OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2 (B)

S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(b)

 

Question 1. Write down the relation shown by the arrow diagram, by listing the ordered pairs. State the domain, co-domain, and the range of the relation.

Answer: The relation R, shown by the arrow diagram, can be written as a set of ordered pairs: \( R = \{ (p, 3), (r, 5), (q, 5), (s, 7) \} \).
The domain of R consists of all the first elements in the ordered pairs: \( \text{Domain of } R = \{p, q, r, s\} \).
The co-domain of R is the entire second set shown in the diagram: \( \text{Co-domain of } R = \{1, 3, 5, 7, 9\} \).
The range of R includes all the second elements in the ordered pairs that are actually mapped to: \( \text{Range of } R = \{3, 5, 7\} \). A relation describes how elements from one set are linked to elements in another set.
In simple words: The relation is a list of pairs where an arrow goes from the first item to the second. The domain is all the starting items. The co-domain is all possible ending items. The range is only the ending items that actually get an arrow.

🎯 Exam Tip: Always remember that the range is a subset of the co-domain. The co-domain is the entire set of possible output values, while the range includes only the actual output values for the given inputs.

 

Question 2. Which of the following are relations from B to A, where \( A = \{a, b, c, d\} \) and \( B = \{x, y, z\} \)?
(i) \( \{(z, x), (z, y), (x, a)\} \)
(ii) \( \{(z, a), (z, b), (z, c), (z, d)\} \)
(iii) \( \{(x, b), (y, a)\} \)
(iv) \( \{(b, y), (c, z), (a, x)\} \)
(v) \( \{(x, d), (y, c), (z, a)\} \)
Answer: For a relation to be from B to A, the first element of each ordered pair must belong to set B, and the second element must belong to set A. This means each pair should be of the form \( (b, a) \) where \( b \in B \) and \( a \in A \).
(i) In \( \{(z, x), (z, y), (x, a)\} \), the element 'x' appears as a second component in the pair \( (z, x) \). However, 'x' is not in set A. Therefore, this is not a relation from B to A.
(ii) In \( \{(z, a), (z, b), (z, c), (z, d)\} \), every first component (z) is from B, and every second component (a, b, c, d) is from A. So, this is a valid relation from B to A.
(iii) In \( \{(x, b), (y, a)\} \), every first component (x, y) is from B, and every second component (b, a) is from A. Thus, this is also a valid relation from B to A.
(iv) In \( \{(b, y), (c, z), (a, x)\} \), the elements 'b', 'c', and 'a' are from set A, but they appear as the first components of the ordered pairs. For a relation from B to A, the first components must be from B. Therefore, this is not a relation from B to A.
(v) In \( \{(x, d), (y, c), (z, a)\} \), every first component (x, y, z) is from B, and every second component (d, c, a) is from A. Hence, this is a valid relation from B to A.
In simple words: A relation from set B to set A means that in every pair, the first item must be from B and the second item must be from A. We check each list of pairs to see if they follow this rule.

🎯 Exam Tip: Pay close attention to the order of sets in "relation from B to A" - it means \( b \in B \) and \( a \in A \) for each pair \( (b, a) \). If it were "relation from A to B", then it would be \( (a, b) \).

 

Question 3. In each of the following, state which of the ordered pairs belong to the given relations?
(i) \( \{(x, y) : x > y + 5\} ; (1, 0), (8, 2), (0,1), (2, 8), (9, 3), (10, 7), (123, 4) \)
(ii) \( \{(x, y) : xy = 12\} ; (3, 4), (4, 3), (12, 0), (0,12), (12, 1), (6, 2), (7, 5) \)
(iii) \( \{(x, y) : y = \frac{x+3}{x-1}, x \neq 1\} ; (0, 1), (2, 5), (5, 2), (3, 3), (7, 5), (7, \frac{5}{3}) \)
Answer: We need to check each ordered pair against the given rule for each relation.
(i) For the relation \( R = \{(x, y) : x > y + 5\} \):
\( (1, 0) \): \( 1 > 0 + 5 \implies 1 > 5 \) (False). So, \( (1, 0) \notin R \).
\( (8, 2) \): \( 8 > 2 + 5 \implies 8 > 7 \) (True). So, \( (8, 2) \in R \).
\( (0, 1) \): \( 0 > 1 + 5 \implies 0 > 6 \) (False). So, \( (0, 1) \notin R \).
\( (2, 8) \): \( 2 > 8 + 5 \implies 2 > 13 \) (False). So, \( (2, 8) \notin R \).
\( (9, 3) \): \( 9 > 3 + 5 \implies 9 > 8 \) (True). So, \( (9, 3) \in R \).
\( (10, 7) \): \( 10 > 7 + 5 \implies 10 > 12 \) (False). So, \( (10, 7) \notin R \).
\( (123, 4) \): \( 123 > 4 + 5 \implies 123 > 9 \) (True). So, \( (123, 4) \in R \).
Hence, the pairs belonging to R are \( (8, 2), (9, 3), (123, 4) \).
(ii) For the relation \( R = \{(x, y) : xy = 12\} \):
\( (3, 4) \): \( 3 \times 4 = 12 \) (True). So, \( (3, 4) \in R \).
\( (4, 3) \): \( 4 \times 3 = 12 \) (True). So, \( (4, 3) \in R \).
\( (12, 0) \): \( 12 \times 0 = 0 \neq 12 \) (False). So, \( (12, 0) \notin R \).
\( (0, 12) \): \( 0 \times 12 = 0 \neq 12 \) (False). So, \( (0, 12) \notin R \).
\( (12, 1) \): \( 12 \times 1 = 12 \) (True). So, \( (12, 1) \in R \).
\( (6, 2) \): \( 6 \times 2 = 12 \) (True). So, \( (6, 2) \in R \).
\( (7, 5) \): \( 7 \times 5 = 35 \neq 12 \) (False). So, \( (7, 5) \notin R \).
Hence, the pairs belonging to R are \( (3, 4), (4, 3), (12, 1), (6, 2) \).
(iii) For the relation \( R = \{(x, y) : y = \frac{x+3}{x-1}, x \neq 1\} \):
\( (0, 1) \): When \( x = 0 \), \( y = \frac{0+3}{0-1} = \frac{3}{-1} = -3 \). Since \( 1 \neq -3 \), \( (0, 1) \notin R \).
\( (2, 5) \): When \( x = 2 \), \( y = \frac{2+3}{2-1} = \frac{5}{1} = 5 \). So, \( (2, 5) \in R \).
\( (5, 2) \): When \( x = 5 \), \( y = \frac{5+3}{5-1} = \frac{8}{4} = 2 \). So, \( (5, 2) \in R \).
\( (3, 3) \): When \( x = 3 \), \( y = \frac{3+3}{3-1} = \frac{6}{2} = 3 \). So, \( (3, 3) \in R \).
\( (7, 5) \): When \( x = 7 \), \( y = \frac{7+3}{7-1} = \frac{10}{6} = \frac{5}{3} \). Since \( 5 \neq \frac{5}{3} \), \( (7, 5) \notin R \).
\( (7, \frac{5}{3}) \): When \( x = 7 \), \( y = \frac{7+3}{7-1} = \frac{10}{6} = \frac{5}{3} \). So, \( (7, \frac{5}{3}) \in R \).
Hence, the pairs belonging to R are \( (2, 5), (5, 2), (3, 3), (7, \frac{5}{3}) \).
In simple words: For each rule, we take each pair of numbers and put them into the rule. If the rule works out to be true, then that pair belongs to the relation. If it's false, the pair does not belong.

🎯 Exam Tip: Always carefully substitute the x and y values from each ordered pair into the given relation rule. Double-check your calculations for inequalities and equations.

 

Question 4. Let N be the set of natural numbers. Describe the following relations in words, giving their domain and the range.
(a) \( \{(2, 1), (4, 2), (10, 5), (18, 9), (20, 10)\} \)
(b) \( \{(3, 1), (6, 2), (15, 5)\} \)
(c) \( \{(1, 4), (5, 16), (7, 22), (12, 37)\} \)
Answer: We need to find the pattern between the first and second components in each ordered pair and then identify the domain and range.
(a) Given relation: \( \{(2, 1), (4, 2), (10, 5), (18, 9), (20, 10)\} \).
We observe that in each ordered pair \( (x, y) \), the first component \( x \) is twice the second component \( y \). So, \( x = 2y \).
In words: The relation is "x is twice y".
Domain (R) = Set of all first components \( = \{2, 4, 10, 18, 20\} \).
Range (R) = Set of all second components \( = \{1, 2, 5, 9, 10\} \).
(b) Given relation: \( \{(3, 1), (6, 2), (15, 5)\} \).
We observe that in each ordered pair \( (x, y) \), the first component \( x \) is three times the second component \( y \). So, \( x = 3y \).
In words: The relation is "x is three times y".
Domain (R) = Set of all first components \( = \{3, 6, 15\} \).
Range (R) = Set of all second components \( = \{1, 2, 5\} \).
(c) Given relation: \( \{(1, 4), (5, 16), (7, 22), (12, 37)\} \).
We observe that in each ordered pair \( (x, y) \), the second component \( y \) is one more than three times the first component \( x \). So, \( y = 3x + 1 \).
In words: The relation is "y is one more than three times x".
Domain (R) = Set of all first components \( = \{1, 5, 7, 12\} \).
Range (R) = Set of all second components \( = \{4, 16, 22, 37\} \).
In simple words: Look for a pattern between the first and second numbers in each pair. Describe this pattern in simple English. The domain is all the first numbers, and the range is all the second numbers in the pairs.

🎯 Exam Tip: When describing relations in words, clearly state the relationship between \( x \) and \( y \) (the first and second components). Always list the distinct elements for the domain and range without repetition.

 

Question 5. Z is the set of integers. Describe the following relation in set builder form, given its domain and range.
\( \{(0, – 7), (2, – 5), (4, – 3), (- 13, – 20), ...\} \)
Answer: Given the relation \( \{(0, – 7), (2, – 5), (4, – 3), (- 13, – 20), ...\} \).
Let's examine the difference between the first and second components for each pair \( (x, y) \):
For \( (0, -7) \): \( 0 - (-7) = 0 + 7 = 7 \).
For \( (2, -5) \): \( 2 - (-5) = 2 + 5 = 7 \).
For \( (4, -3) \): \( 4 - (-3) = 4 + 3 = 7 \).
For \( (-13, -20) \): \( -13 - (-20) = -13 + 20 = 7 \).
We observe a consistent pattern: the difference between the first component and the second component in each ordered pair is 7. So, the rule is \( x - y = 7 \). Since Z is the set of integers, both x and y must be integers.
In set builder form, the relation R can be written as: \( R = \{(x, y) : x - y = 7, x \in Z, y \in Z\} \). This type of relation is often shown on a number line or coordinate plane.
The domain of R is \( \{0, 2, 4, -13, ...\} \), which are the first elements of the pairs.
The range of R is \( \{-7, -5, -3, -20, ...\} \), which are the second elements of the pairs.
In simple words: We look at the numbers in each pair and find a rule that connects them. Here, the first number minus the second number always equals 7. So, we write this rule using symbols, stating that both numbers must be whole numbers (integers).

🎯 Exam Tip: To find the rule for a relation, try simple operations like addition, subtraction, multiplication, or division between the x and y components. Look for a constant value or a simple linear relationship.

 

Question 6. Write down the domain and range of the relation \( (x, y) : x = 3y \) and x and y are natural numbers less than 10.
Answer: Given the relation \( R = \{(x, y) : x = 3y\} \), where x and y are natural numbers less than 10. Natural numbers less than 10 are \( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \).
We need to find all possible pairs \( (x, y) \) that satisfy \( x = 3y \) with x and y from this set.
If \( y = 1 \), then \( x = 3 \times 1 = 3 \). So, \( (3, 1) \in R \).
If \( y = 2 \), then \( x = 3 \times 2 = 6 \). So, \( (6, 2) \in R \).
If \( y = 3 \), then \( x = 3 \times 3 = 9 \). So, \( (9, 3) \in R \).
If \( y = 4 \), then \( x = 3 \times 4 = 12 \). But 12 is not less than 10, so we stop here. Any further values of y would also result in x values greater than or equal to 10.
Therefore, the relation R consists of the ordered pairs: \( R = \{(3, 1), (6, 2), (9, 3)\} \).
The domain of R is the set of all first components: \( \text{Domain of } R = \{3, 6, 9\} \).
The range of R is the set of all second components: \( \text{Range of } R = \{1, 2, 3\} \). This method systematically finds all valid pairs for the given conditions.
In simple words: We need to list all pairs of numbers where the first number is three times the second number, and both numbers are natural numbers smaller than 10. Then, the domain is all the first numbers we found, and the range is all the second numbers we found.

🎯 Exam Tip: When given constraints on the values (e.g., "natural numbers less than 10"), always list out the possible values for x and y first. Then, systematically test values to form the ordered pairs that fit the relation's rule.

 

Question 7. Determine the domain and range of the relation R.
(a) \( R = \{(x + 1, x + 5) | x \in \{0, 1, 2, 3, 4, 5\}\} \). Draw the graph of R.
(b) \( R = \{(x, x^3) | x \text{ is a prime number less than 10}\} \).
Answer:
(a) Given \( R = \{(x + 1, x + 5) | x \in \{0, 1, 2, 3, 4, 5\}\} \).
We calculate the ordered pairs for each value of x:
When \( x = 0 \): \( (0 + 1, 0 + 5) = (1, 5) \).
When \( x = 1 \): \( (1 + 1, 1 + 5) = (2, 6) \).
When \( x = 2 \): \( (2 + 1, 2 + 5) = (3, 7) \).
When \( x = 3 \): \( (3 + 1, 3 + 5) = (4, 8) \).
When \( x = 4 \): \( (4 + 1, 4 + 5) = (5, 9) \).
When \( x = 5 \): \( (5 + 1, 5 + 5) = (6, 10) \).
So, the relation R is \( \{(1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)\} \).
The domain of R is the set of all first components: \( \text{Domain of } R = \{1, 2, 3, 4, 5, 6\} \).
The range of R is the set of all second components: \( \text{Range of } R = \{5, 6, 7, 8, 9, 10\} \).
The graph of R is a set of distinct points on a coordinate plane, showing a linear progression.

(b) Given \( R = \{(x, x^3) | x \text{ is a prime number less than 10}\} \).
First, list the prime numbers less than 10: \( \{2, 3, 5, 7\} \).
Now, calculate the ordered pairs for each prime number:
When \( x = 2 \): \( (2, 2^3) = (2, 8) \).
When \( x = 3 \): \( (3, 3^3) = (3, 27) \).
When \( x = 5 \): \( (5, 5^3) = (5, 125) \).
When \( x = 7 \): \( (7, 7^3) = (7, 343) \).
So, the relation R is \( \{(2, 8), (3, 27), (5, 125), (7, 343)\} \).
The domain of R is the set of all first components: \( \text{Domain of } R = \{2, 3, 5, 7\} \).
The range of R is the set of all second components: \( \text{Range of } R = \{8, 27, 125, 343\} \). This demonstrates how a simple rule can generate specific pairs based on an input set.
In simple words: For part (a), we create pairs by adding 1 to 'x' for the first number and adding 5 to 'x' for the second number. For part (b), we use prime numbers less than 10, then pair each number with its cube. The domain is all the first numbers in the pairs, and the range is all the second numbers.

🎯 Exam Tip: When graphing a relation with a discrete domain (like a finite set of numbers), always plot individual points and do not connect them with a line, as the relation is not continuous.

 

Question 8. Given \( A = \{- 2, – 1, 0, 1, 2\} \), list the ordered pairs determined by each of the following relations applied on A :
(i) R\( _1 \) = "is less than”
(ii) R\( _2 \) = "is the square of”
(iii) R\( _3 \) = "is the additive inverse of”
(iv) R\( _4 \) = "is equal to"
Answer: Given set \( A = \{-2, -1, 0, 1, 2\} \). We need to form ordered pairs \( (a, b) \) where both \( a \) and \( b \) are from set A, based on the given relation rules.
(i) R\( _1 \) = "is less than" \( \implies R_1 = \{(a, b) | a < b, a, b \in A\} \).
\( -2 < -1 \implies (-2, -1) \in R_1 \)
\( -2 < 0 \implies (-2, 0) \in R_1 \)
\( -2 < 1 \implies (-2, 1) \in R_1 \)
\( -2 < 2 \implies (-2, 2) \in R_1 \)
\( -1 < 0 \implies (-1, 0) \in R_1 \)
\( -1 < 1 \implies (-1, 1) \in R_1 \)
\( -1 < 2 \implies (-1, 2) \in R_1 \)
\( 0 < 1 \implies (0, 1) \in R_1 \)
\( 0 < 2 \implies (0, 2) \in R_1 \)
\( 1 < 2 \implies (1, 2) \in R_1 \)
So, \( R_1 = \{(-2, -1), (-2, 0), (-2, 1), (-2, 2), (-1, 0), (-1, 1), (-1, 2), (0, 1), (0, 2), (1, 2)\} \).
(ii) R\( _2 \) = "is the square of" \( \implies R_2 = \{(a, b) | a = b^2, a, b \in A\} \).
We check which elements in A are squares of other elements in A.
\( 1 = (-1)^2 \implies (1, -1) \in R_2 \)
\( 1 = 1^2 \implies (1, 1) \in R_2 \)
\( 0 = 0^2 \implies (0, 0) \in R_2 \)
\( 4 \) is not in A, so we cannot have \( (4, 2) \) or \( (4, -2) \).
So, \( R_2 = \{(1, -1), (1, 1), (0, 0)\} \).
(iii) R\( _3 \) = "is the additive inverse of" \( \implies R_3 = \{(a, b) | a + b = 0, a, b \in A\} \).
This means 'b' is the negative of 'a', or 'a' is the negative of 'b'. The sum of an element and its additive inverse is always zero.
\( 2 + (-2) = 0 \implies (2, -2) \in R_3 \)
\( -2 + 2 = 0 \implies (-2, 2) \in R_3 \)
\( 1 + (-1) = 0 \implies (1, -1) \in R_3 \)
\( -1 + 1 = 0 \implies (-1, 1) \in R_3 \)
\( 0 + 0 = 0 \implies (0, 0) \in R_3 \)
So, \( R_3 = \{(2, -2), (-2, 2), (1, -1), (-1, 1), (0, 0)\} \).
(iv) R\( _4 \) = "is equal to" \( \implies R_4 = \{(a, b) | a = b, a, b \in A\} \).
This means the first and second components of the pair must be the same.
\( -2 = -2 \implies (-2, -2) \in R_4 \)
\( -1 = -1 \implies (-1, -1) \in R_4 \)
\( 0 = 0 \implies (0, 0) \in R_4 \)
\( 1 = 1 \implies (1, 1) \in R_4 \)
\( 2 = 2 \implies (2, 2) \in R_4 \)
So, \( R_4 = \{(-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2)\} \).
In simple words: For each rule, we pick two numbers from set A and see if they fit the rule. "Is less than" means the first number is smaller. "Is the square of" means the first number is the square of the second. "Is the additive inverse of" means they add up to zero. "Is equal to" means they are the same number.

🎯 Exam Tip: When working with relations on a single set, remember that both components of each ordered pair must come from that specific set. Be methodical and test all possible combinations or check if elements satisfy the condition.

 

Question 9. Given \( A = \{2, 3, 4, 5, 6\} \). List the elements of each of the following relations :
(i) \( \{(x, y) \in A \times A : x = y\} \)
(ii) \( \{(x, y) \in A \times A : x > y, \frac{x}{y} \in W\} \)
(iii) \( \{(x, y) \in A \times A : x \text{ is a divisor of y and } x \neq y\} \).
Answer: Given set \( A = \{2, 3, 4, 5, 6\} \). We need to list the ordered pairs \( (x, y) \) where both \( x \) and \( y \) are from set A, based on the given relation rules.
(i) For the relation \( R = \{(x, y) \in A \times A : x = y\} \):
This means the first and second components of the pair must be identical.
\( x = 2, y = 2 \implies (2, 2) \in R \)
\( x = 3, y = 3 \implies (3, 3) \in R \)
\( x = 4, y = 4 \implies (4, 4) \in R \)
\( x = 5, y = 5 \implies (5, 5) \in R \)
\( x = 6, y = 6 \implies (6, 6) \in R \)
So, \( R = \{(2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \).
(ii) For the relation \( R = \{(x, y) \in A \times A : x > y, \frac{x}{y} \in W\} \) (where W is the set of whole numbers, meaning \( \frac{x}{y} \) must be a non-negative integer):
We list pairs \( (x, y) \) from A where \( x > y \) and \( \frac{x}{y} \) is a whole number.
If \( y = 2 \):
\( x = 3 \): \( 3 > 2 \), \( \frac{3}{2} \notin W \).
\( x = 4 \): \( 4 > 2 \), \( \frac{4}{2} = 2 \in W \implies (4, 2) \in R \).
\( x = 5 \): \( 5 > 2 \), \( \frac{5}{2} \notin W \).
\( x = 6 \): \( 6 > 2 \), \( \frac{6}{2} = 3 \in W \implies (6, 2) \in R \).
If \( y = 3 \):
\( x = 4 \): \( 4 > 3 \), \( \frac{4}{3} \notin W \).
\( x = 5 \): \( 5 > 3 \), \( \frac{5}{3} \notin W \).
\( x = 6 \): \( 6 > 3 \), \( \frac{6}{3} = 2 \in W \implies (6, 3) \in R \).
If \( y = 4 \):
\( x = 5 \): \( 5 > 4 \), \( \frac{5}{4} \notin W \).
\( x = 6 \): \( 6 > 4 \), \( \frac{6}{4} \notin W \).
If \( y = 5 \):
\( x = 6 \): \( 6 > 5 \), \( \frac{6}{5} \notin W \).
So, \( R = \{(4, 2), (6, 2), (6, 3)\} \).
(iii) For the relation \( R = \{(x, y) \in A \times A : x \text{ is a divisor of y and } x \neq y\} \):
This means \( y \) must be divisible by \( x \), and \( x \) and \( y \) must be different.
If \( x = 2 \):
\( y = 4 \): 2 divides 4, \( 2 \neq 4 \implies (2, 4) \in R \).
\( y = 6 \): 2 divides 6, \( 2 \neq 6 \implies (2, 6) \in R \).
If \( x = 3 \):
\( y = 6 \): 3 divides 6, \( 3 \neq 6 \implies (3, 6) \in R \).
If \( x = 4 \): (No \( y \in A, y \neq 4 \) such that 4 divides y)
If \( x = 5 \): (No \( y \in A, y \neq 5 \) such that 5 divides y)
If \( x = 6 \): (No \( y \in A, y \neq 6 \) such that 6 divides y)
So, \( R = \{(2, 4), (2, 6), (3, 6)\} \).
In simple words: For each relation, we check pairs from set A. For (i), pairs must have the same number. For (ii), the first number must be bigger than the second, and the first number must divide the second number evenly (no remainder). For (iii), the first number must divide the second number evenly, but the two numbers cannot be the same.

🎯 Exam Tip: When defining relations from a set to itself (like \( A \times A \)), systematically consider all possible pairs or elements, and apply each condition one by one to avoid missing any valid pairs.

 

Question 10.
(i) If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B is
(a) 29
(b) 2\( ^6 \)
(c) 32
(d) 29 – 1
(ii) Let A be a finite set. The number of relations on A where A has 3 elements are :
(a) 9
(b) 0.81
(c) 24.3
(d) 512
Answer:
(i) First, let's find the elements of sets A and B:
Set A = even natural numbers less than 8 \( = \{2, 4, 6\} \). So, the number of elements in A, \( n(A) = 3 \).
Set B = prime numbers less than 7 \( = \{2, 3, 5\} \). So, the number of elements in B, \( n(B) = 3 \).
The number of elements in the Cartesian product \( A \times B \) is \( n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9 \).
The total number of relations from A to B is \( 2^{n(A \times B)} = 2^9 = 512 \). A relation is essentially any subset of the Cartesian product.
Therefore, the correct option is (a) 29, which means \( 2^9 \).
(ii) Given that A is a finite set with 3 elements, so \( n(A) = 3 \).
The number of relations on A means the number of relations from A to A. In this case, the Cartesian product is \( A \times A \).
The number of elements in \( A \times A \) is \( n(A \times A) = n(A) \times n(A) = 3 \times 3 = 9 \).
The total number of relations on A is \( 2^{n(A \times A)} = 2^9 = 512 \).
Thus, the correct option is (d) 512.
In simple words: The number of possible relations between two sets (or a set and itself) is found by taking 2 raised to the power of the total number of pairs you can make between the elements of those sets. If set A has 3 items and set B has 3 items, there are 3 multiplied by 3, which is 9 possible pairs. So, there are 2 to the power of 9 (512) relations.

🎯 Exam Tip: The formula for the total number of relations from set A to set B is \( 2^{n(A) \times n(B)} \). For relations "on A", it's \( 2^{n(A) \times n(A)} \). Remember to first accurately determine the number of elements in each set before applying the formula.

 

Question 11. Let n (A) = p. Then the number of all relations on A is
(a) 2P
(b) 2p\( ^1 \)
(c) 2p\( ^2 \)
(d) None of the options
Answer: (c) 2p\( ^2 \)
Given that the number of elements in set A is \( n(A) = p \).
We need to find the number of all possible relations on A. This means the relation is from set A to set A.
The number of elements in the Cartesian product \( A \times A \) is \( n(A \times A) = n(A) \times n(A) = p \times p = p^2 \).
The total number of relations on set A is \( 2^{n(A \times A)} \).
Substituting the value of \( n(A \times A) \), the number of relations on A is \( 2^{p^2} \). This is a fundamental concept in set theory.
In simple words: If a set has 'p' number of items, then the total number of ways to link these items to themselves (called relations) is found by multiplying 'p' by 'p', and then raising the number 2 to that power.

🎯 Exam Tip: This is a standard formula question. Remember that the number of relations is always \( 2 \) raised to the power of the number of elements in the Cartesian product of the domain and co-domain.