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JEE • MATHEMATICS • SET THEORY

Set theory practice · Mixed difficulty

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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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Showing 181-210 of 273 questions

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181 Easy
181 of 273
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.
182 Medium
182 of 273
Analyzing +4 | -1
A={x/x∈R,∣x∣<1}B={x/x∈R,∣x−1∣≥1}A=\{x / x \in R,|x|<1\} \quad B=\{x / x \in R,|x-1| \geq 1\} and A∪B=R−DA \cup B=R-D Then D=D=
View answer and explanation
Correct option: B
{x/1≤x<2}\{x / 1 \leq x<2\}
The correct answer cannot be determined without the options. Please provide the options to determine the correct answer.
183 Medium
183 of 273
Application +4 | -1
If A={1,2,3}A=\{1,2,3\} and B={3,8}B=\{3,8\}, then (A∪B)×(A∩B)(A \cup B) \times(A \cap B) is equal to
View answer and explanation
Correct option: D
{(1,3),(2,3),(3,3),(8,3)}\{(1,3),(2,3),(3,3),(8,3)\}
Option D ({(1,3),(2,3),(3,3),(8,3)}\left\{(1,3),(2,3),(3,3),(8,3)\right\}) is correct because it represents the Cartesian product of the union of sets A and B with the intersection of sets A and B.
184 Medium
184 of 273
Understanding +4 | -1
Which of the following is the empty set?
View answer and explanation
Correct option: B
{x:x\left\{x: x\right. is a real number and x2+1=0}\left.x^2+1=0\right\}
Option B ({x:x\{x: x is a real number and x2+1=0}\left.x^2+1=0\right\}) is the empty set because there is no real number whose square plus 1 equals 0.
185 Easy
185 of 273
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.
186 Easy
186 of 273
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.
187 Easy
187 of 273
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.
188 Easy
188 of 273
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.
189 Easy
189 of 273
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.
190 Easy
190 of 273
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.
191 Easy
191 of 273
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.
192 Easy
192 of 273
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.
193 Easy
193 of 273
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.
194 Easy
194 of 273
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.
195 Easy
195 of 273
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.
196 Medium
196 of 273
Understanding +4 | -1
Let A={1,2},B={3,4}\mathrm{A}=\{1,2\}, \mathrm{B}=\{3,4\}. Then, number of subsets of A×B\mathrm{A} \times \mathrm{B} is
View answer and explanation
Correct option: D
16
Option D (16) is correct because the number of subsets of a set with n elements is 2^n. A x B has 4 elements.
197 Medium
197 of 273
Understanding +4 | -1
If A={(x,y):x2+y2=4;  x,y∈R}A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} = 4;\;x,y \in R} \right\} and
B={(x,y):x2+y2=9;x,y∈R}B = \left\{ {\left( {x,y} \right):{x^2} + {y^2} = 9;x,y \in R} \right\}, then
View answer and explanation
Correct option: B
B−A=BB - A = B
Option B (B−A=BB - A = B) is correct because A and B represent circles centered at the origin with radii 2 and 3 respectively. Since A is contained within B, their intersection is empty, and removing A from B leaves B.
198 Easy
198 of 273
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.
199 Easy
199 of 273
Understanding +4 | -1
Which one of the following is not true?
View answer and explanation
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.
200 Easy
200 of 273
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 ?
View answer and explanation
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.
201 Medium
201 of 273
Application +4 | -1
Let XX be the universal set for sets AA and BB. If n(A)=200,n(B)=300n(A)=200, n(B)=300 and n(A∩B)=100n(A \cap B)=100, then n(A′∩B′)n\left(A^{\prime} \cap B^{\prime}\right) is equal to 300 , provided (X)(X) is equal to
View answer and explanation
Correct option: B
700
Option B (700) is correct because it is calculated using the formula for the union of two sets and De Morgan's law to find the size of the universal set.
202 Medium
202 of 273
Understanding +4 | -1
If A,BA, B and CC are three finite sets, then what is [(A∪B)∩C][(A \cup B) \cap C] equal to?
View answer and explanation
Correct option: C
A′∩B′∪C′A^{\prime} \cap B^{\prime} \cup C^{\prime}
Option C (A′∩B′∪C′A^{\prime} \cap B^{\prime} \cup C^{\prime}) is correct because it correctly applies De Morgan's law and the complement properties to simplify the expression.
203 Easy
203 of 273
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.
204 Medium
204 of 273
Application +4 | -1
If XX and YY are two sets such that (X∪Y)(X \cup Y) has 60 elements, XX has 38 elements and YY has 42 elements, how many elements does (X∩Y)(\mathrm{X} \cap \mathrm{Y}) have?
View answer and explanation
Correct option: B
20
Option B (20) is correct because it is calculated using the formula for the number of elements in the union of two sets: n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y).
205 Easy
205 of 273
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.
206 Medium
206 of 273
Analyzing +4 | -1
If A∪B=A∪CA \cup B=A \cup C and A∩B=A∩CA \cap B=A \cap C, then
View answer and explanation
Correct option: C
B=CB=C
Option A (A′A^{\prime}) is correct because the expression (A∪B)′∪(A′∩B)(A \cup B)^{\prime} \cup (A^{\prime} \cap B) simplifies to A′A^{\prime} using set identities.
207 Medium
207 of 273
Analyzing +4 | -1
Let P={θ:sin⁡θ−cos⁡θ=2cos⁡θ}P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} and Q={θ:sin⁡θ+Q=\{\theta: \sin \theta+ cos⁡θ=2sin⁡θ}\cos \theta=\sqrt{2} \sin \theta\} be two sets. Then:
View answer and explanation
Correct option: C
P=Q{P}={Q}
Option C (P=Q) is correct because by manipulating the first equation, it can be shown to be equivalent to the second equation, indicating that both sets contain the same elements.
208 Easy
208 of 273
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.
209 Easy
209 of 273
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.
210 Easy
210 of 273
Understanding +4 | -1
If X∪{3,4}={1,2,3,4,5,6}X \cup\{3,4\}=\{1,2,3,4,5,6\} then which of the following is true
View answer and explanation
Correct option: A
Smallest set X={1,2,5,6}X=\{1,2,5,6\}
Option A is correct because the smallest set X is obtained by removing {3,4} from {1,2,3,4,5,6}.
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