NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions

NCERT Solutons for class 11 Maths Chapter 2 Relations and Functions of Class provides detailed step-by-step solutions, ensuring that students can easiy understand the concept and score good marks in exam. It’s important to pay attention to the definitions and properties of relations and functions, as these are frequently tested in exams. This chapter contains 3 exercises along with a miscellaneous exercise.

Every student aims to score excellent marks in the Maths examinations. Practising the NCERT textual questions by referring to the NCERT Solutions of Class 11 Maths can undoubtedly help the students in fulfilling their aim of scoring high marks in the board examination.

NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions Exercise 2.2 (9 Questions)

NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions are based on Relations Exercise 2.2 contains the following topics.

1. A relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. The subset is derived by describing the relationship between the first element and the second element of the ordered pairs in A × B. The second element is called the image of the first element.
2. The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.
3. The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the codomain of the relation R. Note that range ⊂ codomain.
   1. A relation may be represented algebraically either by the Roster method or by the Set-builder method.
   2. An arrow diagram is a visual representation of a relation.

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.

Solution:

The relation R from A to A is given as

R = {(x, y): 3x – y = 0, where x, y ∈ A}

= {(x, y): 3x = y, where x, y ∈ A}

So,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Now,

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

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the relation R.

Hence, Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {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.

Solution:

The relation R is given by

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

The natural numbers less than 4 are 1, 2, and 3.

So,

R = {(1, 6), (2, 7), (3, 8)}

Now,

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

Hence, Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {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.

Solution:

Given,

A = {1, 2, 3, 5} and B = {4, 6, 9}

The relation from A to B is given as

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

Thus,

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

4. The figure shows a relationship between the sets P and Q. Write this relation

(i) in set-builder form (ii) in roster form.

What are their domain and range?

Solution:

From the given figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q

Set-builder form

(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}

Roster form

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {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.

Solution:

Given,

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

Hence,

(i) 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) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of 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}}.

Solution:

Given,

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

Thus,

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

So,

Domain of R = {0, 1, 2, 3, 4, 5} and

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.

Solution:

Given,

Relation R = {(x, x3): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

Therefore,

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.

Solution:

Given, A = {x, y, z} and B = {1, 2}

Now,

A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

As n(A × B) = 6, the number of subsets of A × B will be 26.

Thus, the number of relations from A to B is 26.

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.

Solution:

Given,

Relation R = {(a, b): a, b ∈ Z, a – b is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R = Z and Range of R = Z

NCERT Solutions For Class 11 Maths Chapter 2 Relations and Functions Exercise 2.3 (10 Questions)

NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions are based on Relations Exercise 2.3 contains the following topics.

1. Functions
   1. Some functions and their graphs
      • Identity function
      • Constant function
      • Polynomial function
      • Rational functions
      • The Modulus function
      • Signum function
      • Greatest integer function
   2. Algebra of real functions
      • Addition of two real functions
      • Subtraction of a real function from another
      • Multiplication by a scalar
      • Multiplication of two real functions
      • The quotient of two real functions

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)}

Solution:

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

As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

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

As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

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

It’s seen that the same first element, i.e., 1, corresponds to two different images, i.e., 3 and 5; this relation cannot be called a function.

2. Find the domain and range of the following real function.

(i) f(x) = –|x| (ii) f(x) = √(9 – x2)

Solution:

(i) Given,

f(x) = –|x|, x ∈ R

We know that

As f(x) is defined for x ∈ R, the domain of f is R.

It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.

Therefore, the range of f is given by (–∞, 0].

(ii) f(x) = √(9 – x2)

As √(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0

So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3]

Now,

For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.

Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3]

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)

Solution:

Given,

Function, f(x) = 2x – 5

Therefore,

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) 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.

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

Solution:

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

(i) f(x) = 2 – 3x, x ∈ R, x > 0

(ii) f(x) = x2 + 2, x is a real number.

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

Solution:

(i) Given,

f(x) = 2 – 3x, x ∈ R, x > 0

We have,

x > 0

So,

3x > 0

-3x < 0 [Multiplying by -1 on both sides, the inequality sign changes.]

2 – 3x < 2

Therefore, the value of 2 – 3x is less than 2.

Hence, Range = (–∞, 2)

(ii) Given,

f(x) = x2 + 2, x is a real number

We know that

x2 ≥ 0

So,

x2 + 2 ≥ 2 [Adding 2 on both sides]

Therefore, the value of x2 + 2 is always greater or equal to 2, for x is a real number.

Hence, Range = [2, ∞)

(iii) Given,

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

Clearly, the range of f is the set of all real numbers.

Thus,

Range of f = R