ML Aggarwal Class 11 Maths Solutions Chapter 02 Relations and Functions

Relations and Functions

Introduction

In daily life, we come across many relations such as Teacher and Student, Mother and Daughter, Book and Cost. In mathematics also, we come across many relations such as

• (i) number x is square of number y
• (ii) line l is perpendicular to line m
• (iii) set A is a proper subset of set B
• (iv) area of a circle with radius r is πr²

In each of these, we notice that a relation involves pairs of objects in a certain order. In this chapter, we will learn how to connect pairs of objects from two sets and then introduce relation between two objects of the pair. Finally, we shall learn about special type of relations called functions. From the beginning of modern mathematics in the 17th century, the concept of function has been at the very centre of mathematical thought. It gives the mathematical rule by which one quantity corresponds to another quantity.

2.1 Ordered Pair

An ordered pair is a pair of objects taken in a specific order.

An ordered pair is written by listing its two members in a specific order, separating them by a comma and enclosing the pair in parentheses. In the ordered pair (a, b), a is called the first member (or component) and b is called the second member (or component).

Equality of ordered pairs. Two ordered pairs (a, b) and (c, d) are called equal, written as (a, b) = (c, d), iff a = c and b = d.

Remarks

• 1. The word ordered implies that the order in which the two elements of the pair occur is meaningful. For example, if we have a sock and a shoe, the order in which they are put on matters. In fact, there are situations in which order is very important and essential.
• 2. The ordered pairs (a, b) and (b, a) are different unless a = b.
• 3. The two components of an ordered pair may be equal.
• 4. Note that {a , b} ≠ (a , b), because {a , b} is a set whereas (a , b) is an ordered pair.

2.2 Cartesian Product of Two Sets

Let A and B be any two non-empty sets, then the set of all ordered pairs (a, b) for all a ∈ A and b ∈ B is called the cartesian product of A and B. It is written as A × B (read as 'A cross B').

Symbolically, A × B = {(a, b) : for all a ∈ A, b ∈ B}.

For example, let A = {1, 2, 3} and B = {3, 4}, then

A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} and
B × A = {(3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3)}.

From this example, we observe that

• (i) A × B ≠ B × A.
• (ii) n(A × B) = 6 = n(B × A).
• (iii) n(A × B) = 6 = 3 × 2 = n(A) × n(B).

Remark

• 1. A × B ≠ B × A unless A = B.
• 2. If A and B are finite sets, then n(A × B) = n(A) × n(B) and n(A × B) = n(B × A).
• 3. A × B = φ when one or both of A, B are empty sets.
• 4. A × B ≠ φ iff A ≠ φ and B ≠ φ.
• 5. If A and B are non-empty sets and either A or B is an infinite set, then A × B is also infinite set.

The concept of cartesian product can be extended to more than two sets.

Let A, B and C be any non-empty sets, then the set of all triplets (a, b, c) for all a ∈ A, b ∈ B and c ∈ C is called the cartesian product of A, B and C. It is written as A × B × C. Thus,

A × B × C = {(a, b, c) : for all a ∈ A, b ∈ B, c ∈ C}.

Example 1. If the ordered pairs (x - 1, y + 3) and (2, x + 4) are equal, find x and y.
Answer: Since the ordered pairs are equal, their corresponding components must be equal. From the first component: x - 1 = 2, which gives x = 3. From the second component: y + 3 = x + 4. Substituting x = 3 into this equation: y + 3 = 3 + 4, so y = 4.
In simple words: When two ordered pairs are equal, match each component from the first pair with the matching component from the second pair, then solve for the unknown values.

Exam Tip: Always equate corresponding components of equal ordered pairs - first components with first, second with second. Solve systematically from one equation to find unknowns in the other.

Example 2. If P = {a, b, c} and Q = {d}, form the sets P × Q and Q × P. Are these two cartesian products equal?
Answer: By the definition of cartesian product, P × Q is formed by pairing each element of P with each element of Q: P × Q = {(a, d), (b, d), (c, d)}. Similarly, Q × P pairs each element of Q with each element of P: Q × P = {(d, a), (d, b), (d, c)}. These two sets are not equal because their ordered pairs are different - for instance, (a, d) from the first set does not match (d, a) from the second set, since the components are in different order.
In simple words: The cartesian product P × Q is different from Q × P. The order of the sets matters - switching them changes which values appear first in each pair.

Exam Tip: Remember that P × Q ≠ Q × P in general. Only when P = Q will the two cartesian products be equal. Show both sets explicitly to prove they are different.

Example 3. If A = {1, 2, 3, 4} and x, y ∈ A, form the set of all ordered pairs (x, y) such that x is a divisor of y.
Answer: We need to find all pairs where the first element divides the second element evenly. Going through systematically: 1 divides all four numbers, so (1, 1), (1, 2), (1, 3), (1, 4) are included. 2 divides 2 and 4, giving (2, 2), (2, 4). 3 divides only itself, giving (3, 3). 4 divides only itself, giving (4, 4). Therefore, the required set is {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}.
In simple words: Check each possible pair and keep it only if the first number divides into the second number with no remainder.

Exam Tip: For divisibility relations, start by listing what each element divides. This systematic approach prevents missing pairs and avoids incorrect inclusions.

Example 4. Express {(x , y) : y + 2x = 5, x , y ∈ W} as the set of ordered pairs.
Answer: We need to find all pairs of whole numbers satisfying the equation y + 2x = 5. Testing x = 0: y + 0 = 5 gives y = 5, so (0, 5) works. Testing x = 1: y + 2 = 5 gives y = 3, so (1, 3) works. Testing x = 2: y + 4 = 5 gives y = 1, so (2, 1) works. Testing x = 3: y + 6 = 5 gives y = -1, which is not a whole number, so this fails. For all higher values of x, we get negative values for y, which are not whole numbers. Therefore, the required set of ordered pairs is {(0, 5), (1, 3), (2, 1)}.
In simple words: Substitute values for x starting from zero, find the matching y value from the equation, and include the pair only if both x and y are whole numbers.

Exam Tip: When a relation involves whole numbers or natural numbers, test systematically from the smallest values and stop when further values give results outside the required set.

Example 5. If A = {1, 5}, B = {2, 6}, C = {2, 4}, find A × (B ∪ C).
Answer: First, find the union of B and C. B ∪ C combines all unique elements from both sets: B ∪ C = {2, 4, 6}. Now form the cartesian product of A with this union. Each element of A pairs with each element of B ∪ C: A × (B ∪ C) = {(1, 2), (1, 4), (1, 6), (5, 2), (5, 4), (5, 6)}.
In simple words: Combine the two sets into their union first, then match each element from A with each element in that union to create all pairs.

Exam Tip: When computing A × (B ∪ C), always compute the union before the cartesian product. This follows the order of operations and ensures accuracy.

Example 6. If A = {x | x ∈ W, x < 3}, B = {x | x ∈ N, 2 ≤ x < 4} and C = {3, 4}, then verify that (A ∪ B) × C = (A × C) ∪ (B × C).
Answer: First, identify the sets. A consists of whole numbers less than 3: A = {0, 1, 2}. B consists of natural numbers from 2 up to (but not including) 4: B = {2, 3}. C = {3, 4} is given. So A ∪ B = {0, 1, 2, 3}. Now compute the left side: (A ∪ B) × C = {(0, 3), (0, 4), (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}. For the right side, compute A × C = {(0, 3), (0, 4), (1, 3), (1, 4), (2, 3), (2, 4)} and B × C = {(2, 3), (2, 4), (3, 3), (3, 4)}. Taking the union: (A × C) ∪ (B × C) = {(0, 3), (0, 4), (1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}. Both sides are equal, so the property is verified.
In simple words: When you have a union of two sets multiplied by a third set, you can instead multiply each set separately by the third set, then combine the results - you get the same answer.

Exam Tip: To verify set identities involving unions and cartesian products, explicitly write out both sets and their products. Comparing the ordered pairs directly confirms whether they are equal.

Example 7. If A = {1, 2}, B = {2, 3} and C = {0}, form the set A × B × C.
Answer: The cartesian product of three sets produces triplets where the first component comes from A, the second from B, and the third from C. Since A has 2 elements, B has 2 elements, and C has 1 element, there will be 2 × 2 × 1 = 4 triplets. We get: A × B × C = {(1, 2, 0), (1, 3, 0), (2, 2, 0), (2, 3, 0)}.
In simple words: For a three-way cartesian product, pair each element from the first set with each element from the second set, then pair that result with each element from the third set.

Exam Tip: The number of triplets in A × B × C equals n(A) × n(B) × n(C). This helps verify you have found all triplets.

Example 8. If A = {- 1, 1}, form the set A × A × A.
Answer: We need all triplets where each component is either -1 or 1. Since A has 2 elements, there are 2³ = 8 triplets total. Listing them systematically: A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}.
In simple words: For A × A × A, list every possible combination of the two values, taking three values at a time, where each slot can be filled with either value independently.

Exam Tip: When forming A × A × A, organize systematically: fix the first component and vary the other two, then move to the next first component. This avoids missing or repeating triplets.

Example 9. If R is the set of all real numbers, what do the cartesian products R × R and R × R × R represent?
Answer: The cartesian product R × R represents all ordered pairs (x, y) where both x and y are real numbers. Geometrically, this describes every point in a two-dimensional plane, where x is the horizontal coordinate and y is the vertical coordinate. The cartesian product R × R × R represents all ordered triplets (x, y, z) where each component is a real number. Geometrically, this describes every point in three-dimensional space, with x, y, and z as the three spatial coordinates.
In simple words: R × R gives you all the points on a flat 2D surface, while R × R × R gives you all the points in 3D space.

Exam Tip: Understanding the geometric interpretation of cartesian products helps visualize abstract concepts and is often tested conceptually in exams.

Example 10. If A × B = {(0, 2), (3, - 1), (4, 2), (0, - 1), (3, 2), (4, - 1)}, then find B × A.
Answer: To obtain B × A from A × B, we swap the components of each ordered pair. Each pair (a, b) from A × B becomes (b, a) in B × A. Applying this transformation: (0, 2) becomes (2, 0), (3, -1) becomes (-1, 3), (4, 2) becomes (2, 4), (0, -1) becomes (-1, 0), (3, 2) becomes (2, 3), and (4, -1) becomes (-1, 4). Therefore, B × A = {(2, 0), (-1, 3), (2, 4), (-1, 0), (2, 3), (-1, 4)}.
In simple words: To switch from A × B to B × A, simply reverse the order of the numbers in each pair.

Exam Tip: You can quickly identify the sets A and B from A × B by collecting all first components (which form A) and all second components (which form B).

Example 11. If A × B = {(a, p), (b, q), (c, p), (a, q), (b, p), (c, q)}, find A and B.
Answer: Set A contains all the first components of the ordered pairs in A × B. Collecting these: a, b, c, a, b, c → the unique elements are a, b, c. So A = {a, b, c}. Set B contains all the second components of the ordered pairs in A × B. Collecting these: p, q, p, q, p, q → the unique elements are p, q. So B = {p, q}.
In simple words: To find A and B from A × B, gather all the first parts of the pairs for set A, and all the second parts for set B, removing duplicates.

Exam Tip: Always remove duplicate elements when identifying sets from a cartesian product - each set should contain only distinct elements.

Example 12. Let A and B be two sets such that n(A) = 5 and n(B) = 2. If (a, 1), (b, 5), (c, 5), (d, 1), (e, 5) are in A × B, find A and B, where a, b, c, d, e are distinct elements. Also write the remaining elements of A × B.
Answer: From the given pairs, the first components are a, b, c, d, e (all distinct), and the second components are 1 and 5. Since n(A) = 5 and we have 5 distinct first components, A = {a, b, c, d, e}. Since n(B) = 2 and the second components are 1 and 5, B = {1, 5}. Now, A × B would contain n(A) × n(B) = 5 × 2 = 10 pairs total. The given pairs are (a, 1), (b, 5), (c, 5), (d, 1), (e, 5) - that is 5 pairs. The remaining pairs must pair each element of A with the second component not yet shown for it. These are: (a, 5), (b, 1), (c, 1), (d, 5), (e, 1).
In simple words: Find A from the first parts of the given pairs. Find B from the second parts. Then complete the cartesian product by adding all missing pairs needed to reach n(A) × n(B) total pairs.

Exam Tip: When given partial information about a cartesian product, use the counts n(A) and n(B) to determine how many total pairs should exist, then identify which ones are missing.

Example 13. The cartesian product A × A has 9 elements among which are found (- 1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
Answer: Let n(A) = m. Since A × A has 9 elements, we have m² = 9, which gives m = 3 (taking the positive value). Since (-1, 0) ∈ A × A, both -1 and 0 must be in A. Since (0, 1) ∈ A × A, both 0 and 1 must be in A. So far, we know -1, 0, 1 ∈ A, and since n(A) = 3, these are the only elements: A = {-1, 0, 1}. The cartesian product A × A contains all pairs of elements from A. The complete set is: A × A = {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}. The remaining elements (those not given in the problem) are: {(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)}.
In simple words: Count the elements: 9 elements means A has 3 elements since 3 × 3 = 9. The given pairs tell you which elements belong to A. List all possible pairs from these elements.

Exam Tip: For A × A with n elements, the set A must have √n elements. Use information from given pairs to identify which specific elements belong in A.

Exercise 2.1

Very short answer type questions (1 to 19):

Question 1. Find a and b if
(i) (a + 1, b - 2) = (3, 1)
(ii) \( \left( \frac{a}{3} + 1, b - \frac{2}{3} \right) = \left( \frac{5}{3}, \frac{1}{3} \right) \)
(iii) (2a, a + b) = (6, 2)
(iv) (a + b, 3b - 2) = (7, - 5)
Answer:
(i) Equating first components: a + 1 = 3 → a = 2. Equating second components: b - 2 = 1 → b = 3.
(ii) From the first component: \( \frac{a}{3} + 1 = \frac{5}{3} \) → \( \frac{a}{3} = \frac{5}{3} - 1 = \frac{2}{3} \) → a = 2. From the second component: \( b - \frac{2}{3} = \frac{1}{3} \) → b = 1.
(iii) From the first component: 2a = 6 → a = 3. From the second component: a + b = 2 → 3 + b = 2 → b = -1.
(iv) From the first component: a + b = 7. From the second component: 3b - 2 = -5 → 3b = -3 → b = -1. Substituting into the first: a - 1 = 7 → a = 8.
In simple words: Match up each component from the first ordered pair with the matching component from the second ordered pair, then solve the equations.

Exam Tip: Always start by equating corresponding components. If one variable appears in only one equation, solve that equation first, then substitute into the other.

Question 2. Find x and y if
(i) (4x + 3, y) = (3x + 5, - 2)
(ii) (x - y, x + y) = (6, 10)
(iii) (2x + y, x - y) = (8, 3)
(iv) (x - 2, 2y + 1) = (y - 1, x + 2)
Answer:
(i) First components: 4x + 3 = 3x + 5 → x = 2. Second components: y = -2.
(ii) First components: x - y = 6. Second components: x + y = 10. Adding these equations: 2x = 16 → x = 8. Substituting back: 8 + y = 10 → y = 2.
(iii) First components: 2x + y = 8. Second components: x - y = 3. Adding: 3x = 11 → x = \( \frac{11}{3} \). Substituting: \( \frac{11}{3} - y = 3 \) → y = \( \frac{2}{3} \).
(iv) First components: x - 2 = y - 1 → x = y + 1. Second components: 2y + 1 = x + 2. Substituting x = y + 1: 2y + 1 = (y + 1) + 2 → y = 2. Therefore x = 3.
In simple words: Set up equations by matching components, then solve the system using substitution or elimination.

Exam Tip: When two equations result from matching components, use algebraic techniques like substitution or elimination to find both unknowns efficiently.

Question 3. If the ordered pairs (a, - 1) and (5, b) belong to {(x, y) : y = 2x - 3}, find the values of a and b.
Answer: Since (a, -1) belongs to the set, it must satisfy y = 2x - 3. Substituting x = a and y = -1: -1 = 2a - 3 → 2a = 2 → a = 1. Since (5, b) belongs to the set, it must satisfy y = 2x - 3. Substituting x = 5 and y = b: b = 2(5) - 3 = 10 - 3 = 7.
In simple words: Each point in the set must satisfy the given rule. Substitute the known coordinate value and solve for the unknown.

Exam Tip: When a point belongs to a relation defined by an equation, substitute its coordinates into the equation to find unknown values.

Question 4. If P = {7, 8} and Q = {5, 4, 2}, find P × Q and Q × P.
Answer: P × Q pairs each element of P with each element of Q: P × Q = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}. Q × P pairs each element of Q with each element of P: Q × P = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}.
In simple words: For P × Q, list every combination where a P element comes first and a Q element comes second. For Q × P, reverse the order.

Exam Tip: Count your pairs: if n(P) = 2 and n(Q) = 3, then n(P × Q) should be 2 × 3 = 6. This helps verify completeness.

Question 5. If A = {- 1, 0, 1} and B = {3, 5}, write the following:
(i) A × B
(ii) B × A
(iii) B × B
Answer:
(i) A × B = {(-1, 3), (-1, 5), (0, 3), (0, 5), (1, 3), (1, 5)}
(ii) B × A = {(3, -1), (3, 0), (3, 1), (5, -1), (5, 0), (5, 1)}
(iii) B × B = {(3, 3), (3, 5), (5, 3), (5, 5)}
In simple words: Systematically pair each element from the first set with each element from the second set to form all ordered pairs.

Exam Tip: For B × B, each element from B pairs with every element in B, including itself. Check that you have n(B)² = 4 pairs.

Question 6. If n(A) = 2 and B = {- 1, 0, 3}, then what is number of elements in A × B?
Answer: The number of elements in a cartesian product is found by multiplying the number of elements in each set. n(A × B) = n(A) × n(B) = 2 × 3 = 6.
In simple words: The size of a cartesian product equals the size of the first set times the size of the second set.

Exam Tip: Use the formula n(A × B) = n(A) × n(B) to quickly find the number of ordered pairs without listing them all.

Question 7. If A is a set such that n(A) = 3 and B = {3, 4, 5}, then what is the number of elements in A × B?
Answer: n(A × B) = n(A) × n(B) = 3 × 3 = 9.
In simple words: Multiply the number of elements in A by the number of elements in B to find the total number of pairs.

Exam Tip: This is a straightforward application of the multiplication principle for cartesian products.

Question 8. If A = {- 3, - 1, 0, 4} and B = {- 1, 0, 1, 2, 3}, then write the number of elements in each of the following cartesian products:
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B
Answer:
(i) n(A × B) = n(A) × n(B) = 4 × 5 = 20
(ii) n(B × A) = n(B) × n(A) = 5 × 4 = 20
(iii) n(A × A) = n(A) × n(A) = 4 × 4 = 16
(iv) n(B × B) = n(B) × n(B) = 5 × 5 = 25
In simple words: Count the elements in each set, then multiply appropriately based on which cartesian product you need.

Exam Tip: Note that n(A × B) = n(B × A) even though A × B ≠ B × A. The number of pairs is the same, but the pairs themselves are different.

Question 9. If A = {1, 2} and B = {3, 4}, then how many subsets will A × B have?
Answer: First, find n(A × B) = n(A) × n(B) = 2 × 2 = 4. A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}. The number of subsets of a set with n elements is 2ⁿ. Therefore, the number of subsets of A × B is 2⁴ = 16.
In simple words: Find how many ordered pairs are in A × B, then the number of subsets is 2 raised to that count.

Exam Tip: The power set of A × B has 2^(n(A×B)) subsets, including the empty set and A × B itself.

Question 10. If x ∈ {2, 3, 5} and y ∈ {2, 4, 6}, form the set of all ordered pairs (x, y) such that x < y.
Answer: We need pairs where the first element is less than the second element. Checking systematically: (2, 4), (2, 6), (3, 4), (3, 6), (5, 6) satisfy this condition. Therefore, the required set is {(2, 4), (2, 6), (3, 4), (3, 6), (5, 6)}.
In simple words: Test each possible pair and keep it only if the first value is smaller than the second value.

Exam Tip: When forming restricted cartesian products, systematically check the condition for each potential pair rather than randomly selecting.

Question 11. If x ∈ {- 1, 2, 3, 4, 5} and y ∈ {0, 3, 6}, form the set of all ordered pairs (x, y) such that x + y = 5.
Answer: We need pairs where the sum equals 5. Testing each x value: x = -1 gives y = 6, so (-1, 6) works but 6 ∉ the given y set. x = 2 gives y = 3, so (2, 3) works. x = 3 gives y = 2, but 2 ∉ the given y set. x = 4 gives y = 1, but 1 ∉ the given y set. x = 5 gives y = 0, so (5, 0) works. Therefore, the required set is {(2, 3), (5, 0)}.
In simple words: For each x value, calculate what y must be to satisfy the equation, then check if that y is in the given y set.

Exam Tip: Always verify that both components of your pair belong to their respective given sets before including the pair.

Question 12. If x ∈ {2, 3, 4} and y ∈ {4, 6, 9, 10}, form the set of all ordered pairs (x, y) such that x is a factor of y.
Answer: We need pairs where the first element divides the second element. Testing: x = 2 divides 4, 6, 10 → (2, 4), (2, 6), (2, 10). x = 3 divides 6, 9 → (3, 6), (3, 9). x = 4 divides 4 → (4, 4). Therefore, the required set is {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4)}.
In simple words: Check if each x value divides evenly into each y value. Include the pair if it does.

Exam Tip: For divisibility, a number divides another if the remainder is zero. Check this systematically for each combination.

Question 13. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ B and y ∈ A.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
Answer:
(i) False. Since P = Q = {m, n}, the cartesian product P × Q actually contains all pairs of elements from this set: P × Q = {(m, m), (m, n), (n, m), (n, n)}. The correct statement is: If P = {m, n} and Q = {n, m}, then P × Q = {(m, m), (m, n), (n, m), (n, n)}.
(ii) False. The definition requires x ∈ A and y ∈ B, not the other way around. The correct statement is: If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) True. B ∩ φ = φ (the intersection of any set with the empty set is empty). Therefore, A × (B ∩ φ) = A × φ = φ.
In simple words: Always check set definitions carefully. A × B means first components from A and second from B. Any cartesian product with an empty set is empty.

Exam Tip: For true-false questions, recall the definitions of cartesian products and set operations. A small error in phrasing makes a statement false.

Question 14. If A = {- 1, 0, 1, 2, 3}, write the subset S of A × A such that the second component of the elements of S is 0.
Answer: We need all pairs (x, y) from A × A where y = 0. Since the second component must be 0, we form pairs by taking each element of A as the first component and 0 as the second: S = {(-1, 0), (0, 0), (1, 0), (2, 0), (3, 0)}.
In simple words: List all pairs from the set with 0 as the second coordinate, letting the first coordinate run through all values in A.

Exam Tip: When restricting a cartesian product by fixing one component, let the other component vary through all set values.

Question 15. If A × B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.
Answer: Set A consists of all first components: {p, m}. Set B consists of all second components: {q, r}. Therefore, A = {p, m} and B = {q, r}.
In simple words: Collect all unique first coordinates for set A and all unique second coordinates for set B.

Exam Tip: Given a cartesian product, identifying the original sets is straightforward - just extract unique first and second components.

Question 16. If A × B = {(- 1, 1), (- 1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}, find A and B.
Answer: First components from all pairs: -1, -1, 2, 2, 3, 3 → unique elements: A = {-1, 2, 3}. Second components: 1, 2, 1, 2, 1, 2 → unique elements: B = {1, 2}.
In simple words: Extract the unique first coordinates to get A and unique second coordinates to get B.

Exam Tip: The number of elements: n(A) should equal the number of distinct first components, and n(B) should satisfy n(A) × n(B) = total number of pairs.

Question 17. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Answer: From the given pairs, the first components x, y, z are distinct, and together with the constraint n(A) = 3, we have A = {x, y, z}. The second components are 1 and 2. With n(B) = 2, we have B = {1, 2}.
In simple words: The distinct first components from the given pairs form set A (check that you have the right count). The distinct second components form set B.

Exam Tip: Use the given counts n(A) and n(B) to confirm you have identified the correct sets from the partial information provided.

Question 18. If A = {x, y, z} and some elements of A × B are (x, 1), (y, 2), (z, 1), then write the set B such that n(A × B) = 6.
Answer: We have A = {x, y, z}, so n(A) = 3. We want n(A × B) = 6. Using the formula n(A × B) = n(A) × n(B): 6 = 3 × n(B), so n(B) = 2. From the given pairs, the second components include 1 and 2. These are the only distinct second components mentioned. Since n(B) = 2, we must have B = {1, 2}.
In simple words: Divide the desired total number of pairs by n(A) to find n(B). The second components from the given pairs identify the elements of B.

Exam Tip: The condition n(A × B) = 6 directly tells you the size of set B: divide 6 by n(A) to get n(B).

Question 19. If A × B = {(x, 1), (y, 2), (x, 3), (y, 3), (y, 1), (x, 2)}, then find B × A.
Answer: From A × B, the first components are x, y, x, y, y, x → unique: A = {x, y}. The second components are 1, 2, 3, 3, 1, 2 → unique: B = {1, 2, 3}. Now, B × A reverses the components: B × A = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}.
In simple words: Identify A and B from the given cartesian product, then form B × A by reversing each pair's components.

Exam Tip: To switch from A × B to B × A, identify the sets first, then systematically create all reversed pairs.

Question 20. If A = {2, 3, 4} and B = {0, 1}, find
(i) A × B
(ii) B × A
(iii) A × A
(iv) B × B
(v) n(A × B)
(vi) n(B × A)
(vii) n(A × A)
(viii) n(B × B)
Is A × B = B × A?
Answer:
(i) A × B = {(2, 0), (2, 1), (3, 0), (3, 1), (4, 0), (4, 1)}
(ii) B × A = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
(iii) A × A = {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}
(iv) B × B = {(0, 0), (0, 1), (1, 0), (1, 1)}
(v) n(A × B) = 3 × 2 = 6
(vi) n(B × A) = 2 × 3 = 6
(vii) n(A × A) = 3 × 3 = 9
(viii) n(B × B) = 2 × 2 = 4
No, A × B ≠ B × A because their ordered pairs are different (e.g., (2, 0) ∈ A × B but (2, 0) ∉ B × A).
In simple words: Form each cartesian product by systematically pairing elements. Count them to verify. Note that the cartesian products themselves are different, though they contain the same number of pairs.

Exam Tip: While n(A × B) = n(B × A), the sets themselves are generally different unless A = B. Show specific pairs to prove inequality.

2.3 Relations

In everyday life, we frequently speak of relations between two or more objects. To learn the concept properly, consider the following examples:

(i) Let A = {1, 2, 3, 5} and B = {2, 4}. Then A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4), (5, 2), (5, 4)}. We can obtain a subset of A × B by introducing a relation 'is less than' between the elements of the sets A and B. If we write R for the relation 'is less than', then we get 1 R 2, 1 R 4, 2 R 4, 3 R 4. Omitting the letter R between the above pairs of numbers and writing these pairs of numbers as ordered pairs, the above information can be written as a set of ordered pairs R where R = {(1, 2), (1, 4), (2, 4), (3, 4)} = {(x, y) : x ∈ A, y ∈ B, x < y}. Thus the relation 'is less than' from the set A to the set B gives rise to a subset R of A × B such that (x, y) ∈ R iff x R y i.e. iff x < y.

(ii) Let A = {2, 3, 5, 9} and B = {4, 6, 9, 15, 25}. There is a relation 'is a divisor of' between the elements of the sets A and B. If we write R for the relation 'is a divisor of', then we get 2 R 4, 2 R 6, 3 R 6, 3 R 9, 3 R 15, 5 R 15, 5 R 25, 9 R 9. This can be written as a set of ordered pairs R where R = {(2, 4), (2, 6), (3, 6), (3, 9), (3, 15), (5, 15), (5, 25), (9, 9)} = {(x, y) : x ∈ A, y ∈ B, x is a divisor of y}. Thus, the relation 'is a divisor of' from the set A to the set B gives rise to a subset R of A × B such that (x, y) ∈ R iff x R y i.e. iff x is a divisor of y.

(iii) Let N be the set of natural numbers. There is a relation 'has as its square' from the set N to N. If we write R for the relation 'has as its square', then we get 1 R 1, 2 R 4, 3 R 9, 4 R 16, 5 R 25, …… This can be written as a set of ordered pairs R where R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), ……} = {(x, y) : x , y ∈ N, y = x²}. Thus, R is a subset of N × N such that (x, y) ∈ R iff x R y i.e. iff y = x².

Definition. If A, B are any two (non-empty) sets, then any subset of A × B is called a relation from A to B.

Let R be a relation from A to B. If R = φ, then R is called the empty relation and if R = A × B, then R is called the universal relation.

If R is a relation from A to B and if (a, b) ∈ R, then we write a R b and say that a is related to b and if (a, b) ∉ R, then we write a R̅ b and say that a is not related to b.

In particular, if A is any (non-empty) set, then any subset of A × A is called a relation on A.

2.3.1 Representation of a Relation

1. Roster form. In this form, a relation is represented by the set of all ordered pairs which belong to the given relation.

For example, let A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, …, 20}, and let R be the relation 'has as its square' from A to B, then R = {(1, 1), (2, 4), (3, 9), (4, 16)}.

2. Set-builder form. In this form, the relation is represented as {(x, y) : x ∈ A, y ∈ B, x …y}, the blank is to be replaced by the rule which associates x and y.

For example, let A = {1, 3, 4, 5, 7}, B = {2, 4, 6, 8} and R = {(1, 2), (3, 4), (5, 6), (7, 8)} then R in the builder form can be written as R = {(x, y) : x ∈ A, y ∈ B, x is one less than y}.

3. By arrow diagram. In this form, the relation is represented by drawing arrows from first components to the second components of all ordered pairs which belong to the given relation.

For example, let A = {1, 2, 3, 5}, B = {2, 3, 4} and R be the relation 'is greater than' from A to B, then R = {(3, 2), (5, 2), (5, 3), (5, 4)}. This relation R from A to B can be represented by an arrow diagram.

2.3.2 Domain and Range of a Relation

Let A, B be any two (non-empty) sets and R be a relation from A to B. The domain of the relation R is the set of all first components of the ordered pairs which belong to R. The range of the relation R is the set of all second components of the ordered pairs which belong to R. Thus:

domain of R = {x : x ∈ A, (x, y) ∈ R for some y ∈ B} and
range of R = {y : y ∈ B, (x, y) ∈ R for some x ∈ A}.

If R is a relation from A to B, then B is called codomain of R.

For example, let A = {1, 3, 4, 5, 7}, B = {2, 4, 6, 8} and R be the relation 'is one less than' from A to B, then R = {(1, 2), (3, 4), (5, 6), (7, 8)}. Here, domain of R = {1, 3, 5, 7} and range of R = {2, 4, 6, 8}. In this example, note that range of R = B = codomain of R.

Example 1. If A and B are finite sets such that n(A) = m and n(B) = k, find the number of relations from A to B.
Answer: Given n(A) = m and n(B) = k, we calculate n(A × B) = n(A) × n(B) = mk. Since every subset of A × B is a relation from A to B, and a set with n elements has 2ⁿ subsets, the number of subsets of A × B equals 2^(mk). Therefore, the number of relations from A to B is 2^(mk).
In simple words: The number of possible relations from A to B equals 2 raised to the power of (total pairs in A × B).

Exam Tip: Remember that every subset of A × B, including the empty set and A × B itself, counts as a relation.

Example 2. If a relation R = {(0, 0), (2, 4), (- 1, - 2), (3, 6), (1, 2)}, then
(i) write domain of R.
(ii) write range of R.
(iii) write R in the builder form.
(iv) represent R by an arrow diagram.
Answer:
(i) Domain of R consists of all first components: {0, 2, -1, 3, 1}.
(ii) Range of R consists of all second components: {0, 4, -2, 6, 2}.
(iii) Observing the relationship in each pair: 0 = 2(0), 4 = 2(2), -2 = 2(-1), 6 = 2(3), 2 = 2(1). The pattern is y = 2x. R in builder form: R = {(x, y) : x ∈ I, -1 ≤ x ≤ 3, y = 2x}.
(iv) The arrow diagram shows arrows from each first component to its corresponding second component in B.
In simple words: Domain is all the first values, range is all the second values. Look for a rule connecting them to write builder form.

Exam Tip: When writing a relation in builder form, identify the rule by checking what operation relates each x to its y value.

Example 3. If A = {- 1, 2, 5, 8}, B = {0, 1, 3, 6, 7} and R be the relation 'is one less than' from A to B, then
(i) find R as a set of ordered pairs.
(ii) find domain and range of R.
Answer:
(i) For the relation 'is one less than', we include (x, y) if x is one less than y, i.e., x = y - 1 or y = x + 1. Checking: -1 is one less than 0 → (-1, 0). 2 is one less than 3 → (2, 3). 5 is one less than 6 → (5, 6). 8 is one less than 9, but 9 ∉ B. Therefore, R = {(-1, 0), (2, 3), (5, 6)}.
(ii) Domain of R = {-1, 2, 5} and range of R = {0, 3, 6}.
In simple words: Test each element from A: does it relate to any element in B according to the given rule?

Exam Tip: For relations with verbal descriptions like 'is one less than', translate to an equation and check each potential pair.

Example 4. If A = {1, 2, 3}, B = {1, 2, 3, 4} and R = {(x, y): (x, y) ∈ A × B, y = x + 1}, then
(i) find A × B.
(ii) write R in roster form.
(iii) write domain and range of R.
(iv) represent R by an arrow diagram.
Answer:
(i) A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4)}.
(ii) From A × B, we keep pairs satisfying y = x + 1: (1, 2), (2, 3), (3, 4). So R = {(1, 2), (2, 3), (3, 4)}.
(iii) Domain of R = {1, 2, 3} and range of R = {2, 3, 4}.
(iv) The arrow diagram shows arrows: 1 → 2, 2 → 3, 3 → 4.
In simple words: Filter A × B by keeping only pairs that satisfy the given rule.

Exam Tip: Always generate A × B first if asked, then apply the additional condition to get the final relation.

Example 5. If A = {1, 2, 3, 4, …, 14} and a relation R is defined from A to A by R = {(x, y) : 3x - y = 0, x, y ∈ A}.
(i) Write R in roster form.
(ii) Write its domain, codomain and range.
(iii) Depict this relationship by an arrow diagram.
Answer:
(i) From 3x - y = 0, we get y = 3x. Finding pairs where both x and y are in A: x = 1 → y = 3 ✓, x = 2 → y = 6 ✓, x = 3 → y = 9 ✓, x = 4 → y = 12 ✓, x = 5 → y = 15 (not in A). So R = {(1, 3), (2, 6), (3, 9), (4, 12)}.
(ii) Domain = {1, 2, 3, 4}, codomain = {1, 2, 3, …, 14} = A, range = {3, 6, 9, 12}.
(iii) The arrow diagram shows 1 → 3, 2 → 6, 3 → 9, 4 → 12.
In simple words: Solve the equation for y in terms of x, then test values of x to find which give y values also in the set.

Exam Tip: The codomain is always the set B in "relation from A to B", while range is only the elements of B that actually appear as second components.

Example 6. The adjoining diagram shows a relation between the sets P and Q. Write this relation
(i) in roster form
(ii) in set builder form.
What is its domain and range?
Answer:
(i) Reading from the diagram: 4 relates to 2 and -2; 9 relates to 3 and -3; 25 relates to 5 and -5. In roster form, R = {(4, 2), (4, -2), (9, 3), (9, -3), (25, 5), (25, -5)}.
(ii) Looking at the pattern, each pair (x, y) satisfies x = y². In set builder form, R = {(x, y) : x = y², x ∈ P, y ∈ Q}.
Domain = {4, 9, 25}, range = {2, -2, 3, -3, 5, -5}.
In simple words: From the diagram, read off each arrow. Identify the pattern to write the set-builder form.

Exam Tip: Arrow diagrams are visual - carefully trace each arrow to list all pairs. Then spot the pattern for set-builder form.

Example 7. Find the domain and the range of the relation R defined by R = {(x + 1, x + 3) : x ∈ {0, 1, 2, 3, 4, 5}}.
Answer: For each value of x, we compute x + 1 (first component) and x + 3 (second component):
x = 0: (1, 3)
x = 1: (2, 4)
x = 2: (3, 5)
x = 3: (4, 6)
x = 4: (5, 7)
x = 5: (6, 8)
Therefore, R = {(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8)}. Domain of R = {1, 2, 3, 4, 5, 6} and range of R = {3, 4, 5, 6, 7, 8}.
In simple words: Substitute each x value to find the corresponding pair, then collect all first components for domain and all second components for range.

Exam Tip: When the relation is defined parametrically (using x as a parameter), compute all pairs by substitution, then identify domain and range.

Example 8. If R = {(x, y) : x, y ∈ W, 2x + y = 8}, then
(i) find the domain and the range of R.
(ii) write R as a set of ordered pairs.
Answer:
(i) and (ii): From 2x + y = 8, we get y = 8 - 2x. For x, y ∈ W (whole numbers), we test values: x = 0 → y = 8, so (0, 8); x = 1 → y = 6, so (1, 6); x = 2 → y = 4, so (2, 4); x = 3 → y = 2, so (3, 2); x = 4 → y = 0, so (4, 0); x = 5 → y = -2 (not a whole number). Therefore, R = {(0, 8), (1, 6), (2, 4), (3, 2), (4, 0)}, domain = {0, 1, 2, 3, 4}, range = {8, 6, 4, 2, 0}.
In simple words: Rearrange the equation to express y in terms of x, then substitute whole number values of x to find valid pairs.

Exam Tip: For constraints like "x, y ∈ W", test systematically and stop when further x values produce invalid y values.

Example 9. Find the domain and the range of the relation R given by R = {(x, y) : y = x + \( \frac{6}{x} \), where x, y ∈ N and x < 6}.
Answer: We need both x and y to be natural numbers with x < 6. Testing values: x = 1 → y = 1 + 6 = 7 (natural number) ✓, x = 2 → y = 2 + 3 = 5 (natural number) ✓, x = 3 → y = 3 + 2 = 5 (natural number) ✓, x = 4 → y = 4 + 1.5 (not a natural number), x = 5 → y = 5 + 1.2 (not a natural number). Therefore, R = {(1, 7), (2, 5), (3, 5)}, domain = {1, 2, 3}, range = {7, 5}.
In simple words: Compute y for each valid x value. Include the pair only if y is a natural number.

Exam Tip: When the formula produces fractions or decimals, check whether the result is in the required set before including the pair.

Example 10. Find the linear relation between the components of the ordered pairs of the relation R where R = {(2, 1), (4, 7), (1, - 2), …}.
Answer: Let the linear relation be y = ax + b. Using pair (2, 1): 1 = 2a + b ... (i). Using pair (4, 7): 7 = 4a + b ... (ii). Subtracting (i) from (ii): 6 = 2a, so a = 3. Substituting a = 3 into (i): 1 = 6 + b, so b = -5. Therefore, the linear relation is y = 3x - 5.
In simple words: Use two pairs to set up two equations, solve for the coefficients a and b.

Exam Tip: To find a linear relation, always use exactly two different pairs to avoid under-determined or over-determined systems.

Example 11. If A = {1, 2, 3, 5}, B = {4, 6, 9} and a relation R from A to B is defined by R = {(x, y) : the difference between x and y is odd, x ∈ A, y ∈ B}. Then
(i) write R in the roster form
(ii) represent R by an arrow diagram.
Answer:
(i) We need pairs where x - y is an odd number. Testing: (1, 4): 1 - 4 = -3 (odd) ✓, (1, 6): 1 - 6 = -5 (odd) ✓, (1, 9): 1 - 9 = -8 (even), (2, 9): 2 - 9 = -7 (odd) ✓, (3, 4): 3 - 4 = -1 (odd) ✓, (3, 6): 3 - 6 = -3 (odd) ✓, (5, 4): 5 - 4 = 1 (odd) ✓, (5, 6): 5 - 6 = -1 (odd) ✓. Therefore, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.
(ii) The arrow diagram shows arrows from each x in A to those y in B where the difference is odd.
In simple words: Check each possible pair to see if their difference is odd. A difference is odd when one number is even and the other is odd.

Exam Tip: Remember: odd - even = odd, even - odd = odd, odd - odd = even, even - even = even.

Example 12. Let A = {3, 5} and B = {7, 11}. Let R = {(a, b) : a ∈ A, b ∈ B, a - b is odd}. Show that R is an empty relation from A to B.
Answer: The cartesian product A × B = {(3, 7), (3, 11), (5, 7), (5, 11)}. Checking each pair: 3 - 7 = -4 (even), 3 - 11 = -8 (even), 5 - 7 = -2 (even), 5 - 11 = -6 (even). Since none of the differences is odd, no pair satisfies the condition. Therefore, R = φ (the empty relation).
In simple words: Both 3 and 5 are odd; both 7 and 11 are odd. Odd minus odd is always even, so the condition "difference is odd" can never be satisfied.

Exam Tip: When no pair from A × B satisfies the given condition, the relation is empty. Show the calculation to verify this.

Example 13. If A = {2, 4, 6, 9}, B = {4, 6, 18, 27, 54} and a relation R from A to B is defined by R = {(a, b) : a ∈ A, b ∈ B, a is a factor of b and a < b}, then find R in Roster form. Also find its domain and range.
Answer: We need pairs where a divides b and a < b. Testing each element from A: a = 2 divides 4, 6, 18, 54 (and is less than each) → (2, 4), (2, 6), (2, 18), (2, 54). a = 4 divides only 4, but 4 is not less than 4. a = 6 divides 6, 18, 54 (and 6 is less than 18 and 54 but not itself) → (6, 18), (6, 54). a = 9 divides 18, 27, 54 (and 9 is less than each) → (9, 18), (9, 27), (9, 54). Therefore, R = {(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)}, domain = {2, 6, 9}, range = {4, 6, 18, 27, 54}.
In simple words: Check if a divides b and if a is smaller than b. Include the pair only if both conditions hold.

Exam Tip: For divisibility with an additional inequality constraint, verify both the divisibility and the order before including a pair.

Example 14. Let R be a relation from Q to Q defined by R = {(a, b) : a, b ∈ Q and a - b ∈ Z}. Show that
(i) (a, a) ∈ R for all a ∈ Q
(ii) (a, b) ∈ R implies that (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R.
Answer:
(i) For any a ∈ Q, a - a = 0, and 0 ∈ Z. Therefore, (a, a) ∈ R.
(ii) Assume (a, b) ∈ R. Then a - b ∈ Z. This means -(a - b) = b - a ∈ Z. Therefore, (b, a) ∈ R.
(iii) Assume (a, b) ∈ R and (b, c) ∈ R. Then a - b ∈ Z and b - c ∈ Z. Adding these: (a - b) + (b - c) = a - c ∈ Z. Therefore, (a, c) ∈ R.
In simple words: Verify each property by using the definition of the relation and properties of integers.

Exam Tip: Properties like reflexivity, symmetry, and transitivity are important concepts tested in relation problems.

Example 15. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
(i) (a, a) ∈ R for all a ∈ N
(ii) (a, b) ∈ R implies (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.
Answer:
(i) False. For (a, a) ∈ R, we need a = a², which means a² - a = 0 or a(a - 1) = 0, so a = 0 or a = 1. But 0 is not a natural number. For a = 1: 1 = 1², which is true, but this works only for a = 1. For a = 2: 2 ≠ 4, so (2, 2) ∉ R. The statement is false.
(ii) False. Counterexample: (4, 2) ∈ R because 4 = 2². However, (2, 4) ∉ R because 2 ≠ 4² = 16. So the implication fails.
(iii) False. Counterexample: (16, 4) ∈ R (since 16 = 4²) and (4, 2) ∈ R (since 4 = 2²). However, (16, 2) ∉ R because 16 ≠ 2² = 4. So the implication fails.
In simple words: Test the properties with specific examples. If even one counterexample exists, the statement is false.

Exam Tip: When asked "Are these true?", provide counterexamples when the answer is no. This shows clear understanding.

2.4 Functions

A function is a special case of a relation. To be specific, let X, Y be two non-empty sets and R (or f) be a relation from X to Y. The relation may not associate an element of X with an element of Y, or it may relate an element of X to more than one element of Y. However, a function associates each element of X to a unique element of Y.

Definition. If X, Y are two non-empty sets then a subset f of X × Y is called a function (or mapping or map) from X to Y iff for each x ∈ X, there exists a unique y ∈ Y such that (x, y) ∈ f. It is written as f : X → Y.

Thus, a subset f of X × Y is called a function from X to Y iff

• (i) for each x ∈ X, there exists y ∈ Y such that (x, y) ∈ f and
• (ii) no two different ordered pairs have the same first component.

In other words, a function from X to Y is a rule (or correspondence) which associates to each element x of X, a unique element y of Y.

Image of an element. The unique element y ∈ Y is called the image of the element x of X under the function f : X → Y. It is denoted by f(x) i.e. y = f(x). The element y is also called the value of the function f at x.

2.4.1 Domain and Range of a Function

Let f be a function from X to Y. The set X is called the domain of the function f and the set Y is called the codomain.

The set consisting of all the images of the elements of X under the function f is called the range of f. It is denoted by f(X). Thus, range of f = {f(x) : for all x ∈ X}.

Note that range of f is a subset of Y (codomain) which may or may not be equal to Y.

For example:

(1) Let X = {1, 2, 3, 4, 5}, Y = {0, 1, 2, 3, 5, 7, 9, 11, 13} and

• (i) f = {(1, 1), (2, 0), (3, 7), (4, 9), (5, 13)}, then f is a function from X to Y because each element of X has a unique image in Y. Range of f = {1, 0, 7, 9, 13}. Note that some elements of Y are not associated with any element of X.
• (ii) f = {(1, 3), (2, 3), (3, 5), (4, 7), (5, 5)}, then f is a function from X to Y because each element of X has a unique image in Y. Range of f = {3, 5, 7}. Note that the second components may repeat.
• (iii) f = {(1, 5), (2, 7), (4, 9), (5, 0)}, then f is not a function from X to Y because the element 3 of X has no image in Y.
• (iv) f = {(1, 1), (1, 2), (2, 3), (3, 5), (4, 7), (5, 11)}, then f is not a function because the different pairs (1, 1) and (1, 2) have same first component i.e. the element 1 of X has two different images.

(2) Let X = {a, b, c, d} and Y = {p, q, r, s, t}, then the rule depicted by an arrow diagram represents a function from X to Y because each element of X has a unique image in Y. Range of the function = {p, q, r, t}. Note that the element s of Y is not associated with any element of X.

2.4.2 Main Features of a Function

Let f be a function from X to Y, then

• (i) to every x ∈ X, there exists a unique element y ∈ Y such that y = f(x).
• (ii) no element of X can have more than one images in Y.
• (iii) there may be elements of Y which are not associated with any element of X.
• (iv) distinct elements of X may have same image in Y.
• (v) function f is determined when f(x) is known for all x ∈ X.

Example 1. Which of the following relations are functions? Give reasons.
(i) R = {(2, 1), (3, 1), (4, 2)}
(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}
Answer:
(i) Domain of R = {2, 3, 4}. Each element has exactly one image: 2 maps to 1, 3 maps to 1, 4 maps to 2. Therefore, R is a function.
(ii) Domain of R = {2, 3, 4}. The element 2 has two different images: 2 and 4. Since one element maps to multiple values, R is not a function.
(iii) Domain of R = {1, 2, 3, 4, 5, 6}. Each element maps to exactly one element: 1 → 2, 2 → 3, 3 → 4, 4 → 5, 5 → 6, 6 → 7. Therefore, R is a function.
In simple words: A relation is a function only if every element in the domain maps to exactly one element in the codomain. Check that no first element appears twice.

Exam Tip: To verify if a relation is a function, examine the first components. If any first component appears more than once, it is not a function.

Question 2. If A = {1, 2, 3} and f, g, h and s are relations corresponding to the subsets of A × A indicated against them, which of f, g, h and s are functions? In case of a function, find its domain and range.
(i) f = {(2, 1), (3, 3)}
(ii) g = {(1, 2), (1, 3), (2, 3), (3, 1)}
(iii) h = {(1, 3), (2, 1), (3, 2)}
(iv) s = {(1, 2), (2, 2), (3, 1)}
Answer:
(i) f is not a function because element 1 from A does not appear as the first element of any ordered pair in f. This means 1 has no corresponding value in A.
(ii) g is not a function because two different pairs (1, 2) and (1, 3) have the same first element. This shows that element 1 of A maps to two different values in A.
(iii) h is a function because all elements of A map to exactly one element each in A. Domain of h = {1, 2, 3} = A and range of h = {3, 1, 2} = A.
(iv) s is a function because all elements of A map to exactly one element each in A. Domain of s = {1, 2, 3} = A and range of s = {2, 1}.
In simple words: A function means every input gets exactly one output. If any input has no output or two outputs, it is not a function.

Exam Tip: Check two conditions: (1) every element from set A must appear as a first element in some pair, (2) no element appears as a first element more than once. If both hold, it is a function.

Question 3. Consider the following diagrams carefully and state whether they represent functions. Give reasons for your answer. In case of a function, write its domain and range.
Answer:
(i) The diagram shown represents a function because each element in the set {a, b, c, d} has exactly one image in the set {g, f, m}. Domain = {a, b, c, d} and range = {g, f, m}.
(ii) The diagram shown does not represent a function because element 3 from the set {2, 3, 4} has two different images: 2 and 4, both in the set {1, 2, 4, 5, 7}.
In simple words: In a function diagram, each arrow starting from the left set must go to exactly one element on the right. If any element has two arrows, it fails the function test.

Exam Tip: Always count the arrows leaving each element on the left. If any element has zero or more than one arrow, it is not a function.

Question 4. Let N be the set of natural numbers and the relation R be defined on N by R = {(x, y) : y = 2x, x, y ∈ N}. What is the domain, codomain and range of R? Is this relation a function?
Answer: Given R = {(x, y) : y = 2x, x, y ∈ N}. Domain of R is N (all natural numbers), codomain of R is N. The range of R is the set of all even natural numbers. Since every natural number x gets a unique image 2x, the relation R is a function.
In simple words: For each natural number you pick, doubling it gives one answer. That answer is always an even number. So this is a function.

Exam Tip: Remember that domain is the set of all starting values, range is the set of all actual output values, and codomain is the set where outputs are allowed to belong. Range is always a subset of codomain.

Question 5. A relation 'f' is defined by f : x → x² - 2, where x ∈ {-1, -2, 0, 2}. (i) List the elements of f. (ii) Is f a function?
Answer:
(i) The relation f is defined by f(x) = x² - 2 where x ∈ {-1, -2, 0, 2}. Calculate f for each value:
\( f(-1) = (-1)^2 - 2 = 1 - 2 = -1 \)
\( f(-2) = (-2)^2 - 2 = 4 - 2 = 2 \)
\( f(0) = 0^2 - 2 = 0 - 2 = -2 \)
\( f(2) = 2^2 - 2 = 4 - 2 = 2 \)

Therefore, f = {(-1, -1), (-2, 2), (0, -2), (2, 2)}

(ii) We observe that all elements from the domain have exactly one corresponding image. Hence, the relation f is a function.
In simple words: Plug each x value into the formula. Each input gives one output number. When all inputs give one output each, it is a function.

Exam Tip: For function verification, list all pairs and check that no first element repeats. If the first element never repeats, it passes the function test.