Basic of Set theory Questions in English

(A) The given equation is $x^2 + x - 2 = 0$.
Factorizing the quadratic equation:
$x^2 + 2x - x - 2 = 0$
$x(x + 2) - 1(x + 2) = 0$
$(x - 1)(x + 2) = 0$
Thus,the roots are $x = 1$ and $x = -2$.
Therefore,the solution set in roster form is $\{1, -2\}$.

(A) The condition given is that $x$ is a positive integer and $x^2 < 40$.
We test the positive integers:
$1^2 = 1 < 40$ (True)
$2^2 = 4 < 40$ (True)
$3^2 = 9 < 40$ (True)
$4^2 = 16 < 40$ (True)
$5^2 = 25 < 40$ (True)
$6^2 = 36 < 40$ (True)
$7^2 = 49 > 40$ (False)
Thus,the set in roster form is $\{1, 2, 3, 4, 5, 6\}$.

(A) The given set is $A = \{ 1, 4, 9, 16, 25, \dots \}$.
Observing the elements,we see that they are squares of natural numbers:
$1 = 1^2$
$4 = 2^2$
$9 = 3^2$
$16 = 4^2$
$25 = 5^2$
Thus,the set $A$ can be written in set-builder form as:
$A = \{ x : x = n^2, n \in \mathbb{N} \}$.

(A) We observe that each element in the set follows the pattern $\frac{n}{n+1}$,where $n$ is the numerator.
The numerator $n$ starts from $1$ and goes up to $6$.
Thus,the set-builder form is $\{ x : x = \frac{n}{n+1}, n \in \mathbb{N}, 1 \le n \le 6 \}$.

(A) In $(d)$,the word $PRINCIPAL$ consists of the letters $P, R, I, N, C, A, L$ (since $P$ and $I$ are repeated,they are written once in the set). Thus,$(i)$ matches $(d)$.
In $(c)$,$x + 1 = 1$ implies $x = 0$. Thus,$(ii)$ matches $(c)$.
In $(a)$,the positive divisors of $18$ are $1, 2, 3, 6, 9, 18$. Thus,$(iii)$ matches $(a)$.
In $(b)$,$x^2 - 9 = 0$ implies $x^2 = 9$,so $x = 3$ or $x = -3$. Thus,$(iv)$ matches $(b)$.
The correct matching is $(i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)$.

(N/A) The collection of all months of a year beginning with the letter $J$ is a well-defined collection of objects because one can definitely identify a month that belongs to this collection.
Specifically,the months are January,June,and July.
Since the collection is well-defined,it is a set.

(N/A) set is a well-defined collection of objects.
The collection of the ten most talented writers of India is not a well-defined collection because the criteria for determining a writer's talent are subjective and vary from person to person.
Therefore,this collection is not a set.

(B) set is defined as a well-defined collection of objects.
$A$ team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman's talent (such as average,strike rate,or consistency) may vary from person to person.
Since the collection is subjective and not well-defined,it is not a set.

(A) set is a well-defined collection of objects.
The collection of all boys in your class is a well-defined collection because you can definitely identify whether a specific person belongs to this collection or not.
Since the criteria for inclusion are clear and objective,this collection is a set.

(N/A) The collection of all natural numbers less than $100$ is a well-defined collection because one can definitely identify a number that belongs to this collection.
Hence,this collection is a set.

(N/A) collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify whether a specific book belongs to this collection or not.
Since the collection is well-defined,it is a set.

(A) The collection of all even integers is a well-defined collection because one can definitely identify whether any given integer is even or not.
Since the criteria for membership in this collection are clear and unambiguous,this collection is a set.

(N/A) The collection of questions in this chapter is a well-defined collection because one can definitely identify whether a specific question belongs to this chapter or not.
Since the collection is well-defined,it is a set.

(N/A) set is defined as a well-defined collection of objects.
The collection of the most dangerous animals of the world is not a well-defined collection because the criteria for determining the 'dangerousness' of an animal can vary from person to person.
Since the criteria are subjective,this collection is not a set.

(A) The set $A$ is defined as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $5$ is present in the set $A$,we use the symbol $\in$ to denote that $5$ is an element of $A$.
Therefore,$5 \in A$.

(B) The set $A$ is given as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $8$ is not present in the set $A$,we use the symbol $\notin$.
Therefore,$8 \notin A$.

(B) The set $A$ is defined as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $0$ is not present in the set $A$,we use the symbol $\notin$ to denote that $0$ is not an element of $A$.
Therefore,$0 \notin A$.

(A) The set $A$ is given as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $4$ is present in the set $A$,we use the symbol $\in$ to denote that $4$ belongs to $A$.
Therefore,$4 \in A$.

(A) The set $A$ is defined as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $2$ is present in the set $A$,the correct symbol to use is $\in$.
Therefore,$2 \in A$.

(B) The set $A$ is defined as $A = \{1, 2, 3, 4, 5, 6\}$.
Since the element $10$ is not present in the set $A$,we use the symbol $\notin$ (does not belong to).
Therefore,$10 \notin A$.

(N/A) The set $A$ is defined as $A = \{ x : x \text{ is an integer and } -3 < x < 7 \}$.
The integers strictly between $-3$ and $7$ are $-2, -1, 0, 1, 2, 3, 4, 5, 6$.
Therefore,the roster form of the set is $A = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}$.

(N/A) The set $B$ is defined as the collection of all natural numbers less than $6$.
Natural numbers are positive integers starting from $1$.
Therefore,the natural numbers less than $6$ are $1, 2, 3, 4, \text{ and } 5$.
In roster form,we list these elements within curly braces:
$B = \{1, 2, 3, 4, 5\}$

(N/A) The set $C$ consists of all two-digit natural numbers where the sum of the digits is $8$.
Let the two-digit number be $xy$,where $x \in \{1, 2, \dots, 9\}$ and $y \in \{0, 1, \dots, 9\}$.
We require $x + y = 8$.
Possible pairs $(x, y)$ are:
If $x=1, y=7 \implies 17$
If $x=2, y=6 \implies 26$
If $x=3, y=5 \implies 35$
If $x=4, y=4 \implies 44$
If $x=5, y=3 \implies 53$
If $x=6, y=2 \implies 62$
If $x=7, y=1 \implies 71$
If $x=8, y=0 \implies 80$
Thus,the set in roster form is $C = \{17, 26, 35, 44, 53, 62, 71, 80\}$.

(A) To find the elements of the set $D$,we first find the prime factorization of $60$.

$2$	$60$
$2$	$30$
$3$	$15$
$5$	$5$

$\therefore 60 = 2 \times 2 \times 3 \times 5$.
The prime divisors of $60$ are $2, 3,$ and $5$.
Therefore,the set in roster form is $D = \{ 2, 3, 5 \}$.

(A) The word is $TRIGONOMETRY$.
To write the set in roster form,we list each distinct letter present in the word exactly once.
The letters in $TRIGONOMETRY$ are $T, R, I, G, O, N, O, M, E, T, R, Y$.
Identifying the distinct letters: $T, R, I, G, O, N, M, E, Y$.
Therefore,the set in roster form is $E = \{ T, R, I, G, O, N, M, E, Y \}$.

(N/A) $F =$ The set of all letters in the word $BETTER$.
There are $6$ letters in the word $BETTER,$ out of which the letters $E$ and $T$ are repeated.
In roster form,each distinct element is listed only once.
Therefore,the set $F$ in roster form is $F = \{B, E, T, R\}$.

(N/A) The given set is $A = \{ 3, 6, 9, 12 \}$.
We observe that each element is a multiple of $3$ and can be written as $3n$,where $n$ is a natural number.
For $n=1$,$3(1) = 3$.
For $n=2$,$3(2) = 6$.
For $n=3$,$3(3) = 9$.
For $n=4$,$3(4) = 12$.
Thus,the set can be written in set-builder form as $\{ x : x = 3n, n \in \mathbb{N} \text{ and } 1 \le n \le 4 \}$.

(N/A) The given set is $A = \{ 2, 4, 8, 16, 32 \}$.
We observe that each element is a power of $2$:
$2 = 2^{1}$
$4 = 2^{2}$
$8 = 2^{3}$
$16 = 2^{4}$
$32 = 2^{5}$
Thus,the set can be represented in set-builder form as:
$A = \{ x : x = 2^{n}, n \in \mathbb{N} \text{ and } 1 \le n \le 5 \}$

(N/A) The given set is $A = \{ 5, 25, 125, 625 \}$.
Observe the elements of the set:
$5 = 5^{1}$
$25 = 5^{2}$
$125 = 5^{3}$
$625 = 5^{4}$
It can be seen that each element is of the form $5^{n}$,where $n$ is a natural number such that $1 \le n \le 4$.
Therefore,the set-builder form is $\{ x : x = 5^{n}, n \in N \text{ and } 1 \le n \le 4 \}$.

(N/A) The given set is $\{ 2, 4, 6, \dots \}$.
This set consists of all positive even integers.
Therefore,the set-builder form is $\{ x : x \text{ is an even natural number} \}$.

(N/A) The given set is $A = \{ 1, 4, 9, \ldots, 100 \}$.
We observe that the elements are squares of natural numbers:
$1 = 1^2$
$4 = 2^2$
$9 = 3^2$
$\vdots$
$100 = 10^2$
Thus,the set can be written in set-builder form as:
$A = \{ x : x = n^2, n \in \mathbb{N} \text{ and } 1 \le n \le 10 \}$

(A) The set $A$ is defined as the set of all odd natural numbers.
Natural numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, \dots$.
Odd natural numbers are those natural numbers which are not divisible by $2$.
Therefore,the elements of set $A$ are $1, 3, 5, 7, 9, \dots$.
Thus,$A = \{ 1, 3, 5, 7, 9, \dots \}$.

(N/A) The set is given by $B = \{ x : x \text{ is an integer}, -\frac{1}{2} < x < \frac{9}{2} \}$.
We know that $-\frac{1}{2} = -0.5$ and $\frac{9}{2} = 4.5$.
The integers $x$ such that $-0.5 < x < 4.5$ are $0, 1, 2, 3, 4$.
Therefore,$B = \{0, 1, 2, 3, 4\}$.

The set $C$ is defined as $C = \{ x : x \text{ is an integer; } x^2 \le 4 \}$.
We need to find all integers $x$ such that their square is less than or equal to $4$.
Testing integers:
$(-3)^2 = 9 > 4$
$(-2)^2 = 4 \le 4$
$(-1)^2 = 1 \le 4$
$0^2 = 0 \le 4$
$1^2 = 1 \le 4$
$2^2 = 4 \le 4$
$3^2 = 9 > 4$
Thus,the integers satisfying the condition are $\{-2, -1, 0, 1, 2\}$.
Therefore,$C = \{-2, -1, 0, 1, 2\}$.

(D) To list the elements of the set $D$,we identify the distinct letters in the word $\text{"LOYAL"}$.
The letters are $L, O, Y, A, L$.
Since elements in a set must be distinct,we list each letter only once.
Therefore,$D = \{ L, O, Y, A \}$.

(A) The set $E$ consists of all months in a year that have fewer than $31$ days.
Months with $31$ days are: January,March,May,July,August,October,and December.
Months with $30$ days or fewer are: February ($28$ or $29$ days),April,June,September,and November.
Therefore,$E = \{ \text{February, April, June, September, November} \}$.

(N/A) The consonants in the English alphabet that precede the letter $k$ are $b, c, d, f, g, h, j$.
Therefore,the set $F$ is given by:
$F = \{b, c, d, f, g, h, j\}$

(A) $(i)$ All the elements of this set are natural numbers as well as the divisors of $6$. Therefore,$(i)$ matches with $(c)$.
$(ii)$ It can be seen that $2$ and $3$ are prime numbers. They are also the divisors of $6$. Therefore,$(ii)$ matches with $(a)$.
$(iii)$ All the elements of this set are letters of the word $MATHEMATICS$. Therefore,$(iii)$ matches with $(d)$.
$(iv)$ All the elements of this set are odd natural numbers less than $10$. Therefore,$(iv)$ matches with $(b)$.

(A) The given equation is $(x - 1)(x - 2) = 0$.
Solving this,we get $x = 1$ or $x = 2$.
Since both $1$ and $2$ are natural numbers $(x \in \mathbb{N})$,the set is $\{ 1, 2 \}$.
Since the number of elements in the set is countable and limited,it is a finite set.

(A) The given set is defined as $\{ x:x \in N \text{ and } x^2 = 4 \}$.
Solving the equation $x^2 = 4$,we get $x = 2$ or $x = -2$.
Since $x \in N$ (the set of natural numbers),only $x = 2$ is a valid element.
Thus,the set is $\{ 2 \}$.
Since the number of elements in the set is $1$,which is a finite number,the set is finite.

(A) The given set is defined by the condition $2x - 1 = 0$,which implies $2x = 1$,or $x = \frac{1}{2}$.
Since $x$ must be a natural number $(x \in N)$,and $\frac{1}{2}$ is not a natural number,there are no elements that satisfy this condition.
Therefore,the set is the empty set,denoted by $\phi$.
Since the empty set has $0$ elements,it is a finite set.

(B) The given set $A$ consists of all prime numbers,i.e.,$A = \{2, 3, 5, 7, 11, \dots\}$.
Since the set of prime numbers is infinite,the given set is an infinite set.

(B) The set is defined as the collection of all odd natural numbers.
Since the set of natural numbers $N = \{1, 2, 3, 4, 5, \dots\}$ is infinite,the subset of odd natural numbers $\{1, 3, 5, 7, \dots\}$ also contains an infinite number of elements.
Therefore,the given set is infinite.

(C) First,we determine the elements of each set:
$A = \{ 0 \}$
$B = \phi$ (since no number is both greater than $15$ and less than $5$)
$C = \{ 5 \}$ (since $x - 5 = 0 \implies x = 5$)
$D = \{ -5, 5 \}$ (since $x^2 = 25 \implies x = \pm 5$)
$E = \{ 5 \}$ (since $x^2 - 2x - 15 = 0 \implies (x - 5)(x + 3) = 0$,so $x = 5$ or $x = -3$. The positive integral root is $5$)
Comparing the sets:
$A = \{ 0 \}, B = \phi, C = \{ 5 \}, D = \{ -5, 5 \}, E = \{ 5 \}$.
We observe that $C = E$ because they contain the same elements.
Thus,the only pair of equal sets is $(C, E)$.

(A) We have,$X = \{A, L, L, O, Y\} = \{A, L, O, Y\}$ and $B = \{L, O, Y, A, L\} = \{L, O, Y, A\}$.
Since the repetition of elements in a set does not change the set,we have $X = \{A, L, O, Y\}$ and $B = \{A, L, O, Y\}$.
Since all elements of $X$ are in $B$ and all elements of $B$ are in $X$,the sets are equal.
Thus,$X = B$.

(A) For set $A$,the condition is $n \in \mathbb{Z}$ and $n^2 \le 4$. The integers satisfying this are $\{-2, -1, 0, 1, 2\}$.
For set $B$,the condition is $x \in \mathbb{R}$ and $x^2 - 3x + 2 = 0$. Solving the quadratic equation: $(x-1)(x-2) = 0$,which gives $x = 1$ or $x = 2$. Thus,$B = \{1, 2\}$.
Since $A = \{-2, -1, 0, 1, 2\}$ and $B = \{1, 2\}$,the elements of $A$ and $B$ are not the same.
Therefore,$A \neq B$.

(A) null set (or empty set) is a set that contains no elements.
For the set of odd natural numbers divisible by $2$:
An odd number is of the form $2n + 1$,where $n$ is an integer.
By definition,an odd number is not divisible by $2$.
Therefore,there are no odd natural numbers divisible by $2$.
Thus,the set is a null set (denoted by $\emptyset$ or $\{\}$).

(B, C, D) null set (or empty set) is a set that contains no elements.
$1$. The set of even prime numbers is $\{2\}$,which is not a null set.
$2$. The set of odd natural numbers divisible by $2$ is a null set,as no odd number is divisible by $2$.
$3$. The set of even prime numbers greater than $2$ is a null set,as $2$ is the only even prime number.
$4$. The set of natural numbers $x$ such that $x < 5$ and $x > 7$ is a null set,as no number can satisfy both conditions simultaneously.

(B) null set (or empty set) is a set that contains no elements.
Consider the set $A = \{ x : x \in \mathbb{N}, x < 5 \text{ and } x > 7 \}$.
For a natural number $x$ to be in this set,it must satisfy both conditions: $x < 5$ and $x > 7$ simultaneously.
There is no natural number that is both less than $5$ and greater than $7$.
Therefore,the set $A$ contains no elements,which makes it a null set.

(A) The set $\{ y : y \text{ is a point common to any two parallel lines} \}$ is a null set because parallel lines do not intersect. Therefore,they have no common point.