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📐 Mathematics  ·  Class 12  ·  JEE

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

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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.

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
  • C
  • 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
  • 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
  • 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
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.