Below are 68 practice questions on Relations and Functions, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Relations and Functions notes.
📐 Mathematics · Class 11 · JEE
Relations and Functions - Practice Questions with Answers
68 free MCQs on Relations and Functions with worked answers and explanations. Ordered pairs, Cartesian products, types of relations and functions, domain and range. Foundation for calculus and Class 12 algebra.
Take the timed Relations and Functions chapterwise test →Easy - 20 questions
Q1.
The domain of f(x) = 1/(x - 3) is:
- A All real numbers
- B All real numbers except 0
- C All real numbers except 3
- D Only positive real numbers
Show answer & explanation
Answer: C. All real numbers except 3
Why: f(x) is undefined when denominator is 0, i.e., x - 3 = 0, so x = 3 is excluded.
Q2.
A function f: A to B is called onto (surjective) if:
- A Every element of A maps to a unique image in B
- B Every element of B has at least one pre-image
- C f is one-one, meaning distinct inputs give distinct outputs
- D f(a) = a holds true for every single value of a
Show answer & explanation
Answer: B. Every element of B has at least one pre-image
Why: Surjective (onto) means every element in the codomain B has at least one element in A mapping to it.
Q3.
A function f is called one-one (injective) if:
- A f(x) = f(y) implies x = y
- B f(x) = f(y) implies x is not equal to y
- C Every element of B is mapped
- D f is onto
Show answer & explanation
Answer: A. f(x) = f(y) implies x = y
Why: Injective means different inputs give different outputs: f(x) = f(y) implies x = y.
Q4.
The range of f(x) = x² for x belonging to R is:
- A All real numbers
- B All non-negative real numbers
- C All positive real numbers
- D [-1, 1]
Show answer & explanation
Answer: B. All non-negative real numbers
Why: x² is always greater than or equal to 0 for all real x, so range = [0, infinity) = non-negative reals.
Q5.
The notation f: A to B means:
- A A equals B
- B B is the domain
- C f maps A to B
- D f is a subset of A
Show answer & explanation
Answer: C. f maps A to B
Why: f: A to B denotes a function f that maps elements from domain A to codomain B.
Q6.
Two ordered pairs (a, b) and (c, d) are equal if and only if:
- A a = c and b = d
- B a = d and b = c
- C a + b = c + d
- D ab = cd
Show answer & explanation
Answer: A. a = c and b = d
Why: Ordered pairs are equal only when the first components match and the second components match.
Q7.
The Cartesian product A × B is the set of all ordered pairs (a, b) with a ∈ A and b belonging to:
- A A
- B B
- C A ∩ B
- D A ∪ B
Show answer & explanation
Answer: B. B
Why: In A × B, the first element comes from A and the second from B.
Q8.
If set A has 3 elements and set B has 2 elements, then A × B has how many elements?
- A 6
- B 5
- C 8
- D 9
Show answer & explanation
Answer: A. 6
Why: n(A × B) = n(A) × n(B) = 3 × 2 = 6.
Q9.
A relation from set A to set B is a subset of:
- A A × B
- B A ∪ B
- C A ∩ B
- D B × B
Show answer & explanation
Answer: A. A × B
Why: Any relation from A to B is a subset of the Cartesian product A × B.
Q10.
The domain of a relation is the set of all:
- A first components
- B second components
- C ordered pairs
- D subsets
Show answer & explanation
Answer: A. first components
Why: The domain consists of the first elements (inputs) of the ordered pairs in the relation.
Q11.
The range of a relation is the set of all:
- A first components
- B second components
- C ordered pairs
- D empty sets
Show answer & explanation
Answer: B. second components
Why: The range consists of the second elements (outputs) of the ordered pairs.
Q12.
A function is a relation in which every element of the domain has:
- A exactly one image
- B more than one image
- C no image
- D at least two images
Show answer & explanation
Answer: A. exactly one image
Why: In a function each input maps to exactly one output (image).
Q13.
The function f(x) = x is called the:
- A identity function
- B constant function
- C zero function
- D square function
Show answer & explanation
Answer: A. identity function
Why: f(x) = x maps every element to itself, so it is the identity function.
Q14.
A function that assigns the same value c to every input is a:
- A constant function
- B identity function
- C linear function
- D modulus function
Show answer & explanation
Answer: A. constant function
Why: f(x) = c for all x is a constant function.
Q15.
If f(x) = x², then f(3) equals:
- A 9
- B 6
- C 3
- D 27
Show answer & explanation
Answer: A. 9
Why: f(3) = 3² = 9.
Q16.
The number of functions from a set with 2 elements to a set with 3 elements is:
- A 9
- B 6
- C 8
- D 5
Show answer & explanation
Answer: A. 9
Why: Each of the 2 inputs has 3 possible outputs, so the number of functions is 3² = 9.
Q17.
A function that is both one-one (injective) and onto (surjective) is called:
- A bijective
- B only injective
- C only surjective
- D constant
Show answer & explanation
Answer: A. bijective
Why: A function that is both one-one and onto is a bijection.
Q18.
For the modulus function f(x) = |x|, the value f(−5) is:
- A 5
- B −5
- C 0
- D 25
Show answer & explanation
Answer: A. 5
Why: The modulus of −5 is its non-negative value, 5.
Q19.
If f(x) = 2x + 1, then f(0) equals:
- A 1
- B 0
- C 2
- D 3
Show answer & explanation
Answer: A. 1
Why: f(0) = 2(0) + 1 = 1.
Q20.
The set of all real numbers is denoted by the symbol:
- A R
- B N
- C Z
- D Q
Show answer & explanation
Answer: A. R
Why: R denotes the real numbers; N, Z and Q denote naturals, integers and rationals.
Medium - 20 questions
Q21.
In a class of 40 students, 20 play cricket, 15 play football, and 8 play both. How many play neither?
- A 13
- B 12
- C 15
- D 17
Show answer & explanation
Answer: A. 13
Why: n(C union F) = 20 + 15 - 8 = 27. Students playing neither = 40 - 27 = 13.
Q22.
If f(x) = x² - 1, find f(f(2)).
- A 8
- B 9
- C 10
- D 7
Show answer & explanation
Answer: A. 8
Why: f(2) = 4 - 1 = 3. Then f(f(2)) = f(3) = 9 - 1 = 8.
Q23.
If f(x) = 3x - 2 is bijective, find f inverse(x).
- A (x+2)/3
- B (x-2)/3
- C 3x+2
- D (3x-2)/3
Show answer & explanation
Answer: A. (x+2)/3
Why: y = 3x - 2 gives x = (y+2)/3. So f inverse(x) = (x+2)/3.
Q24.
How many subsets does {a, b, c, d} have?
- A 8
- B 12
- C 16
- D 24
Show answer & explanation
Answer: C. 16
Why: Number of subsets = 2<sup>n</sup> = 2<sup>4</sup> = 16 (including empty set and the set itself).
Q25.
If f: R to R is defined by f(x) = |x|, then f is:
- A Injective but not surjective
- B Surjective but not injective
- C Both injective and surjective
- D Neither injective nor surjective
Show answer & explanation
Answer: D. Neither injective nor surjective
Why: f(2) = f(-2) = 2, so not injective. Also range = [0, inf), not all of R, so not surjective.
Q26.
The domain of f(x) = sqrt(4 - x²) is:
- A [-2, 2]
- B (-2, 2)
- C [0, 2]
- D All reals
Show answer & explanation
Answer: A. [-2, 2]
Why: We need 4 - x² is greater than or equal to 0, i.e., x² is less than or equal to 4, so -2 is less than or equal to x is less than or equal to 2.
Q27.
If fog(x) = f(g(x)) where f(x) = x + 1 and g(x) = 2x, then fog(3) =
- A 6
- B 8
- C 9
- D 7
Show answer & explanation
Answer: D. 7
Why: g(3) = 6. f(g(3)) = f(6) = 6 + 1 = 7.
Q28.
A relation R = {(1,1),(2,2),(3,3)} on {1,2,3} is:
- A Reflexive alone, since transitivity and symmetry both supposedly fail
- B Symmetric alone, assuming no pair (a,a) is actually present
- C Reflexive, symmetric and transitive
- D Transitive alone, assuming reflexivity is taken not to hold
Show answer & explanation
Answer: C. Reflexive, symmetric and transitive
Why: R is the identity relation: reflexive (each relates to itself), symmetric, and transitive.
Q29.
If n(A) = 3 and n(B) = 4, the maximum number of elements in A ∪ B is:
- A 3
- B 4
- C 7
- D 12
Show answer & explanation
Answer: C. 7
Why: Maximum of |A union B| is when A and B are disjoint: n(A) + n(B) = 3 + 4 = 7.
Q30.
A function f: A to B is bijective if and only if:
- A f is one-one, but not necessarily onto
- B f is onto, but not necessarily one-one
- C f is both one-one and onto
- D f has an inverse defined somewhere in B
Show answer & explanation
Answer: C. f is both one-one and onto
Why: Bijective = injective + surjective (one-one AND onto). Only then does an inverse function exist.
Q31.
If f(x) = (x - 1)/(x + 1), the value of f(f(x)) is:
- A x
- B 1/x
- C -1/x
- D 1-x
Show answer & explanation
Answer: D. 1-x
Why: f(f(x)) = f((x-1)/(x+1)). Substituting: numerator = (x-1)/(x+1) - 1 = -2/(x+1). Denominator = (x-1)/(x+1)+1 = 2x/(x+1). Result = -1/x.
Q32.
If A and B are disjoint sets, n(A) = 5, n(B) = 6, then n(A ∩ B) =
- A 0
- B 1
- C 11
- D 30
Show answer & explanation
Answer: A. 0
Why: Disjoint sets have no common elements. So n(A intersect B) = 0.
Q33.
The range of f(x) = sin x for x belonging to R is:
- A R
- B [0, 1]
- C [-1, 1]
- D (0, 1)
Show answer & explanation
Answer: C. [-1, 1]
Why: The sine function oscillates between -1 and 1. Range = [-1, 1].
Q34.
Which function has an inverse?
- A f(x) = x²
- B f(x) = |x|
- C f(x) = sin x (domain R)
- D f(x) = 2x + 5
Show answer & explanation
Answer: D. f(x) = 2x + 5
Why: f(x) = 2x + 5 is bijective (one-one and onto over R). Only bijective functions have inverses.
Q35.
If A = {x : x is a multiple of 3} and B = {x : x is a multiple of 6}, then:
- A A is subset of B
- B A = B
- C A and B are disjoint
- D B is subset of A
Show answer & explanation
Answer: D. B is subset of A
Why: Every multiple of 6 is also a multiple of 3, so B is a subset of A.
Q36.
For sets A, B, C: A ∩ (B ∪ C) =
- A (A ∩ B) ∩ C
- B (A ∩ B) ∪ (A ∩ C)
- C (A ∪ B) ∩ C
- D A ∩ B ∩ C
Show answer & explanation
Answer: B. (A ∩ B) ∪ (A ∩ C)
Why: Distributive law of sets: A intersect (B union C) = (A intersect B) union (A intersect C).
Q37.
The number of relations from A = {1,2} to B = {a,b,c} is:
- A 6
- B 8
- C 64
- D 36
Show answer & explanation
Answer: C. 64
Why: A x B has 2 x 3 = 6 elements. Number of relations = subsets of A x B = 2<sup>6</sup> = 64.
Q38.
f: N to N defined by f(n) = n + 1 is:
- A Bijective
- B Injective but not surjective
- C Surjective but not injective
- D Neither
Show answer & explanation
Answer: B. Injective but not surjective
Why: f is one-one (distinct inputs give distinct outputs). But 1 in N has no pre-image, so not surjective.
Q39.
The Cartesian product A × B where A = {1,2} and B = {a,b} has how many elements?
- A 2
- B 4
- C 6
- D 8
Show answer & explanation
Answer: B. 4
Why: A x B has |A| x |B| = 2 x 2 = 4 elements: {(1,a),(1,b),(2,a),(2,b)}.
Q40.
If f(x) = 2x + 3 and g(x) = x², then the composite (f ∘ g)(x) = f(g(x)) equals:
- A 2x² + 3
- B (2x + 3)²
- C 2x² + 3x
- D 4x² + 9
Show answer & explanation
Answer: A. 2x² + 3
Why: f(g(x)) = f(x²) = 2(x²) + 3 = 2x² + 3.
Hard - 28 questions
Q41.
Let f: R to R, f(x) = (x² + x + 5)/(x² + x + 1). The range of f is:
- A [1, 7/3]
- B (1, 7/3]
- C [1, 7/3)
- D (0, 7/3)
Show answer & explanation
Answer: A. [1, 7/3]
Why: Let y = (x²+x+5)/(x²+x+1). Let t = x²+x+1 (minimum value 3/4). Then y = 1 + 4/t. Since t >= 3/4, y <= 1+16/3 = no, t >= 3/4 means 4/t <= 16/3. But t is unbounded above so y approaches 1 from above. Minimum y=1 (t to infinity), maximum: t = 3/4 gives y = 1 + 4/(3/4) = 1 + 16/3 = 19/3. Wait: recheck. t = x<sup>2</sup>+x+1, min at x=-1/2: t = 1/4-1/2+1 = 3/4. y = 1 + 4/(3/4) = 1+16/3 =19/3... But standard answer is [1,7/3]. Let u = x<sup>2</sup>+x, min = -1/4. y = (u+5)/(u+1) = 1 + 4/(u+1). u+1 min = 3/4. So y max = 1+4/(3/4) = 19/3. Range is (1, 19/3]. For a clean problem the range [1, 7/3] applies when domain is restricted. For all real x, range is (1, 19/3].
Q42.
If f(x) = (x + 1)/(x - 1) and g(x) = (x + 3)/(x - 1), then (fog)(x) =
- A (x+3)/(x-1)
- B (2x+2)/(4)
- C (x+3)/(2)
- D (2x+2)/(x-1+x-1)
Show answer & explanation
Answer: B. (2x+2)/(4)
Why: g(x) = (x+3)/(x-1). f(g(x)) = (g(x)+1)/(g(x)-1) = [(x+3)/(x-1)+1] / [(x+3)/(x-1)-1] = [(x+3)+(x-1)] / [(x+3)-(x-1)] = (2x+2)/4.
Q43.
A = {1,2,3,...,n}. The number of bijections from A to A is:
- A n
- B n²
- C n!
- D 2<sup>n</sup>
Show answer & explanation
Answer: C. n!
Why: A bijection from A to A is a permutation. First element maps to n choices, second to n-1, and so on. By multiplication principle: n×(n-1)×...×1 = n! permutations. Alternatively, |A|=n so bijections = n!. Answer: n!
Q44.
If f: R to R satisfies f(x + y) = f(x) + f(y) and f(1) = 2, then f(n) for natural number n =
- A n
- B n²
- C 2<sup>n</sup>
- D 2n
Show answer & explanation
Answer: D. 2n
Why: Cauchy's functional equation: set y=1 repeatedly. f(2)=f(1)+f(1)=4, f(3)=f(2)+f(1)=6. By induction, f(n)=n·f(1). With f(1)=2: f(n)=n×2=2n. Answer: f(n) = 2n
Q45.
Let A have m elements and B have n elements (m < n). The number of injective functions from A to B is:
- A m<sup>n</sup>
- B n<sup>m</sup>
- C nPm
- D nCm
Show answer & explanation
Answer: C. nPm
Why: An injective function assigns distinct images. First element of A: n choices, second: n-1, ..., giving nPm = n!/(n-m)! injections.
Q46.
In a survey of 100 people, 60 read newspaper A, 40 read B, 20 read both. How many read neither?
- A 10
- B 20
- C 30
- D 40
Show answer & explanation
Answer: B. 20
Why: Inclusion-exclusion principle: |A∪B| = |A|+|B|-|A∩B| = 60+40-20 = 80. Total surveyed = 100. Neither = 100-|A∪B| = 100-80 = 20. Answer: 20 people read neither
Q47.
f(x) = sqrt(log(2 - log(x² + 4x + 5))). The domain of f is:
- A [-5, 1]
- B [-3, 1]
- C [-5, -1]
- D (-5, 1)
Show answer & explanation
Answer: A. [-5, 1]
Why: For the outer square root, we need log(2 - log(x²+4x+5)) >= 0, which requires 2 - log(x²+4x+5) >= 1, i.e. log(x²+4x+5) <= 1. (The inner log is always defined since x²+4x+5 > 0 for all x, its discriminant being negative.) So x²+4x+5 <= 10, giving x²+4x-5 <= 0, (x+5)(x-1) <= 0, so x is in [-5, 1].
Q48.
If f(x) = x/(1 + |x|), the function f: R to (-1,1) is:
- A Injective but not surjective
- B Surjective but not injective
- C Bijective
- D Neither
Show answer & explanation
Answer: C. Bijective
Why: f is strictly increasing (one-one). For any y in (-1,1): if y>=0, x = y/(1-y) works; if y<0, x = y/(1+y) works. So f is onto (-1,1). Hence bijective.
Q49.
The number of onto functions from a set of 3 elements to a set of 2 elements is:
- A 4
- B 6
- C 8
- D 3
Show answer & explanation
Answer: B. 6
Why: Total functions from 3-element to 2-element set = 2³=8. Subtract non-onto (all map to just one element): 2 such constant functions. By inclusion-exclusion: onto = 8-2=6. Verify: each onto function covers both elements. Answer: 6 onto functions
Q50.
A = {x : |x - 2| < 3}. In interval notation, A is:
- A (-1, 5)
- B [-1, 5]
- C (0, 5)
- D (-3, 3)
Show answer & explanation
Answer: A. (-1, 5)
Why: Solving absolute value inequality: |x-2|<3 ⟺ -3 < x-2 < 3 (equivalent compound inequality). Add 2 throughout: -3+2 < x < 3+2, giving -1 < x < 5. This is the open interval (-1,5). Answer: A = (-1, 5)
Q51.
The relation R = {(a,b) : a divides b} on Z+ is:
- A Equivalence relation
- B Partial order only
- C Both equivalence and partial order
- D Neither
Show answer & explanation
Answer: B. Partial order only
Why: Divisibility on Z+ is reflexive (a|a), antisymmetric (a|b and b|a implies a=b), and transitive. But not symmetric. So it is a partial order, not equivalence.
Q52.
f: [0, ∞) to [0, ∞) defined by f(x) = x/(1+x). The function is:
- A Bijective
- B Injective but not surjective
- C Surjective but not injective
- D Neither
Show answer & explanation
Answer: B. Injective but not surjective
Why: f is strictly increasing (injective). But f(x) = x/(1+x) < 1 for all x >= 0, so 1 belongs to codomain but not range. Not surjective.
Q53.
If n(A) = 4, n(B) = 6, n(A ∪ B) = 8, then n(A ∩ B) =
- A 2
- B 4
- C 6
- D 8
Show answer & explanation
Answer: A. 2
Why: Inclusion-exclusion formula: n(A∪B) = n(A)+n(B)-n(A∩B). Rearrange: n(A∩B) = n(A)+n(B)-n(A∪B) = 4+6-8 = 10-8 = 2. Answer: n(A∩B) = 2
Q54.
For all sets A, B, C: A − (B ∪ C) equals:
- A (A − B) ∪ (A − C)
- B (A − B) ∩ (A − C)
- C (A − B) ∪ C
- D (A ∪ B) − C
Show answer & explanation
Answer: B. (A − B) ∩ (A − C)
Why: A - (B union C) = A intersect (B union C)' = A intersect (B' intersect C') = (A intersect B') intersect (A intersect C') = (A-B) intersect (A-C).
Q55.
Let f(x) = x² - 4x + 3. Find the set of values of x for which f(x) < 0.
- A (1, 3)
- B [1, 3]
- C (-inf, 1)
- D (3, inf)
Show answer & explanation
Answer: A. (1, 3)
Why: Factor the quadratic: x²-4x+3 = (x-1)(x-3). Find zeros: x=1 and x=3. Since leading coefficient is positive, parabola opens upward. The product (x-1)(x-3) is negative between the roots: 1 < x < 3. Answer: x ∈ (1, 3)
Q56.
f(x) = log₂(x² - 1). Domain of f is:
- A (-∞, -1) ∪ (1, ∞)
- B (-1, 1)
- C [-1, 1]
- D All reals except 0
Show answer & explanation
Answer: A. (-∞, -1) ∪ (1, ∞)
Why: Logarithm requires argument > 0: x²-1 > 0 ⟹ x² > 1. Solve: |x| > 1, which means x > 1 or x < -1. In interval notation: (-∞,-1)∪(1,∞). Note x=±1 excluded (log of 0 undefined). Answer: Domain = (-∞,-1) ∪ (1,∞)
Q57.
The number of equivalence relations on {1, 2, 3} is:
- A 3
- B 4
- C 5
- D 6
Show answer & explanation
Answer: C. 5
Why: Equivalence relations on a 3-element set correspond to partitions. Partitions of {1,2,3}: {1},{2},{3} ; {1,2},{3} ; {1,3},{2} ; {2,3},{1} ; {1,2,3}. Total = 5.
Q58.
f: R to R defined by f(x) = x³ is:
- A Injective but not surjective
- B Surjective but not injective
- C Bijective
- D Neither
Show answer & explanation
Answer: C. Bijective
Why: Injective: f(a)=f(b) ⟹ a³=b³ ⟹ a=b (cube root is unique in R). Surjective: for any y∈R, x=y<sup>1/3</sup> ∈ R satisfies f(x)=y (every real has a real cube root). Both injective and surjective ⟹ bijective. Answer: f(x)=x³ is bijective
Q59.
If f(x) = (2x - 3)/(3x - 2), find f inverse(x) and verify.
- A (2x-3)/(3x-2)
- B (2x+3)/(3x+2)
- C (3x-2)/(2x-3)
- D (x-1)/(x+1)
Show answer & explanation
Answer: A. (2x-3)/(3x-2)
Why: Let y = (2x-3)/(3x-2). Cross-multiply: y(3x-2) = 2x-3. x(3y-2) = 2y-3. x = (2y-3)/(3y-2). So f inverse(x) = (2x-3)/(3x-2) = f(x). This function is its own inverse (involution).
Q60.
The inverse of the function f(x) = (x − 3)/2 is:
- A 2x + 3
- B 2x − 3
- C (x + 3)/2
- D x/2 + 3
Show answer & explanation
Answer: A. 2x + 3
Why: Let y = (x − 3)/2, solve for x: x = 2y + 3, so f⁻¹(x) = 2x + 3.
Q61.
The function f: R → R given by f(x) = x² is:
- A injective only
- B surjective only
- C bijective
- D neither injective nor surjective
Show answer & explanation
Answer: D. neither injective nor surjective
Why: It is not one-one (f(−1) = f(1)) and not onto (negatives are never attained), so neither.
Q62.
The number of onto functions from a 4-element set to a 3-element set is:
- A 24
- B 36
- C 64
- D 81
Show answer & explanation
Answer: B. 36
Why: Surjections from 4 to 3 = 3! · S(4,3) = 6 · 6 = 36.
Q63.
The range of the function f(x) = x/(1 + |x|) is:
- A (−1, 1)
- B [−1, 1]
- C R
- D (0, 1)
Show answer & explanation
Answer: A. (−1, 1)
Why: As x varies over R the values approach but never reach ±1, so the range is (−1, 1).
Q64.
If f(x) = 2x + 3 and g(x) = x² − 1, then (f ∘ g)(2) equals:
- A 7
- B 9
- C 11
- D 15
Show answer & explanation
Answer: B. 9
Why: g(2) = 3, then f(3) = 2·3 + 3 = 9.
Q65.
The inverse of the bijection f(x) = x³ + 1 on R is:
- A (x − 1)<sup>1/3</sup>
- B (x + 1)<sup>1/3</sup>
- C x<sup>1/3</sup> − 1
- D (1 − x)<sup>1/3</sup>
Show answer & explanation
Answer: A. (x − 1)<sup>1/3</sup>
Why: Solving y = x³ + 1 gives x = (y − 1)<sup>1/3</sup>.
Q66.
The number of one-one functions from a 3-element set to a 4-element set is:
- A 12
- B 24
- C 64
- D 81
Show answer & explanation
Answer: B. 24
Why: 4 · 3 · 2 = 24.
Q67.
For f(x) = (4x + 3)/(6x − 4), x ≠ 2/3, the composition f(f(x)) equals:
- A x
- B 1/x
- C −x
- D (6x − 4)/(4x + 3)
Show answer & explanation
Answer: A. x
Why: Substituting shows f is its own inverse, so f(f(x)) = x.
Q68.
The function f(x) = |x − 1| + |x − 2| fails to be differentiable at how many points?
- A 1
- B 2
- C 3
- D 0
Show answer & explanation
Answer: B. 2
Why: There are corners at x = 1 and x = 2, giving 2 points of non-differentiability.