**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)**