Relations And Functions

Chapter 2.1: Relations

• Definition: A relation between two sets A and B is a subset of their Cartesian product A × B.
• Representation: Relations can be represented using:
  • Roster method (listing ordered pairs)
  • Set-builder form (defining the condition for ordered pairs)
  • Arrow diagrams
  • Graphs

Chapter 2.2: Functions

• Definition: A relation between two sets A and B is called a function if every element of A has exactly one element associated with it in B.
• Domain: The set of all elements in A.
• Codomain: The set of all elements in B.
• Range: The set of all elements in B that are associated with some element in A.
• Types of Functions:
  • Identity function
  • Constant function
  • Polynomial function
  • Rational function
  • Modulus function
  • Greatest integer function
  • Signum function

Chapter 2.3: Real-Valued Functions

• Definition: Functions with domain and codomain subsets of real numbers.
• Graphs of Real-Valued Functions: Visual representation of the relationship between the input (domain) and output (range).
• Properties of Graphs:
  • Increasing, decreasing, constant intervals
  • Even and odd functions

Chapter 2.4: Algebra of Real-Valued Functions

• Operations on Functions: Addition, subtraction, multiplication, division, composition.
• Properties of Function Operations:
  • Commutative, associative, distributive

Key Concepts:

• Relations and their representations
• Functions and their properties
• Types of functions
• Graphs of real-valued functions
• Algebra of real-valued functions

Exercise 2.1

1.

Ans :

2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).

Ans : 

The Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

Since set A has 3 elements and set B has 3 elements, the number of possible ordered pairs in A × B is 3 × 3 = 9.

Therefore, the number of elements in (A × B) is 9.

3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G

Ans : 

Given:

• G = {7, 8}
• H = {5, 4, 2}

G × H

• This represents the Cartesian product of G and H, which is the set of all ordered pairs (g, h) where g ∈ G and h ∈ H.
• G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G

• This represents the Cartesian product of H and G, which is the set of all ordered pairs (h, g) where h ∈ H and g ∈ G.
• H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

4.  State whether each of the following statements are 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 ∈ A and y ∈ B. 

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

Ans : 

(i) True

(ii) True

(iii) True

5. If A = {–1, 1}, find A × A × A.

Ans : 

A × A × A represents the Cartesian product of set A with itself three times. It consists of all possible ordered triples (a, b, c) where a, b, and c are elements of A.

Since A = {-1, 1}, the elements of A × A × A will be all possible combinations of -1 and 1 taken three at a time:

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)}

6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Ans : 

To find sets A and B from their Cartesian product A × B, we can identify the unique elements in the first and second positions of the ordered pairs.

Identifying elements in the first position:

• a, a, b, b

Identifying elements in the second position:

• x, y, x, y

Therefore, we can deduce that:

• A = {a, b}
• B = {x, y}

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that 

(i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D

Ans : 

8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Ans : 

• Empty set: ∅
• Single-element subsets: {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}
• Two-element subsets: {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}
• Three-element subsets: {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}
• Four-element subset (A × B itself): {(1, 3), (1, 4), (2, 3), (2, 4)}

9. 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 and z are distinct elements. 

Ans : 

Given:

• n(A) = 3 (A has 3 elements)
• n(B) = 2 (B has 2 elements)
• (x, 1), (y, 2), (z, 1) ∈ A × B

Analysis:

• Since (x, 1) and (z, 1) are in A × B, this implies that 1 is an element of B.
• Similarly, since (y, 2) is in A × B, this implies that 2 is an element of B.
• Therefore, B = {1, 2} (since we know B has 2 elements)

Now, we need to find A. We know that A has 3 elements and contains x, y, and z. Since x and y are paired with 1 and 2 in the Cartesian product, they must be distinct from each other.

Therefore, we can conclude:

• A = {x, y, z} (where x, y, and z are distinct)
• B = {1, 2}

10. 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.

Ans : 

Given:

• A × A has 9 elements.
• (-1, 0) and (0, 1) are elements of A × A.

Analysis:

• Since (-1, 0) and (0, 1) are in A × A, it implies that -1, 0, and 1 are elements of A.
• The Cartesian product A × A will have 9 elements if A has 3 elements.

Therefore, A = {-1, 0, 1}.

Now, let’s find the remaining elements of A × A:

• A × A = {(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)}

The remaining elements of A × A are: 

(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), and (1, 1).

Exercise 2.2

1. Let A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range

Ans : 

1. Domain:

The domain is the set of all first elements of the ordered pairs in R.

Since 3x – y = 0 implies y = 3x, the domain consists of all elements x in A such that 3x is also in A.

Domain = {1, 2, 3, 4, 5} 

(because 3 * 6, 3 * 7, 3 * 8, 3 * 9, 3 * 10, 3 * 11, 3 * 12, 3 * 13, and 3 * 14 are not in A)

2. Codomain:

The codomain is the set of all possible second elements of the ordered pairs in the relation.

Since the relation is defined from A to A, the codomain is also A.

Codomain = {1, 2, 3, …, 14}

3. Range:

The range is the set of all second elements that actually appear in the relation.

Since y = 3x for all (x, y) in R, the range is the set of all multiples of 3 that are in A.

Range = {3, 6, 9, 12}

2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range.

Ans : 

Given:

• R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}

Understanding R:

• The ordered pairs in R are of the form (x, y) where y is 5 more than x, and x is a natural number less than 4.

Roster Form:

• Listing all the possible ordered pairs based on the given condition:
  • R 
  • = {(1, 6), (2, 7), (3, 8)}

Domain and Range:

• Domain: The set of all first elements of the ordered pairs.
• Domain = {1, 2, 3}
• Range = {6, 7, 8}

3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Ans : 

Given:

• A = {1, 2, 3, 5}
• B = {4, 6, 9}
• R = {(x, y) : the difference between x and y is odd; x ∈ A, y ∈ B}

Understanding R:

• R is a relation from A to B where the difference between the elements in each ordered pair (x, y) is odd.

Finding the ordered pairs in R:

• By checking the differences between elements in A and B, we can identify the pairs where the difference is odd:
  • (1, 4) (difference: 4 – 1 = 3)
  • (1, 6) (difference: 6 – 1 = 5)
  • (2, 9) (difference: 9 – 2 = 7)
  • (3, 4) (difference: 4 – 3 = 1)
  • (3, 6) (difference: 6 – 3 = 3)
  • (5, 4) (difference: 4 – 5 = -1)
  • (5, 6) (difference: 6 – 5 = 1)

Therefore, R in roster form is:

R =

= {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

4. The Fig2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?

Ans : 

From the given figure, we can observe the following relationships between the elements of sets P and Q:

• 5 is related to 3
• 6 is related to 4
• 7 is related to 5

Representation of the Relation:

(i) Set-Builder Form:

• R = {(x, y) : x ∈ P, y ∈ Q, and y = x – 2}

This notation means that the relation R consists of all ordered pairs (x, y) where x belongs to set P, y belongs to set Q, and y is 2 less than x.

(ii) Roster Form:

• R 
• = {(5, 3), (6, 4), (7, 5)}

This notation lists all the ordered pairs that satisfy the relation.

Domain and Range:

• Domain = {5, 6, 7}
• Range = {3, 4, 5}

5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}. 

(i) Write R in roster form (ii) Find the domain of R (iii) Find the range of R.

Ans : 

Given:

• A = {1, 2, 3, 4, 6}
• R 
• = {(a, b): a, b ∈ A, b is exactly divisible by a}

(i) Writing R in Roster Form

To write R in roster form, we need to list all the ordered pairs (a, b) that satisfy the given condition.

• R 
• = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Finding the Domain of R

• Domain(R) = {1, 2, 3, 4, 6}

(iii) Finding the Range of R

• Range(R) = {1, 2, 3, 4, 6}

6. Determine the domain and range of the relation R defined by 

R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Ans : 

Given:

• R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}

Domain

• The domain is the set of all first elements of the ordered pairs in R.
• From the definition of R, the first element of each ordered pair is x, which is taken from the set {0, 1, 2, 3, 4, 5}.
• Therefore, the domain of R is {0, 1, 2, 3, 4, 5}.

Range

• The range is the set of all second elements of the ordered pairs in R.
• Since the second element is x + 5, we can find the range by adding 5 to each element in the domain.
• Range 
• = {5, 6, 7, 8, 9, 10}

In summary:

• Domain of R
•  = {0, 1, 2, 3, 4, 5}
• Range of R 
• = {5, 6, 7, 8, 9, 10}

7. Write the relation R = {(x, x3 ) : x is a prime number less than 10} in roster form.

Ans : 

To write the relation R = {(x, x^3) : x is a prime number less than 10} in roster form, we need to list all the ordered pairs (x, x^3) 

Let’s find the corresponding cubes:

• 2^3 = 8
• 3^3 = 27
• 5^3 = 125
• 7^3 = 343

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Ans : 

Step 1: Find A × B

A × B 

= {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Step 2: Count the elements in A × B

6 elements in A × B.

Step 3: Find the number of subsets of A × B

The number of subsets of a set with n elements is 2^n.

Therefore, the number of relations from A to B is 2^6 = 64.

9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R. 

Ans : 

Exercise 2.3

1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. (i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)} 

(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)} (iii) {(1,3), (1,5), (2,5)}.

Ans : 

(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}

• Every element in the domain (2, 5, 8, 11, 14, 17) is paired with a unique element in the codomain (1).
• This relation is a function.
• Domain: {2, 5, 8, 11, 14, 17}
• Range: {1}

(ii)

• Every element in the domain (2, 4, 6, 8, 10, 12, 14) is paired with a unique element in the codomain (1, 2, 3, 4, 5, 6, 7).
• This relation is a function.
• Domain: 
• {2, 4, 6, 8, 10, 12, 14}
• Range: {1, 2, 3, 4, 5, 6, 7}

(iii) {(1,3), (1,5), (2,5)}

• The element 1 in the domain is paired with both 3 and 5 in the codomain.
• This relation is not a function.

2. Find the domain and range of the following real functions: (i) f(x) = – x 

(ii) f(x) = 2 9 − x .

Ans : 

(i) f(x) = -x

Domain:

• The function f(x) =
•  -x is defined for all real numbers.
• Therefore, the domain is R (all real numbers).

Range:

• The function f(x) = -x can take on any real number value.

(ii) f(x) = 2/(9-x)

Domain:

• The function is undefined when the denominator is zero.
• So, we need to find the values of x that make 9 – x = 0.
• 9 – x = 0 implies x = 9.
• Therefore, the domain is R except for x = 9, or in interval notation: (-∞, 9) U (9, ∞).

Range:

• The function is a rational function with a vertical asymptote at x = 9.
• As x approaches 9 from the left or right, the function approaches positive or negative infinity, respectively.
• The function can take on all real values except 0 (since the numerator is a constant).
• Therefore, the range is R except for y = 0, or in interval notation: (-∞, 0) U (0, ∞).

3. A function f is defined by f(x) = 2x –5. Write down the values of (i) f (0), (ii) f (7), (iii) f (–3).

Ans : 

To find the values of f(0), f(7), and f(-3), we simply substitute these values into the function f(x) = 2x – 5.

(i) f(0): f(0) = 2(0) – 5 = 0 – 5 = -5

(ii) f(7): f(7) = 2(7) – 5 = 14 – 5 = 9

(iii) f(-3): f(-3) = 2(-3) – 5 = -6 – 5 = -11

4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C) = 9C /5 + 32.

 Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212

Ans : 

The given function is:

t(C) = 9C/5 + 32

(i) t(0) = 9*0/5 + 32 = 0 + 32 = 32

(ii) t(28) = 9*28/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4

(iii) t(-10) = 9*(-10)/5 + 32 = -18 + 32 = 14

(iv) t(C) = 212 9C/5 + 32 = 212 9C/5 = 212 – 32 9C/5 = 180 C = 180*5/9 C = 100

Therefore, the values are:

(i) t(0) = 32 

(ii) t(28) = 82.4 

(iii) t(-10) = 14 

(iv) C = 100

5. Find the range of each of the following functions. 

(i) f (x) = 2 – 3x, x ∈ R, x > 0. (ii) f (x) = x 2 + 2, x is a real number. (iii) f (x) = x, x is a real number.

Ans :

(i) f(x) = 2 – 3x, x ∈ R, x > 0

Analysis:

• As x increases (since x > 0), -3x decreases.
• Therefore, 2 – 3x decreases as x increases.
• The maximum value of f(x) occurs when x is at its minimum (x = 0).

Range:

• f(x) is decreasing and has a maximum value of 2 when x = 0.
• Range = (-∞, 2]

(ii) f(x) = x^2 + 2, x is a real number

Analysis:

• For any real number x, x^2 is always non-negative (x^2 ≥ 0).
• So, x^2 + 2 is always greater than or equal to 2.

Range:

• Range = [2, ∞)

(iii) f(x) = x, x is a real number

Analysis:

• This function is simply the identity function.
• For any real number x, f(x) will also be that same real number.

Range:

• Range = R (all real numbers)