Relations and Functions: 50 Practice Questions for Competitive Exams
Below are 50 questions on Relations and Functions for UP TGT/PGT, NDA, IAS, and KVS exams. Click “Show Answer” to reveal the answer and explanation after attempting each question.
1. If R is a relation on set A = {1, 2, 3} defined by R = {(1, 1), (2, 2), (3, 3)}, then R is:
a) Reflexive only
b) Symmetric only
c) Transitive only
d) Equivalence relation
Explanation: R is reflexive (each element is related to itself), symmetric ((a, a) implies (a, a)), and transitive (no counterexample exists). Hence, it’s an equivalence relation.
Year: UP TGT 2016
2. The function f: R → R defined by f(x) = x² is:
a) One-one
b) Onto
c) Neither one-one nor onto
d) Both one-one and onto
Explanation: f(x) = x² is not one-one (e.g., f(2) = f(-2) = 4) and not onto (no real x satisfies f(x) = -1).
Year: KVS PGT 2018
3. The domain of the function f(x) = 1/(x – 2) is:
a) R
b) R – {2}
c) R – {0}
d) R⁺
Explanation: The function is undefined when x – 2 = 0, i.e., x = 2. Thus, the domain is all real numbers except 2.
Year: NDA 2019
4. If f(x) = 2x + 3 and g(x) = x – 1, then (f ∘ g)(x) is:
a) 2x + 1
b) 2x – 1
c) 2x + 5
d) 2x – 3
Explanation: (f ∘ g)(x) = f(g(x)) = f(x – 1) = 2(x – 1) + 3 = 2x – 2 + 3 = 2x + 1.
Year: UP PGT 2020
5. A relation R on set A = {1, 2, 3, 4} is defined by R = {(1, 2), (2, 3), (3, 4)}. Is R transitive?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: For transitivity, if (1, 2) and (2, 3) are in R, then (1, 3) must be in R. Since (1, 3) is not in R, it is not transitive.
Year: IAS Prelims 2017
6. The range of the function f(x) = |x| is:
a) R
b) R⁺ ∪ {0}
c) R⁻
d) {0}
Explanation: The absolute value function gives non-negative outputs for all real inputs.
Year: KVS TGT 2014
7. If f: A → B is a bijective function, then:
a) f is one-one only
b) f is onto only
c) f is both one-one and onto
d) f is neither one-one nor onto
Explanation: A bijective function is both injective (one-one) and surjective (onto).
Year: UP TGT 2019
8. The inverse of the function f(x) = 3x + 5 is:
a) (x – 5)/3
b) (x + 5)/3
c) 3/(x – 5)
d) 5/(x – 3)
Explanation: For f(x) = 3x + 5, solve y = 3x + 5 for x: x = (y – 5)/3. Thus, f⁻¹(x) = (x – 5)/3.
Year: NDA 2020
9. A relation R on Z defined by (a, b) ∈ R if a – b is divisible by 3 is:
a) Reflexive only
b) Symmetric only
c) Transitive only
d) Equivalence relation
Explanation: R is reflexive (a – a = 0 is divisible by 3), symmetric (if a – b is divisible by 3, so is b – a), and transitive (if a – b and b – c are divisible by 3, so is a – c).
Year: UP PGT 2018
10. The function f(x) = sin(x) is:
a) One-one
b) Onto
c)
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Relations and Functions: 50 Practice Questions for Competitive Exams
Below are 50 questions on Relations and Functions for UP TGT/PGT, NDA, IAS, and KVS exams. Click “Show Answer” to reveal the answer and explanation after attempting each question.
1. If R is a relation on set A = {1, 2, 3} defined by R = {(1, 1), (2, 2), (3, 3)}, then R is:
a) Reflexive only
b) Symmetric only
c) Transitive only
d) Equivalence relation
Explanation: R is reflexive (each element is related to itself), symmetric ((a, a) implies (a, a)), and transitive (no counterexample exists). Hence, it’s an equivalence relation.
Year: UP TGT 2016
2. The function f: R → R defined by f(x) = x² is:
a) One-one
b) Onto
c) Neither one-one nor onto
d) Both one-one and onto
Explanation: f(x) = x² is not one-one (e.g., f(2) = f(-2) = 4) and not onto (no real x satisfies f(x) = -1).
Year: KVS PGT 2018
3. The domain of the function f(x) = 1/(x – 2) is:
a) R
b) R – {2}
c) R – {0}
d) R⁺
Explanation: The function is undefined when x – 2 = 0, i.e., x = 2. Thus, the domain is all real numbers except 2.
Year: NDA 2019
4. If f(x) = 2x + 3 and g(x) = x – 1, then (f ∘ g)(x) is:
a) 2x + 1
b) 2x – 1
c) 2x + 5
d) 2x – 3
Explanation: (f ∘ g)(x) = f(g(x)) = f(x – 1) = 2(x – 1) + 3 = 2x – 2 + 3 = 2x + 1.
Year: UP PGT 2020
5. A relation R on set A = {1, 2, 3, 4} is defined by R = {(1, 2), (2, 3), (3, 4)}. Is R transitive?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: For transitivity, if (1, 2) and (2, 3) are in R, then (1, 3) must be in R. Since (1, 3) is not in R, it is not transitive.
Year: IAS Prelims 2017
6. The range of the function f(x) = |x| is:
a) R
b) R⁺ ∪ {0}
c) R⁻
d) {0}
Explanation: The absolute value function gives non-negative outputs for all real inputs.
Year: KVS TGT 2014
7. If f: A → B is a bijective function, then:
a) f is one-one only
b) f is onto only
c) f is both one-one and onto
d) f is neither one-one nor onto
Explanation: A bijective function is both injective (one-one) and surjective (onto).
Year: UP TGT 2019
8. The inverse of the function f(x) = 3x + 5 is:
a) (x – 5)/3
b) (x + 5)/3
c) 3/(x – 5)
d) 5/(x – 3)
Explanation: For f(x) = 3x + 5, solve y = 3x + 5 for x: x = (y – 5)/3. Thus, f⁻¹(x) = (x – 5)/3.
Year: NDA 2020
9. A relation R on Z defined by (a, b) ∈ R if a – b is divisible by 3 is:
a) Reflexive only
b) Symmetric only
c) Transitive only
d) Equivalence relation
Explanation: R is reflexive (a – a = 0 is divisible by 3), symmetric (if a – b is divisible by 3, so is b – a), and transitive (if a – b and b – c are divisible by 3, so is a – c).
Year: UP PGT 2018
10. The function f(x) = sin(x) is:
a) One-one
b) Onto
c) Neither one-one nor onto
d) Both one-one and onto
Explanation: sin(x) is not one-one (e.g., sin(0) = sin(2π)) and not onto (range is [-1, 1], not all of R).
Year: KVS PGT 2020
11. If f(x) = x + 1 and g(x) = x², then (g ∘ f)(x) is:
a) x² + 1
b) (x + 1)²
c) x² + 2x + 1
d) x + 1
Explanation: (g ∘ f)(x) = g(f(x)) = g(x + 1) = (x + 1)².
Year: NDA 2018
12. The number of onto functions from a set with 3 elements to a set with 2 elements is:
a) 6
b) 8
c) 0
d) 9
Explanation: The number of surjective functions from a set of m elements to a set of n elements (m ≥ n) is n! × S(m, n), where S(m, n) is the Stirling number of the second kind. For m = 3, n = 2, S(3, 2) = 3, so 2! × 3 = 6.
Year: IAS Prelims 2019
13. The function f(x) = e^x is:
a) One-one but not onto
b) Onto but not one-one
c) Both one-one and onto
d) Neither one-one nor onto
Explanation: e^x is strictly increasing (hence one-one), but its range is (0, ∞), not all of R.
Year: UP TGT 2021
14. The domain of f(x) = √(4 – x²) is:
a) [-2, 2]
b) (-2, 2)
c) R
d) [0, 2]
Explanation: For the square root to be defined, 4 – x² ≥ 0, i.e., x² ≤ 4, so x ∈ [-2, 2].
Year: KVS TGT 2017
15. If R = {(a, b) ∈ Z × Z : a² = b²}, then R is:
a) Reflexive only
b) Symmetric only
c) Equivalence relation
d) Transitive only
Explanation: R is reflexive (a² = a²), symmetric (if a² = b², then b² = a²), and transitive (if a² = b² and b² = c², then a² = c²).
Year: UP PGT 2016
16. The function f(x) = log(x) is defined for:
a) R
b) R⁺
c) R – {0}
d) [0, ∞)
Explanation: The logarithm is defined only for positive real numbers (x > 0).
Year: NDA 2021
17. If f(x) = x³, then f is:
a) One-one but not onto
b) Onto but not one-one
c) Both one-one and onto
d) Neither one-one nor onto
Explanation: f(x) = x³ is strictly increasing (one-one) and its range is all of R (onto).
Year: IAS Prelims 2018
18. The number of reflexive relations on a set with 3 elements is:
a) 2⁹
b) 2⁶
c) 2³
d) 2⁸
Explanation: For a set with n elements, the number of reflexive relations is 2^(n² – n). For n = 3, it’s 2^(9 – 3) = 2⁶.
Year: UP TGT 2020
19. If f(x) = 2x and g(x) = x/2, then (f ∘ g)(x) is:
a) x
b) 2x
c) x²
d) 1/x
Explanation: (f ∘ g)(x) = f(g(x)) = f(x/2) = 2(x/2) = x.
Year: KVS PGT 2017
20. The range of f(x) = cos(x) is:
a) [-1, 1]
b) [0, 1]
c) R
d) (-1, 1)
Explanation: The cosine function outputs values between -1 and 1 inclusive.
Year: NDA 2017
21. A relation R on {1, 2, 3} defined by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} is:
a) Reflexive only
b) Symmetric only
c) Equivalence relation
d) Transitive only
Explanation: R is reflexive (contains (1, 1), (2, 2), (3, 3)), symmetric ((1, 2) implies (2, 1)), and transitive ((1, 2) and (2, 1) imply (1, 1)).
Year: UP TGT 2017
22. The inverse of f(x) = 2x – 4 is:
a) (x + 4)/2
b) (x – 4)/2
c) 2x + 4
d) x/2
Explanation: Solve y = 2x – 4 for x: x = (y + 4)/2. Thus, f⁻¹(x) = (x + 4)/2.
Year: KVS TGT 2016
23. The function f(x) = [x] (greatest integer function) is:
a) One-one
b) Onto
c) Both one-one and onto
d) Neither one-one nor onto
Explanation: f(x) = [x] is not one-one (e.g., [1.2] = [1.8] = 1) and not onto (range is Z, not R).
Year: NDA 2019
24. If R = {(a, b) ∈ Z × Z : |a – b| ≤ 2}, is R transitive?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: If (1, 3) ∈ R (|1 – 3| = 2 ≤ 2) and (3, 5) ∈ R (|3 – 5| = 2 ≤ 2), but (1, 5) ∉ R (|1 – 5| = 4 > 2), so not transitive.
Year: UP PGT 2019
25. The period of the function f(x) = sin(2x) is:
a) π
b) 2π
c) π/2
d) π/4
Explanation: For f(x) = sin(kx), period = 2π/k. Here, k = 2, so period = 2π/2 = π/2.
Year: IAS Prelims 2019
26. If f(x) = x² – 4x + 3, then f is:
a) One-one
b) Onto
c) Neither one-one nor onto
d) Both one-one and onto
Explanation: f(x) is a parabola opening upwards, not one-one (e.g., f(1) = f(3) = 0). Minimum value is -1, so not onto R.
Year: KVS PGT 2020
27. The number of symmetric relations on a set with 3 elements is:
a) 2⁹
b) 2⁶
c) 2³
d) 2⁸
Explanation: For n elements, the number of symmetric relations is 2^(n(n+1)/2). For n = 3, it’s 2^(3×4/2) = 2⁶.
Year: UP TGT 2018
28. If f(x) = 1/(x² – 3x + 2), the domain is:
a) R
b) R – {1, 2}
c) R – {0}
d) R⁺
Explanation: Denominator x² – 3x + 2 = (x – 1)(x – 2) ≠ 0, so x ≠ 1, 2.
Year: NDA 2020
29. If f(x) = 2x + 1 and g(x) = x², then (f ∘ g)(2) is:
a) 5
b) 9
c) 10
d) 13
Explanation: (f ∘ g)(x) = f(g(x)) = f(x²) = 2x² + 1. Thus, (f ∘ g)(2) = 2(2²) + 1 = 9.
Year: UP PGT 2020
30. The function f(x) = tan(x) has domain:
a) R
b) R – {nπ}
c) R – {(2n+1)π/2}
d) R⁺
Explanation: tan(x) is undefined at x = (2n+1)π/2, where n ∈ Z.
Year: KVS TGT 2018
31. If R = {(a, b) ∈ N × N : a divides b}, is R transitive?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: If a divides b and b divides c, then a divides c (e.g., a = 2, b = 4, c = 8). Thus, R is transitive.
Year: IAS Prelims 2017
32. The range of f(x) = 1/(1 + x²) is:
a) (0, 1]
b) [0, 1]
c) (0, ∞)
d) [1, ∞)
Explanation: Since x² ≥ 0, 1 + x² ≥ 1, so f(x) = 1/(1 + x²) ∈ (0, 1]. Maximum is 1 when x = 0.
Year: UP TGT 2019
33. If f(x) = |x – 1|, then f is:
a) One-one
b) Onto
c) Neither one-one nor onto
d) Both one-one and onto
Explanation: f(x) = |x – 1| is not one-one (e.g., f(0) = f(2) = 1) and not onto (range is [0, ∞), not R).
Year: NDA 2018
34. The number of equivalence relations on a set with 2 elements is:
a) 1
b) 2
c) 3
d) 4
Explanation: For n elements, the number of equivalence relations is the Bell number B(n). For n = 2, B(2) = 2 (partitions: {{1, 2}, {1},{2}}).
Year: UP PGT 2018
35. If f(x) = x/(x + 1), the inverse is:
a) x/(1 – x)
b) (x + 1)/x
c) x/(x – 1)
d) (x – 1)/x
Explanation: Solve y = x/(x + 1) for x: xy + y = x, x(1 – y) = y, x = y/(1 – y). Thus, f⁻¹(x) = x/(1 – x).
Year: KVS PGT 2019
36. If f(x) = {(x, y) ∈ R × R : y = x + 2}, is f a function?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: For each x ∈ R, there is exactly one y = x + 2, satisfying the definition of a function.
Year: NDA 2016
37. The function f(x) = x² + 2x + 2 has minimum value:
a) 0
b) 1
c) 2
d) 3
Explanation: Complete the square: f(x) = (x + 1)² + 1. Minimum value is 1 when x = -1.
Year: UP TGT 2020
38. If R = {(a, b) ∈ Z × Z : a + b is even}, then R is:
a) Reflexive only
b) Symmetric only
c) Equivalence relation
d) Transitive only
Explanation: R is reflexive (a + a = 2a is even), symmetric (a + b even implies b + a even), and transitive (a + b and b + c even imply a + c even).
Year: IAS Prelims 2018
39. The domain of f(x) = 1/√(x – 3) is:
a) (3, ∞)
b) [3, ∞)
c) R – {3}
d) R
Explanation: For 1/√(x – 3) to be defined, x – 3 > 0, i.e., x > 3.
Year: KVS TGT 2019
40. If f(x) = 3x and g(x) = x/3, then (g ∘ f)(x) is:
a) x
b) 3x
c) x³
d) 1/x
Explanation: (g ∘ f)(x) = g(f(x)) = g(3x) = (3x)/3 = x.
Year: NDA 2017
41. The range of f(x) = e^(x – |x|) is:
a) (0, ∞)
b) [0, ∞)
c) (0, 1]
d) [1, e³]
Explanation: Since |x| ≥ 0, x – |x| ≤ 0, so e^(x – |x|) ≤ 1. Maximum is 1 when x = 0, and as x → ±∞, f(x) → 0.
Year: UP PGT 2020
42. If R = {(a, b) ∈ N × N : a² = b}, is R a function?
a) Yes
b) No
c) Partially
d) Cannot be determined
Explanation: For a = 2, there is no b ∈ N such that 2² = b (b = 4, but 4 is not a square of a natural number). R is not a function as it’s not defined for all a ∈ N.
Year: KVS PGT 2018
43. The function f(x) = x² is one-one if restricted to:
a) R
b) [0, ∞)
c) (-∞, 0]
d) Both b and c
Explanation: f(x) = x² is strictly increasing on [0, ∞) and strictly decreasing on (-∞, 0], making it one-one on both intervals.
Year: UP TGT 2018
44. If f(x) = x + 1/x, the range is:
a) R
b) (-∞, -2] ∪ [2, ∞)
c) [-2, 2]
d) [0, ∞)
Explanation: By AM-GM, |x + 1/x| ≥ 2 for x ≠ 0. Thus, f(x) ≤ -2 or f(x) ≥ 2. All values in these intervals are achievable.
Year: NDA 2019
45. If R = {(a, b) ∈ Z × Z : a ≡ b (mod 5)}, then R is:
a) Reflexive only
b) Symmetric only
c) Equivalence relation
d) Transitive only
Explanation: R is reflexive (a ≡ a (mod 5)), symmetric (a ≡ b implies b ≡ a), and transitive (a ≡ b and b ≡ c imply a ≡ c).
Year: IAS Prelims 2019
46. The period of f(x) = cos(3x) is:
a) π/3
b) 2π/3
c) π
d) 2π
Explanation: For f(x) = cos(kx), period = 2π/k. Here, k = 3, so period = 2π/3.
Year: KVS TGT 2017
47. The domain of f(x) = 1/(x² – 1) is:
a) R
b) R – {1, -1}
c) R – {0}
d) R⁺
Explanation: The denominator x² – 1 = (x – 1)(x + 1) ≠ 0, so x ≠ 1, -1.
Year: UP PGT 2017
48. If f(x) = x³ – x is one-one, then:
a) True for all x ∈ R
b) False
c) True for x > 0
d) True for x < 0
Explanation: f'(x) = 3x² – 1 has critical points, but f is strictly increasing (check via second derivative or graph), so one-one.
Year: NDA 2018
49. If R = {(x, y) ∈ R × R : x² + y² = 1}, is R a function?
a) Yes
b) No
c) Yes, one-to-one
d) Cannot be determined
Explanation: For x = 0, y = ±1, so (0, 1) and (0, -1) ∈ R, violating the function definition.
Year: KVS PGT 2019
50. The range of f(x) = sin(x) + cos(x) is:
a) [-1, 1]
b) [-√2, √2]
c) [0, 2]
d) [-2, 2]
Explanation: Rewrite f(x) = √2 sin(x + π/4). Since sin θ ∈ [-1, 1], f(x) ∈ [-√2, √2].
Year: UP TGT 2019
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