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JEE • MATHEMATICS • SET THEORY

Set theory practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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121 Easy
121 of 194
Analyzing +4 | -1
Let P={1,2,3,…,100}P=\{1,2,3, \ldots, 100\} be the number of non-empty subsets AA of SS such that the product of elements in AA is even is
View answer and explanation
Correct option: A
250(250−1)2^{50}\left(2^{50}-1\right)
Option A (250(250−1)2^{50}(2^{50}-1)) is correct because it calculates the total number of non-empty subsets minus the number of subsets containing only odd numbers.
122 Easy
122 of 194
Understanding +4 | -1
A set S\mathrm{S} contains 3 elements, the number of subsets of which of following sets is 256 .
View answer and explanation
Correct option: C
P(P(S))\mathrm{P}(\mathrm{P}(\mathrm{S}))
Option C (P(P(S))\mathrm{P}(\mathrm{P}(\mathrm{S}))) is correct because the power set of the power set of S has 256 subsets.
123 Easy
123 of 194
Analyzing +4 | -1
Let RR be a relation on N\mathbf{N} defined by R={(m,n):m,n∈N and m=n2}.R=\left\{(m, n): m, n \in N \text { and } m=n^2\right\}.
Which of the following is true.
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because none of the given properties (reflexivity, symmetry, transitivity) hold true for the relation R on the set of natural numbers.
124 Easy
124 of 194
Analyzing +4 | -1
If A={θ:2cos⁡2θ+sin⁡θ≤2}A=\left\{\theta: 2 \cos ^2 \theta+\sin \theta \leq 2\right\} and B={θ:π2≤θ≤3π2}B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}, then A∩BA \cap B is equal to
View answer and explanation
Correct option: C
{θ:π/2≤θ≤5π/6}∪{θ:π≤θ≤3π/2}\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup\{\theta: \pi \leq \theta \leq 3 \pi / 2\}
Option C ({θ:π/2≤θ≤5π/6}∪{θ:π≤θ≤3π/2}\{\theta: \pi / 2 \leq \theta \leq 5 \pi / 6\} \cup \{\theta: \pi \leq \theta \leq 3 \pi / 2\}) is correct because it represents the intersection of the solution sets for the trigonometric inequality and the given interval.
125 Easy
125 of 194
Understanding +4 | -1
The value of n[P(P(P(ϕ)))]n\left[ {P\left( {P\left( {P\left( \phi \right)} \right)} \right)} \right] is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because it represents the cardinality of the power set of the power set of the power set of the empty set.
126 Easy
126 of 194
Analyzing +4 | -1
Let S={1,  2,  3,  4}S = \left\{ {1,\;2,\;3,\;4} \right\}. The total number of unordered pairs of disjoint subsets of SS is equal to
View answer and explanation
Correct option: D
41
Option D (41) is correct because the total number of unordered pairs of disjoint subsets of S is calculated using the formula (3^n + 1)/2 where n is the number of elements in S.
127 Easy
127 of 194
Application +4 | -1
Let X={1,2,3,4,5}\mathrm{X=}\{1,2,3,4,5\}. The number of different ordered pairs (Y,Z)(\mathrm{Y}, \mathrm{Z}) that can formed such that Y⊆X,Z⊆X\mathrm{Y} \subseteq \mathrm{X}, \mathrm{Z} \subseteq \mathrm{X} and Y∩Z\mathrm{Y} \cap \mathrm{Z} is empty is:
View answer and explanation
Correct option: B
353^5
Option B (353^5) is correct because each element in X has three possibilities: belonging to Y, belonging to Z, or belonging to neither, given that Y and Z are disjoint subsets of X.
128 Easy
128 of 194
Application +4 | -1
If A={1,  2,  3,  4},A = \left\{ {1,\;2,\;3,\;4} \right\}, then the number of subsets of AA that contain the element 2 but not 3, is
View answer and explanation
Correct option: B
4
Option C (4) is correct because there are 22 = 4 subsets that contain the element 2 but not 3, formed from the remaining elements 1 and 4.
129 Easy
129 of 194
Understanding +4 | -1
If A={x:x=3n,n∈Z}A=\{x: x=3 n, n \in Z\} and B={x:x=4n,n∈Z}B=\{x: x=4 n, n \in Z\} then A∩BA \cap B is
View answer and explanation
Correct option: D
None of these
Option D ({x:x=12n,n∈Z}\{x: x=12 n, n \in Z\}) is correct because the intersection of multiples of 3 and 4 are multiples of 12.
130 Easy
130 of 194
Application +4 | -1
If A={2,4}A=\{2,4\} and B={3,4,5}B=\{3,4,5\}, then (A∩B)×(A∪B)(A \cap B) \times(A \cup B) is
View answer and explanation
Correct option: D
{(4,2),(4,3),(4,4),(4,5)}\{(4,2),(4,3),(4,4),(4,5)\}
Option D ({ (4,2),(4,3),(4,4),(4,5) }) is correct because it accurately represents the Cartesian product of the intersection of A and B with the union of A and B.
131 Easy
131 of 194
Application +4 | -1
If AA and BB are two sets such that n(A)=7,n(B)=6n\left( A \right) = 7,n\left( B \right) = 6 and (A∩B)≠ϕ.\left( {A \cap B} \right) \ne \phi . Then the greatest possible value of n(A  Δ  B),n\left( {A\;\mathrm{\Delta \;}B} \right), is
View answer and explanation
Correct option: A
11
Option A (11) is correct because the greatest possible value of n(AΔB)n(A \Delta B) occurs when n(A∩B)n(A \cap B) is minimized, which is 1 in this case.
132 Easy
132 of 194
Analyzing +4 | -1
If a  N={a  x  :x∈N}a\;N = \left\{ {a\;x\;:x \in N} \right\} and b  N  ∩c  N=d  N,b\;N\; \cap c\;N = d\;N, where b,c∈Nb,c \in N are relatively prime, then
View answer and explanation
Correct option: A
d=bcd = bc
Option A (d=bcd = bc) is correct because the intersection of multiples of two relatively prime numbers b and c is the set of multiples of their product bc.
133 Easy
133 of 194
Analyzing +4 | -1
If X={8n−7n−1:n∈N}X=\left\{8^n-7 n-1: n \in N\right\} and Y={49(n−1):n∈N}Y=\{49(n-1): n \in N\}, then
View answer and explanation
Correct option: A
X⊂YX \subset Y
Option A (X⊂YX \subset Y) is correct because every element in X is also an element in Y, but not vice versa.
134 Easy
134 of 194
Analyzing +4 | -1
Let NN be the set of natural numbers and for a∈N,aNa \in N, a N denotes the set {ax:x∈N}.\{a x: x \in N\}. If bN∩cN=dN,b N \cap c N=d N, where b,c,db, c, d are natural numbers greater than 1 and the greatest common divisor (GCD) of bb and cc is 1, then dd equals
View answer and explanation
Correct option: D
bcbc
The correct answer is D (bcbc). Since GCD(b, c) = 1, the least common multiple is bc, and bN ∩ cN represents multiples of both b and c, hence multiples of bc.
135 Easy
135 of 194
Analyzing +4 | -1
If X={4n−3n−1  :n∈N}X = \left\{ {{4^n} - 3n - 1\;:n \in N} \right\} and Y={9(n−1):n∈N},Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}, then X∪YX \cup Y is equal to
View answer and explanation
Correct option: B
YY
Option B (Y) is correct because every element of X is also an element of Y, meaning X is a subset of Y, and therefore the union of X and Y is Y.
136 Easy
136 of 194
Understanding +4 | -1
If A={x  :x  A = \{ x\;:x\; is a multiple of   3}\;3\} and,
B={x  :x  B = \{ x\;:x\; is a multiple of 5}, then A−BA - B is
View answer and explanation
Correct option: B
A∩BˉA \cap \bar B
Option B (A∩BˉA \cap \bar B) is correct because A - B contains elements that are in A but not in B, which is the definition of A∩BˉA \cap \bar B.
137 Easy
137 of 194
Understanding +4 | -1
Let AA and BB be two sets such that A∪B=AA \cup B=A. Then, A∩BA \cap B is equal to
View answer and explanation
Correct option: B
BB
Option B (BB) is correct because if A∪B=AA \cup B = A, it means that B is a subset of A, and therefore A∩BA \cap B is equal to B.
138 Easy
138 of 194
Application +4 | -1
If A1⊂A2⊂A3⊂…⊂A50{A_1} \subset {A_2} \subset {A_3} \subset \ldots \subset {A_{50}} and n(Ai)=i−1,n\left( {{A_i}} \right) = i - 1, then n(∩i=1150Ai)=n\left( { \cap _{i = 11}^{50}{A_i}} \right) =
View answer and explanation
Correct option: D
10
The correct answer is D (10) because the intersection of nested sets is the smallest set, and n(A_11) = 10.
139 Easy
139 of 194
Analyzing +4 | -1
The number of elements in the set S={x∈R:2cos⁡(x2+x6)=4x+4−x}S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\right\} is :
View answer and explanation
Correct option: A
1
Option A (1) is correct because the equation holds true only when both sides are equal to 2, which occurs only at x=0.
140 Easy
140 of 194
Understanding +4 | -1
{n(n+1)(2n+1):n∈Z}⊂\left\{ {n\left( {n + 1} \right)\left( {2n + 1} \right):n \in Z} \right\} \subset subset
View answer and explanation
Correct option: A
{6k  :k∈Z}\left\{ {6k\;:k \in Z} \right\}
Option A ({6k  :  k∈Z}\{ {6k\;:\;k \in Z} \}) is correct because the expression n(n+1)(2n+1)n(n+1)(2n+1) is always divisible by 6 for any integer n.
141 Easy
141 of 194
Application +4 | -1
In a group of 70 people, 37 like coffee, 52 like tea and each person like atleast one of the two drinks. The number of persons liking both coffee and tea is
View answer and explanation
Correct option: C
19
Option C (19) is correct because it uses the principle of inclusion-exclusion to find the number of people liking both coffee and tea.
142 Easy
142 of 194
Understanding +4 | -1
If A={x  :xA = \{ x\;:x is a multiple of 4} and,
B={x  :x  B = \{ x\;:x\; is a multiple of   6}\;6\}, then A∩BA \cap B consists of multiples of
View answer and explanation
Correct option: B
12
Option B (12) is correct because the least common multiple of 4 and 6 is 12, so any number that is a multiple of both 4 and 6 must be a multiple of 12.
143 Easy
143 of 194
Understanding +4 | -1
If A={x∈R/f(x)=0}{A}=\{x \in {R} / {f}(x)=0\} and B={x∈R/g(x)=0}{B}=\{x \in {R} / {g}(x)=0\} then A∩B={A} \cap {B}=
View answer and explanation
Correct option: D
{x∈R/f(x)2+g(x)2=0}\left\{x \in {R} / {f}(x)^2+{g}(x)^2=0\right\}
Option D is correct because A ∩ B represents the intersection of sets A and B, meaning elements that satisfy both f(x) = 0 and g(x) = 0, which is equivalent to f(x)² + g(x)² = 0.
144 Easy
144 of 194
Analyzing +4 | -1
Consider the set AA of all determinants of order 3 with entries 0 or 1 only. Let BB be the subset of AA consisting of all determinants with value 1. Let CC be the subset of the set of all determinants with value −1.- 1.Then
View answer and explanation
Correct option: B
BB has as many elements as CC
Option B (BB has as many elements as CC) is correct because interchanging two rows or columns in a determinant changes its sign, creating a one-to-one correspondence between determinants with value 1 and -1.
145 Easy
145 of 194
Understanding +4 | -1
Which one of the following is not true?
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Correct option: C
A⊆A−BA \subseteq A-B
Option C (A⊆A−BA \subseteq A-B) is not true because A−BA - B contains elements in A but not in B, so A cannot be a subset of A-B.
146 Easy
146 of 194
Understanding +4 | -1
If the cardinality of a set AA is 4 and that of a set BB is 3 , then what is the cardinality of the set A△BA \triangle B ?
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Correct option: D
Cannot be determined as the sets AA and BB
The correct answer is D (Cannot be determined as the sets AA and BB) because without knowing the elements of sets A and B, we cannot determine their symmetric difference.
147 Easy
147 of 194
Understanding +4 | -1
Let Y={1,2,3,4,5,6,7},A={3,4},B={1,2,5}Y=\{1,2,3,4,5,6,7\}, A=\{3,4\}, B=\{1,2,5\}, then (Y×A)∩(Y×B)(Y \times A) \cap(Y \times B) is
View answer and explanation
Correct option: D
ϕ\phi
Option D (ϕ\phi) is correct because the intersection of A and B is an empty set, leading to an empty Cartesian product.
148 Easy
148 of 194
Application +4 | -1
In an election, two contestants AA and BB contested x %x\,\% of the total voters voted for AA and (x+20)%\left( {x + 20} \right)\% for BB. If 20% of the voters did not vote, then x=x =
View answer and explanation
Correct option: A
30
Option A (30) is correct because it is the value of x that satisfies the equation derived from the given percentages of voters for A, B, and those who did not vote.
149 Easy
149 of 194
Application +4 | -1
If A={(x,y)  :y=4x,x≠0}A = \left\{ {\left( {x,y} \right)\;:y = \frac{4}{x},x \ne 0} \right\} and
B={(x,y):x2+y2=8,x,y∈R}B = \left\{ {\left( {x,y} \right):{x^2} + {y^2} = 8,x,y \in R} \right\}, then
View answer and explanation
Correct option: C
A∩BA \cap B contains two points only
The correct answer is C (A∩BA \cap B contains two points only) because solving the system of equations y = 4/x and x2 + y2 = 8 yields two intersection points.
150 Easy
150 of 194
Understanding +4 | -1
Consider the following relations:

I. A−B=A−(A∩B)A-B=A-(A \cap B)

II. A=(A∩B)∪(A−B){A}=({A} \cap {B}) \cup({A}-{B})

III. A−(B∪C)=(A−B)∪(A−C)A-(B \cup C)=(A-B) \cup(A-C)

Which of these is/are correct?
View answer and explanation
Correct option: D
Both I and II
Option D (Both I and II) is correct because the given set operations are valid set identities.
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