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Class 11 Maths NCERT Solutions

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Relations and Functions Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 2, Relations and Functions — 24 questions from Ex 2.1 to Ex 2.3, each worked through step by step in the CBSE marking pattern. Cartesian products, relations, domain and range, and one-one and onto functions.

Class:11Subject:MathsChapter:2
3 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 2?

Chapter 2 carries 3 exercise questions, numbered Ex 2.1 to Ex 2.3. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

This chapter introduces ordered pairs, Cartesian products, relations between two sets, and functions as special relations where each input has exactly one output. You will practise writing relations in roster and set-builder form, finding domains and ranges, and classifying functions. Every question below is from the NCERT Class 11 textbook (rationalised edition), worked line by line.

Board pattern

A relation from A to B is any subset of A × B; a function is a relation in which no first coordinate repeats. Board questions often ask for domain (all first coordinates) and range (all second coordinates) — right the roster form before attempting these. When A has m elements and B has n elements, the number of relations from A to B is 2ᵐⁿ.
02

Exercise 2.1 — Ordered Pairs and Cartesian Products

10Exercise questions

Step-by-step solution

  1. 1Two ordered pairs are equal iff their first coordinates match and their second coordinates match.
  2. 2x/3 + 1 = 5/3 → x/3 = 2/3 → x = 2.
  3. 3y − 2/3 = 1/3 → y = 1.

Final answer

x = 2, y = 1.

Step-by-step solution

  1. 1A has 3 elements; B has 3 elements.
  2. 2n(A × B) = n(A) × n(B) = 3 × 3 = 9.

Final answer

A × B has 9 elements.

Step-by-step solution

  1. 1G × H: first coordinate from G, second from H → {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}.
  2. 2H × G: first from H, second from G → {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}.

Final answer

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)}.

Step-by-step solution

  1. 1(i) False. P × Q has 2 × 2 = 4 elements: P × Q = {(m,n), (m,m), (n,n), (n,m)}.
  2. 2(ii) True — this is exactly the definition of the Cartesian product.
  3. 3(iii) True — B ∩ ∅ = ∅, so A × ∅ = ∅.

Final answer

(i) False → P × Q = {(m,n),(m,m),(n,n),(n,m)}. (ii) True. (iii) True.

Step-by-step solution

  1. 1A × A × A = ordered triples with each coordinate in {−1, 1}.
  2. 2{(-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 answer

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)}.

Step-by-step solution

  1. 1The first coordinates are a and b → A = {a, b}.
  2. 2The second coordinates are x and y → B = {x, y}.

Final answer

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

Step-by-step solution

  1. 1B ∩ C = ∅ (no common elements).
  2. 2So A × (B ∩ C) = A × ∅ = ∅.
  3. 3A × B = 8 pairs, A × C = 4 pairs, with no ordered pair in common since seconds are disjoint → (A × B) ∩ (A × C) = ∅.
  4. 4Both sides equal ∅, hence verified.

Final answer

Verified: both sides equal ∅.

Step-by-step solution

  1. 1A × B = {(1,3), (1,4), (2,3), (2,4)} — 4 elements.
  2. 2Number of subsets = 2⁴ = 16.
  3. 3They are: ∅; singletons {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}; three pairs sharing one element; the 6 two-element subsets; 4 three-element subsets; and A × B itself.

Final answer

A × B = {(1,3),(1,4),(2,3),(2,4)}; it has 2⁴ = 16 subsets.

Step-by-step solution

  1. 1First coordinates give A; they are x, y, z (distinct) → A = {x, y, z}.
  2. 2Second coordinates are 1 and 2 → B = {1, 2}.

Final answer

A = {x, y, z}, B = {1, 2}.

Step-by-step solution

  1. 1n(A × A) = n(A)² = 9 → n(A) = 3.
  2. 2−1, 0 appear as first coordinates, and 0, 1 appear as second coordinates → A = {−1, 0, 1}.
  3. 3A × A = {(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)}.
  4. 4Remaining elements besides (−1, 0) and (0, 1): (-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1).

Final answer

A = {−1, 0, 1}; remaining: (-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1).

03

Exercise 2.2 — Relations, Domain, Codomain and Range

9Exercise questions

Step-by-step solution

  1. 13x − y = 0 → y = 3x. For x = 1, 2, 3, 4: y = 3, 6, 9, 12, all in A.
  2. 2x = 5 gives y = 15 ∉ A, so stop there. R = {(1,3), (2,6), (3,9), (4,12)}.
  3. 3Domain = {1, 2, 3, 4}.
  4. 4Codomain = A = {1, 2, 3, …, 14}.
  5. 5Range = {3, 6, 9, 12}.

Final answer

R = {(1,3),(2,6),(3,9),(4,12)}; domain {1,2,3,4}, codomain {1,…,14}, range {3,6,9,12}.

Step-by-step solution

  1. 1x < 4, x ∈ N → x = 1, 2, 3. Then y = x + 5 → 6, 7, 8.
  2. 2R = {(1,6), (2,7), (3,8)}.
  3. 3Domain = {1, 2, 3}; range = {6, 7, 8}.

Final answer

R = {(1,6),(2,7),(3,8)}; domain {1,2,3}, range {6,7,8}.

Step-by-step solution

  1. 1Check every pair: x − y odd when x and y have opposite parity.
  2. 2x = 1: y = 4, 6 give odd differences → (1,4), (1,6).
  3. 3x = 2: y = 9 gives odd difference → (2,9).
  4. 4x = 3: y = 4, 6 → (3,4), (3,6).
  5. 5x = 5: y = 4, 6 → (5,4), (5,6).
  6. 6R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}.

Final answer

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

Step-by-step solution

  1. 1(i) Set-builder form: R = {(x, y) : y = x − 2, x ∈ P, y ∈ Q}.
  2. 2(ii) Roster form: R = {(5,3), (6,4), (7,5)}.
  3. 3Domain = {5, 6, 7}; range = {3, 4, 5}.

Final answer

R = {(x,y) : y = x − 2, x ∈ P, y ∈ Q} = {(5,3),(6,4),(7,5)}; domain {5,6,7}, range {3,4,5}.

Step-by-step solution

  1. 1b divisible by a means a | b, so b ≥ a.
  2. 2a = 1: all b → (1,1),(1,2),(1,3),(1,4),(1,6).
  3. 3a = 2: (2,2),(2,4),(2,6). a = 3: (3,3),(3,6). a = 4: (4,4). a = 6: (6,6).
  4. 4R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}.
  5. 5Domain = {1, 2, 3, 4, 6}; range = {1, 2, 3, 4, 6}.

Final answer

R as listed above; domain = {1,2,3,4,6}, range = {1,2,3,4,6}.

Step-by-step solution

  1. 1Roster form: R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}.
  2. 2Domain = {0, 1, 2, 3, 4, 5}.
  3. 3Range = {5, 6, 7, 8, 9, 10}.

Final answer

Domain = {0,1,2,3,4,5}, range = {5,6,7,8,9,10}.

Step-by-step solution

  1. 1Primes less than 10: 2, 3, 5, 7.
  2. 2x³ for each: 8, 27, 125, 343.
  3. 3R = {(2,8), (3,27), (5,125), (7,343)}.

Final answer

R = {(2,8),(3,27),(5,125),(7,343)}.

Step-by-step solution

  1. 1n(A) = 3, n(B) = 2 → n(A × B) = 6.
  2. 2Every subset of A × B is a relation from A to B.
  3. 3Number of relations = 2⁶ = 64.

Final answer

2⁶ = 64 relations.

Step-by-step solution

  1. 1a and b are any integers, and the difference of two integers is always an integer.
  2. 2So R contains every (a, b) with a, b ∈ Z.
  3. 3Domain = Z; range = Z.

Final answer

Domain = Z, range = Z.

04

Exercise 2.3 — Functions

5Exercise questions

Step-by-step solution

  1. 1(i) Each first coordinate appears once → function. Domain = {2,5,8,11,14,17}, range = {1}.
  2. 2(ii) Each first coordinate distinct → function. Domain = {2,4,6,8,10,12,14}, range = {1,2,3,4,5,6,7}.
  3. 3(iii) First coordinate 1 repeats with different images → NOT a function.

Final answer

(i) and (ii) are functions; (iii) is not a function.

Step-by-step solution

  1. 1(i) |x| is defined for all real x → domain = R.
  2. 2|x| ≥ 0 so −|x| ≤ 0, and every non-positive value is attained → range = (−∞, 0].
  3. 3(ii) 9 − x² ≥ 0 → x² ≤ 9 → −3 ≤ x ≤ 3 → domain = [−3, 3].
  4. 4x² ∈ [0, 9] → 9 − x² ∈ [0, 9] → √(9 − x²) ∈ [0, 3] → range = [0, 3].

Final answer

(i) Domain R, range (−∞, 0]. (ii) Domain [−3, 3], range [0, 3].

Step-by-step solution

  1. 1(i) f(0) = 2(0) − 5 = −5.
  2. 2(ii) f(7) = 2(7) − 5 = 9.
  3. 3(iii) f(−3) = 2(−3) − 5 = −11.

Final answer

f(0) = −5, f(7) = 9, f(−3) = −11.

Step-by-step solution

  1. 1(i) t(0) = 0 + 32 = 32.
  2. 2(ii) t(28) = 9(28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4.
  3. 3(iii) t(−10) = 9(−10)/5 + 32 = −18 + 32 = 14.
  4. 4(iv) 212 = 9C/5 + 32 → 180 = 9C/5 → C = 100.

Final answer

t(0) = 32, t(28) = 82.4, t(−10) = 14, and C = 100 when t(C) = 212.

Step-by-step solution

  1. 1(i) For x > 0, −3x < 0 so 2 − 3x < 2; as x → 0+, f → 2 and as x → ∞, f → −∞. Range = (−∞, 2).
  2. 2(ii) x² ≥ 0 → x² + 2 ≥ 2, and every value ≥ 2 occurs. Range = [2, ∞).
  3. 3(iii) f(x) = x takes every real value → range = R.

Final answer

(i) (−∞, 2). (ii) [2, ∞). (iii) R.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Size of a cartesian product

Composite function

Inverse function

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • A function from A to B needs every element of A mapped to exactly one element of B — mapping two different inputs to the same output is allowed, mapping one input to two outputs is not.
  • For an onto function, n(A) cannot exceed n(B); the pigeonhole principle settles every such existence question immediately.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 2 (Relations and Functions)?

There are 3 exercise questions in this chapter, numbered Ex 2.1 to Ex 2.3. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Relations and Functions Class 11 Maths?

The formulas this chapter's questions actually turn on are: Size of a cartesian product, Composite function, Inverse function. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Relations and Functions important for JEE Main?

Important — domain, range, one-one and onto classification carries easy marks, and the JEE Main function questions often build directly on these definitions.

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