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

These Class 11 Maths Chapter 2 solutions cover Relations and Functions from the NCERT textbook (updated for 2026–27). Every question of Exercise 2.1, 2.2, 2.3 and the Miscellaneous Exercise on Chapter 2 is solved step by step — Cartesian products, relations, domains, ranges and real functions — with answers verified against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 2 Exercises: 2.1, 2.2, 2.3, Miscellaneous Session: 2026–27
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Chapter 2 Overview

Chapter 2 of Class 11 Maths, Relations and Functions, builds the language used everywhere in higher mathematics. It begins with the ordered pair and the Cartesian product A × B, then defines a relation from a set A to a set B as a subset of A × B, together with its domain, codomain and range. Finally it introduces the function — a special relation in which every element of the domain has exactly one image — and studies the standard real functions (identity, constant, polynomial, rational, modulus, signum and greatest-integer) and the algebra of real functions (sum, difference, product, scalar multiple and quotient). The solutions below work through every textbook exercise question in order.

Key Concepts & Definitions

Ordered pair: a pair (p, q) in which order matters; (p, q) = (r, s) only when p = r and q = s.

Cartesian product: A × B = {(a, b) : a ∈ A, b ∈ B}. If n(A) = p and n(B) = q, then n(A × B) = pq.

Relation: a relation R from A to B is any subset of A × B. The set of first elements is the domain, the set of second elements is the range, and the whole set B is the codomain (range ⊆ codomain).

Number of relations: from a set A to a set B there are 2pq possible relations, where p = n(A), q = n(B).

Function: a relation f from A to B in which every element of A has one and only one image in B. We write f : A → B and f(a) = b.

Real function: a function whose domain and range are both R or subsets of R.

Important Formulas (Chapter 2)

Cartesian product size: n(A × B) = n(A) × n(B).

Number of relations: from A to B = 2n(A)·n(B).

Algebra of real functions (for f, g : X → R): (f + g)(x) = f(x) + g(x); (f − g)(x) = f(x) − g(x); (fg)(x) = f(x)g(x); (αf)(x) = α·f(x); (f/g)(x) = f(x)/g(x), provided g(x) ≠ 0.

Domain rules: a rational function f/g is defined where g(x) ≠ 0; an even root √(expr) is defined where expr ≥ 0.

Exercise 2.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

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

SOLUTION Equal ordered pairs ⇒ corresponding components are equal. First components: x/3 + 1 = 5/3 ⇒ x/3 = 5/3 − 1 = 2/3 ⇒ x = 2. Second components: y − 2/3 = 1/3 ⇒ y = 1/3 + 2/3 = 1. ∴ x = 2 and y = 1.

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

SOLUTION n(A) = 3 and n(B) = 3. n(A × B) = n(A) × n(B) = 3 × 3 = 9 elements.

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

SOLUTION 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)}. (Note G × H ≠ H × G, though both have 6 elements.)

4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. (i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, 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) such that x ∈ A and y ∈ B. (iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.

SOLUTION (i) False. Here P = {m, n}, Q = {n, m} both have 2 elements, so P × Q must have 4 pairs. Correct statement: P × Q = {(m, n), (m, m), (n, n), (n, m)}. (ii) True. By definition A × B = {(x, y) : x ∈ A, y ∈ B}, and if A, B are non-empty so is A × B. (iii) True. B ∩ φ = φ, and A × φ = φ.

5. If A = {−1, 1}, find A × A × A.

SOLUTION A × A × A is the set of all ordered triplets (a, b, c) with a, b, c ∈ {−1, 1}; there are 2 × 2 × 2 = 8 such triplets. 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 A = set of all first elements = {a, b}. B = set of all second elements = {x, y}. ∴ A = {a, b}, B = {x, y}.

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (i) A × (B ∩ C) = (A × B) ∩ (A × C). (ii) A × C is a subset of B × D.

SOLUTION (i) B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = φ, so A × (B ∩ C) = A × φ = φ. A × B and A × C share no pair (second elements 1–4 vs 5–6), so (A × B) ∩ (A × C) = φ. Both sides equal φ, hence A × (B ∩ C) = (A × B) ∩ (A × C). Verified. (ii) A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}. Each first element (1, 2) ∈ B and each second element (5, 6) ∈ D, so every pair of A × C lies in B × D. Hence A × C ⊆ B × D. Verified.

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

SOLUTION A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}, so n(A × B) = 4. Number of subsets = 24 = 16. They are: 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)}; and {(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 First elements x, y, z are the elements of A; since n(A) = 3, A = {x, y, z}. Second elements 1, 2 are the elements of B; since n(B) = 2, B = {1, 2}. ∴ A = {x, y, z} and 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 n(A × A) = n(A)2 = 9 ⇒ n(A) = 3. From the given pairs, −1, 0 and 1 are elements of A, and that is already 3 elements, so A = {−1, 0, 1}. A × A has all 9 ordered pairs; removing the two given ones, the remaining elements are: (−1, −1), (−1, 1), (0, −1), (0, 0), (1, −1), (1, 0), (1, 1).

Exercise 2.2 Solutions

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 3x − y = 0 ⇒ y = 3x. Both x and y must lie in A = {1, …, 14}, so 3x ≤ 14 ⇒ x = 1, 2, 3, 4. R = {(1, 3), (2, 6), (3, 9), (4, 12)}. Domain of R = {1, 2, 3, 4}; Range of R = {3, 6, 9, 12}; Codomain of R = A = {1, 2, …, 14}.

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 x is a natural number less than 4, so x = 1, 2, 3, and y = x + 5. R = {(1, 6), (2, 7), (3, 8)}. Domain of R = {1, 2, 3}; 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 For each x ∈ A and y ∈ B, keep the pair when |x − y| is odd (i.e. one of x, y is even and the other odd). x = 1: 1−4 = −3 (odd) ✓, 1−6 = −5 (odd) ✓, 1−9 = −8 (even) ✗. x = 2: 2−4 even, 2−6 even, 2−9 = −7 odd ✓. x = 3: 3−4 odd ✓, 3−6 odd ✓, 3−9 even. x = 5: 5−4 odd ✓, 5−6 odd ✓, 5−9 even. R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

4. The Fig 2.7 shows a relationship between the sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?

SOLUTION In Fig 2.7, P = {5, 6, 7} and Q = {3, 4, 5}, and each element of P is mapped to a value 2 less than itself (5→3, 6→4, 7→5). (i) Set-builder form: R = {(x, y) : y = x − 2 for x = 5, 6, 7}. (ii) Roster form: 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 Keep (a, b) when a divides b, with a, b ∈ {1, 2, 3, 4, 6}. (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 For x = 0, 1, 2, 3, 4, 5 the pairs are (0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10). Domain of R = {0, 1, 2, 3, 4, 5}; 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 Prime numbers less than 10 are 2, 3, 5, 7; their cubes are 8, 27, 125, 343. 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 n(A) = 3, n(B) = 2, so n(A × B) = 6. Number of relations = number of subsets of A × B = 26 = 64.

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 For any a, b ∈ Z, a − b is always an integer, so every ordered pair of integers belongs to R. Domain of R = Z and Range of R = Z.

Exercise 2.3 Solutions

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 A relation is a function when no first element is repeated (each input has exactly one output). (i) Yes, a function. All first elements are distinct. Domain = {2, 5, 8, 11, 14, 17}, Range = {1}. (ii) Yes, a function. All first elements are distinct. Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}. (iii) Not a function, because the first element 1 is repeated with two different images (3 and 5).

2. Find the domain and range of the following real functions: (i) f(x) = −|x| (ii) f(x) = √(9 − x2).

SOLUTION (i) f(x) = −|x| is defined for every real x, so Domain = R. Since |x| ≥ 0, −|x| ≤ 0, so Range = (−∞, 0]. (ii) For the square root, 9 − x2 ≥ 0 ⇒ x2 ≤ 9 ⇒ −3 ≤ x ≤ 3, so Domain = {x : −3 ≤ x ≤ 3} = [−3, 3]. As x runs over [−3, 3], 9 − x2 ranges from 0 to 9, so √(9 − x2) ranges from 0 to 3. Range = {y : 0 ≤ y ≤ 3} = [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 (i) f(0) = 2(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 t(C) = 9C/5 + 32. Find (i) t(0) (ii) t(28) (iii) t(−10) (iv) The value of C, when t(C) = 212.

SOLUTION (i) t(0) = 9(0)/5 + 32 = 32. (ii) t(28) = 9(28)/5 + 32 = 252/5 + 160/5 = 412/5 = 412/5 (= 82.4). (iii) t(−10) = 9(−10)/5 + 32 = −90/5 + 32 = −18 + 32 = 14. (iv) 212 = 9C/5 + 32 ⇒ 9C/5 = 180 ⇒ C = 180 × 5/9 = 100.

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) For x > 0, 3x > 0, so 2 − 3x < 2; as x → 0+, f → 2 (not attained) and as x → ∞, f → −∞. Range = (−∞, 2). (ii) x2 ≥ 0, so x2 + 2 ≥ 2, with value 2 reached at x = 0. Range = [2, ∞). (iii) f(x) = x takes every real value, so Range = R.

Miscellaneous Exercise on Chapter 2 — Solutions

1. The relation f is defined by f(x) = x2, 0 ≤ x ≤ 3; f(x) = 3x, 3 ≤ x ≤ 10. The relation g is defined by g(x) = x2, 0 ≤ x ≤ 2; g(x) = 3x, 2 ≤ x ≤ 10. Show that f is a function and g is not a function.

SOLUTION For f, the two pieces overlap only at x = 3. First piece gives f(3) = 32 = 9; second piece gives f(3) = 3(3) = 9. The two rules agree at x = 3, so every x in [0, 10] has a single image. Hence f is a function. For g, the pieces overlap at x = 2. First piece gives g(2) = 22 = 4; second piece gives g(2) = 3(2) = 6. The element 2 has two different images (4 and 6), so g is not a function.

2. If f(x) = x2, find [f(1.1) − f(1)] / (1.1 − 1).

SOLUTION f(1.1) = (1.1)2 = 1.21 and f(1) = 12 = 1. [f(1.1) − f(1)] / (1.1 − 1) = (1.21 − 1) / (0.1) = 0.21 / 0.1 = 2.1.

3. Find the domain of the function f(x) = (x2 + 2x + 1) / (x2 − 8x + 12).

SOLUTION The function is undefined where the denominator is zero. x2 − 8x + 12 = (x − 2)(x − 6) = 0 ⇒ x = 2 or x = 6. ∴ Domain = R − {2, 6}, i.e. the set of all real numbers except 2 and 6.

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

SOLUTION Square root requires x − 1 ≥ 0 ⇒ x ≥ 1, so Domain = [1, ∞). For x ≥ 1, √(x − 1) ≥ 0 and grows without bound, so Range = [0, ∞).

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

SOLUTION The modulus is defined for every real x, so Domain = R. |x − 1| ≥ 0 for all x and takes every non-negative value, so Range = [0, ∞) (all non-negative real numbers).

6. Let f = {(x, x2/(1 + x2)) : x ∈ R} be a function from R into R. Determine the range of f.

SOLUTION Let y = x2/(1 + x2). Then y(1 + x2) = x2 ⇒ x2(1 − y) = y ⇒ x2 = y/(1 − y). Since x2 ≥ 0, we need y/(1 − y) ≥ 0, which holds for 0 ≤ y < 1. Also y ≠ 1 (denominator 1 − y ≠ 0), and y = 0 is attained at x = 0. ∴ Range = [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 (f + g)(x) = (x + 1) + (2x − 3) = 3x − 2. (f − g)(x) = (x + 1) − (2x − 3) = −x + 4. (f/g)(x) = (x + 1)/(2x − 3), provided 2x − 3 ≠ 0, i.e. 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 Using (0, −1): f(0) = a(0) + b = b = −1, so b = −1. Using (1, 1): f(1) = a + b = 1 ⇒ a − 1 = 1 ⇒ a = 2. Check (2, 3): 2(2) − 1 = 3 ✓ and (−1, −3): 2(−1) − 1 = −3 ✓. ∴ a = 2, b = −1.

9. Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Are the following true? (i) (a, a) ∈ R, for all a ∈ N (ii) (a, b) ∈ R, implies (b, a) ∈ R (iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R. Justify your answer in each case.

SOLUTION (i) No. (a, a) ∈ R needs a = a2, which fails for a = 2 (2 ≠ 4). So it is not true for all a ∈ N. (ii) No. Take (4, 2) ∈ R since 4 = 22. Then (2, 4) would need 2 = 42 = 16, which is false. So (a, b) ∈ R does not imply (b, a) ∈ R. (iii) No. Take (16, 4) ∈ R (16 = 42) and (4, 2) ∈ R (4 = 22). Then (16, 2) would need 16 = 22 = 4, which is false. So it is not transitive.

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)}. Are the following true? (i) f is a relation from A to B (ii) f is a function from A to B. Justify your answer in each case.

SOLUTION (i) Yes. Every ordered pair of f has its first element in A and second element in B, so f ⊆ A × B, i.e. f is a relation from A to B. (ii) No. The element 2 of A has two images (9 and 11), so f 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 No, f is not a function. The same first element ab can come from different (a, b) giving different sums a + b. Example: with a = 1, b = 6, we get (6, 7); with a = 2, b = 3, we get (6, 5). The first element 6 has two different images (7 and 5), so 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 f(9) = 3 (9 = 3 × 3); f(10) = 5 (10 = 2 × 5); f(11) = 11 (prime); f(12) = 3 (12 = 22 × 3); f(13) = 13 (prime). Collecting the distinct images: Range of f = {3, 5, 11, 13}.

Common Mistakes to Avoid

Watch out for these

  • Treating A × B and B × A as equal — ordered pairs reverse, so in general A × B ≠ B × A.
  • Confusing range with codomain: the range is the set of images actually attained; range ⊆ codomain.
  • Declaring a relation a function without checking that no first element repeats — one input with two images breaks it.
  • Forgetting domain restrictions: a rational function needs denominator ≠ 0, and an even root needs the radicand ≥ 0.
  • Counting subsets/relations wrongly — the number of relations from A to B is 2n(A)·n(B), not n(A) × n(B).
  • For piecewise definitions, not checking the overlap point: if the two rules disagree there, the relation is not a function.

Practice MCQs & Assertion–Reason

1. If n(A) = 3 and n(B) = 4, then n(A × B) is:

(a) 7    (b) 12    (c) 81    (d) 64

2. The number of relations from a set of 2 elements to a set of 3 elements is:

(a) 6    (b) 32    (c) 64    (d) 8

3. The domain of the function f(x) = √(9 − x2) is:

(a) (−3, 3)    (b) [−3, 3]    (c) [0, 3]    (d) R

4. The range of f(x) = x2 + 2, x ∈ R, is:

(a) R    (b) [0, ∞)    (c) [2, ∞)    (d) (2, ∞)

5. If f(x) = 2x − 5, then f(−3) equals:

(a) −1    (b) −11    (c) 11    (d) 1

6. Which of these relations is a function?

(a) {(1, 2), (1, 3)}    (b) {(1, 2), (2, 2)}    (c) {(1, 2), (1, 4), (2, 5)}    (d) {(2, 1), (2, 3)}

7. The domain of f(x) = 1 / (x2 − 8x + 12) is:

(a) R    (b) R − {2, 6}    (c) R − {−2, −6}    (d) R − {4}

8. The range of the modulus function f(x) = |x − 1| is:

(a) R    (b) (−∞, 0]    (c) [0, ∞)    (d) (0, ∞)

9. If f(x) = x2, the value of [f(1.1) − f(1)] / (1.1 − 1) is:

(a) 1.1    (b) 2    (c) 2.1    (d) 0.21

10. The range of f(x) = x2/(1 + x2), x ∈ R, is:

(a) [0, 1]    (b) [0, 1)    (c) (0, 1)    (d) R

Answer key: 1-(b), 2-(c), 3-(b), 4-(c), 5-(b), 6-(b), 7-(b), 8-(c), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: If n(A) = 3 and n(B) = 3, then n(A × B) = 9.

Reason: n(A × B) = n(A) × n(B).

A-R 2. Assertion: The relation {(1, 3), (1, 5), (2, 5)} is a function.

Reason: In a function every element of the domain must have one and only one image.

A-R 3. Assertion: The domain of f(x) = √(x − 1) is [1, ∞).

Reason: A square root is real only when the expression under it is non-negative.

A-R 4. Assertion: For all sets A and B, A × B = B × A.

Reason: The Cartesian product A × B and B × A always contain the same number of elements.

A-R 5. Assertion: The number of relations from A to B, where n(A) = 2 and n(B) = 3, is 64.

Reason: The number of relations from A to B equals the number of subsets of A × B, i.e. 2n(A)·n(B).

Answer key: 1-(A), 2-(D), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • An ordered pair (a, b) records order; (a, b) = (c, d) iff a = c and b = d.
  • Cartesian product A × B = {(a, b) : a ∈ A, b ∈ B}; n(A × B) = n(A) × n(B); in general A × B ≠ B × A.
  • A relation from A to B is a subset of A × B, with a domain (first elements), range (second elements) and codomain B; range ⊆ codomain.
  • The number of relations from A to B is 2n(A)·n(B).
  • A function assigns to each element of its domain exactly one image; no first element repeats.
  • Standard real functions: identity, constant, polynomial, rational, modulus, signum and greatest-integer functions.
  • Algebra of functions: (f ± g)(x), (fg)(x), (αf)(x) and (f/g)(x) with g(x) ≠ 0.

How to score full marks in this chapter

Write Cartesian products in proper roster order and remember A × B ≠ B × A. For domain questions, state the rule you use (denominator ≠ 0, radicand ≥ 0) before solving. For range questions, set y = f(x), solve for x, and read off the values of y that keep x real. When checking “is this a function?”, look only for a repeated first element — one is enough to disqualify it. Keep every step shown so each earns its mark.

Frequently Asked Questions

What is Class 11 Maths Chapter 2 Relations and Functions about?

Chapter 2 introduces ordered pairs and the Cartesian product of sets, then defines relations (with their domain, codomain and range) and functions as a special kind of relation. It also covers standard real functions and the algebra of real functions (sum, difference, product, scalar multiple and quotient).

How many exercises are there in Class 11 Maths Chapter 2?

There are three numbered exercises — Exercise 2.1, 2.2 and 2.3 — plus a Miscellaneous Exercise on Chapter 2. Every question of all four exercises is solved step by step on this page.

What is the difference between range and codomain?

The codomain is the whole set B in a relation or function f : A → B, while the range is the set of elements of B that are actually images of some element of A. The range is always a subset of the codomain.

Are these Class 11 Maths Chapter 2 solutions free?

Yes. All solutions are free and follow the official NCERT Class 11 Mathematics textbook for the 2026–27 session, with answers verified against the book’s answer key.

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