Sets, Relation, and Function: Previous Year Questions with Solutions

1.0Introduction

Sets, Relations, and Functions is a fundamental topic in the JEE syllabus, forming the basis for understanding more advanced mathematical concepts. Previous year questions from this chapter typically focus on key concepts like the definition and representation of sets, types of sets (finite, infinite, equal, null, etc.), Venn diagrams, Cartesian products, types of relations (reflexive, symmetric, transitive, equivalence), and various types of functions (one-one, onto, bijective, identity, inverse, composition of functions).

Examples include identifying and proving the type of a given relation, determining the domain and range of a function, checking if a function is invertible, or simplifying expressions using set operations like union, intersection, and complement.

Solutions generally involve using set identities, understanding mappings and their properties, analyzing graphs of functions, and applying logical reasoning to determine the nature of relations and functions.

Practicing these previous year questions builds a strong conceptual base, enhances your ability to visualize abstract mathematical structures, and sharpens problem-solving skills crucial for success in JEE.

2.0Sets, Relations, and Functions – Key Concept and Formula

Sets 

• Set: A collection of well-defined, distinct objects.
• Represented by capital letters: A, B, C.
• Element: An object in a set.
• Roster Method: Listing all elements, e.g., A = {1, 2, 3}.
• Set Builder Method: Describing properties, e.g., A = {x: x is an odd number < 10}.

Relations and Functions:

• Relation: A relation from set A to set B is a subset of A × B.
• Function: A special relation where every element of A is associated with exactly one element of B.

Types of Functions:

• One-One (Injective)
• Onto (Surjective)
• Bijective (One-One and Onto)

Sets: Key Concepts & Formulas

1. Representation of Sets:

• Roster form: {a, b, c}
• Set-builder form: {x | condition on x}

2. Types of Sets:

• Empty set: ∅
• Finite/Infinite set
• Equal sets: A = B if all elements are same
• Subset: A ⊆ B ⇔ every element of A is in B
• Proper subset: A ⊂ B and A ≠ B
• Power set: P(A) = set of all subsets of A, including ∅ and A itself
• If n(A) = n, then n(P(A)) = 2ⁿ

3. Operations on Sets:

• Union: A ∪ B = {x | x ∈ A or x ∈ B}
• Intersection: A ∩ B = {x | x ∈ A and x ∈ B}
• Difference: A - B = {x | x ∈ A and x ∉ B}
• Complement: A' = {x ∈ U | x ∉ A}, where U is universal set

4. Laws of Set Algebra:

• Commutative: A ∪ B = B ∪ A, A ∩ B = B ∩ A
• Associative: (A ∪ B) ∪ C = A ∪ (B ∪ C)
• Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
• De Morgan’s Laws:

(A ∪ B)' = A' ∩ B'

(A ∩ B)' = A' ∪ B'

Cardinality Formulas:

If A and B are finite sets:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

For three sets:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C)

Relations: Key Concepts

Cartesian Product:

• A × B = {(a, b) | a ∈ A, b ∈ B}
• If n(A) = m, n(B) = n, then n(A × B) = m × n

Relation:

A relation R from A to B is a subset of A × B

Types of Relations on a Set A:

Reflexive: (a, a) ∈ R ∀ a ∈ A

Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R

Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Equivalence Relation: A relation that is reflexive, symmetric, and transitive

Functions: Key Concepts & Formulas

1. Definition:
   A function f: A → B assigns every element of A to a unique element of B.
2. Domain, Co-domain, Range:

• Domain: input set A
• Co-domain: set B
• Range: actual outputs f(A) ⊆ B
• Types of Functions:
• One-One (Injective): f(a1) = f(a2) ⇒ a1 = a2
• Onto (Surjective): Range(f) = Co-domain
• Bijective: Both one-one and onto
• Identity Function: f(x) = x
• Constant Function: f(x) = c
• Polynomial, Rational, Modulus, Greatest Integer, etc.

3. Operations on Functions:

Sum: (f + g)(x) = f(x) + g(x)

Product: (f × g)(x) = f(x) × g(x)

Quotient: (f/g)(x) = f(x)/g(x), g(x) ≠ 0

4. Composition of Functions:

(f ∘ g)(x) = f(g(x))

Associative: (f ∘ g) ∘ h = f ∘ (g ∘ h)

5. Inverse of a Function:

If f: A → B is bijective, then its inverse f⁻¹: B → A exists such that:

f(f⁻¹(x)) = x and f⁻¹(f(x)) = x

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3.0JEE Mains Past Year Questions with Solutions on Sets, Relation and Functions

Previous Year Questions From Sets

1. In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let m and n respectively, be the least and the most number of students who studied all the three subjects. Then m + n is equal to _____

Ans. (45)

Sol.

125 ≤ m + 90 – x ≤ 130

85 ≤ P + 70 – x ≤ 95

75 ≤ C + 80 – x ≤ 90

m + P + C + 120 – 2x = 210

⇒ 15 ≤ x ≤ 45 & 30 – x ≥ 0

⇒ 15 ≤ x ≤ 30

30 + 15 = 45

2. Let $S=\left\{x\in \mathbb{R}:{\left(\sqrt{3}+\sqrt{2}\right)}^x+{\left(\sqrt{3}-\sqrt{2}\right)}^x=10\right\}$

Then the number of elements in S is :

(1) 4

(2) 0

(3) 2

(4) 1

Ans. (3)

Sol.

Let

t² – 10t + 1 = 0 

x = 2 or x = –2

$$\begin{gathered} {\left(\sqrt{3}+\sqrt{2}\right)}^x+{\left(\sqrt{3}-\sqrt{2}\right)}^x=10 \\ Let{\left(\sqrt{3}+\sqrt{2}\right)}^x=t,then: \\ t+\frac{1}{t}=10 \\ {t}^2-10t+1=0 \\ t=\frac{10\pm \sqrt{100-4}}{2}=\frac{10\pm \sqrt{96}}{2}=5\pm 2\sqrt{6} \\ {\left(\sqrt{3}+\sqrt{2}\right)}^x={\left(\sqrt{3}\pm \sqrt{2}\right)}^2 \\ x=2\text{or}x=-2 \\ \text{Number of solutions = 2} \end{gathered}$$

3. Let A and B be two finite sets with m and n elements respectively. The total number of subsets of the set A is 56 more than the total number of subsets of B. Then the distance of the point P(m, n) from the point Q(–2, –3) is

(1) 10

(2) 6

(3) 4

(4) 8

Ans. (1)

Sol.

$$\begin{gathered} 2^m-2^n=56 \\ {2}^n(2^{m-n}-1)=2^3\times 7 \\ {2}^n=2^3\text{and}{2}^{m-n}-1=7 \\ \Rightarrow n=3\text{and}{2}^{m-n}=8 \\ \Rightarrow n=3\text{and}m-n=3 \\ \Rightarrow n=3\text{and}m=6P(6,3)andQ(-2,-3) \\ PQ=\sqrt{(6-(-2){)}^2+(3-(-3){)}^2}=\sqrt{8^2+6^2} \\ =\sqrt{100}=10 \end{gathered}$$

Hence, option (1) is correct

4. Let the set $C=\left\{(x,y)\mid x^2-2y^2=2023,x,y\in \mathbb{Z}\right\}$. Then $\sum _{(x,y)\in C}(x+y)$ is equal to _______.

Ans. 46

Sol.

$$\begin{gathered} x^2-2y^2=2023 \\ \Rightarrow x=45,y=1 \\ {\sum }_{(x,y)\in C}(x+y)=46 \end{gathered}$$

4.0Previous Year Questions on Relation 

1. Let a relation R on be defined as: (x₁, y₁) R(x₂, y₂) if and only if x₁ < x₂ or y₁ < y₂ 

Consider the two statements: 

(I) R is reflexive but not symmetric.

(II) R is transitive 

Then, which one of the following is true? 

(1) Only (II) is correct.

(2) Only (I) is correct.

(3) Both (I) and (II) are correct.

(4) Neither (I) nor (II) is correct.

Ans. (3)

Sol.

For Reflexive: 

All ((x₁, y₁), (x₁, y₁)) are in R where 

x₁, y₁ ∈ N ∴ R is reflexive

For Symmetric: 

((1, 1), (2, 3)) ∈ R but ((2, 3), (1, 1)) ∉ R

∴ R is not symmetric 

For Reflexive: 

If (x₁, y₁) R (x₂, y₂) ⇒ x₁ ≤ x₂ & y₁ ≤ y₂  

& (x₂, y₂) R (x₃, y3) ⇒ x₂ ≤ x₃ & y₂ ≤ y₃  

For (x₁, y₁) R (x₃, y₃) ⇒ x₁ ≤ x₃ & y₁ ≤ y₃  

So, R is transitive 

2. Let the relations R₁ and R₂ on the set 

X = {1, 2, 3, ..., 20} be given by 

R₁ = {(x, y) : 2x – 3y = 2} and 

R₂ = {(x, y) : –5x + 4y = 0}. If M and N be the minimum number of elements required to be added in R₁ and R₂, respectively, in order to make the relations symmetric, then M + N equals

(1) 8

(2) 16

(3) 12

(4) 10

Ans. (4)

Sol. x = {1, 2, 3, .......20}

R₁ = {(x, y) : 2x – 3y = 2}

R₂ = {(x, y) : –5x + 4y = 0}

R₁ = {(4, 2), (7, 4), (10, 6), (13, 8), (16, 10), (19, 12)}

R₂ = {(4, 5), (8, 10), (12, 15), (16, 20)}

in R₁ 6 element needed

in R₂ 4 element needed

So, total 6 + 4 = 10 element

3. If a function f satisfies f(m + n) = f(m) + f(n) for all m, n ∈ N and  f(1) = 1, then the largest natural number λ such that ${\sum }_{k=1}^{2022}f(\lambda +k)\le (2022{)}^2$ is equal to  ________.

Ans. (1010)

Sol. f (m + n) = f(m) + f(n)

$$\begin{gathered} \Rightarrow f(x)=kx \\ \Rightarrow f(1)=1 \\ \Rightarrow k=1 \\ f(x)=x \\ Now \\ {\sum }_{k=1}^{2022}f(\lambda +k)\le (2022{)}^2 \\ \Rightarrow {\sum }_{k=1}^{2022}(\lambda +k)\le (2022{)}^2 \\ \Rightarrow 2022\lambda +\frac{2022\times 2023}{2}\le (2022{)}^2 \\ \Rightarrow \lambda \le 2022-\frac{2023}{2} \\ \Rightarrow \lambda \le 1010.5 \\ \therefore \text{largest natural number }\lambda \text{ is }\boxed{1010}. \end{gathered}$$

4. Let A = {2, 3, 6, 7} and B = {4, 5, 6, 8}. Let R be a relation defined on A × B by (a₁, b₁) R (a₂, b₂) is and only if a₁ + a₂ = b₁ + b₂. Then the number of elements in R is ________.

Ans. (25)

Sol. A = {2, 3, 6, 7}

B = { 2, 5, 6, 8}

(a₁, b₁) R (a₂, b₂)

a₁ + a₂ = b₁ + b₂

Total 24 + 1 = 25

5. Let S = {l, 2, 3, ... , 10}. Suppose M is the set of all the subsets of S, then the relation

R = {(A, B): A ∩ B ≠ φ; A, B ∈ M} is : 

(1) symmetric and reflexive only

(2) reflexive only

(3) symmetric and transitive only

(4) symmetric only

Ans. (4)

Sol. Let S = {1, 2, 3, …, 10}

R = {(A, B): A ∩ B ≠ φ; A, B ∈ M}

For Reflexive,

M is subset of ‘S’

So φ ∈ M

for φ ∩ φ = φ

⇒ but relation is A ∩ B ≠ φ

So, it is not reflexive.

For symmetric, 

ARB A ∩ B ≠ φ,

⇒ BRA ⇒ B ∩ A ≠ φ,

So, it is symmetric.

For transitive,

If A = {(1, 2), (2, 3)}

B = {(2, 3), (3, 4)}

C = {(3, 4), (5, 6)}

ARB & BRC but A does not relate to C 

So it not transitive

6. The number of symmetric relations defined on the set {1, 2, 3, 4} which are not reflexive is _____ .

Ans. (960)

Sol. Total number of relation both symmetric and reflexive $=2^{\frac{n^2-n}{2}}$

Total number of symmetric relation $=2^{\left(\frac{n^2+n}{2}\right)}$

⇒ Then number of symmetric relation which are not reflexive

$=2^{\frac{n(n+1)}{2}}-2^{\frac{n(n-1)}{2}}$

⇒ 2¹⁰ – 2⁶

⇒ 1024 – 64

= 960

7. Let A = {1, 2, 3, ………100}. Let R be a relation on A defined by (x, y) $\in R$ if and only if 2x = 3y. Let R₁ be a symmetric relation on A such that $R\subseteq R_1$ and the number of elements in R₁ is n. Then, the minimum value of n is ___________.

Ans. (66)

Sol.-

$$\begin{gathered} R=\{(3,2),(6,4),(9,6),(12,8),\dots ,(99,66)\} \\ n(R)=33 \\ \therefore \boxed{66} \end{gathered}$$

5.0Previous Year Questions on Functions

1. Let A = {1, 3, 7, 9, 11} and B = {2, 4, 5, 7, 8, 10, 12}. Then the total number of one-one maps
f : A → B, such that f (1) + f(3) = 14, is: 

(1) 180

(2) 120 

(3) 480

(4) 240 

Ans. (4)

Sol.

A = {1, 3, 7, 9, 11}

B = {2, 4, 5, 7, 8, 10, 12}

f(1) + f(3) = 14

(i) 2 + 12

(ii) 4 + 10

2 × (2 × 5 × 4 × 3) = 240 

2. The number of distinct real roots of the equation |x| |x + 2| – 5|x + l| – 1 = 0 is __________.

Ans. (3)

Sol.

Case-1

x ≥ 0

x² + 2x – 5x – 5 – 1 = 0

x² – 3x – 6 = 0

$x=\frac{3\pm \sqrt{9+24}}{2}=\frac{3\pm \sqrt{33}}{2}$

One positive root

Case-2

–1 ≤ x < 0

–x² – 2x – 5x – 5 – 1 = 0

x² + 7x + 6 = 0

(x + 6) (x + 1) = 0

x = –1

one root in range

Case-3

–2 ≤ x < –1

x² – 2x + 5x + 5 – 1 = 0

x² – 3x – 4 = 0

(x – 4) (x + 1) = 0

No root in range

Case-4

x < –2

x² + 7x + 4 = 0

$x=\frac{-7\pm \sqrt{49-16}}{2}=\frac{-7\pm \sqrt{33}}{2}$

one root in range

Total number of distinct roots are 3

3. Let  

$f(x)=\{\begin{array}{l}-a\text{if }-a\le x\le 0\\ x+a\text{if }0<x\le a\end{array}$

where a > 0 and g(x) = (f |x| ) – | f (x)| )/2. 

Then the function g : [ –a, a] → [ –a, a] is

(1) neither one-one nor onto.

(2) both one-one and onto.

(3) one-one.

(4) onto

Ans. (1)

Sol. y = f(x)

y = f|x|

y = |f(x)|

g(x) = $\frac{f(|x|)-|f(x)|}{2}$

4. If the domain of the function $f(x)=\frac{\sqrt{x^2-25}}{4-x^2}$ +log₁₀ (x² + 2x – 15) is (– ∞, α) U [β,∞), then α² + β³ is equal to : 

(1) 140

(2) 175

(3) 150

(4) 125

Ans. (3)

Sol.

$ƒ(x)=\frac{\sqrt{x^2-25}}{4-x^2}$ + log₁₀(x² + 2x - 15)

Domain : x² – 25 ≥ 0  ⇒ x ∈ (–∞, -5] ∪ [5, ∞)

4 – x² ≠ 0 ⇒ x ≠{–2, 2}

x² + 2x – 15 > 0 ⇒  (x + 5) (x – 3) > 0 

⇒ x ∈ (–∞, –5) ∪ (3, ∞)

∴ x ∈ (–∞, –5) ∪ [5, ∞)

α = –5; β = 5

∴ α2 + β3 = 150

5. Let f : R → R  and g : R → R be defined as

f(x) = $\{\begin{array}{l}{\log }_{e}x,x>0\\ e^{-x},x\le 0\end{array}andg(x)=\{\begin{array}{l}x,x\ge 0\\ e^x,x<0\end{array}$ . Then, go f : R → R is :

(1) one-one but not onto

(2) neither one-one nor onto

(3) onto but not one-one

(4) both one-one and onto

Ans. (2)

Sol.

$$\begin{gathered} g(f(x))=\{\begin{array}{l}f(x),f^{\prime }(x)\ge 0\\ e^{f(x)},f^{\prime }(x)<0\end{array} \\ g(f(x))=\{\begin{array}{l}{e}^{-x},(-\infty ,0]\\ e^{\ln x},(0,1)\\ \ln x,[1,\infty )\end{array} \end{gathered}$$

Graph of g(f(x))

g(f(x)) $\Rightarrow$ Many one into 

6. The function f : N – {1} → N; defined by f(n) = the highest prime factor of n, is :

(1) both one-one and onto

(2) one-one only

(3) onto only

(4) neither one-one nor onto

Ans. (4)

Sol. f : N – {1} → N

f(n) = The highest prime factor of n.

f(2) = 2

f(4) = 2

⇒ many one

4 is not image of any element 

⇒ into

Hence many one and into

Neither one-one nor onto.

7. If the function $f:(-\infty ,-1]\to (a,b],\text{defined by}f(x)=e^{x^3-3x+1}$ is one-one and onto, then the distance of the point P(2b + 4, a + 2) from the line x + e^–3y = 4 is : 

$(1)\sqrt{\frac{2}{1+e^6}}(2)\sqrt{\frac{4}{1+e^6}}(3)\sqrt{\frac{3}{1+e^6}}(4)\sqrt{\frac{1}{1+e^6}}$

Ans. (1)

Sol.-

$$\begin{gathered} f(x)=e^{x^3-3x+1} \\ {f}^{\prime }(x)=e^{x^3-3x+1}\cdot (3x^2-3) \\ =e^{x^3-3x+1}\cdot 3(x-1)(x+1) \\ For{f}^{\prime }(x)\ge 0 \\ \therefore f(x)\text{ is increasing function} \\ \therefore a=e^{-\infty }=0=f(-\infty ) \\ b=e^{-1^3+3\cdot 1+1}=e^3=f(-1) \\ Given:P(2b+4,a+2) \\ \therefore P(2e^3+4,2) \\ d=\frac{(2e^3+4)+2e^3-4}{\sqrt{1+e^{-6}}}=\frac{4e^3}{\sqrt{1+e^{-6}}}=2\sqrt{1+e^6} \end{gathered}$$

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