Set theory practice · Mixed difficulty

The set $A=\left\{x: x^4-x^3-x^2=0\right.$ and $\left.x \in N\right\}$ represents

If $A = \left\{ {p:p = \frac{{\left( {n + 2} \right)\left( {2{n^5} + 3{n^4} + 4{n^3} + 5{n^2} + 6} \right)}}{{{n^2} + 2n}},n,p \in {{\text{Z}}^ + }} \right\}$ then the number of elements in the set $A$, is

In a class of 35 students, 17 have taken Mathematics, 10 have taken Mathematics but not Economics. If each student has taken either Mathematics or Economics or both, then the number of students who have taken Economics but not Mathematics is

If $S$ is the set of squares and $R$ is the set of rectangles, then $\left( {S \cup R} \right) - \left( {S \cap S} \right)$ is

If a set $A$ contains n elements, then which of the following cannot be the number of reflexive relations on the set $A$?

Let $S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}}}\theta + {8^{2{{\cos }^2}\theta }} = 16} \right\}.$ Then $n(S)+$ $\sum\limits_{\theta \in S} {\left( {\sec \left( {\frac{\pi }{4} + 2\theta } \right)\cos ec\left( {\frac{\pi }{4} + 2\theta } \right)} \right)}$ is equal to :

If the number of elements in $A$ is $m$ and number of element in $B$ is $n$ then the number of elements in the power set of $A \times B$ is

If $A=\{1,2,3\}, B=\{3,4,5\},$ then $(A \cap B) \times A$ is

Given the sets $A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}$, then: $A \cup(B \cap C)$ is

If $X$ and $Y$ are two sets, then $X \cap(Y \cap X)^{\prime}$ equals

Let $\mathcal{U}$ be the universal set for sets A and B such that $n\left( A \right) = 200,n\left( B \right) = 300$ and $n\left( {A \cap B} \right) = 100.$ Then, $n\left( {A' \cap B'} \right)$ is equal to 300, provided that $n\left( \mathcal{U} \right)$ is equal to

Let $P(S)$ denote the power set of $S=\{1,2,3, \ldots, 10\}$. Define the relations $R_{1}$ and $R_{2}$ on $P(S)$ as $A R_{1} B$ if $\left(A \cap B^{C}\right) \cup\left(B \cap A^{C}\right)=\phi$ and $A R_{2} B$ if $A \cup B^{C}=B \cup A^{C}, \forall A, B \in P(S)$. Then

Let $S=\{x \in R: x \geq 0$ and $2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0$. Then $S$:

Two finite sets have $m$ and $n$ elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of $m$ and $n$ are

Let $A = \left\{ {{\text{ONGC}},{\text{BHEL}},{\text{SAIL}},{\text{GAIL}},{\text{IOCL}}} \right\}$ and $R$ be a relation defined as “two elements of $A$ are related if they share exactly one letter”. The relation $R$ is

If $A = \left\{ {1,\;2,\;3} \right\},\;B\left\{ {3,\;4} \right\},\;C\left\{ {4,\;5,\;6} \right\}$. Then, $A \cup \left( {B \cap C} \right)$ is

The value of $n\{P[P(\phi)]\}$ is equal to

For any two sets $A$ and $B$, $A - \left( {A - B} \right)$ equals

If $A$ and $B$ are two given sets, then $A \cap {\left( {A \cap B} \right)^c}$ is equal to

In a class of 55 students, the number of students studying different subjects are 23 in mathematics, 24 in physics, 19 in chemistry, 12 in mathematics and physics, 9 in mathematics and chemistry, 7 in physics and chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is:

The set $(A \backslash B) \cup(B \backslash A)$ is equal to

If $A = \left\{ {x,\;y} \right\}$, then the power set of $A$ is

Let $\mathrm{A}=\{\mathrm{x} \in \mathrm{W}$, the set of whole numbers and $x < 3\}, B=\{x \in N$, the set of natural numbers and $2 \leq x < 4\}$ and $C=\{3,4\}$, then how manyelements will $(\mathrm{A} \cup \mathrm{B}) \times \mathrm{C}$ contain?

Let $S=\left\{\theta \in[0,2 \pi]: 2 \cos ^2 \theta+3 \sin \theta=0\right\}$ then number of elements of $S$ is

If $A,B$ and $C$ are three sets such that $A \cap B = A \cap C$ and $A \cup B = A \cup C$, then

Statement-I: The set $\left\{x\right.$ : $x$ is a real number and $\left.x^2-1=0\right\}$ is the empty set.

Statement-II: The set ${A}=\left\{{x}: {x} \in {R}, {x}^2=16\right.$ and $\left.2 {x}=6\right\}$ is an empty set.

If $A=\left\{\theta: 2 \cos ^2 \theta+\sin \theta \leq 2\right\}$ and $B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}$, then $A \cap B$ is equal to

Let $f(x)=(x+1)^2-1,(x \geq-1)$. Then, the set $S=\left\{x: f(x)=f^{-1}(x)\right\}$ is

If $A = \left\{ {p \in N:p\;\mathrm{is\;}a\mathrm{\;prime\;and\;}p = \frac{{7{n^2} + 3n + 3}}{n}\mathrm{\;for\;some}\;n \in N} \right\},$ then the number of elements in the set $A$, is

If $A$ and $B$ are two sets, then $(A-B) \cup(B-A) \cup(A \cap B)$ is equal to