JEE Main Previous Year Questions (2021): Sets and Relations | Chapter-wise Tests for JEE Main & Advanced PDF Download

[JEE Mains MCQs]

Q1: Which of the following is not correct for relation R on the set of real numbers ?
(a) (x, y) ∈ R ⇔ 0 < |x| − |y| ≤ 1 is neither transitive nor symmetric.
(b) (x, y) ∈ R ⇔ 0 < |x − y| ≤ 1 is symmetric and transitive.
(c) (x, y) ∈ R ⇔ |x| − |y| ≤ 1 is reflexive but not symmetric.
(d) (x, y) ∈ R ⇔ |x − y| ≤ 1 is reflexive nd symmetric.
Ans: (b)
Note that (a, b) and (b, c) satisfy 0 < |x − y| ≤ 1 but (a, c) does not satisfy it so 0 ≤ |x − y| ≤ 1 is symmetric but not transitive.
For example,
x = 0.2, y = 0.9, z = 1.5
0 ≤ |x – y| = 0.7 ≤ 1
0 ≤ |y – z| = 0.6 ≤ 1
But |x – z| = 1.3 > 1
So, (b) is correct. 

Q2: Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If K% of them are suffering from both ailments, then K can not belong to the set :
(a) {80, 83, 86, 89}
(b) {84, 86, 88, 90}
(c) {79, 81, 83, 85}
(d) {84, 87, 90, 93}
Ans: (c)
This solution begins by applying the principle of inclusion and exclusion, which in the context of this problem, is represented by the formula:
n(A ∪ B) ≥ n(A) + n(B) - n(A ∩ B)
Here, n(A ∪ B) represents the total number of patients in the hospital, which is 100%. n(A) represents the proportion of patients with a heart ailment (89%), and n(B) represents the proportion of patients with a lung infection (98%).
By rearranging this formula, the solution establishes an inequality for n(A ∩ B), the proportion of patients suffering from both ailments:
100% ≥ 89% + 98% - n(A ∩ B)
Therefore,
n(A ∩ B) ≥ 87%
Next, the solution notes that n(A ∩ B) cannot be greater than the smaller of n(A) and n(B), since it cannot be larger than the smallest group. Thus, we have another inequality:
n(A ∩ B) ≤ 89%
Combining these two inequalities gives:
87% ≤ n(A ∩ B) ≤ 89%
Hence, the proportion of patients suffering from both ailments must be a value between 87% and 89% inclusive. So, the set of values {79,81,83,85} which are all less than 87% are values that n(A ∩ B) cannot belong to. Therefore, option C is the correct answer.

Q3: Let N be the set of natural numbers and a relation R on N be defined by R = {(x , y) ∈ N × N : x³ − 3x²y − xy² + 3y³ = 0}. Then the relation R is :
(a) symmetric but neither reflexive nor transitive
(b) reflexive but neither symmetric nor transitive
(c) reflexive and symmetric, but not transitive
(d) an equivalence relation
Ans: (b)

Now, x = y ∀ (x, y) ∈ N × N so reflexive but not symmetric & transitive. See, (3, 1) satisfies but (1, 3) does not. Also (3, 1) & (1, − 1) satisfies but (3, − 1) does not.

Q4: Define a relation R over a class of n × n real matrices A and B as "ARB iff there exists a non-singular matrix P such that PAP⁻¹ = B". Then which of the following is true?
(a) R is reflexive, transitive but not symmetric
(b) R is symmetric, transitive but not reflexive.
(c) R is reflexive, symmetric but not transitive
(d) R is an equivalence relation
Ans: (d)
For reflexive relation,

So, R is reflexive relation.
For symmetric relation,
Let (A , B) ∈ R for matrix P.
⇒ A = PBP⁻¹ 
After pre-multiply by P⁻¹ and post-multiply by P , we get
P⁻¹ AP = B
So, (B , A) ∈ R for matrix P⁻¹.
So, R is a symmetric relation.
For transitive relation,
Let ARB and BRC

∴ R is transitive relation.
Hence, R is an equivalence relation.

Q5: In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?
(a) Q and R
(b) P and Q
(c) P and R
(d) None of these
Ans: (d)
As none play all three games the intersection of all three circles must be zero.
Hence none of P, Q, R justify the given statement

Q6: Let A = {2, 3, 4, 5, ....., 30} and '≃' be an equivalence relation on A × A, defined by (a, b) ≃ (c, d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to:
(a) 5
(b) 6
(c) 8
(d) 7
Ans: (d)
ad = bc
(a, b) R (4, 3)
⇒ 3a = 4b
a = 4/3 b
b must be multiple of 3
b = {3, 6, 9 ..... 30}
(a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)}
⇒ 7 ordered pair

Q7: The number of elements in the set {x ∈ R : (|x| − 3) |x + 4| = 6} is equal to:
(a) 4
(b) 2
(c) 3
(d) 1
Ans: (b)
Case 1:
x ≤ −4
(−x − 3)(−x − 4) = 6
⇒ (x + 3)(x + 4) = 6
⇒ x² + 7x + 6 = 0
⇒ x = −1 or −6
but x ≤ −4
x = −6

Case 2:
x ∈ (−4, 0)
(−x − 3)(x + 4) = 6
⇒ −x² − 7x − 12 − 6 = 0
⇒ x² + 7x + 18 = 0
D < 0 No solution

Case 3:
x ≥ 0
(x − 3)(x + 4) = 6
⇒ x² + x − 12 − 6 = 0
⇒ x² + x − 18 = 0


Q8: Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, − 1) is the set : 
(a) S = {(x , y) | x² + y² = √2} 
(b) S = {(x , y) | x² + y² = 2} 
(c) S = {(x , y) | x² + y² = 1} 
(d) S = {(x , y) | x² + y² = 4}
Ans: (b)
Given R = {(P, Q) | P and Q are at the same distance from the origin}.
Then equivalence class of (1, −1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, −1).
i.e., radius of circle
∴ Required equivalence class of (S)

[JEE Mains Numericals]

Q9: If A = {x ∈ R : |x − 2| > 1},
B = {x ∈ R : },
C = {x ∈ R : |x − 4| ≥ 2} and Z is the set of all integers, then the number of subsets of the set (A ∩ B ∩ C)^c ∩ Z is ________________.
Ans: 256
A = (− ∞ , 1) ∪ (3, ∞)
B = (− ∞ , − 2) ∪ (2, ∞)
C = (− ∞ , 2] ∪ [6, ∞)
So, A ∩ B ∩ C = (− ∞ , − 2) ∪ [6, ∞)
z ∩ (A ∩ B ∩ C)' = {− 2, − 1, 0, − 1, 2, 3, 4, 5}
Hence, no. of its subsets = 2⁸ = 256.

Q10: Let A = {n ∈ N | n² ≤ n + 10,000}, B = {3k + 1 | k ∈ N} an dC = {2k | k ∈ N}, then the sum of all the elements of the set A ∩ (B − C) is equal to _____________.
Ans: 832
B − C ≡ {7, 13, 19, ......, 97, .......}
Now, n² − n ≤ 100 × 100
⇒ n(n − 1) ≤ 100 × 100
⇒ A = {1, 2, ......., 100}.
So, A∩(B − C) = {7, 13, 19, ......., 97} 
Hence, sum

Q11: Let  A = {n ∈ N: n is a 3-digit number}
B = {9k + 2: k ∈ N} and C = {9k + l : k ∈ N} for some l (0 < l < 9)
If the sum of all the elements of the set A ∩ (B ∪ C) is 274 × 400, then l is equal to ________.
Ans: 5
In this problem, we're dealing with 3-digit numbers in set A , and subsets B and C which represent numbers of specific forms.
1. First, we consider the numbers of the form 9k + 2 (Set B) within the 3-digit range, which starts at 101 and ends at 992.
2. We calculate the sum of these numbers, denoted as s₁. To calculate s₁ , you use the formula for the sum of an arithmetic series: {first term}{last term} (n/2) × ( {first term} + {last term})
Here, n is the total count of such numbers. These are 3-digit numbers of the form 9k + 2 , and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.
The sum s₁ is calculated as follows:
(100 / 2) × (101 + 992) = 54650
3. According to the problem, the sum of all elements of the set A ∩ (B ∪ C) is 274 × 400 = 109600.
Since the set A ∩ (B ∪ C) is the union of two disjoint sets (the set of all three-digit numbers of form 9k + 2 and the set of all three-digit numbers of form 9k + l), we can write this sum as :
s₁ (for numbers of the form 9k + 2) + s₂ (for numbers of the form 9k + l) = 109600
4. Solving this equation for s₂ (the sum of numbers of the form 9k + l), we get :
s₂ = 109600 − s₁ = 109600 − 54650 = 54950
5. The sum s₂ can be expressed as (n / 2) × ({first term} + {last term}) , where n is the count of numbers of the form 9k + l. The first term here is the smallest 3-digit number of this form, which is 99 + l, and the last term is the largest such number, which is 990 + l.
We equate this to s₂ to solve for l:
54950 = (100 / 2) [(99 + l) + ( 990 + l)]
6. Simplifying this equation, we get:
2l + 1089 = 1099
Solving for l , we find:
l = 5
So, the correct answer is 5.

Q12: Set A has m elements and set B has n elements. If the total number of subsets of A is 112 more than the total number of subsets of B, then the value of m.n is ______.
Ans: 28
Number of subsets of A = 2^m 
Number of subsets of B = 2ⁿ 
Given = 2^m – 2ⁿ = 112
∴ m = 7, n = 4 (2⁷ – 24 = 112)
∴ m × n = 7 × 4 = 28