Relations And Functions

NCERT Solutions for Class 11 Maths Chapter 2

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

NCERT Solutions for Class 11 Maths Chapter 2

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}

NCERT Solutions for Class 11 Maths Chapter 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)

NCERT Solutions for Class 11 Maths Chapter 2

FAQ’s

What is covered in Class 11 Maths Chapter 2?

Class 11 Maths Chapter 2 focuses on Relations and Functions, including types of relations, domain, range, and different types of functions.

Why is Relations and Functions important in Class 11 Maths?

It builds the foundation for advanced concepts in calculus, algebra, and higher-level mathematics in Class 12 and competitive exams.

How can NCERT solutions help in Chapter 2?

NCERT solutions provide step-by-step explanations that help students understand definitions, solve problems easily, and prepare better for exams.

What are the key topics in Relations and Functions?

Key topics include ordered pairs, Cartesian products, relations, functions, domain, range, and types of functions.

Where can I find free solutions for Class 11 Maths Chapter 2?

You can find free, detailed solutions for Class 11 Maths Chapter 2 – Relations and Functions – on trusted educational websites and learning platforms.