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# Relations and Functions NCERT Solutions Class 11 Maths Chapter 2 Exercise 2.1 2.2 2.3 Miscellaneous Free pdf Study Material Notes download-Anand Classes

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions thoroughly covers the most fundamental concepts used in algebra. A relation is a set of ordered pairs, whereas a function expresses the correspondence between input and output.

NCERT Solutions Class 11 Maths Chapter 2 comprehensively explains all the key concepts related to relations and functions with examples. With the help of these solutions, students will easily acquire detailed knowledge of this topic and its practical applications.

# , find the values of x and y.

Solution:

Given,

As the ordered pairs are equal, the corresponding elements should also be equal.

Thus,

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]

x = 2 and 3y = 3

Therefore, x = 2 and y = 1

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

Solution:

Given, set A has 3 elements, and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B) = 3 × 3 = 9

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

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

Solution:

Given, G = {7, 8} and H = {5, 4, 2}

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P × Q = {(pq): ∈ P, q ∈ Q}

So,

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

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

Solution:

(i) The statement is false. The correct statement is

If P = {mn} and Q = {nm}, then

P × Q = {(mm), (mn), (n, m), (nn)}

(ii) True

(iii) True

# 5. If A = {–1, 1}, find A × A × A.

Solution:

The A × A × A for a non-empty set A is given by

A × A × A = {(abc): ab∈ A}

Here, it is given A = {–1, 1}

So,

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.

Solution:

Given,

A × B = {(ax), (a, y), (bx), (by)}

We know that the Cartesian product of two non-empty sets, P and Q is given by:

P × Q = {(pq): p ∈ P, q ∈ Q}

Hence, A is the set of all first elements, and B is the set of all second elements.

Therefore, A = {ab} and B = {xy}

# (ii) A × C is a subset of B × D

Solution:

Given,

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

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

Thus,

L.H.S. = A × (B ∩ C) = A × Φ = Φ

Next,

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

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

Thus,

R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) To verify: A × C is a subset of B × D

First,

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

And,

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

Hence verified

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

Solution:

Given,

A = {1, 2} and B = {3, 4}

So,

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

Number of elements in A × B is n(A × B) = 4

We know that,

If C is a set with n(C) = m, then n[P(C)] = 2m.

Thus, the set A × B has 24 = 16 subsets.

And these subsets are as given below:

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(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.

Solution:

Given,

n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that,

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B

So, clearly, xy, and z are the elements of A; and

1 and 2 are the elements of B.

As n(A) = 3 and n(B) = 2, it is clear that set A = {xyz} and set 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.

Solution:

We know that,

If n(A) = and n(B) = q, then n(A × B) = pq.

Also, n(A × A) = n(A) × n(A)

Given,

n(A × A) = 9

So, n(A) × n(A) = 9

Thus, n(A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(a, a): a ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As n(A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

# 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 = {(xy): 3x – y = 0, where xy ∈ A}

= {(xy): 3x = y, where xy ∈ 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 = {(xy): y = x + 5, x is a natural number less than 4, xy ∈ 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 = {(xy): the difference between x and y is odd; x ∈ A, ∈ B}

Thus,

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

# What is its 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}

# (iii) Find the range of R

Solution:

Given,

A = {1, 2, 3, 4, 6} and relation R = {(ab): ab ∈ 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 = {(xx + 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 = {(xx3): 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 = {xy, 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 = {(ab): ab ∈ Z, – 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

# (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.

# (ii) f(x) = √(9 – x2)

Solution:

(i)

Given,

Since | x | is defined i.e. is real and finite for all real numbers, – | x | is also defined for all real numbers.
∴ Domain of f = R
Also, for all x ∈ R, | x | ≥ 0
⇒ – | x | ≤ 0 ⇒ f (x) ≤ 0
∴ Range of f = (– ∞, 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].

# (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:

Given,

(i) Replacing C by 0, we have

# (iii) f(x) = x, x is a real number

Solution:

(i)

Given,  f(x) = 2 – 3xx ∈ 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

# Show that f is a function and g is not a function.

Solution:

The given relation f is defined as:

It is seen that for 0 ≤ x < 3,

f(x) = xand for 3 < x ≤ 10,

f(x) = 3x

Also, at x = 3

f(x) = 32 = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9 [Single image]

Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Therefore, the given relation is a function.

Now,

In the given relation, g is defined as

It is seen that, for x = 2

g(x) = 22 = 4 and g(x) = 3 × 2 = 6

Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.

Therefore, this relation is not a function.

Solution:

Given,

f(x) = x2

Hence,

# 3. Find the domain of the function

Solution:

Given function, .f is a rational function.
∴ Domain of f = R – { x : x2 – 8x + 12 = 0 }

∴ Domain of f = R – { x : (x – 2)(x – 6) = 0 }

It’s clearly seen that the function f is defined for all real numbers except at x = 6 and x = 2, as the denominator becomes zero otherwise.

∴ Domain of f = R – { 2, 6 }.

Therefore, the domain of f is R – {2, 6}.

# 4. Find the domain and the range of the real function f defined by f(x) = √(x – 1).

Solution:

Given real function,

f(x) = √(x – 1)

Clearly, √(x – 1) is defined for (x – 1) ≥ 0

So, the function f(x) = √(x – 1) is defined for x ≥ 1

Thus, the domain of f is the set of all real numbers greater than or equal to 1.

Domain of f = [1, ∞)

Now,

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0

Thus, the range of f is the set of all real numbers greater than or equal to 0.

Range of f = [0, ∞)

# 5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.

Solution:

Given a real function,

f (x) = |x – 1|

Clearly, the function |x – 1| is defined for all real numbers.

Hence,

Domain of f = R

Also, for x ∈ R, |x – 1| assumes all real numbers.

Therefore, the range of f is the set of all non-negative real numbers.

# be a function from R into R. Determine the range of f.

Solution:

Given function,

We know that, for x ∈ R,

x≥ 0

Then,

x2 + 1 ≥ x2

1 ≥ x/ (x+ 1)

Therefore, the range of f = [0, 1)

# 7. Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.

Solution:

Given the functions fg: R → R is defined as

f(x) = + 1, g(x) = 2x – 3

Now,

(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

Thus, (f + g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

Thus, (f – g) (x) = –x + 4

f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R

f/g(x) = + 1/ 2x – 3, 2x – 3 ≠ 0

Thus, f/g(x) = + 1/ 2x – 3, x ≠ 3/2

# 8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Solution:

Given, = {(1, 1), (2, 3), (0, –1), (–1, –3)}

And the function defined as,  f(x) = ax + b

For (1, 1) ∈ f

We have,  f(1) = 1

So, a × 1 + b = 1

a + b = 1 …. (i)

And for (0, –1) ∈ f

We have f(0) = –1

a × 0 + b = –1

b = –1

On substituting b = –1 in (i), we get

a + (–1) = 1 ⇒ a = 1 + 1 = 2.

Therefore, the values of a and b are 2 and –1, respectively.

# (iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R

Solution:

Given relation R = {(ab): ab ∈ N and a = b2}

(i) It can be seen that 2 ∈ N; however, 2 ≠ 22 = 4.

Thus, the statement “(aa) ∈ R, for all a ∈ N” is not true.

(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

Thus, the statement “(ab) ∈ R, implies (ba) ∈ R” is not true.

(iii) It’s clearly seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.

Now, 16 ≠ 22 = 4; therefore, (16, 2) ∉ N

Thus, the statement “(ab) ∈ R, (bc) ∈ R implies (ac) ∈ R” is not true.

# (i) f is a relation from A to B (ii) f is a function from A to B

Solution:

Given,

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

So,

A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}

Also, given that,

= {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It’s clearly seen that f is a subset of A × B.

Therefore, f is a relation from A to B.

(ii) As the same first element, i.e., 2 corresponds to two different images (9 and 11), relation is not a function.

# 11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.

Solution:

Given relation, f is defined as

= {(aba + b): ab ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.

As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It’s clearly seen that the same first element, 12, corresponds to two different images (8 and –8).

Therefore, the relation f is not a function.

# 12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Solution:

Given,

A = {9, 10, 11, 12, 13}

Now, f: A → N is defined as

f(n) = The highest prime factor of n

So,

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

Thus, it can be expressed as

f(9) = The highest prime factor of 9 = 3

f(10) = The highest prime factor of 10 = 5

f(11) = The highest prime factor of 11 = 11

f(12) = The highest prime factor of 12 = 3

f(13) = The highest prime factor of 13 = 13

The range of f is the set of all f(n), where n ∈ A.

Therefore,

Range of f = {3, 5, 11, 13}

# Important topics and subtopics of NCERT Solutions for Class 11 Maths Chapter 2 – Relations and Functions

The following ideas from Chapter 2 Relations and Functions for Class 11, are given elaborately.

### 2.1 Introduction

This section introduces the concepts covered in the chapter Relations and Functions.

The combination of the enrolled number of the student and their corresponding marks is a relationship, which can be written as a set of ordered-pair numbers. Ordered-pair numbers are expressed as (x, y). The set of all elements of x is called the domain of the relation, and the set of all elements of y is called the range of the relation.

### 2.2 Cartesian Product of Sets

This section defines the Cartesian product and ordered pairs by giving a real-life model, its representation, and some worked examples.

Lindt chocolates come in five shapes, three flavours and six colours.

C :={circle, triangle, rectangle, rhombus, square}

N :={orange, vanilla, peach}

S :={red, blue, pink, white, yellow, purple}

C:={circle, triangle, rectangle, rhombus, square}, N:={orange, vanilla, peach}, S:={red, blue, pink, white, yellow, purple}

be the five shapes, three flavours and six colours, respectively. Then the set of all chocolates to be manufactured in the triple Cartesian product C×N×S and consists of 5⋅3⋅6=90 elements. As a manager, to sell this set of chocolates would have to make room for 90 heaps.

### 2.3 Relations

This section explains the mapping of set A to set B with a few solved problems. Definitions of domain and codomain are also introduced.

The idea of mapping a particular phone number to the respective person to whom the number belongs. That’s a relation – from phone number to person.

### 2.4 Functions

This section covers functions, the visualisation of functions, and how a relation is said to be a function, with a few examples. Meaning of image and preimage.

The height of a person can be determined by the length of his femur bone. Hence, it is an example of a function.

### 2.4.1 Some functions and their graphs

This section talks about different types of functions and their graphical representations. Some of the types of functions are listed below.

• Identity function
• Constant function
• Polynomial function
• Rational functions
• The Modulus function
• Signum function
• Greatest integer function

### 2.4.2 Algebra of real functions

This section includes the algebraic operations on 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

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

• The two elements grouped in a particular order are called an ordered pair.
• Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}
• Relation R from a set A to a set B is a subset of the cartesian product A × B obtained by explaining the relationship between the first element x and the second element y of the ordered pairs in A × B.
• The image of an element x under a relation R is given by y, where (x, y) ∈ R.
• The domain of R is the set of all first elements of the ordered pairs in a relation R.
• The range of a relation R is the set of all second elements of the ordered pairs in a relation R.
• Function A from a set A to a set B is a specific type of relation for which every element x of set A has one and only one image y in set B. We write f: A→B, where f(x) = y.
• The range of the function is the set of images.
• A real function has a set of real numbers or one of its subsets both as to its domain and as its range.

Basic concepts covered in the NCERT Solutions for Class 11 Maths modules aid students in moving ahead in their studies. The latest update of the CBSE Syllabus ensures that the content covered is apt for the students to move ahead in their respective streams in the future. A student needs to understand the concept of Relations and Functions as it covers the main part of the question paper. Before solving real-world applications and problems, the concept has to be learned thoroughly.

# Frequently Asked Questions (FAQs) on NCERT Solutions for Class 11 Maths Chapter 2

Q1

### How to find which relation is a function in Chapter 2 of NCERT Solutions for Class 11 Maths?

According to the definition, a function can relate every element which is present in a domain to only one element, which is found in the range. It means that any vertical line drawn by a student on a graph can pass through the x-axis only once. A relation from a function can be found by using vertical line tests or with the help of different formulas.
Q2

### Explain the basic steps for the Cartesian product of sets in NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions.

In order to understand the basic steps for solving a question regarding the Cartesian product of sets, students must comprehend the first exercise of the chapter thoroughly. Students are provided with solved examples before each exercise-wise problem to help them understand the method of solving problems in a shorter duration. By solving the problems from the NCERT textbook, students will improve their conceptual understanding, which is necessary to perform well in the exams.
Q3

### What is the meaning of relations in Chapter 2 of NCERT Solutions for Class 11 Maths?

Relations are nothing but the collection of ordered pairs which has one object from every set. A function can also be considered as a relation, but the conceptual ideas of both of them are completely different. The NCERT Solutions for Class 11 Maths Chapter 2 provides the students with a proper definition and analysis of relations as per the CBSE Syllabus. Several examples present in the solutions will help students solve problems related to relations without difficulty.

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