NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions 2026-27

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions 2026-27 provide clear, step-by-step answers to all questions given in the latest NCERT Mathematics textbook. These solutions help students understand the basic ideas of relations and functions in simple words and prepare well for school exams, assignments, and future competitive exams.

In this chapter, students learn how two sets can be connected through ordered pairs. The chapter begins with the Cartesian product of sets and explains how ordered pairs are formed. It then introduces relations, their meaning, and their representation. Students also study functions, domain, codomain, and range. Easy examples make it simpler to understand how every function is a relation, but every relation is not always a function.

The NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions explain each exercise question in a logical and student-friendly way. Important definitions, formulas, symbols, and methods are presented clearly so that learners can solve problems without confusion. The solutions are useful for quick revision because they follow the same order and method as the NCERT textbook.

By using these chapter solutions, students can improve their understanding of concepts, check their answers, correct mistakes, and learn the right steps for solving questions. They can also build a strong base for later topics such as inverse trigonometric functions, calculus, and algebra.

Download NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions PDF

Download NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions PDF to study the chapter easily anytime and anywhere. The PDF includes clear, step-by-step solutions to NCERT questions on Cartesian products, relations, functions, domain, codomain, and range. Students can use the Class 11 Maths Chapter 2 PDF for quick revision, homework help, and self-study during the 2026-27 academic session. It is a useful study resource for learners searching for accurate and easy NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions.

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions - Question with answers

Exercise 2.1

Question 1: If ((x/3) + 1, y - (2/3)) = (5/3, 1/3), find the values of x and y.

Solution:

Two ordered pairs are equal only when their corresponding components are equal.

Therefore:

(x/3) + 1 = 5/3

and

y - (2/3) = 1/3

Finding x:

(x/3) + 1 = 5/3

x/3 = (5/3) - 1

x/3 = (5/3) - (3/3)

x/3 = 2/3

x = 2

Finding y:

y - (2/3) = 1/3

y = (1/3) + (2/3)

y = 3/3

y = 1

Final answer: x = 2 and y = 1.

Question 2: A set A has 3 elements and B = {3, 4, 5}. Find the number of elements in A × B.

Solution:

The number of elements in a Cartesian product is found by multiplying the number of elements in the two sets.

n(A × B) = n(A) × n(B)

Here, n(A) = 3 and n(B) = 3.

Therefore:

n(A × B) = 3 × 3 = 9

Final answer: A × B has 9 elements.

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

Solution:

In G × H, the first component of every ordered pair comes from G and the second component comes from H.

G × H:

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

In H × G, the first component comes from H and the second component comes from G.

H × G:

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

Question 4: State whether each statement is true or false. Correct the statement when it is false.

(i): If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.

Solution:

The statement is false.

Each element of P must be paired with every element of Q.

Correct statement:

P × Q = {(m, n), (m, m), (n, 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), where x belongs to A and y belongs to B.

Solution:

The statement is true. Since both sets contain at least one element, at least one ordered pair can be formed.

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

Solution:

The statement is true.

The intersection of any set with the empty set is empty.

B ∩ ∅ = ∅

Therefore:

A × (B ∩ ∅) = A × ∅ = ∅

Question 5: If A = {-1, 1}, find A × A × A.

Solution:

Every ordered triple in A × A × A has the form (p, q, r), where p, q and r are selected from A.

Since A has 2 elements, the number of ordered triples is:

2 × 2 × 2 = 8

Therefore:

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)

}

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

Solution:

The first components of the ordered pairs belong to A, while the second components belong to B.

Final answer: A = {a, b} and B = {x, y}.

Question 7: Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify the following statements.

(i): A × (B ∩ C) = (A × B) ∩ (A × C)

Solution:

The sets B and C have no common element.

Therefore:

B ∩ C = ∅

A × (B ∩ C) = A × ∅ = ∅

Now calculate the sets on the right-hand side.

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)}

These Cartesian products have no common ordered pair. Hence:

(A × B) ∩ (A × C) = ∅

Thus, both sides are equal:

A × (B ∩ C) = (A × B) ∩ (A × C) = ∅

(ii) Verify that A × C is a subset of B × D.

Solution:

First calculate A × C:

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

Now calculate B × D:

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)

}

Every ordered pair of A × C is also present in B × D.

Therefore, A × C is a subset of B × D.

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

Solution:

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

Let:

p = (1, 3), q = (1, 4), r = (2, 3) and s = (2, 4).

The set A × B has 4 elements. A set with n elements has 2ⁿ subsets.

Number of subsets = 2⁴ = 16

Replacing p, q, r and s by the ordered pairs, the 16 subsets are:

∅,

{(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)}.

Question 9: Let A and B be sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2) and (z, 1) belong to A × B, find A and B, where x, y and z are distinct.

Solution:

The first components x, y and z belong to A. Since they are distinct and A has exactly 3 elements:

A = {x, y, z}

The second components 1 and 2 belong to B. Since B has exactly 2 elements:

B = {1, 2}

Final answer: A = {x, y, z} and B = {1, 2}.

Question 10: The Cartesian product A × A contains 9 elements, including (-1, 0) and (0, 1). Find A and the remaining elements of A × A.

Solution:

For a finite set A:

n(A × A) = n(A) × n(A) = [n(A)]²

Given:

[n(A)]² = 9

Therefore, n(A) = 3.

Since (-1, 0) belongs to A × A, both -1 and 0 belong to A.

Since (0, 1) belongs to A × A, both 0 and 1 belong to A.

Thus, A has the three elements -1, 0 and 1:

A = {-1, 0, 1}

Therefore:

A × A = {

(-1, -1), (-1, 0), (-1, 1),

(0, -1), (0, 0), (0, 1),

(1, -1), (1, 0), (1, 1)

}

The two given pairs are (-1, 0) and (0, 1). Hence, the remaining elements are:

{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)}

Exercise 2.2

Question 1: Let A = {1, 2, 3, ..., 14}. A relation R from A to A is defined by:

R = {(x, y): 3x - y = 0, where x and y belong to A}

Write the domain, codomain and range of R.

Solution:

The condition is:

3x - y = 0

Therefore, y = 3x.

For x ≥ 5, the value of 3x is greater than 14 and does not belong to A.

Therefore:

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

Domain: {1, 2, 3, 4}

Codomain: A = {1, 2, 3, ..., 14}

Range: {3, 6, 9, 12}

Question 2: A relation R on the set N of natural numbers is defined by:

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

Write R in roster form and find its domain and range.

Solution:

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

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

Domain: {1, 2, 3}

Range: {6, 7, 8}

Question 3: Let 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}

Write R in roster form.

Solution:

The difference of two integers is odd when one integer is odd and the other is even.

In A, the odd elements are 1, 3 and 5, while 2 is even.

In B, 4 and 6 are even, while 9 is odd.

Final answer:

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

Question 4: A relation between P = {5, 6, 7} and Q = {3, 4, 5} pairs each element x of P with x - 2 in Q.

(i) Write the relation in set-builder form.

Solution:

For every related pair, the second component is 2 less than the first component.

Therefore:

R = {(x, y): y = x - 2, x belongs to P and y belongs to Q}

(ii) Write the relation in roster form. Also find its domain and range.

Solution:

Roster form: R = {(5, 3), (6, 4), (7, 5)}

Domain: {5, 6, 7}

Range: {3, 4, 5}

Question 5: Let A = {1, 2, 3, 4, 6}. A relation R on A is defined by:

R = {(a, b): b is exactly divisible by a}

(i) Write R in roster form.

Solution:

Therefore:

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) Find the domain of R.

Solution:

Every element of A occurs as the first component of at least one ordered pair.

Domain of R: {1, 2, 3, 4, 6}

(iii) Find the range of R.

Solution:

Every element of A occurs as the second component of at least one ordered pair.

Range of R: {1, 2, 3, 4, 6}

Question 6: Find the domain and range of:

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

Solution:

Thus:

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

Domain: {0, 1, 2, 3, 4, 5}

Range: {5, 6, 7, 8, 9, 10}

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

Solution:

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

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

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

Solution:

A relation from A to B is any subset of A × B.

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

Thus, A × B contains 6 elements.

A set with 6 elements has 2⁶ subsets.

Number of relations = 2⁶ = 64

Final answer: There are 64 relations from A to B.

Question 9: Let R be a relation on Z defined by:

R = {(a, b): a and b are integers and a - b is an integer}

Find the domain and range of R.

Solution:

The difference of any two integers is always an integer. Therefore, every pair of integers satisfies the given condition.

This means R = Z × Z.

Domain of R: Z

Range of R: Z

Exercise 2.3

Question 1: Determine which of the following relations are functions. Give a reason and find the domain and range whenever the relation is a function.

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

Solution:

The first components are 2, 5, 8, 11, 14 and 17. Each one is paired with exactly one value.

Different inputs may have the same output in a function. Here, every input has the output 1.

Therefore, the relation is a function.

Domain: {2, 5, 8, 11, 14, 17}

Range: {1}

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

Solution:

Every first component occurs only once and has exactly one image.

Therefore, the 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)}

Solution:

The input 1 is paired with two different outputs, 3 and 5.

A function cannot assign two different images to the same input.

Therefore, this relation is not a function.

Question 2: Find the domain and range of the following real functions.

(i) f(x) = -|x|

Solution:

The absolute value |x| is defined for every real number. Therefore, the domain is R.

The function can be written as:

f(x) = -x when x ≥ 0

f(x) = x when x < 0

Since |x| is always non-negative, -|x| is always less than or equal to 0.

The maximum value is 0, obtained when x = 0. As |x| becomes larger, -|x| can become as negative as required.

Domain: R

Range: (-∞, 0]

(ii) f(x) = √(9 - x²)

Solution:

For a real square root, the expression inside the square root must be non-negative.

9 - x² ≥ 0

x² ≤ 9

-3 ≤ x ≤ 3

Therefore, the domain is [-3, 3].

The smallest value of f(x) is 0. It occurs when x = -3 or x = 3.

The largest value is:

f(0) = √(9 - 0) = 3

Domain: [-3, 3]

Range: [0, 3]

Question 3: A function is defined by f(x) = 2x - 5. Find the following values.

(i) Find f(0).

Solution:

Substitute x = 0:

f(0) = 2 × 0 - 5

f(0) = -5

(ii) Find f(7).

Solution:

Substitute x = 7:

f(7) = 2 × 7 - 5

f(7) = 14 - 5 = 9

(iii) Find f(-3).

Solution:

Substitute x = -3:

f(-3) = 2 × (-3) - 5

f(-3) = -6 - 5 = -11

Question 4: The function t converts a temperature in degrees Celsius into degrees Fahrenheit and is defined by:

t(C) = (9C/5) + 32

(i) Find t(0).

Solution:

t(0) = (9 × 0/5) + 32

t(0) = 32

Final answer: 0°C = 32°F.

(ii) Find t(28).

Solution:

t(28) = (9 × 28/5) + 32

t(28) = 252/5 + 32

t(28) = 50.4 + 32

t(28) = 82.4

Final answer: 28°C = 82.4°F.

(iii) Find t(-10).

Solution:

t(-10) = (9 × (-10)/5) + 32

t(-10) = -18 + 32

t(-10) = 14

Final answer: -10°C = 14°F.

(iv) Find C when t(C) = 212.

Solution:

Substitute t(C) = 212 into the conversion formula:

(9C/5) + 32 = 212

9C/5 = 212 - 32

9C/5 = 180

9C = 900

C = 900/9

C = 100

Final answer: 212°F = 100°C.

Question 5: Find the range of each function.

(i) f(x) = 2 - 3x, where x is a real number and x > 0.

Solution:

Since x > 0:

3x > 0

-3x < 0

2 - 3x < 2

Therefore, f(x) is always less than 2.

As x approaches 0 from the positive side, f(x) approaches 2, but it never becomes 2 because x cannot equal 0.

As x increases without limit, 2 - 3x decreases without limit.

Range: (-∞, 2)

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

Solution:

For every real number x:

x² ≥ 0

Therefore:

x² + 2 ≥ 2

Thus, f(x) ≥ 2.

The minimum value 2 occurs when x = 0. There is no upper limit because x² can become as large as required.

Range: [2, ∞)

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

Solution:

This is the identity function. Every real input is returned as the output.

Therefore, every real number occurs as a value of f.

Range: R

Miscellaneous Exercise

Question 1

The relation f is defined by:

f(x) = x² for 0 ≤ x ≤ 3

f(x) = 3x for 3 ≤ x ≤ 10

The relation g is defined by:

g(x) = x² for 0 ≤ x ≤ 2

g(x) = 3x for 2 ≤ x ≤ 10

Show that f is a function but g is not a function.

Solution:

For f, the two rules overlap only at x = 3.

Both rules give the same output at x = 3. Every input in the domain therefore has exactly one image. Hence, f is a function.

For g, the two rules overlap at x = 2.

The same input x = 2 has two different images, 4 and 6. Therefore, g is not a function.

Question 2

If f(x) = x², find:

(f(1.1) - f(1))/(1.1 - 1)

Solution:

First calculate the function values:

f(1.1) = (1.1)² = 1.21

f(1) = 1² = 1

Now substitute:

(f(1.1) - f(1))/(1.1 - 1)

= (1.21 - 1)/(0.1)

= 0.21/0.1

= 2.1

Final answer: 2.1

Question 3

Find the domain of:

f(x) = (x² + 2x + 1)/(x² - 8x + 12)

Solution:

A rational function is undefined when its denominator is zero.

Factor the denominator:

x² - 8x + 12

= x² - 6x - 2x + 12

= x(x - 6) - 2(x - 6)

= (x - 2)(x - 6)

The denominator becomes zero when:

x - 2 = 0 or x - 6 = 0

Therefore, x = 2 or x = 6.

These two values must be excluded.

Domain: R - {2, 6}

Question 4

Find the domain and range of the real function f(x) = √(x - 1).

Solution:

For the square root to be real, the expression inside it must be non-negative.

x - 1 ≥ 0

x ≥ 1

Therefore, the domain is [1, ∞).

A square root is always non-negative. The smallest value is 0, obtained when x = 1. As x increases, √(x - 1) can become as large as required.

Domain: [1, ∞)

Range: [0, ∞)

Question 5

Find the domain and range of the real function f(x) = |x - 1|.

Solution:

The absolute-value expression |x - 1| is defined for every real value of x.

Therefore, the domain is R.

An absolute value cannot be negative. Its minimum value is 0, which occurs when:

x - 1 = 0

x = 1

There is no maximum value.

Domain: R

Range: [0, ∞)

Question 6

Let f be the function from R to R defined by:

f(x) = x²/(1 + x²)

Determine the range of f.

Solution:

Since x² ≥ 0 and 1 + x² is always positive:

x²/(1 + x²) ≥ 0

Also:

f(x) = x²/(1 + x²)

f(x) = (1 + x² - 1)/(1 + x²)

f(x) = 1 - 1/(1 + x²)

Because 1/(1 + x²) is always positive, f(x) is always less than 1.

At x = 0:

f(0) = 0/(1 + 0) = 0

Thus, 0 is included, but 1 is not included.

To check that every number between 0 and 1 can occur, let:

y = x²/(1 + x²)

y + yx² = x²

y = x²(1 - y)

x² = y/(1 - y)

For every y satisfying 0 ≤ y < 1, the quantity y/(1 - y) is non-negative, so a real value of x exists.

Range: [0, 1)

Question 7

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

Solution:

Sum of the functions:

(f + g)(x) = f(x) + g(x)

= (x + 1) + (2x - 3)

= 3x - 2

Therefore, (f + g)(x) = 3x - 2 for all real x.

Difference of the functions:

(f - g)(x) = f(x) - g(x)

= (x + 1) - (2x - 3)

= x + 1 - 2x + 3

= 4 - x

Therefore, (f - g)(x) = 4 - x for all real x.

Quotient of the functions:

(f/g)(x) = f(x)/g(x)

= (x + 1)/(2x - 3)

The denominator must not be zero:

2x - 3 ≠ 0

2x ≠ 3

x ≠ 3/2

Therefore:

(f/g)(x) = (x + 1)/(2x - 3), where x ≠ 3/2.

Question 8

Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be defined by f(x) = ax + b, where a and b are integers. Determine a and b.

Solution:

Since (0, -1) belongs to f:

f(0) = -1

a × 0 + b = -1

b = -1

Since (1, 1) belongs to f:

f(1) = 1

a × 1 + b = 1

a + b = 1

Substitute b = -1:

a - 1 = 1

a = 2

The rule is therefore f(x) = 2x - 1.

A quick check gives:

Final answer: a = 2 and b = -1.

Question 9

Let R be a relation from N to N defined by:

R = {(a, b): a = b²}

Decide whether each statement is true or false and justify the answer.

(i) (a, a) belongs to R for every a belonging to N.

Solution:

For (a, a) to belong to R, the condition a = a² must hold.

This does not hold for every natural number. For example, take a = 2:

2² = 4 ≠ 2

Therefore, (2, 2) does not belong to R.

The statement is false.

(ii) If (a, b) belongs to R, then (b, a) also belongs to R.

Solution:

Take a = 9 and b = 3.

Since 9 = 3², the pair (9, 3) belongs to R.

For (3, 9) to belong to R, we would need:

3 = 9² = 81

This is not true.

Therefore, (3, 9) does not belong to R.

The statement is false.

(iii) If (a, b) belongs to R and (b, c) belongs to R, then (a, c) belongs to R.

Solution:

Consider a = 16, b = 4 and c = 2.

Since 16 = 4², (16, 4) belongs to R.

Since 4 = 2², (4, 2) belongs to R.

For (16, 2) to belong to R, we would need:

16 = 2² = 4

This is false.

Thus, (16, 2) does not belong to R.

The statement is false.

Question 10

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

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

Determine whether the following statements are true.

(i) f is a relation from A to B.

Solution:

A relation from A to B is any subset of A × B.

In every ordered pair of f, the first component belongs to A and the second component belongs to B.

Therefore, f is a subset of A × B.

The statement is true.

(ii) f is a function from A to B.

Solution:

For f to be a function, every element of A must have exactly one image in B.

The element 2 is paired with both 9 and 11:

(2, 9) belongs to f and (2, 11) belongs to f.

Thus, one input has two different outputs.

The statement is false. The relation f is not a function from A to B.

Question 11

Let f be the subset of Z × Z defined by:

f = {(ab, a + b): a and b belong to Z}

Is f a function from Z to Z? Justify the answer.

Solution:

A function must assign exactly one output to each input.

Take a = 2 and b = 6:

ab = 2 × 6 = 12

a + b = 2 + 6 = 8

Therefore, (12, 8) belongs to f.

Now take a = -2 and b = -6:

ab = (-2) × (-6) = 12

a + b = -2 + (-6) = -8

Therefore, (12, -8) belongs to f.

The same input 12 has two different outputs, 8 and -8.

Therefore, f is not a function from Z to Z.

Question 12

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

Solution:

The outputs are 3, 5, 11, 3 and 13. Repeated values are written only once in a set.

Range of f: {3, 5, 11, 13}

How to Solve NCERT Class 11 Relations and Functions Questions?

To solve NCERT Class 11 Relations and Functions questions, first understand the basic terms used in the chapter. These include ordered pairs, Cartesian product, relation, function, domain, codomain, and range. Learning these definitions will make the exercises easier.

1. Start every question by writing the given sets clearly. If the question asks for a Cartesian product, form all possible ordered pairs. Remember that the order of elements matters. For example, ((a,b)) and ((b,a)) are usually different ordered pairs.
2. While solving questions on relations, check which elements of one set are connected to the elements of another set. Write the relation in roster form or set-builder form, as required. To find the domain, write all the first elements of the ordered pairs. To find the range, write all the second elements that are actually used.
3. For function-based questions, check whether every element of the domain has exactly one image in the codomain. If one input has two different outputs, the relation is not a function.
4. Read each question carefully, solve it step by step, and compare your answer with the NCERT solution. Regular practice will help you avoid common mistakes and improve your understanding.

Best Study Guide for Class 11 Maths Chapter 2 2026-27

The best study guide for Class 11 Maths Chapter 2 Relations and Functions 2026-27 should include simple explanations, important definitions, solved examples, exercise-wise NCERT solutions, and practice questions.

• Begin by studying sets and ordered pairs because they form the base of the chapter. Next, understand Cartesian products and learn how to list ordered pairs correctly. After that, study relations, domain, codomain, and range with examples.
• Give special attention to functions. Learn the main condition of a function: every input must have only one output. Practice identifying whether a given relation is a function or not.
• Use diagrams, tables, and real-life examples to understand concepts better. Solve all NCERT examples before attempting the exercises. Revise important symbols and definitions regularly.
• A good study routine is to read the theory, solve examples, complete NCERT exercises, check mistakes, and then attempt extra questions. This method will help students build strong concepts, prepare for exams, and score better marks in Class 11 Maths.