Class 12 Mathematics - NCERT Mathematics Part I
NCERT Solutions for Class 12 Maths Chapter 1: Relations and Functions
Chapter 1 extends the concepts of relations and functions introduced in Class 11. Students learn to classify relations as reflexive, symmetric, transitive, and equivalence relations. The chapter covers types of functions (one-one, onto, bijective), composition of functions, and inverse of a function. This chapter is part of the Relations and Functions unit which carries 8 marks in the CBSE Class 12 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 1 Relations and Functions has 3 exercises plus a Miscellaneous Exercise. Exercise 1.1 covers types of relations, Exercise 1.2 covers types of functions, and Exercise 1.3 covers composition and invertible functions. The Miscellaneous Exercise combines all concepts.
Table of Contents
1.1Exercise 1.1 - Types of Relations
Q1Determine whether each of the following relations are reflexive, symmetric and transitive: (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x − y = 0} (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} (iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x} (iv) Relation R in the set Z of all integers defined as R = {(x, y) : x − y is an integer} (v) Relation R in the set A of human beings in a town at a particular time given by (a) R = {(x, y) : x and y work at the same place} (b) R = {(x, y) : x and y live in the same locality} (c) R = {(x, y) : x is exactly 7 cm taller than y} (d) R = {(x, y) : x is wife of y} (e) R = {(x, y) : x is father of y}
(i) R = {(x, y) : 3x − y = 0} i.e., R = {(x, y) : y = 3x}
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Reflexive: (1, 1) ∉ R (since 3(1) ≠ 1). Not reflexive.
Symmetric: (1, 3) ∈ R but (3, 1) ∉ R (since 3(3) = 9 ≠ 1). Not symmetric.
Transitive: (1, 3) ∈ R and (3, 9) ∈ R but (1, 9) ∉ R (since 3(1) = 3 ≠ 9). Not transitive.
(ii) R = {(x, y) : y = x + 5, x < 4}
R = {(1, 6), (2, 7), (3, 8)}
Reflexive: (1, 1) ∉ R. Not reflexive.
Symmetric: (1, 6) ∈ R but (6, 1) ∉ R. Not symmetric.
Transitive: Since no pair (a, b) and (b, c) both belong to R (no element appears as both second and first component), the condition is vacuously true. Transitive.
(iii) R = {(x, y) : y is divisible by x} in A = {1, 2, 3, 4, 5, 6}
Reflexive: Every number divides itself, so (x, x) ∈ R for all x ∈ A. Reflexive.
Symmetric: (1, 2) ∈ R (2 is divisible by 1) but (2, 1) ∉ R (1 is not divisible by 2). Not symmetric.
Transitive: If (x, y) ∈ R and (y, z) ∈ R, then y is divisible by x and z is divisible by y. So z is divisible by x, giving (x, z) ∈ R. Transitive.
(iv) R = {(x, y) : x − y is an integer} in Z
Reflexive: x − x = 0, which is an integer. So (x, x) ∈ R. Reflexive.
Symmetric: If x − y is an integer, then y − x = −(x − y) is also an integer. Symmetric.
Transitive: If x − y is an integer and y − z is an integer, then x − z = (x − y) + (y − z) is an integer. Transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
(v)(a) R = {(x, y) : x and y work at the same place}
Reflexive: Yes (x works at the same place as x). Symmetric: Yes (if x works with y, y works with x). Transitive: Yes (if x works with y and y works with z at the same place, then x works with z). Equivalence relation.
(v)(b) R = {(x, y) : x and y live in the same locality}
Same reasoning as (a). Equivalence relation.
(v)(c) R = {(x, y) : x is exactly 7 cm taller than y}
Reflexive: No (x cannot be 7 cm taller than x). Not reflexive. Hence not an equivalence relation.
Symmetric: No (if x is 7 cm taller than y, y is 7 cm shorter, not taller). Not symmetric.
Transitive: No (if x is 7 cm taller than y, and y is 7 cm taller than z, then x is 14 cm taller than z, not 7 cm). Not transitive.
(v)(d) R = {(x, y) : x is wife of y}
Reflexive: No (x cannot be wife of x). Not reflexive.
Symmetric: No (if x is wife of y, y is husband of x, not wife). Not symmetric.
Transitive: No (if x is wife of y, y cannot be wife of anyone in the usual sense). Not transitive.
(v)(e) R = {(x, y) : x is father of y}
Reflexive: No. Not reflexive.
Symmetric: No (if x is father of y, y is not father of x). Not symmetric.
Transitive: No (if x is father of y and y is father of z, x is grandfather of z, not father). Not transitive.
Q2Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²} is neither reflexive, nor symmetric, nor transitive.
Reflexive: Take a = 1/2. Then a² = 1/4. We need a ≤ a², i.e., 1/2 ≤ 1/4, which is false.
So (1/2, 1/2) ∉ R. R is not reflexive.
Symmetric: Take a = 1, b = 4. Then a ≤ b² gives 1 ≤ 16 (true), so (1, 4) ∈ R.
But b ≤ a² gives 4 ≤ 1 (false), so (4, 1) ∉ R. R is not symmetric.
Transitive: Take a = 3, b = 2, c = 1.5.
(3, 2) ∈ R since 3 ≤ 4 ✓
(2, 1.5) ∈ R since 2 ≤ 2.25 ✓
But (3, 1.5): 3 ≤ (1.5)² = 2.25? No. (3, 1.5) ∉ R. R is not transitive.
Q3Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Reflexive: (1, 1) ∉ R since 1 ≠ 1 + 1. Not reflexive.
Symmetric: (1, 2) ∈ R but (2, 1) ∉ R since 1 ≠ 2 + 1. Not symmetric.
Transitive: (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R since 3 ≠ 1 + 1. Not transitive.
Q4Show that the relation R in R defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.
Reflexive: For any a ∈ R, a ≤ a is true. So (a, a) ∈ R for all a. R is reflexive.
Transitive: If (a, b) ∈ R and (b, c) ∈ R, then a ≤ b and b ≤ c.
Since a ≤ b ≤ c, we have a ≤ c, so (a, c) ∈ R. R is transitive.
Symmetric: Take a = 3, b = 5. Then a ≤ b (3 ≤ 5) is true, so (3, 5) ∈ R.
But b ≤ a (5 ≤ 3) is false, so (5, 3) ∉ R. R is not symmetric.
Q5Check whether the relation R in R defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
Reflexive: Take a = −1. Then a ≤ a³ means −1 ≤ (−1)³ = −1. This is true.
But take a = 1/2. Then 1/2 ≤ (1/2)³ = 1/8. This is false.
So (1/2, 1/2) ∉ R. R is not reflexive.
Symmetric: Take a = 1, b = 2. Then 1 ≤ 8 (true), so (1, 2) ∈ R.
But 2 ≤ 1 (false), so (2, 1) ∉ R. R is not symmetric.
Transitive: Take a = 3, b = 3/2, c = 6/5.
(3, 3/2): 3 ≤ (3/2)³ = 27/8 = 3.375 ✓
(3/2, 6/5): 3/2 ≤ (6/5)³ = 216/125 = 1.728 ✓
(3, 6/5): 3 ≤ (6/5)³ = 1.728? No. R is not transitive.
Q6Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Symmetric: (1, 2) ∈ R and (2, 1) ∈ R. For every (a, b) ∈ R, (b, a) ∈ R. R is symmetric.
Reflexive: (1, 1) ∉ R, (2, 2) ∉ R, (3, 3) ∉ R. R is not reflexive.
Transitive: (1, 2) ∈ R and (2, 1) ∈ R, but (1, 1) ∉ R. R is not transitive.
Q7Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have the same number of pages} is an equivalence relation.
Reflexive: Every book has the same number of pages as itself. So (x, x) ∈ R for all x ∈ A. R is reflexive.
Symmetric: If x and y have the same number of pages, then y and x have the same number of pages. So if (x, y) ∈ R, then (y, x) ∈ R. R is symmetric.
Transitive: If x and y have the same number of pages, and y and z have the same number of pages, then x and z have the same number of pages. So if (x, y) ∈ R and (y, z) ∈ R, then (x, z) ∈ R. R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Q8Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a − b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Reflexive: |a − a| = 0, which is even. So (a, a) ∈ R for all a. R is reflexive.
Symmetric: If |a − b| is even, then |b − a| = |a − b| is also even. R is symmetric.
Transitive: If |a − b| is even and |b − c| is even, then a − b and b − c are both even.
So a − c = (a − b) + (b − c) is even, meaning |a − c| is even. R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Elements of {1, 3, 5}:
|1 − 3| = 2 (even), |1 − 5| = 4 (even), |3 − 5| = 2 (even). All pairs are related. ✓
Elements of {2, 4}:
|2 − 4| = 2 (even). Related. ✓
Cross-check {1, 3, 5} with {2, 4}:
|1 − 2| = 1 (odd), |1 − 4| = 3 (odd), |3 − 2| = 1 (odd), |3 − 4| = 1 (odd), |5 − 2| = 3 (odd), |5 − 4| = 1 (odd). No pair is related. ✓
Q9Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by (i) R = {(a, b) : |a − b| is a multiple of 4} (ii) R = {(a, b) : a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
(i) R = {(a, b) : |a − b| is a multiple of 4}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Reflexive: |a − a| = 0 = 4 × 0, a multiple of 4. Reflexive.
Symmetric: If |a − b| is a multiple of 4, then |b − a| = |a − b| is also a multiple of 4. Symmetric.
Transitive: If |a − b| = 4k and |b − c| = 4m, then a − b = ±4k and b − c = ±4m.
a − c = (a − b) + (b − c) = ±4k ± 4m = 4(±k ± m), which is a multiple of 4. Transitive.
R is an equivalence relation.
Set of elements related to 1: {a ∈ A : |a − 1| is a multiple of 4}
|a − 1| = 0 → a = 1 ✓
|a − 1| = 4 → a = 5 ✓
|a − 1| = 8 → a = 9 ✓
|a − 1| = 12 → a = 13 ∉ A
Elements related to 1 = {1, 5, 9}
(ii) R = {(a, b) : a = b}
This is the identity relation.
Reflexive: a = a. Reflexive.
Symmetric: If a = b, then b = a. Symmetric.
Transitive: If a = b and b = c, then a = c. Transitive.
R is an equivalence relation.
Set of elements related to 1: {a ∈ A : a = 1} = {1}
Q10Give an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive.
Let A = {1, 2, 3}
(i) Symmetric but neither reflexive nor transitive:
R = {(1, 2), (2, 1)}
Symmetric: (1, 2) ∈ R ⟹ (2, 1) ∈ R ✓
Not reflexive: (1, 1) ∉ R ✗
Not transitive: (1, 2) ∈ R, (2, 1) ∈ R but (1, 1) ∉ R ✗
(ii) Transitive but neither reflexive nor symmetric:
R = {(1, 2), (2, 3), (1, 3)}
Transitive: (1, 2) ∈ R, (2, 3) ∈ R ⟹ (1, 3) ∈ R ✓
Not reflexive: (1, 1) ∉ R ✗
Not symmetric: (1, 2) ∈ R but (2, 1) ∉ R ✗
(iii) Reflexive and symmetric but not transitive:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
Reflexive: (1, 1), (2, 2), (3, 3) ∈ R ✓
Symmetric: (1, 2) ↔ (2, 1), (2, 3) ↔ (3, 2) ✓
Not transitive: (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∉ R ✗
(iv) Reflexive and transitive but not symmetric:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)}
Reflexive ✓, Transitive ✓ (check all pairs)
Not symmetric: (1, 2) ∈ R but (2, 1) ∉ R ✗
(v) Symmetric and transitive but not reflexive:
R = {(1, 2), (2, 1), (1, 1), (2, 2)}
Symmetric ✓, Transitive ✓
Not reflexive: (3, 3) ∉ R ✗
Q11Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Let d(P) denote the distance of point P from the origin.
R = {(P, Q) : d(P) = d(Q)}
Reflexive: d(P) = d(P) for every point P. So (P, P) ∈ R. Reflexive.
Symmetric: If d(P) = d(Q), then d(Q) = d(P). So if (P, Q) ∈ R, then (Q, P) ∈ R. Symmetric.
Transitive: If d(P) = d(Q) and d(Q) = d(S), then d(P) = d(S). So if (P, Q) ∈ R and (Q, S) ∈ R, then (P, S) ∈ R. Transitive.
R is an equivalence relation.
The set of all points related to P = {Q : d(Q) = d(P)} = set of all points at distance d(P) from the origin.
This is the circle with centre at the origin and radius = d(P), passing through P.
Q12Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂}, is an equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13 and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂ and T₃ are related?
Reflexive: Every triangle is similar to itself (same angles, ratio 1:1). Reflexive.
Symmetric: If T₁ is similar to T₂, then T₂ is similar to T₁ (similarity is mutual). Symmetric.
Transitive: If T₁ is similar to T₂ and T₂ is similar to T₃, then T₁ is similar to T₃ (corresponding angles remain equal). Transitive.
R is an equivalence relation.
Checking T₁ (3, 4, 5), T₂ (5, 12, 13), T₃ (6, 8, 10):
T₁ and T₃: sides 3:6 = 4:8 = 5:10 = 1:2. T₁ is similar to T₃. ✓
T₁ and T₂: sides 3:5 ≠ 4:12. Not similar. ✗
T₂ and T₃: sides 5:6 ≠ 12:8. Not similar. ✗
T₁ and T₃ are related.
Q13Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Reflexive: Every polygon has the same number of sides as itself. Reflexive.
Symmetric: If P₁ and P₂ have the same number of sides, then P₂ and P₁ have the same number of sides. Symmetric.
Transitive: If P₁ and P₂ have the same number of sides, and P₂ and P₃ have the same number of sides, then P₁ and P₃ have the same number of sides. Transitive.
R is an equivalence relation.
The right triangle T with sides 3, 4, 5 is a triangle (3 sides).
The set of all elements related to T = set of all triangles in A = the set of all triangles.
Q14Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Reflexive: Every line is parallel to itself. Reflexive.
Symmetric: If L₁ ∥ L₂, then L₂ ∥ L₁. Symmetric.
Transitive: If L₁ ∥ L₂ and L₂ ∥ L₃, then L₁ ∥ L₃ (lines parallel to the same line are parallel to each other). Transitive.
R is an equivalence relation.
The line y = 2x + 4 has slope = 2.
The set of all lines related to it = set of all lines parallel to y = 2x + 4 = {y = 2x + c : c ∈ R}
i.e., the set of all lines with slope 2.
Q15Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer. (A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) R is an equivalence relation.
A = {1, 2, 3, 4}
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
Reflexive: (1, 1) ∈ R ✓, (2, 2) ∈ R ✓, (3, 3) ∈ R ✓, (4, 4) ∈ R ✓. Reflexive.
Symmetric: (1, 2) ∈ R but (2, 1) ∉ R. Not symmetric.
Transitive: Check: (1, 3) ∈ R, (3, 2) ∈ R ⟹ need (1, 2) ∈ R ✓
(1, 2) ∈ R, (2, 2) ∈ R ⟹ need (1, 2) ∈ R ✓
All required pairs are present. Transitive.
Answer: (B) R is reflexive and transitive but not symmetric.
Q16Let R be the relation in the set N given by R = {(a, b) : a = b − 2, b > 6}. Choose the correct answer. (A) (2, 4) ∈ R (B) (3, 8) ∈ R (C) (6, 8) ∈ R (D) (8, 7) ∈ R
R = {(a, b) : a = b − 2, b > 6}
(A) (2, 4): a = 2, b = 4. b > 6? No (4 < 6). (2, 4) ∉ R ✗
(B) (3, 8): a = 3, b = 8. b > 6? Yes. a = b − 2? 3 = 8 − 2 = 6? No. (3, 8) ∉ R ✗
(C) (6, 8): a = 6, b = 8. b > 6? Yes. a = b − 2? 6 = 8 − 2 = 6? Yes. (6, 8) ∈ R ✓
(D) (8, 7): a = 8, b = 7. b > 6? Yes. a = b − 2? 8 = 7 − 2 = 5? No. (8, 7) ∉ R ✗
Answer: (C) (6, 8) ∈ R
Stuck on a question? AI explains any step in detail - in English or Hindi.
Ask AI TutorPrevious Year Board Exam Questions
Let A = {1, 2, 3, ..., 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c. Prove that R is an equivalence relation. Also obtain the equivalence class [(2, 5)].
Reflexive: (a, b) R (a, b) requires a + b = b + a, which is always true. Reflexive. ✓
Symmetric: If (a, b) R (c, d), then a + d = b + c. This gives c + b = d + a, so (c, d) R (a, b). Symmetric. ✓
Transitive: If (a, b) R (c, d) and (c, d) R (e, f), then a + d = b + c and c + f = d + e.
Adding: a + d + c + f = b + c + d + e → a + f = b + e. So (a, b) R (e, f). Transitive. ✓
R is an equivalence relation.
Equivalence class [(2, 5)]: All (a, b) such that a + 5 = b + 2, i.e., b − a = 3.
[(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}
Check if the relation R in the set of real numbers defined by R = {(a, b) : a ≤ b³} is reflexive, symmetric or transitive.
Reflexive: Take a = 1/2. Then 1/2 ≤ (1/2)³ = 1/8 is false. (1/2, 1/2) ∉ R. Not reflexive.
Symmetric: (1, 2) ∈ R since 1 ≤ 8. But (2, 1) ∉ R since 2 > 1. Not symmetric.
Transitive: (3, 3/2) ∈ R since 3 ≤ 3.375. (3/2, 6/5) ∈ R since 1.5 ≤ 1.728. But (3, 6/5) ∉ R since 3 > 1.728. Not transitive.
Show that the function f: R → R defined by f(x) = x/(x² + 1) is neither one-one nor onto.
Not one-one: f(2) = 2/5 and f(1/2) = (1/2)/(1/4 + 1) = (1/2)/(5/4) = 2/5. Since f(2) = f(1/2) but 2 ≠ 1/2, f is not one-one.
Not onto: Let y = x/(x² + 1). Then yx² − x + y = 0. For real x, discriminant ≥ 0: 1 − 4y² ≥ 0, so y² ≤ 1/4, meaning −1/2 ≤ y ≤ 1/2. Since the range is [−1/2, 1/2] ≠ R, f is not onto.
A relation R is defined on the set A = {1, 2, 3, 4, 5} by R = {(x, y) : |x² − y²| < 8}. Check whether R is reflexive, symmetric, and transitive.
Reflexive: |x² − x²| = 0 < 8 for all x. Reflexive. ✓
Symmetric: If |x² − y²| < 8, then |y² − x²| = |x² − y²| < 8. Symmetric. ✓
Transitive: (1, 2) ∈ R since |1 − 4| = 3 < 8 ✓
(2, 3) ∈ R since |4 − 9| = 5 < 8 ✓
But (1, 3): |1 − 9| = 8, which is NOT < 8. So (1, 3) ∉ R.
Since (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R, R is not transitive. ✗
Let f: R − {−4/3} → R be a function defined as f(x) = 4x/(3x + 4). Show that f is one-one. Also, find the inverse of f.
One-one: Let f(x₁) = f(x₂).
4x₁/(3x₁ + 4) = 4x₂/(3x₂ + 4)
4x₁(3x₂ + 4) = 4x₂(3x₁ + 4)
12x₁x₂ + 16x₁ = 12x₁x₂ + 16x₂
16x₁ = 16x₂
x₁ = x₂. f is one-one. ✓
Inverse: Let y = 4x/(3x + 4)
y(3x + 4) = 4x
3xy + 4y = 4x
3xy − 4x = −4y
x(3y − 4) = −4y
x = −4y/(3y − 4) = 4y/(4 − 3y)
f⁻¹(y) = 4y/(4 − 3y)
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides (a − b)}, is an equivalence relation.
Reflexive: a − a = 0, and 2 divides 0. So (a, a) ∈ R. Reflexive. ✓
Symmetric: If 2 | (a − b), then a − b = 2k. So b − a = −2k, and 2 | (b − a). Symmetric. ✓
Transitive: If 2 | (a − b) and 2 | (b − c), then a − b = 2k and b − c = 2m. So a − c = 2k + 2m = 2(k + m), and 2 | (a − c). Transitive. ✓
R is an equivalence relation.
Let f: N → N be defined by f(n) = (n + 1)/2 if n is odd, and f(n) = n/2 if n is even. Show that f is onto but not one-one.
Not one-one: f(1) = (1 + 1)/2 = 1 and f(2) = 2/2 = 1. Since f(1) = f(2) but 1 ≠ 2, f is not one-one.
Onto: For any m ∈ N, we need n ∈ N such that f(n) = m.
Take n = 2m (even). Then f(2m) = 2m/2 = m. ✓
Since every m ∈ N has a pre-image 2m ∈ N, f is onto. ✓
Check whether the function f: R → R defined by f(x) = 1 + x² is one-one, onto, or bijective.
Not one-one: f(1) = 2 and f(−1) = 2. Since f(1) = f(−1) but 1 ≠ −1, f is not one-one.
Not onto: Range of f = [1, ∞) since x² ≥ 0 gives f(x) ≥ 1. There is no x with f(x) = 0. Since range ≠ R, f is not onto.
Not bijective (since it is neither one-one nor onto).
A relation R is defined on a set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Show that R is an equivalence relation. Further, show that all the elements of the subset {1, 3, 5, 7} are related to each other.
Reflexive: For any a ∈ A, both a and a are the same (both odd or both even). (a, a) ∈ R. Reflexive. ✓
Symmetric: If both a and b are odd (or both even), then both b and a are odd (or both even). Symmetric. ✓
Transitive: If both a and b are odd/even, and both b and c are odd/even, then since b's parity determines the group, a and c must also be in the same group. Transitive. ✓
R is an equivalence relation.
All elements of {1, 3, 5, 7} are odd. Any two odd numbers satisfy the relation (both are odd). So every pair from {1, 3, 5, 7} is in R. ✓
Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f(x) = (x − 2)/(x − 3). Is f one-one and onto? Justify your answer.
One-one: Let f(x₁) = f(x₂).
(x₁ − 2)/(x₁ − 3) = (x₂ − 2)/(x₂ − 3)
(x₁ − 2)(x₂ − 3) = (x₂ − 2)(x₁ − 3)
x₁x₂ − 3x₁ − 2x₂ + 6 = x₁x₂ − 3x₂ − 2x₁ + 6
−3x₁ − 2x₂ = −3x₂ − 2x₁
−x₁ = −x₂
x₁ = x₂. f is one-one. ✓
Onto: Let y = (x − 2)/(x − 3). Then y(x − 3) = x − 2, yx − 3y = x − 2, x(y − 1) = 3y − 2, x = (3y − 2)/(y − 1).
For any y ∈ B (y ≠ 1), x = (3y − 2)/(y − 1) ∈ A (x ≠ 3, which can be verified). f is onto. ✓
f is bijective.
Let f: W → W be defined by f(n) = n − 1 if n is odd, and f(n) = n + 1 if n is even. Show that f is a bijection.
One-one: Case 1: Both n₁, n₂ odd. f(n₁) = f(n₂) ⟹ n₁ − 1 = n₂ − 1 ⟹ n₁ = n₂.
Case 2: Both even. f(n₁) = f(n₂) ⟹ n₁ + 1 = n₂ + 1 ⟹ n₁ = n₂.
Case 3: n₁ odd, n₂ even. f(n₁) = n₁ − 1 (even), f(n₂) = n₂ + 1 (odd). Even ≠ odd, so they can't be equal. ✓
f is one-one.
Onto: f maps: 0→1, 1→0, 2→3, 3→2, 4→5, 5→4, ... (swaps consecutive pairs).
For any m ∈ W:
If m is even, take n = m + 1 (odd). Then f(m + 1) = (m + 1) − 1 = m. ✓
If m is odd, take n = m − 1 (even). Then f(m − 1) = (m − 1) + 1 = m. ✓
Every m ∈ W has a pre-image, so f is onto.
Since f is both one-one and onto, f is a bijection.
If f: R → R is defined by f(x) = 3x + 2, find f(f(x)).
f(f(x)) = f(3x + 2) = 3(3x + 2) + 2 = 9x + 6 + 2 = 9x + 8
Show that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by R = {(a, b) : |a − b| is a multiple of 4} is an equivalence relation. Find the set of all elements related to 1.
Reflexive: |a − a| = 0 = 4 × 0 (multiple of 4). Reflexive. ✓
Symmetric: |a − b| = |b − a|. If one is a multiple of 4, so is the other. Symmetric. ✓
Transitive: If a − b = 4k and b − c = 4m, then a − c = 4(k + m), a multiple of 4. Transitive. ✓
R is an equivalence relation.
Elements related to 1: |a − 1| = 0, 4, 8, 12 → a = 1, 5, 9 (13 ∉ A).
Set = {1, 5, 9}
If f(x) = (4x + 3)/(6x − 4) where x ≠ 2/3, show that fof(x) = x for all x ≠ 2/3.
f(f(x)) = f((4x + 3)/(6x − 4))
Let u = (4x + 3)/(6x − 4).
f(u) = (4u + 3)/(6u − 4)
= (4 · (4x+3)/(6x−4) + 3) / (6 · (4x+3)/(6x−4) − 4)
= ((16x + 12 + 18x − 12)/(6x − 4)) / ((24x + 18 − 24x + 16)/(6x − 4))
= (34x)/(34)
= x
Hence fof(x) = x. ✓
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The relation R = {(1, 1), (2, 2), (3, 3)} on set A = {1, 2, 3} is reflexive and symmetric.
Reason (R): A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R.
Answer: (A)
Both true. R contains all (a, a), so it is reflexive. For symmetry, the only pairs are of the form (a, a), and (a, a) ∈ R trivially implies (a, a) ∈ R. So R is symmetric. The reason correctly explains this.
AR2
Assertion (A): The function f: N → N defined by f(x) = 2x is one-one but not onto.
Reason (R): A function is onto if every element of the codomain has a pre-image.
Answer: (A)
Both true. f(x₁) = f(x₂) ⟹ 2x₁ = 2x₂ ⟹ x₁ = x₂ (one-one). But odd numbers like 1, 3, 5 have no pre-image (not onto). The reason correctly defines onto.
AR3
Assertion (A): Every function is a relation but every relation is not a function.
Reason (R): In a function, each element of the domain maps to exactly one element of the codomain.
Answer: (A)
Both true. A function is a special case of a relation where each input has exactly one output. A relation can map one input to multiple outputs, which a function cannot. The reason correctly explains the distinction.
AR4
Assertion (A): If f: A → B is a bijection and g: B → C is a bijection, then gof: A → C is a bijection.
Reason (R): Composition of two bijections is always a bijection.
Answer: (A)
Both true. If f and g are one-one, gof is one-one. If f and g are onto, gof is onto. Hence the composition of two bijections is a bijection.
AR5
Assertion (A): The relation R on set {1, 2, 3} defined by R = {(1, 2), (2, 1)} is transitive.
Reason (R): A relation R is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.
Answer: (D)
Assertion is FALSE: (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R, so R is not transitive. Reason is TRUE as a definition.
AR6
Assertion (A): The function f: R → R defined by f(x) = x³ is a bijection.
Reason (R): f(x) = x³ is strictly increasing on R, so it is one-one, and its range is all of R, so it is onto.
Answer: (A)
Both true. x³ is strictly increasing (if x₁ < x₂ then x₁³ < x₂³), so one-one. For any y ∈ R, x = y^(1/3) gives f(x) = y, so onto. The reason correctly explains both properties.
Case Study Questions
The government assigns a unique 12-digit Aadhaar number to every resident of India. Consider a function f that maps each resident to their Aadhaar number. Let A be the set of all residents of India and B be the set of all 12-digit numbers.
(i) Is the function f one-one? Justify. (1 mark)
Yes, f is one-one. Each resident is assigned a unique Aadhaar number, meaning no two different residents share the same number. If f(x₁) = f(x₂), then x₁ and x₂ have the same Aadhaar number, which means x₁ = x₂.
(ii) Is the function f onto? Justify. (1 mark)
No, f is not onto. The codomain B is the set of all 12-digit numbers (from 000000000000 to 999999999999). Not every 12-digit number is assigned to a resident. Many numbers in B have no pre-image in A.
(iii)(a) What type of function is f? (1 mark)
Since f is one-one but not onto, f is an injective (one-one) function but not surjective (onto). It is not bijective.
(iii)(b) If the government also assigns a unique PAN number to each taxpayer, and g maps taxpayers to PAN numbers, what is the nature of the composition gof (for residents who are taxpayers)? (1 mark)
If both f and g are one-one (unique mappings), then gof maps each taxpayer-resident to a unique PAN number via their Aadhaar. Since both are one-one, gof is also one-one.
In a school, 50 students of Class 12 are assigned roll numbers 1 to 50. A relation R is defined on the set A = {1, 2, 3, ..., 50} by R = {(a, b) : a and b have the same remainder when divided by 5}.
(i) Show that R is an equivalence relation. (2 marks)
Reflexive: a divided by 5 has the same remainder as a divided by 5. (a, a) ∈ R. ✓
Symmetric: If a and b have the same remainder when divided by 5, then b and a do too. ✓
Transitive: If a and b have the same remainder r when divided by 5, and b and c have the same remainder (which must also be r), then a and c have the same remainder r. ✓
R is an equivalence relation.
(ii) Find the equivalence class of roll number 7, i.e., [7]. (1 mark)
7 ÷ 5 gives remainder 2. So [7] = all roll numbers with remainder 2 when divided by 5.
[7] = {2, 7, 12, 17, 22, 27, 32, 37, 42, 47}
(iii) How many distinct equivalence classes does R create? (1 mark)
The possible remainders when dividing by 5 are 0, 1, 2, 3, 4. Each remainder creates one equivalence class.
R creates 5 distinct equivalence classes.
MCQ Practice
MCQ1. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is:
(A) Reflexive and symmetric
(B) Transitive and symmetric
(C) Equivalence relation
(D) Reflexive, transitive but not symmetric
Answer: (D) - Reflexive: n divides n ✓. Transitive: if n|m and m|p then n|p ✓. Not symmetric: 2|4 but 4 does not divide 2.
MCQ2. The maximum number of equivalence relations on the set A = {1, 2, 3} is:
(A) 1
(B) 2
(C) 3
(D) 5
Answer: (D) - The equivalence relations correspond to partitions of {1, 2, 3}: {{1,2,3}}, {{1},{2,3}}, {{2},{1,3}}, {{3},{1,2}}, {{1},{2},{3}}. Total = 5.
MCQ3. If f: R → R is given by f(x) = (3 − x³)^(1/3), then fof(x) is:
(A) x^(1/3)
(B) x³
(C) x
(D) (3 − x³)
Answer: (C) - fof(x) = f(f(x)) = f((3 − x³)^(1/3)) = (3 − ((3 − x³)^(1/3))³)^(1/3) = (3 − (3 − x³))^(1/3) = (x³)^(1/3) = x.
MCQ4. Let f: R → R be defined by f(x) = x². The function f is:
(A) One-one and onto
(B) One-one but not onto
(C) Not one-one but onto
(D) Neither one-one nor onto
Answer: (D) - Not one-one: f(1) = f(−1) = 1. Not onto: negative numbers like −1 have no pre-image since x² ≥ 0.
MCQ5. If A = {1, 2, 3} and relation R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3), (3, 2)}, then R is:
(A) Reflexive only
(B) Symmetric only
(C) Transitive only
(D) Equivalence relation
Answer: (D) - R = A × A (all possible pairs). Reflexive ✓, Symmetric ✓, Transitive ✓. This is the universal relation, which is always an equivalence relation.
MCQ6. The function f: N → N given by f(x) = x + 1 is:
(A) One-one and onto
(B) One-one but not onto
(C) Not one-one but onto
(D) Neither one-one nor onto
Answer: (B) - One-one: f(x₁) = f(x₂) ⟹ x₁ + 1 = x₂ + 1 ⟹ x₁ = x₂. Not onto: 1 ∈ N has no pre-image (f(x) = 1 gives x = 0 ∉ N).
MCQ7. The number of all one-one functions from set A = {a, b, c} to itself is:
(A) 3
(B) 6
(C) 9
(D) 27
Answer: (B) - One-one functions from a set to itself are permutations. Number of permutations of 3 elements = 3! = 6.
MCQ8. Let R be the relation in the set Z defined by R = {(a, b) : (a − b) is divisible by 5}. Then R is:
(A) Reflexive but not symmetric
(B) Symmetric but not transitive
(C) Equivalence relation
(D) None of these
Answer: (C) - Reflexive: 5|(a−a)=0 ✓. Symmetric: 5|(a−b) ⟹ 5|(b−a) ✓. Transitive: 5|(a−b) and 5|(b−c) ⟹ 5|(a−c) ✓. Equivalence relation.
MCQ9. If f: A → B is a bijection and n(A) = 5, then n(B) is:
(A) 10
(B) 5
(C) 1
(D) Cannot be determined
Answer: (B) - A bijection establishes a one-to-one correspondence, so domain and codomain must have the same number of elements. n(B) = 5.
MCQ10. The relation R on set A = {1, 2, 3, 4, 5} defined by R = {(a, b) : |a − b| is even} has the equivalence classes:
(A) {1, 3, 5} and {2, 4}
(B) {1, 2, 3} and {4, 5}
(C) {1, 2} and {3, 4, 5}
(D) {1, 4} and {2, 3, 5}
Answer: (A) - Elements with even difference must have the same parity. Odd numbers {1, 3, 5} form one class, even numbers {2, 4} form another.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
If f: R → R is defined by f(x) = 3x + 2, find f(f(x)).
f(f(x)) = f(3x + 2) = 3(3x + 2) + 2 = 9x + 8
State whether the function f: N → N given by f(x) = x + 1 is onto or not.
Not onto, because 1 ∈ N has no pre-image (f(x) = 1 gives x = 0 ∉ N).
What is the maximum number of equivalence relations on set {1, 2, 3}?
5 (corresponding to the 5 partitions of a 3-element set).
If f(x) = x² + 1, is f: R → R one-one?
No. f(1) = f(−1) = 2, but 1 ≠ −1.
What is the number of bijections from set A = {1, 2, 3, 4} to itself?
4! = 24
2-3 Mark Questions
Show that the relation R on Z defined by R = {(a, b) : a − b is divisible by 3} is an equivalence relation.
Reflexive: 3|(a−a)=0 ✓. Symmetric: 3|(a−b) ⟹ 3|(b−a) ✓. Transitive: 3|(a−b) and 3|(b−c) ⟹ 3|((a−b)+(b−c)) = 3|(a−c) ✓. Equivalence relation.
Check whether the function f: Z → Z defined by f(x) = x² is injective, surjective, or bijective.
Not injective: f(1) = f(−1) = 1. Not surjective: −1 has no pre-image (x² ≥ 0). Neither injective, surjective, nor bijective.
If f: R → R is defined by f(x) = 3x − 4, find f⁻¹.
Let y = 3x − 4. Then x = (y + 4)/3. So f⁻¹(x) = (x + 4)/3.
Give an example of a relation which is reflexive and transitive but not symmetric.
On A = {1, 2, 3}: R = {(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)}. Reflexive ✓, Transitive ✓, Not symmetric: (1,2) ∈ R but (2,1) ∉ R.
3-4 Mark Questions
Let A = R − {3} and B = R − {1}. Show that f: A → B defined by f(x) = (x − 2)/(x − 3) is bijective.
One-one: f(x₁) = f(x₂) ⟹ (x₁−2)(x₂−3) = (x₂−2)(x₁−3). Expanding and simplifying: x₁ = x₂. ✓
Onto: For y ∈ B, x = (3y−2)/(y−1) ∈ A gives f(x) = y. ✓
f is bijective.
Show that the function f: R → R defined by f(x) = x/(x² + 1) is neither one-one nor onto.
Not one-one: f(2) = 2/5, f(1/2) = (1/2)/(5/4) = 2/5. Same output, different inputs.
Not onto: From y = x/(x²+1), yx² − x + y = 0. Discriminant 1−4y² ≥ 0 gives −1/2 ≤ y ≤ 1/2. Range ≠ R.
If f: R → R is given by f(x) = (4x + 3)/(6x − 4), x ≠ 2/3, show that fof(x) = x.
f(f(x)) = f((4x+3)/(6x−4)) = (4·(4x+3)/(6x−4) + 3)/(6·(4x+3)/(6x−4) − 4) = (16x+12+18x−12)/(24x+18−24x+16) = 34x/34 = x. ✓
Key Concepts
Important Formulas
Study Plan - Relations and Functions
Types of Relations
Read NCERT Sections 1.2-1.3, solve Exercise 1.1 Q1-Q8
Types of Relations (continued)
Solve Exercise 1.1 Q9-Q16, focus on equivalence relation proofs
Types of Functions
Read Section 1.4, solve Exercise 1.2
Composition and Inverse
Read Sections 1.5-1.6, solve Exercise 1.3
Mixed Practice
Solve Miscellaneous Exercise + 3 PYQs
Common Mistakes to Avoid
- ✗Confusing reflexive with symmetric - reflexive means (a, a) for all a, symmetric means (a, b) ⟹ (b, a)
- ✗Forgetting to check ALL elements for reflexivity - one missing (a, a) makes it non-reflexive
- ✗Vacuous truth in transitivity - if there's no pair (a, b) and (b, c) both in R, transitivity holds vacuously
- ✗Confusing one-one with onto - one-one means distinct inputs give distinct outputs, onto means every element of codomain is mapped to
- ✗Not checking domain restrictions when finding inverse functions
- ✗Assuming gof = fog - composition of functions is NOT commutative in general
Tips for Board Exam
- ✓For equivalence relation proofs, always check all three properties separately and state each conclusion clearly
- ✓When showing a relation is NOT reflexive/symmetric/transitive, one counterexample is enough
- ✓For function questions, always specify domain and codomain clearly
- ✓To prove a function is bijective, prove one-one first, then onto
- ✓When finding inverse, replace f(x) with y, solve for x in terms of y, then write f⁻¹(y)
Practice Relations and Functions with AI Tutor
Stuck on a question? Ask our AI tutor to explain step-by-step in English or Hindi. Choose from 10 learning modes - Lecture, Quiz, Flashcards, and more. Free for 3 chapters per day.
Start Relations and Functions - Choose Your ModeAbout Relations and Functions - Class 12 Mathematics
Relations and Functions is the foundational chapter of Class 12 Mathematics. It builds on the Class 11 introduction to relations and functions by introducing formal classifications. A relation on a set can be reflexive (every element is related to itself), symmetric (if a is related to b, then b is related to a), and transitive (if a is related to b and b is related to c, then a is related to c). When a relation satisfies all three properties, it is called an equivalence relation.
The chapter then classifies functions into injective (one-one), surjective (onto), and bijective (both one-one and onto). A function is one-one if different inputs always produce different outputs, and onto if every element of the codomain has a pre-image. The concepts of composition of functions (applying one function after another) and invertible functions (functions that can be reversed) are introduced. A key result is that a function is invertible if and only if it is bijective.
This chapter is important not just for the board exam but also as a prerequisite for later chapters on Matrices, Determinants, and Calculus. Understanding function properties helps in analyzing continuity, differentiability, and integration in subsequent chapters.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 1?
Chapter 1 Relations and Functions has 3 exercises plus a Miscellaneous Exercise. Exercise 1.1 has 16 questions on types of relations, Exercise 1.2 has 12 questions on types of functions, Exercise 1.3 has 14 questions on composition and invertible functions, and the Miscellaneous Exercise has 19 questions.
What is the difference between a relation and a function?
A relation from set A to set B is any subset of A × B (any collection of ordered pairs). A function is a special type of relation where every element of A is associated with exactly one element of B. All functions are relations, but not all relations are functions.
How many marks does Relations and Functions carry in CBSE Class 12 board exam?
Relations and Functions is part of the unit that carries 8 marks in the CBSE Class 12 Maths board exam 2027. You can expect 1 MCQ (1 mark) and 1 long answer question (5 marks) from this chapter, plus possibly a 2-mark short answer question.
What is an equivalence relation?
A relation R on a set A is an equivalence relation if it is reflexive (aRa for all a), symmetric (aRb implies bRa), and transitive (aRb and bRc implies aRc). Examples include 'has the same number of elements as' on sets, and 'is parallel to' on lines.
Is NCERT enough for Relations and Functions in CBSE Class 12?
Yes, NCERT is sufficient for the board exam. Solve all exercises thoroughly - especially Exercise 1.1 (equivalence relations) and Exercise 1.3 (composition and inverse). The board questions closely follow NCERT patterns. For JEE, supplement with exemplar problems.
What is the difference between one-one and onto functions?
A one-one (injective) function means that different inputs always give different outputs - if f(x₁) = f(x₂), then x₁ = x₂. An onto (surjective) function means that every element of the codomain has at least one pre-image in the domain. A function can be one-one but not onto (e.g., f: N → N, f(x) = x + 1), onto but not one-one (e.g., f: Z → N, f(x) = |x|), both (bijective), or neither (e.g., f: R → R, f(x) = x²).
How to find the inverse of a function?
To find the inverse of a function f: Step 1: Write y = f(x). Step 2: Solve for x in terms of y. Step 3: Replace x with f⁻¹(y). Important: A function must be bijective (one-one and onto) to have an inverse. For example, if f(x) = 3x + 5, then y = 3x + 5 → x = (y − 5)/3 → f⁻¹(y) = (y − 5)/3. Always verify by checking that fof⁻¹(y) = y and f⁻¹of(x) = x.
What are equivalence classes?
An equivalence class is the set of all elements related to a given element under an equivalence relation. If R is an equivalence relation on set A, then the equivalence class of element a, denoted [a], is {x ∈ A : (x, a) ∈ R}. For example, if R on Z is defined by 'a − b is divisible by 3', then [0] = {..., −6, −3, 0, 3, 6, ...}, [1] = {..., −5, −2, 1, 4, 7, ...}, and [2] = {..., −4, −1, 2, 5, 8, ...}. The equivalence classes partition the set A into disjoint subsets.