subset of $A \times B$.
The total number of subsets of a set with $n$ elements is $2^n$.
Therefore,the total number of relations from set $A$ to set $B$ is $2^{pq}$.

(A) We need to find the number of integer pairs $(x, y)$ satisfying $4x^{2} + y^{2} < 52$.
We test values of $x$ such that $4x^{2} < 52$,i.e.,$x^{2} < 13$. Thus,$x \in \{0, \pm 1, \pm 2, \pm 3\}$.
Case $1$: If $x = 0$,then $y^{2} < 52$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 7\}$. Total $15$ values.
Case $2$: If $x = \pm 1$,then $4(1)^{2} + y^{2} < 52 \implies y^{2} < 48$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 6\}$. Total $2 \times 13 = 26$ values.
Case $3$: If $x = \pm 2$,then $4(4) + y^{2} < 52 \implies y^{2} < 36$. So $y \in \{0, \pm 1, \pm 2, \dots, \pm 5\}$. Total $2 \times 11 = 22$ values.
Case $4$: If $x = \pm 3$,then $4(9) + y^{2} < 52 \implies y^{2} < 16$. So $y \in \{0, \pm 1, \pm 2, \pm 3\}$. Total $2 \times 7 = 14$ values.
Total number of elements $= 15 + 26 + 22 + 14 = 77$.

(C) The relation is defined as $R = \{(a, b) : a = b - 2, b > 6\}$.
Given $b > 6$,let us test the options.
For option $(A)$,$(8, 6)$: Here $b = 6$,which does not satisfy $b > 6$.
For option $(B)$,$(3, 8)$: Here $a = 3$ and $b = 8$. Checking the condition $a = b - 2$,we get $3 = 8 - 2 = 6$,which is false.
For option $(C)$,$(6, 8)$: Here $a = 6$ and $b = 8$. Checking the condition $a = b - 2$,we get $6 = 8 - 2 = 6$,which is true. Also,$b = 8 > 6$ is satisfied.
For option $(D)$,$(8, 7)$: Here $a = 8$ and $b = 7$. Checking the condition $a = b - 2$,we get $8 = 7 - 2 = 5$,which is false.
Therefore,$(6, 8) \in R$ is the correct answer.