Below are 68 practice questions on Relations and Functions (Class 12), sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Relations and Functions (Class 12) notes.
Relations and Functions (Class 12) - Practice Questions with Answers
68 free MCQs on Relations and Functions (Class 12) with worked answers and explanations. Types of relations, equivalence classes, one-one and onto functions, composition, and invertible functions
Take the timed Relations and Functions (Class 12) chapterwise test →Easy - 20 questions
Q1.
A relation R on a set A is reflexive if:
- A (a, a) ∈ R for every a ∈ A
- B (a, b) ∈ R implies (b, a) ∈ R
- C (a, b) and (b, c) ∈ R imply (a, c) ∈ R
- D R is the empty set
Show answer & explanation
Answer: A. (a, a) ∈ R for every a ∈ A
Why: Reflexivity requires every element to be related to itself.
Q2.
A relation R is symmetric if:
- A (a, b) ∈ R implies a = b
- B (a, b) ∈ R implies (b, a) ∈ R
- C (a, a) ∈ R for all a
- D R contains every ordered pair
Show answer & explanation
Answer: B. (a, b) ∈ R implies (b, a) ∈ R
Why: Symmetry means the relation holds in both directions whenever it holds at all.
Q3.
A relation R is transitive if:
- A (a, a) ∈ R for all a
- B R has exactly three elements
- C (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R
- D (a, b) ∈ R implies (b, a) ∈ R
Show answer & explanation
Answer: C. (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R
Why: Transitivity lets related pairs chain together.
Q4.
An equivalence relation is one that is:
- A Reflexive and symmetric only
- B Symmetric and transitive only
- C One-one and onto
- D Reflexive, symmetric and transitive
Show answer & explanation
Answer: D. Reflexive, symmetric and transitive
Why: All three properties together define an equivalence relation.
Q5.
A function f : A → B is one-one (injective) if:
- A f(x₁) = f(x₂) implies x₁ = x₂
- B Every element of B has a preimage
- C The range equals the codomain
- D A and B have the same number of elements
Show answer & explanation
Answer: A. f(x₁) = f(x₂) implies x₁ = x₂
Why: Injectivity means distinct inputs never share an output.
Q6.
A function f : A → B is onto (surjective) if:
- A A has more elements than B
- B Range of f equals B
- C f(x₁) = f(x₂) implies x₁ = x₂
- D f is its own inverse
Show answer & explanation
Answer: B. Range of f equals B
Why: Surjectivity means every element of the codomain is actually attained.
Q7.
A function that is both one-one and onto is called:
- A Transitive
- B Constant
- C Bijective
- D Reflexive
Show answer & explanation
Answer: C. Bijective
Why: A bijection is injective and surjective at the same time.
Q8.
The composite function (g∘f)(x) is defined as:
- A f(g(x))
- B g(x)·f(x)
- C g(x) + f(x)
- D g(f(x))
Show answer & explanation
Answer: D. g(f(x))
Why: In g∘f the inner function f acts first, and g is applied to its output.
Q9.
A function f is invertible if and only if it is:
- A Bijective
- B One-one only
- C Onto only
- D Reflexive
Show answer & explanation
Answer: A. Bijective
Why: An inverse exists exactly when f is both one-one and onto.
Q10.
The identity function on a set A is defined by:
- A I(x) = −x for all x ∈ A
- B I(x) = x for all x ∈ A
- C I(x) = 0 for all x ∈ A
- D I(x) = 1 for all x ∈ A
Show answer & explanation
Answer: B. I(x) = x for all x ∈ A
Why: The identity map leaves every element unchanged.
Q11.
If f : A → B is invertible, then f⁻¹ is a function from:
- A A to A
- B B to B
- C B to A
- D A to B
Show answer & explanation
Answer: C. B to A
Why: The inverse reverses the direction of the mapping.
Q12.
For a relation R on set A, the universal relation is:
- A R = ∅
- B R = {(a, a) : a ∈ A}
- C R = A
- D R = A × A
Show answer & explanation
Answer: D. R = A × A
Why: The universal relation contains every possible ordered pair, so it is all of A × A.
Q13.
The empty relation on a non-empty set A is:
- A Symmetric and transitive but not reflexive
- B Reflexive but not symmetric
- C An equivalence relation
- D Reflexive and transitive
Show answer & explanation
Answer: A. Symmetric and transitive but not reflexive
Why: With no pairs at all, symmetry and transitivity hold vacuously, but reflexivity fails since (a, a) is missing.
Q14.
If f(x) = 2x and g(x) = x + 3, then (f∘g)(1) equals:
- A 4
- B 8
- C 5
- D 7
Show answer & explanation
Answer: B. 8
Why: g(1) = 4, then f(4) = 8.
Q15.
The number of elements in A × B when n(A) = 3 and n(B) = 4 is:
- A 81
- B 64
- C 12
- D 7
Show answer & explanation
Answer: C. 12
Why: The Cartesian product has n(A) × n(B) = 3 × 4 = 12 ordered pairs.
Q16.
If f(x) = x + 5, then f⁻¹(x) equals:
- A 5 − x
- B 1/(x + 5)
- C 5x
- D x − 5
Show answer & explanation
Answer: D. x − 5
Why: Solving y = x + 5 for x gives x = y − 5, so f⁻¹(x) = x − 5.
Q17.
The relation 'is equal to' on any set of numbers is:
- A An equivalence relation
- B Symmetric only
- C Transitive only
- D Not a relation at all
Show answer & explanation
Answer: A. An equivalence relation
Why: Equality is reflexive, symmetric and transitive, making it the standard example of an equivalence relation.
Q18.
If f : A → B and g : B → C, then g∘f is a function from:
- A A to B
- B A to C
- C C to A
- D B to C
Show answer & explanation
Answer: B. A to C
Why: Composition takes an input from A through f into B, then through g into C.
Q19.
A constant function f(x) = c on a set with more than one element is:
- A Bijective
- B An equivalence relation
- C Not one-one
- D One-one but not onto
Show answer & explanation
Answer: C. Not one-one
Why: Every input maps to the same output, so distinct inputs share an image and injectivity fails.
Q20.
If f(x) = 3x − 2, then f(2) equals:
- A 8
- B 1
- C 6
- D 4
Show answer & explanation
Answer: D. 4
Why: Substituting gives 3(2) − 2 = 4.
Medium - 20 questions
Q21.
The relation R = {(1,1), (2,2), (3,3), (1,2)} on A = {1,2,3} is:
- A Reflexive and transitive but not symmetric
- B An equivalence relation
- C Symmetric but not reflexive
- D Neither reflexive nor transitive
Show answer & explanation
Answer: A. Reflexive and transitive but not symmetric
Why: All three (a,a) pairs are present and no chain is broken, but (1,2) is present while (2,1) is not.
Q22.
On the set of lines in a plane, the relation 'is perpendicular to' is:
- A Transitive but not symmetric
- B Symmetric but neither reflexive nor transitive
- C An equivalence relation
- D Reflexive and transitive
Show answer & explanation
Answer: B. Symmetric but neither reflexive nor transitive
Why: If l ⊥ m then m ⊥ l, but no line is perpendicular to itself, and two lines perpendicular to the same line are parallel, not perpendicular.
Q23.
On ℤ, the relation aRb if a − b is divisible by 5 produces how many equivalence classes?
- A 10
- B Infinitely many
- C 5
- D 2
Show answer & explanation
Answer: C. 5
Why: The classes correspond to the possible remainders 0, 1, 2, 3 and 4 on division by 5.
Q24.
The function f : ℝ → ℝ given by f(x) = x² is:
- A One-one but not onto
- B Onto but not one-one
- C Bijective
- D Neither one-one nor onto
Show answer & explanation
Answer: D. Neither one-one nor onto
Why: f(2) = f(−2) breaks injectivity, and no negative number is attained so it is not onto ℝ.
Q25.
The function f : ℝ → ℝ given by f(x) = 3x + 5 is:
- A Bijective
- B One-one but not onto
- C Onto but not one-one
- D Neither one-one nor onto
Show answer & explanation
Answer: A. Bijective
Why: It is strictly increasing so injective, and every real y is attained by x = (y − 5)/3, so surjective.
Q26.
If f(x) = x + 1 and g(x) = x², then (g∘f)(2) equals:
- A 4
- B 9
- C 5
- D 6
Show answer & explanation
Answer: B. 9
Why: f(2) = 3, then g(3) = 9.
Q27.
If f(x) = x + 1 and g(x) = x², then (f∘g)(2) equals:
- A 6
- B 4
- C 5
- D 9
Show answer & explanation
Answer: C. 5
Why: g(2) = 4, then f(4) = 5 - different from (g∘f)(2), showing composition is not commutative.
Q28.
The number of one-one functions from a set with 3 elements to a set with 5 elements is:
- A 125
- B 243
- C 10
- D 60
Show answer & explanation
Answer: D. 60
Why: Choose distinct images in order: 5 × 4 × 3 = 60.
Q29.
The number of relations on a set with 3 elements is:
- A 2⁹
- B 3³
- C 2³
- D 9
Show answer & explanation
Answer: A. 2⁹
Why: A relation is any subset of A × A, which has 9 elements, so there are 2⁹ = 512 relations.
Q30.
If f : ℝ → ℝ is f(x) = 2x − 3, then f⁻¹(x) is:
- A 1/(2x − 3)
- B (x + 3)/2
- C (x − 3)/2
- D 2x + 3
Show answer & explanation
Answer: B. (x + 3)/2
Why: Solving y = 2x − 3 gives x = (y + 3)/2.
Q31.
The function f : ℕ → ℕ defined by f(n) = 2n is:
- A Bijective
- B Neither one-one nor onto
- C One-one but not onto
- D Onto but not one-one
Show answer & explanation
Answer: C. One-one but not onto
Why: Distinct n give distinct 2n, but no odd natural number is in the range.
Q32.
For an equivalence relation on a set A, the equivalence classes are:
- A Always of equal size in every case
- B Allowed to overlap partially
- C Always exactly two in number
- D Pairwise disjoint and their union is A
Show answer & explanation
Answer: D. Pairwise disjoint and their union is A
Why: Equivalence classes partition the set: every element lies in exactly one class.
Q33.
If f : A → B is bijective with n(A) = 5, then n(B) equals:
- A 5
- B 10
- C 25
- D 1
Show answer & explanation
Answer: A. 5
Why: A bijection pairs elements one-to-one with none left over, forcing the sets to be the same size.
Q34.
The relation R on A = {1,2,3} given by R = {(1,2), (2,1)} is:
- A Transitive only
- B Symmetric but not reflexive or transitive
- C An equivalence relation
- D Reflexive and symmetric
Show answer & explanation
Answer: B. Symmetric but not reflexive or transitive
Why: Both directions of the pair are present, but (1,1) is missing and (1,2) with (2,1) does not give (1,1).
Q35.
If f(x) = 1/x for x ≠ 0, then (f∘f)(x) equals:
- A 1/x²
- B x²
- C x
- D 1/x
Show answer & explanation
Answer: C. x
Why: f(f(x)) = 1/(1/x) = x, so f is its own inverse.
Q36.
The function f : ℝ → [0, ∞) given by f(x) = x² is:
- A One-one but not onto
- B Bijective
- C Neither one-one nor onto
- D Onto but not one-one
Show answer & explanation
Answer: D. Onto but not one-one
Why: Restricting the codomain to [0, ∞) makes every value attainable, but f(2) = f(−2) still breaks injectivity.
Q37.
If g∘f is defined, which condition must hold?
- A The range of f must be contained in the domain of g
- B The domain of f must equal the range of g
- C f and g must both be one-one functions
- D f and g must have identical formulas
Show answer & explanation
Answer: A. The range of f must be contained in the domain of g
Why: Outputs of f become inputs of g, so they must lie inside g's domain.
Q38.
The number of equivalence relations on the set {1, 2} is:
- A 3
- B 2
- C 1
- D 4
Show answer & explanation
Answer: B. 2
Why: Either both elements form a single class, or each sits alone - giving exactly 2 partitions and hence 2 equivalence relations.
Q39.
If f(x) = x³, then f : ℝ → ℝ is:
- A Onto but not one-one
- B Neither
- C Bijective
- D One-one but not onto
Show answer & explanation
Answer: C. Bijective
Why: The cube function is strictly increasing and covers all reals, so it is both injective and surjective.
Q40.
The relation 'is a sibling of' on a set of people (excluding oneself) is:
- A Reflexive and symmetric
- B Transitive and reflexive
- C Not symmetric
- D Symmetric but not reflexive
Show answer & explanation
Answer: D. Symmetric but not reflexive
Why: If A is a sibling of B then B is a sibling of A, but nobody is their own sibling.
Hard - 28 questions
Q41.
For a function between two finite sets of the same size, one-one implies onto. Why does this fail for infinite sets?
- A An infinite set can be placed in bijection with a proper subset of itself, leaving room for injective maps that miss elements
- B Infinite sets never admit one-one functions in the majority of documented cases
- C Onto functions cannot be defined on infinite sets under standard conventions
- D The pigeonhole principle applies only when the codomain is uncountable
Show answer & explanation
Answer: A. An infinite set can be placed in bijection with a proper subset of itself, leaving room for injective maps that miss elements
Why: The finite argument relies on counting; with infinite sets f(n) = 2n on ℕ is injective yet misses every odd number.
Q42.
If g∘f is one-one, what can be concluded?
- A Neither f nor g need be one-one
- B f must be one-one, but g need not be
- C g must be one-one, but f need not be
- D Both f and g must be one-one
Show answer & explanation
Answer: B. f must be one-one, but g need not be
Why: If f(x₁) = f(x₂) then g(f(x₁)) = g(f(x₂)), forcing x₁ = x₂; g can still collapse points outside the range of f.
Q43.
For invertible functions f and g, (g∘f)⁻¹ equals:
- A f∘g
- B (f∘g)⁻¹
- C f⁻¹∘g⁻¹
- D g⁻¹∘f⁻¹
Show answer & explanation
Answer: C. f⁻¹∘g⁻¹
Why: Undoing a composite reverses the order - like removing socks then shoes in the opposite order they were put on.
Q44.
Why is the empty relation on a non-empty set not reflexive, even though it is symmetric and transitive?
- A The empty relation is not a valid relation on any set by definition
- B Reflexivity holds vacuously as well, so the relation is in fact an equivalence relation
- C Reflexivity requires the set itself to be empty in all cases
- D Symmetry and transitivity hold vacuously with no pairs to check, but reflexivity actively demands that (a, a) be present
Show answer & explanation
Answer: D. Symmetry and transitivity hold vacuously with no pairs to check, but reflexivity actively demands that (a, a) be present
Why: Symmetry and transitivity are conditional statements satisfied when the hypothesis never occurs; reflexivity is an existence requirement that fails.
Q45.
The number of equivalence relations on the set {1, 2, 3} is:
- A 5
- B 3
- C 6
- D 8
Show answer & explanation
Answer: A. 5
Why: Equivalence relations correspond to partitions; the Bell number for a 3-element set is 5.
Q46.
The number of onto functions from a set of 3 elements to a set of 2 elements is:
- A 9
- B 6
- C 8
- D 2
Show answer & explanation
Answer: B. 6
Why: Total functions 2³ = 8, minus the 2 constant maps that miss an element, gives 6.
Q47.
The number of reflexive relations on a set with n elements is:
- A 2<sup>n</sup>
- B n²
- C 2<sup>n² − n</sup>
- D 2<sup>n²</sup>
Show answer & explanation
Answer: C. 2<sup>n² − n</sup>
Why: The n diagonal pairs are forced to be present; the remaining n² − n off-diagonal pairs are free, giving 2<sup>n² − n</sup>.
Q48.
Why is restricting the domain of f(x) = x² to [0, ∞) significant?
- A It makes the function onto ℝ, which the unrestricted version fails to be
- B It converts the function into a linear map on that interval
- C It removes the need for the function to be continuous
- D It makes the function one-one, allowing the square root to be defined as its inverse
Show answer & explanation
Answer: D. It makes the function one-one, allowing the square root to be defined as its inverse
Why: On [0, ∞) the squaring map is strictly increasing hence injective, and with codomain [0, ∞) it is bijective - exactly what defines √x.
Q49.
If f : ℝ → ℝ is f(x) = x/(1 + |x|), then f is:
- A One-one with range (−1, 1), so not onto ℝ
- B Onto ℝ but not one-one
- C Bijective from ℝ to ℝ
- D Neither one-one nor onto
Show answer & explanation
Answer: A. One-one with range (−1, 1), so not onto ℝ
Why: The map is strictly increasing hence injective, but |f(x)| < 1 always, so its range is the open interval (−1, 1).
Q50.
A relation that is symmetric and transitive but fails reflexivity on some element a indicates that:
- A Symmetry and transitivity are incompatible properties
- B The element a is not related to anything at all
- C The relation must be the universal relation
- D The set must be infinite in such cases
Show answer & explanation
Answer: B. The element a is not related to anything at all
Why: If a were related to some b, symmetry gives (b, a) and transitivity then forces (a, a) - so a must be isolated.
Q51.
The number of bijective functions from a set with n elements to itself is:
- A 2<sup>n</sup>
- B n<sup>n</sup>
- C n!
- D n²
Show answer & explanation
Answer: C. n!
Why: A bijection from a set to itself is a permutation, and there are n! of them.
Q52.
If f : A → B and g : B → C are both onto, then g∘f is:
- A One-one but not necessarily onto
- B Neither one-one nor onto
- C Bijective in every case
- D Onto
Show answer & explanation
Answer: D. Onto
Why: Each c ∈ C has a preimage b under g, and that b has a preimage a under f, so g(f(a)) = c.
Q53.
On the set A = {1, 2, 3}, how many relations are both reflexive and symmetric?
- A 8
- B 64
- C 512
- D 27
Show answer & explanation
Answer: A. 8
Why: The 3 diagonal pairs are forced, and the 3 off-diagonal unordered pairs are each independently in or out, giving 2³ = 8.
Q54.
Why does f invertible require f to be onto, not just one-one?
- A The inverse is only defined for continuous functions in most cases
- B f⁻¹ must be defined at every element of the codomain, which requires each to have a preimage
- C One-one functions never possess inverses in standard practice
- D Ontoness alone is sufficient for invertibility without injectivity
Show answer & explanation
Answer: B. f⁻¹ must be defined at every element of the codomain, which requires each to have a preimage
Why: f⁻¹ : B → A must assign a value to every b ∈ B; if some b has no preimage the inverse is undefined there.
Q55.
The relation R on ℝ given by aRb if a ≤ b is:
- A Symmetric and transitive only
- B Neither reflexive nor transitive
- C Reflexive and transitive but not symmetric
- D An equivalence relation
Show answer & explanation
Answer: C. Reflexive and transitive but not symmetric
Why: a ≤ a holds and the relation chains, but a ≤ b does not give b ≤ a unless they are equal - this is a partial order.
Q56.
If f(x) = (x − 1)/(x + 1) for x ≠ −1, then f(f(x)) equals:
- A x
- B 1/x
- C (x + 1)/(x − 1)
- D −1/x
Show answer & explanation
Answer: D. −1/x
Why: Substituting gives ((x−1)/(x+1) − 1)/((x−1)/(x+1) + 1) = (−2)/(2x) = −1/x.
Q57.
The number of symmetric relations on a set with n elements is:
- A 2^(n(n+1)/2)
- B 2<sup>n² − n</sup>
- C 2<sup>n²</sup>
- D n!
Show answer & explanation
Answer: A. 2^(n(n+1)/2)
Why: Choices are made on the n diagonal entries plus the n(n−1)/2 unordered off-diagonal pairs, totalling n(n+1)/2 free binary choices.
Q58.
Composition of functions is associative but not commutative. This means:
- A Composition is undefined unless the functions commute
- B (h∘g)∘f = h∘(g∘f) always, while g∘f = f∘g may fail
- C g∘f = f∘g always, while grouping matters
- D Both grouping and order are irrelevant to the result
Show answer & explanation
Answer: B. (h∘g)∘f = h∘(g∘f) always, while g∘f = f∘g may fail
Why: Regrouping never changes the result, but swapping the order generally does - f(x) = x + 1 and g(x) = x² already differ.
Q59.
If A has m elements and B has n elements with m > n, the number of one-one functions from A to B is:
- A mⁿ
- B nᵐ
- C 0
- D n!
Show answer & explanation
Answer: C. 0
Why: By the pigeonhole principle two elements of A must share an image, so no injective function exists.
Q60.
On ℤ, the relation aRb if a + b is even is:
- A Not reflexive, since a + a may be odd
- B Symmetric but not transitive
- C An equivalence relation with infinitely many classes
- D An equivalence relation with 2 classes
Show answer & explanation
Answer: D. An equivalence relation with 2 classes
Why: a + a = 2a is always even so it is reflexive, addition is symmetric, and transitivity holds - the classes are the evens and the odds.
Q61.
The number of equivalence relations on a 3-element set is:
- A 3
- B 5
- C 8
- D 15
Show answer & explanation
Answer: B. 5
Why: Equivalence relations correspond to partitions; the Bell number B₃ = 5.
Q62.
The inverse of the bijection f(x) = 3x − 7 on R is:
- A (x + 7)/3
- B (x − 7)/3
- C 3x + 7
- D (7 − x)/3
Show answer & explanation
Answer: A. (x + 7)/3
Why: Solving y = 3x − 7 gives x = (y + 7)/3.
Q63.
For the binary operation a * b = a + b − ab on R, the identity element is:
- A 1
- B 0
- C −1
- D 2
Show answer & explanation
Answer: B. 0
Why: a * e = a requires a + e − ae = a, so e(1 − a) = 0 for all a, giving e = 0.
Q64.
The number of bijections from {1, 2, 3} to {1, 2, 3} is:
- A 3
- B 6
- C 9
- D 27
Show answer & explanation
Answer: B. 6
Why: The number of bijections on a 3-element set is 3! = 6.
Q65.
The binary operation a * b = |a − b| on the reals is:
- A associative but not commutative
- B commutative but not associative
- C both associative and commutative
- D neither
Show answer & explanation
Answer: B. commutative but not associative
Why: |a − b| = |b − a| is commutative, but ||a−b|−c| ≠ |a−|b−c|| in general, so it is not associative.
Q66.
On the positive integers, the relation aRb defined by 'a divides b' is:
- A an equivalence relation
- B a partial order but not an equivalence relation
- C symmetric only
- D not transitive
Show answer & explanation
Answer: B. a partial order but not an equivalence relation
Why: It is reflexive, antisymmetric and transitive but not symmetric, hence a partial order.
Q67.
The number of onto functions from {1, 2, 3, 4} to {1, 2} is:
- A 8
- B 14
- C 16
- D 2
Show answer & explanation
Answer: B. 14
Why: Total functions 2⁴ = 16 minus the 2 constant ones = 14.
Q68.
For f(x) = x/(x − 1), x ≠ 1, the composition f(f(x)) equals:
- A x
- B 1/x
- C −x
- D (x − 1)/x
Show answer & explanation
Answer: A. x
Why: Substituting shows f is its own inverse, so f(f(x)) = x.