Important Questions Class 11 Maths Chapter 2 Relations and Functions

A relation connects ordered pairs of elements from two sets through a defined rule. A function is a relation where every domain element has exactly one image.

Pairs become meaningful in mathematics only when their order and rule are clear. Important Questions Class 11 Maths Chapter 2 help students practise Cartesian products, ordered pairs, relations, domain, range, codomain, functions, real functions, standard function graphs, and algebra of real functions. The 2026 NCERT chapter Relations and Functions builds the base for coordinate geometry, calculus, mappings, and mathematical modelling.

Key Takeaways

• Cartesian Product: A × B contains all ordered pairs (a, b) where a ∈ A and b ∈ B.
• Number Of Ordered Pairs: If n(A) = p and n(B) = q, then n(A × B) = pq.
• Relation: A relation from A to B is any subset of A × B.
• Function: A function maps every element of its domain to one and only one image.

Important Questions Class 11 Maths Chapter 2 Structure 2026

Relations and Functions Class 11 Chapter Overview

A relation links objects in a fixed order. NCERT begins with colour-object pairs and vehicle code pairs to show why order matters.

A function is a special relation with one image for every domain element. The chapter also covers real functions, standard graphs, and operations on functions.

Q1. What Are Relations and Functions Class 11?

Relations and Functions Class 11 studies ordered pairs, set mappings, and special relations called functions.

A relation connects elements from two sets through ordered pairs. A function gives every input exactly one output.

Example: f(x) = 2x + 1 gives one value for each natural number x.

Q2. Why Are Important Questions Class 11 Maths Chapter 2 Useful For Exams?

Important Questions Class 11 Maths Chapter 2 are useful because the chapter tests definitions, set notation, and function checks.

Students must identify domains, ranges, codomains, Cartesian products, and valid functions. Many NCERT questions require roster form and set-builder form.

Example: R = {(x, y): y = x + 1} can be written in roster form.

Q3. What Is The Difference Between A Relation And A Function?

A relation may give one element many images, but a function gives each domain element exactly one image.

Every function is a relation. Every relation need not be a function.

Example: {(2, 2), (2, 4)} is not a function because 2 has two images.

Cartesian Product Class 11 Important Questions

Cartesian products create ordered pairs from two sets. The first element must come from the first set.

The order of sets matters because (a, b) and (b, a) are different ordered pairs.

Q4. What Is Cartesian Product Class 11?

The Cartesian product of two non-empty sets P and Q is the set of all ordered pairs (p, q).

Here, p ∈ P and q ∈ Q. It is written as P × Q.

Formula:
P × Q = {(p, q): p ∈ P, q ∈ Q}

Q5. If A = {red, blue} And B = {b, c, s}, Find A × B.

A × B has 6 ordered pairs.

1. Given Data:
   A = {red, blue}
   B = {b, c, s}
2. Formula Used:
   A × B = {(a, b): a ∈ A, b ∈ B}
3. Write all pairs:
   A × B = {(red, b), (red, c), (red, s), (blue, b), (blue, c), (blue, s)}

Final Result: A × B has 6 ordered pairs

Q6. If P = {a, b, c} And Q = {r}, Find P × Q And Q × P.

P × Q and Q × P have the same number of elements, but they are not equal.

1. Given Data:
   P = {a, b, c}
   Q = {r}
2. Cartesian product:
   P × Q = {(a, r), (b, r), (c, r)}
3. Reverse product:
   Q × P = {(r, a), (r, b), (r, c)}
4. Since (a, r) ≠ (r, a):
   P × Q ≠ Q × P

Final Result: P × Q ≠ Q × P

Q7. If n(A) = 3 And n(B) = 2, Find n(A × B).

The value of n(A × B) is 6.

1. Given Data:
   n(A) = 3
   n(B) = 2
2. Formula Used:
   n(A × B) = n(A) × n(B)
3. Calculation:
   n(A × B) = 3 × 2 = 6

Final Result: 6 ordered pairs

Q8. What Happens If A Or B Is An Empty Set In A × B?

If either A or B is empty, then A × B = φ.

A Cartesian product needs one element from each set. If one set has no element, no ordered pair can form.

Example: A × φ = φ.

Ordered Pairs Class 11 Questions

An ordered pair keeps elements in a fixed order. Equality of ordered pairs depends on matching first and second components.

This rule helps solve unknown values inside pairs.

Q9. When Are Two Ordered Pairs Equal?

Two ordered pairs are equal when their corresponding first elements and second elements are equal.

If (a, b) = (x, y), then a = x and b = y. Both equalities must hold.

Example: (2, 5) ≠ (5, 2).

Q10. If (x + 1, y − 2) = (3, 1), Find x And y.

The values are x = 2 and y = 3.

1. Given Data:
   (x + 1, y − 2) = (3, 1)
2. Equate first components:
   x + 1 = 3
   x = 2
3. Equate second components:
   y − 2 = 1
   y = 3

Final Result: x = 2, y = 3

Q11. If ((2x + 5)/3, y − 1/3) = (1/3, 1/3), Find x And y.

The values are x = −2 and y = 2/3.

1. Equate first components:
   (2x + 5)/3 = 1/3
2. Solve for x:
   2x + 5 = 1
   2x = −4
   x = −2
3. Equate second components:
   y − 1/3 = 1/3
4. Solve for y:
   y = 2/3

Final Result: x = −2, y = 2/3

Q12. If A = {−1, 1}, Find A × A × A.

A × A × A has 8 ordered triplets.

1. Given Data:
   A = {−1, 1}
2. Formula Used:
   n(A × A × A) = 2 × 2 × 2 = 8
3. Roster form:
   {(−1, −1, −1), (−1, −1, 1), (−1, 1, −1), (−1, 1, 1), (1, −1, −1), (1, −1, 1), (1, 1, −1), (1, 1, 1)}

Final Result: 8 ordered triplets

Important Questions Class 11 Maths Chapter 2 On Cartesian Product Properties

Cartesian products also interact with union, intersection, and subsets. NCERT uses these forms to test set operations with ordered pairs.

Students should form the needed product before comparing sets.

Q13. If A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, Find A × (B ∩ C).

The set A × (B ∩ C) is {(1, 4), (2, 4), (3, 4)}.

1. Find intersection:
   B ∩ C = {4}
2. Form product:
   A × (B ∩ C) = A × {4}
3. Write ordered pairs:
   {(1, 4), (2, 4), (3, 4)}

Final Result: {(1, 4), (2, 4), (3, 4)}

Q14. Verify A × (B ∩ C) = (A × B) ∩ (A × C) For A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}.

Both sides are empty sets.

1. Find B ∩ C:
   B = {1, 2, 3, 4}
   C = {5, 6}
   B ∩ C = φ
2. Left side:
   A × (B ∩ C) = A × φ = φ
3. Since B and C have no common element:
   (A × B) ∩ (A × C) = φ

Final Result: Both sides are φ

Q15. If A × B = {(a, x), (a, y), (b, x), (b, y)}, Find A And B.

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

1. First elements form A:
   a, b
2. Second elements form B:
   x, y
3. Therefore:
   A = {a, b}
   B = {x, y}

Final Result: A = {a, b}, B = {x, y}

Q16. If A × B Has 4 Elements, How Many Subsets Will A × B Have?

A × B will have 16 subsets.

1. Given Data:
   n(A × B) = 4
2. Formula Used:
   Number of subsets of a set with n elements = 2ⁿ
3. Calculation:
   2⁴ = 16

Final Result: 16 subsets

Relations Class 11 Maths Important Questions

A relation selects some ordered pairs from a Cartesian product. The selected pairs must follow a stated relationship.

Relations can appear in roster form, set-builder form, or arrow diagrams.

Q17. What Is A Relation Class 11 Maths?

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

It connects the first element of an ordered pair with the second element through a rule. The second element is called the image.

Example: R = {(x, y): y = x + 1}.

Q18. If A = {1, 2, 3, 4, 5, 6} And R = {(x, y): y = x + 1}, Write R In Roster Form.

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

1. Given Data:
   A = {1, 2, 3, 4, 5, 6}
2. Relation rule:
   y = x + 1
3. Valid ordered pairs:
   (1, 2), (2, 3), (3, 4), (4, 5), (5, 6)

Final Result: R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Q19. Let A = {1, 2, 3, ..., 14}. R = {(x, y): 3x − y = 0}. Find Domain And Range.

The domain is {1, 2, 3, 4} and range is {3, 6, 9, 12}.

1. Relation rule:
   3x − y = 0
   y = 3x
2. Since y ∈ A, y must be at most 14.
3. Possible x-values:
   x = 1, 2, 3, 4
4. Corresponding y-values:
   3, 6, 9, 12

Final Result: Domain = {1, 2, 3, 4}, Range = {3, 6, 9, 12}

Q20. Write Roster Form Of R = {(x, y): y = x + 5, x < 4, x ∈ N}.

The roster form is {(1, 6), (2, 7), (3, 8)}.

1. Given condition:
   x ∈ N and x < 4
2. Possible x-values:
   1, 2, 3
3. Use rule y = x + 5:
   y = 6, 7, 8

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

Domain And Range Class 11 Questions

Domain and range come directly from ordered pairs in a relation. The codomain is the full second set given in the mapping.

NCERT clearly states that range is a subset of codomain.

Q21. What Is Domain And Range Class 11?

Domain is the set of first elements, and range is the set of second elements in a relation.

The codomain is the full target set. The range contains only actual images.

Example: For {(1, 2), (2, 3)}, domain is {1, 2} and range is {2, 3}.

Q22. What Is Codomain Class 11 Maths?

Codomain is the full set into which the relation or function maps its elements.

Range may be smaller than codomain. Every range element belongs to the codomain.

Example: If R is from A to B, then B is the codomain.

Q23. For R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}, Find Domain, Range And Codomain If R Is From A To A.

Domain = {1, 2, 3, 4, 5}, range = {2, 3, 4, 5, 6}, and codomain = A.

1. First elements:
   1, 2, 3, 4, 5
2. Second elements:
   2, 3, 4, 5, 6
3. Since relation is from A to A:
   Codomain = A

Final Result: Domain = {1, 2, 3, 4, 5}; Range = {2, 3, 4, 5, 6}

Q24. If R = {(9, 3), (9, −3), (4, 2), (4, −2), (25, 5), (25, −5)}, Find Domain And Range.

The domain is {4, 9, 25}, and the range is {−5, −3, −2, 2, 3, 5}.

1. Domain from first elements:
   9, 4, 25
2. Range from second elements:
   3, −3, 2, −2, 5, −5
3. Remove repeated values.

Final Result: Domain = {4, 9, 25}, Range = {−5, −3, −2, 2, 3, 5}

Functions Class 11 Maths Important Questions

A function is stricter than a relation. Every element of the domain must have exactly one image.

If one input has two outputs, the relation fails the function test.

Q25. What Is A Function Class 11 Maths?

A function from A to B is a relation where every element of A has one and only one image in B.

It is written as f: A → B. If (a, b) ∈ f, then f(a) = b.

Example: f(x) = 2x + 1 gives one value for each x.

Q26. Is R = {(2, 1), (3, 1), (4, 2)} A Function?

Yes, R is a function.

1. First elements:
   2, 3, 4
2. Each first element appears once.
3. Each input has exactly one image.

Final Result: R is a function

Q27. Is R = {(2, 2), (2, 4), (3, 3), (4, 4)} A Function?

No, R is not a function.

1. First element 2 appears in two pairs:
   (2, 2) and (2, 4)
2. Input 2 has two images:
   2 and 4
3. A function cannot assign two images to one input.

Final Result: R is not a function

Q28. Is R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)} A Function?

Yes, R is a function.

1. First elements are:
   1, 2, 3, 4, 5, 6
2. Every first element has one image.
3. No first element repeats with a different image.

Final Result: R is a function

Q29. Why Is The Relation y = 2x On N A Function?

The relation y = 2x on N is a function because every natural number has one image.

For each x ∈ N, the value 2x is fixed. The range is the set of even natural numbers.

Example: 1 → 2, 2 → 4, 3 → 6.

Class 11 Maths Chapter 2 Relations and Functions Questions On Number Of Relations

The number of relations depends on the number of ordered pairs in A × B. Every subset of A × B defines a relation.

This counting idea often appears in one-mark and short-answer questions.

Q30. How Many Relations Are Possible From A To B If n(A) = p And n(B) = q?

The number of relations from A to B is 2^(pq).

Since n(A × B) = pq, every subset of A × B is a relation. A set with pq elements has 2^(pq) subsets.

Formula:
Number of relations = 2^(pq)

Q31. Let A = {1, 2} And B = {3, 4}. Find The Number Of Relations From A To B.

The number of relations from A to B is 16.

1. Find Cartesian product:
   A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
2. Number of ordered pairs:
   n(A × B) = 4
3. Number of relations:
   2⁴ = 16

Final Result: 16 relations

Q32. Let A = {x, y, z} And B = {1, 2}. Find The Number Of Relations From A To B.

The number of relations from A to B is 64.

1. Given Data:
   n(A) = 3
   n(B) = 2
2. Find ordered pairs:
   n(A × B) = 3 × 2 = 6
3. Number of relations:
   2⁶ = 64

Final Result: 64 relations

Real Function Class 11 And Standard Graph Questions

Real functions use real numbers or their subsets as domain and range. NCERT includes identity, constant, polynomial, rational, modulus, signum, and greatest integer functions.

These graphs help students recognise function behaviour before calculus.

Q33. What Is A Real Function Class 11?

A real function has domain and range as R or subsets of R.

It assigns real inputs to real outputs. NCERT calls it a real function when both domain and range are real-number sets or subsets.

Example: f(x) = x² is a real function from R to R.

Q34. What Is Identity Function Class 11?

The identity function is defined by f(x) = x for every real number x.

Its domain and range are both R. Its graph is a straight line passing through the origin.

Example: f(−3) = −3.

Q35. What Is Constant Function Class 11?

A constant function is defined by f(x) = c for every real number x.

Its domain is R and range is {c}. Its graph is a horizontal line parallel to the x-axis.

Example: f(x) = 3 has range {3}.

Q36. What Is Modulus Function Class 11?

The modulus function is defined by f(x) = |x|.

For x ≥ 0, f(x) = x. For x < 0, f(x) = −x.

Example: f(−5) = 5.

Q37. What Is Signum Function Class 11?

The signum function gives −1, 0, or 1 based on the sign of x.

It is defined as f(x) = 1 if x > 0, f(x) = 0 if x = 0, and f(x) = −1 if x < 0.

Example: f(−4) = −1.

Q38. What Is Greatest Integer Function Class 11?

The greatest integer function is defined by f(x) = [x].

It gives the greatest integer less than or equal to x. Its graph has step-like horizontal pieces.

Example: [2.7] = 2 and [−1.2] = −2.

Algebra Of Real Functions Class 11 Questions

Algebra of functions combines two real functions using addition, subtraction, multiplication, scalar multiplication, and division. The input set must support the operation.

For quotient functions, the denominator function cannot be zero.

Q39. If f(x) = x² And g(x) = 2x + 1, Find (f + g)(x), (f − g)(x) And (fg)(x).

The results are x² + 2x + 1, x² − 2x − 1, and 2x³ + x².

1. Given Data:
   f(x) = x²
   g(x) = 2x + 1
2. Addition:
   (f + g)(x) = f(x) + g(x)
   = x² + 2x + 1
3. Subtraction:
   (f − g)(x) = f(x) − g(x)
   = x² − 2x − 1
4. Product:
   (fg)(x) = x²(2x + 1)
   = 2x³ + x²

Final Result: x² + 2x + 1; x² − 2x − 1; 2x³ + x²

Q40. If f(x) = x² And g(x) = 2x + 1, Find (f/g)(x).

The quotient is x²/(2x + 1), where x ≠ −1/2.

1. Given Data:
   f(x) = x²
   g(x) = 2x + 1
2. Formula Used:
   (f/g)(x) = f(x)/g(x)
3. Substitute:
   (f/g)(x) = x²/(2x + 1)
4. Restriction:
   2x + 1 ≠ 0
   x ≠ −1/2

Final Result: x²/(2x + 1), x ≠ −1/2

Q41. If f(x) = x + 1 And g(x) = 2x − 3, Find f + g, f − g And f/g.

The results are 3x − 2, 4 − x, and (x + 1)/(2x − 3).

1. Given Data:
   f(x) = x + 1
   g(x) = 2x − 3
2. Addition:
   (f + g)(x) = x + 1 + 2x − 3 = 3x − 2
3. Subtraction:
   (f − g)(x) = x + 1 − (2x − 3) = 4 − x
4. Quotient:
   (f/g)(x) = (x + 1)/(2x − 3)
5. Restriction:
   2x − 3 ≠ 0, so x ≠ 3/2

Final Result: 3x − 2; 4 − x; (x + 1)/(2x − 3), x ≠ 3/2

NCERT Class 11 Maths Chapter 2 Questions On Domain And Range

Domain restrictions usually come from denominators, square roots, and given intervals. Range depends on the output values.

NCERT asks students to find domain and range of real functions using these basic restrictions.

Q42. Find The Domain Of f(x) = (x² + 3x + 5)/(x² − 5x + 4).

The domain is R − {1, 4}.

1. Denominator:
   x² − 5x + 4
2. Factorise:
   x² − 5x + 4 = (x − 1)(x − 4)
3. Denominator cannot be zero:
   x ≠ 1 and x ≠ 4

Final Result: Domain = R − {1, 4}

Q43. Find The Domain And Range Of f(x) = √(9 − x²).

The domain is [−3, 3], and range is [0, 3].

1. Square root condition:
   9 − x² ≥ 0
2. Solve:
   x² ≤ 9
   −3 ≤ x ≤ 3
3. Since square root values are non-negative:
   Minimum value = 0
   Maximum value = 3

Final Result: Domain = [−3, 3], Range = [0, 3]

Q44. Find The Range Of f(x) = x² + 2, Where x Is Real.

The range is [2, ∞).

1. Since x is real:
   x² ≥ 0
2. Add 2:
   x² + 2 ≥ 2
3. Minimum value occurs at x = 0:
   f(0) = 2

Final Result: Range = [2, ∞)

Q45. If f = {(1, 1), (2, 3), (0, −1), (−1, −3)} Is A Linear Function, Find f(x).

The function is f(x) = 2x − 1.

1. Let linear function be:
   f(x) = mx + c
2. Use point (0, −1):
   f(0) = c = −1
3. Use point (1, 1):
   f(1) = m + c = 1
4. Substitute c = −1:
   m − 1 = 1
   m = 2

Final Result: f(x) = 2x − 1

Important Questions Class 11 Maths Chapter-Wise