50 Extra Questions with Step-by-Step Solutions · PYQ
Class
12
Board
CBSE
Subject
Mathematics
Chapter
Relations and Functions
Category
Extra Questions
Language
English
Questions
50
Academic Session
Not specified
Chapter Introduction
This practice collection covers Class 12 Mathematics Chapter 1: Relations and Functions. Topics include types of relations (reflexive, symmetric, transitive and equivalence relations), types of functions (one-one/injective, onto/surjective and bijective), composition of functions and invertible functions.
The collection contains 50 Extra Questions organised as multiple-choice, short-answer, long-answer and case-based items, with answers and step-by-step solutions. It is intended as a self-study resource aligned with the Class 12 Mathematics syllabus for this chapter. Official reference sites: NCERT and CBSE.
Part A: Multiple Choice Questions (MCQs)
Question 1MCQ[1 Mark]
Let R be a relation on the set N of natural numbers defined by (a, b) in R if and only if a divides b. Then the relation R is:
(A) Equivalence relation
(B) Reflexive and transitive, but not symmetric
(C) Symmetric and transitive, but not reflexive
(D) An empty relation
Correct Option: (B)
Answer: (B)
Explanation: Reflexive since a divides a for all a in N. Not symmetric since 2 divides 4, but 4 does not divide 2. Transitive since if a divides b and b divides c, then a divides c.
Question 2MCQ[1 Mark]
Let A = {1, 2, 3}. The number of equivalence relations containing (1, 2) on set A is:
(A) 1
(B) 2
(C) 3
(D) 4
Correct Option: (B)
Answer: (B)
Explanation: The minimum equivalence relation containing (1, 2) must also contain (2, 1) by symmetry, and all diagonal elements (1,1), (2,2), (3,3) by reflexivity. This gives R1 = {(1,1), (2,2), (3,3), (1,2), (2,1)}. Adding (2,3) and its symmetric pair requires adding (1,3) and (3,1) for transitivity, yielding the universal relation R2 = A × A. Both are valid equivalence relations, totaling 2.
Question 3MCQ[1 Mark]
Consider the function f: R → R defined by f(x) = 3x − 2. Then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (A)
Answer: (A)
Explanation: For injectivity, 3x₁ − 2 = 3x₂ − 2 implies x₁ = x₂ (one-one). For surjectivity, for any y in R, x = (y+2)/3 in R such that f(x) = y (onto).
Question 4MCQ[1 Mark]
Let f: R → R be defined as f(x) = x4. Choose the correct option:
(A) f is one-one and onto
(B) f is many-one and onto
(C) f is one-one but not onto
(D) f is neither one-one nor onto
Correct Option: (D)
Answer: (D)
Explanation: f(−1) = f(1) = 1, so it is not one-one. Also, for negative real numbers like −2, there is no real x such that x4 = −2, so it is not onto.
Question 5MCQ[1 Mark]
If f: R → R is given by f(x) = cos x, then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (D)
Answer: (D)
Explanation: f(0) = f(2π) = 1, hence not one-one. The range of cos x is [−1, 1] not equal to R, hence not onto.
Question 6MCQ[1 Mark]
Let f: {1, 2, 3} → {1, 2, 3} be a function such that f(1) = 1, f(2) = 3, f(3) = 2. The value of (f ∘ f)(3) is:
(A) 1
(B) 2
(C) 3
(D) Undefined
Correct Option: (C)
Answer: (C)
Explanation: (f ∘ f)(3) = f(f(3)) = f(2) = 3.
Question 7MCQ[1 Mark]
Let f: R → R be defined by f(x) = 1 + x2. Then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (D)
Answer: (D)
Explanation: f(−1) = f(1) = 2 (not one-one), and the range is [1, ∞) not equal to R (not onto).
Question 8MCQ[1 Mark]
Let R be the relation in the set A = {1, 2, 3, 4} given by R = {(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2)}. Then R is:
(A) Reflexive and symmetric, but not transitive
(B) Reflexive and transitive, but not symmetric
(C) An equivalence relation
(D) Symmetric and transitive, but not reflexive
Correct Option: (B)
Answer: (B)
Explanation: All elements (a,a) for a in A are present (reflexive). (1,2) in R but (2,1) not in R (not symmetric). Transitivity holds across all pairs.
Question 9MCQ[1 Mark]
If f: R → R is defined by f(x) = x3, then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (A)
Answer: (A)
Explanation: Cubic polynomials over real numbers are strictly increasing, making them bijective (one-one and onto).
Question 10MCQ[1 Mark]
Let A = {1, 2, 3}. The number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is:
(A) 0
(B) 1
(C) 2
(D) 3
Correct Option: (A)
Answer: (A)
Explanation: Symmetric property requires (2,1) and (3,1) to be present. Reflexive requires (1,1), (2,2), (3,3). Transitivity of these mandatory pairs forces (2,3) and (3,2) as well, turning it into an equivalence relation. Thus, no such relation exists.
Question 11MCQ[1 Mark]
Let f: {1, 2, 3} → {4, 5, 6, 7} and g = {(1, 4), (2, 5), (3, 6)}. Then g is:
(A) One-one function
(B) Onto function
(C) Bijective function
(D) Not a function
Correct Option: (A)
Answer: (A)
Explanation: Distinct elements in domain have distinct images in codomain, but element 7 in codomain has no pre-image, making it one-one but not onto.
Question 12MCQ[1 Mark]
Let R be a relation on set A = {1, 2, 3} given by R = {(1,1), (2,2), (3,3), (1,2)}. Then R is:
(A) Reflexive and symmetric, but not transitive
(B) Reflexive and transitive, but not symmetric
(C) Symmetric and transitive, but not reflexive
(D) An equivalence relation
Correct Option: (B)
Answer: (B)
Explanation: Contains all (a,a) (reflexive). Contains (1,2) but not (2,1) (not symmetric). Transitivity is trivially satisfied.
Question 13MCQ[1 Mark]
If f: R → R is defined by f(x) = |x|, then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (D)
Answer: (D)
Explanation: f(−1) = f(1) = 1 (not one-one), and range is [0, ∞) not equal to R (not onto).
Question 14MCQ[1 Mark]
Let f(x) = (x − 2) / (x − 3). Then f: R − {3} → R − {1} is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (A)
Answer: (A)
Explanation: Solving y = (x − 2) / (x − 3) gives x = (3y − 2) / (y − 1), which is unique and well-defined for all y in R − {1}, proving bijectivity.
Question 15MCQ[1 Mark]
The number of all bijective functions from set A to itself, where A contains n elements, is:
(A) n2
(B) 2n
(C) n!
(D) n
Correct Option: (C)
Answer: (C)
Explanation: Total number of bijections (permutations) on a set with n elements is n!.
Question 16MCQ[1 Mark]
Let f: R → R be defined by f(x) = x2 − x + 1. Then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (D)
Answer: (D)
Explanation: f(0) = 1 and f(1) = 1 (not one-one). Range is [3/4, ∞) not equal to R (not onto).
Question 17MCQ[1 Mark]
Let R be the relation in the set Z of integers given by R = {(a, b) : 2 divides a − b}. Then R is:
(A) Reflexive only
(B) Symmetric only
(C) Transitive only
(D) An equivalence relation
Correct Option: (D)
Answer: (D)
Explanation: a − a = 0 is divisible by 2 (reflexive). If 2 divides (a−b), then 2 divides (b−a) (symmetric). If 2 divides (a−b) and 2 divides (b−c), then 2 divides (a−c) (transitive).
Question 18MCQ[1 Mark]
Let f: N → N be defined by f(n) = (n+1)/2 if n is odd, and n/2 if n is even. Then f is:
(A) One-one and onto
(B) One-one but not onto
(C) Onto but not one-one
(D) Neither one-one nor onto
Correct Option: (C)
Answer: (C)
Explanation: f(1) = 1 and f(2) = 1, so it is not one-one. For any natural number y, f(2y) = y, so it is onto.
Question 19MCQ[1 Mark]
Let f: R → R be defined by f(x) = 3 − 4x. Then inverse of f(x) is given by:
(A) (3 − x) / 4
(B) (x − 3) / 4
(C) 3 − 4x
(D) 1 / (3 − 4x)
Correct Option: (A)
Answer: (A)
Explanation: Let y = 3 − 4x implies 4x = 3 − y implies x = (3 − y) / 4, so inverse of f(x) = (3 − x) / 4.
Question 20MCQ[1 Mark]
Let A = {1, 2, 3}. The total number of reflexive relations on set A is:
(A) 23
(B) 26
(C) 29
(D) 2(n2 − n)
Correct Option: (B)
Answer: (B)
Explanation: Total pairs in A × A is 9. Out of these, 3 pairs are diagonal elements (1,1), (2,2), (3,3) which must be present. The remaining 6 pairs can either be chosen or omitted, giving 26 reflexive relations.
Part B: Short Answer Questions (Q21 – Q35)
Question 21Short Answer[2 Marks]
Check whether the relation R defined on the set R of real numbers as R = {(a, b) : a ≤ b3} is reflexive, symmetric, or transitive.
Answer: Neither reflexive, nor symmetric, nor transitive.
Explanation: 1. Reflexive: Take a = 1/2. Then 1/2 ≤ (1/2)3 = 1/8 is false. Not reflexive. 2. Symmetric: Take 1 ≤ 23 = 8 is true, but 2 ≤ 13 = 1 is false. Not symmetric. 3. Transitive: Take 3 ≤ 23 (8) and 2 ≤ (1.5)3 (3.375), but 3 ≤ 1.53 is false. Not transitive.
Question 22Short Answer[2 Marks]
Prove that the greatest integer function f: R → R given by f(x) = [x] is neither one-one nor onto.
Answer: Proved.
Explanation: f(1.2) = 1 and f(1.8) = 1, but 1.2 not equal to 1.8, so f is not one-one. Also, for 1.5 in R (codomain), there is no x in R such that [x] = 1.5 since [x] is always an integer. Thus, f is not onto.
Question 23Short Answer[2 Marks]
Let f: R → R be defined by f(x) = x2 + 1. Find pre-images of 10 and −3.
Answer: Pre-image of 10 is {3, −3}, and pre-image of −3 is empty set.
Explanation: For 10, x2 + 1 = 10 implies x2 = 9 implies x = ±3. For −3, x2 + 1 = −3 implies x2 = −4, which has no real solution.
Question 24Short Answer[2 Marks]
Show that the relation R in the set A of all polygons defined by R = {(P1, P2) : P1 and P2 have the same number of sides} is an equivalence relation.
Answer: Proved.
Explanation: 1. Reflexive: Polygon P1 has the same number of sides as itself implies (P1, P1) in R. 2. Symmetric: If (P1, P2) in R, P1 and P2 have same number of sides implies P2 and P1 have same number of sides implies (P2, P1) in R. 3. Transitive: If (P1, P2) in R and (P2, P3) in R, then P1, P2, P3 all have the same number of sides implies (P1, P3) in R. Hence, an equivalence relation.
Question 25Short Answer[2 Marks]
Let f: R → R and g: R → R be defined by f(x) = sin x and g(x) = x2. Find (f ∘ g)(x) and (g ∘ f)(x). Are they equal?
Determine whether the relation R on the set of all integers Z defined by R = {(x, y) : x − y is an integer} is an equivalence relation.
Answer: It is an equivalence relation.
Explanation: x − x = 0 is an integer (reflexive). If x − y is an integer, y − x = −(x − y) is also an integer (symmetric). If x − y and y − z are integers, their sum (x − y) + (y − z) = x − z is an integer (transitive).
Question 27Short Answer[2 Marks]
Let f: N → Y be a function defined as f(n) = 4n + 3, where Y = {y in N : y = 4n + 3 for some n in N}. Show that f is invertible and find its inverse.
Answer: Inverse of f(y) = (y − 3) / 4.
Explanation: Let y = 4x + 3 implies x = (y − 3) / 4. Define g: Y → N by g(y) = (y − 3) / 4. Then (g ∘ f)(x) = x and (f ∘ g)(y) = y. Thus, f is invertible.
Question 28Short Answer[2 Marks]
If f: R → R is defined by f(x) = (2x − 1) / 3, find the inverse of f evaluated at 5.
Answer: 8
Explanation: Let f(x) = y implies (2x − 1) / 3 = 5 implies 2x − 1 = 15 implies 2x = 16 implies x = 8.
Question 29Short Answer[2 Marks]
Show that the function f: R → R defined by f(x) = x / (x2 + 1) is neither one-one nor onto.
Answer: Proved.
Explanation: For one-one, f(2) = 2/5 and f(1/2) = (1/2)/(5/4) = 2/5, but 2 not equal to 1/2 (not one-one). For onto, range of f(x) is [−1/2, 1/2] not equal to R (not onto).
Question 30Short Answer[2 Marks]
Let A = {1, 2, 3}. Give an example of a relation which is symmetric but neither reflexive nor transitive.
Answer: R = {(1,2), (2,1)}.
Explanation: (1,1) not in R (not reflexive). If (1,2) in R, then (2,1) in R (symmetric). However, (1,2) and (2,1) in R does not imply (1,1) in R (not transitive).
Question 31Short Answer[2 Marks]
Find the number of binary operations on the set {a, b}.
Answer: 16
Explanation: A binary operation on set A is a function from A × A to A. Here |A × A| = 2 × 2 = 4 and |A| = 2. Total number of functions is 24 = 16.
Question 32Short Answer[2 Marks]
Let f, g: R → R be defined by f(x) = |x| and g(x) = [x]. Evaluate (g ∘ f)(−5/3) − (f ∘ g)(−5/3).
Explanation: For a, b in N and a = 10 − 2b: b=1 gives a=8; b=2 gives a=6; b=3 gives a=4; b=4 gives a=2. Thus domain is {2, 4, 6, 8} and range is {1, 2, 3, 4}.
Question 34Short Answer[2 Marks]
Prove that the signum function sgn(x) is neither one-one nor onto.
Answer: Proved.
Explanation: sgn(2) = 1 and sgn(5) = 1, but 2 not equal to 5 (not one-one). The range is {−1, 0, 1} not equal to R (not onto).
Question 35Short Answer[2 Marks]
Let f(x) = x2 − 3x + 2. Find f(f(x)).
Answer: x4 − 6x3 + 10x2 − 3x
Explanation: Substituting f(x) into itself: (x2 − 3x + 2)2 − 3(x2 − 3x + 2) + 2 which simplifies to x4 − 6x3 + 10x2 − 3x.
Part C: Long Answer Questions (Q36 – Q45)
Question 36Long Answer[4 Marks]
Show that the relation R in A = {x in 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.
Answer: Equivalence relation; Equivalence class of 1 is {1, 5, 9}.
Explanation: 1. Reflexive: |a − a| = 0 (multiple of 4) implies (a,a) in R. 2. Symmetric: |a − b| = 4k implies |b − a| = 4k. 3. Transitive: Sum of multiples is a multiple. Equivalence class of 1: {x in A : |x − 1| is multiple of 4} = {1, 5, 9}.
Question 37Long Answer[4 Marks]
Let f(x) = (5x + 3) / (6x − 5). Show that f is invertible and find its inverse.
Answer: Inverse of f(y) = (5y + 3) / (6y − 5).
Explanation: Let y = (5x + 3) / (6x − 5). Cross multiplying and rearranging for x gives x = (5y + 3) / (6y − 5), showing bijectivity and giving the inverse function.
Question 38Long Answer[4 Marks]
Prove that the relation R on R defined by R = {(a, b) : a ≤ b2} is neither reflexive, nor symmetric, nor transitive.
Answer: Proved with counterexamples.
Explanation: For reflexive, take a = 1/2 (1/2 ≤ 1/4 is false). For symmetric, take a=1, b=2 (1 ≤ 4 true, but 2 ≤ 1 false). For transitive, counterexamples verify failure.
Question 39Long Answer[4 Marks]
Let f: N → N be defined as f(x) = 4x2 + 12x + 15. Show that f is invertible and find its inverse.
Answer: Inverse of f(y) = (√(y − 6) − 3) / 2.
Explanation: Rewriting as y = (2x + 3)2 + 6, solving for x in terms of y gives the unique inverse expression.
Question 40Long Answer[4 Marks]
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible and find its inverse.
Answer: Inverse of f(y) = (√(y + 6) − 1) / 3.
Explanation: Complete the square to get f(x) = (3x + 1)2 − 6, then invert by solving for x.
Question 41Long Answer[4 Marks]
Let A = R − {3} and B = R − {1}. Consider f(x) = (x − 2) / (x − 3). Is f bijective?
Answer: Yes, f is bijective.
Explanation: Injectivity and surjectivity are verified algebraically by showing unique pre-images for every element in the codomain.
Question 42Long Answer[4 Marks]
Let R be a relation on lines in a plane defined by parallel lines. Show it is an equivalence relation and find lines related to y = 2x + 4.
Answer: Family of lines y = 2x + c where c is a real constant.
Explanation: Parallelism is reflexive, symmetric, and transitive. All parallel lines share the same slope 2.
Question 43Long Answer[4 Marks]
Let f: R → R be defined by f(x) = x3 + 5. Show that f is a bijection.
Answer: Proved.
Explanation: Strict monotonicity ensures one-one, and for any real y, x = (y − 5)1/3 provides a valid pre-image ensuring surjectivity.
Question 44Long Answer[4 Marks]
Determine whether the relation R on N × N defined by (a, b) R (c, d) iff a + d = b + c is an equivalence relation.
Answer: Proved to be an equivalence relation.
Explanation: Verified using reflexivity (a+b=b+a), symmetry, and transitivity properties of integer addition.
Question 45Long Answer[5 Marks]
Let A = {1, 2, …, 9} and R on A × A be defined by (a, b) R (c, d) iff a + d = b + c. Find equivalence class [(2, 5)].
This collection contains 50 Extra Questions on Class 12 Mathematics Chapter 1: Relations and Functions, including multiple-choice, short-answer, long-answer and case-based items, each with an answer and explanation.
How should students use these questions?
Attempt each question independently, then use the “View Answer / Solution” button on that question to check your working. Use the content index to jump to any question number.
Are solutions included for every question?
Yes. Every question in this collection includes an answer and an explanation in the page source. Solutions are hidden by default and revealed per question when you choose to view them.
Conclusion & Study Guidance
Regular practice on relations, types of functions, composition and invertibility strengthens preparation for Class 12 Mathematics. Work through the questions in this collection and revisit any topic that needs more attention. More resources are available at pyqly.com.