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CBSE - Class 11 Mathematics Sets Worksheet
EXERCISE 1.1
Worksheet Answers
Solution:
The set $C$ is defined by the set-builder notation: $C = \{x : x \in \mathbb{N}, 10 \le x \le 99, \text{sum of digits of } x = 8\}$.
Let the two-digit number be represented as $10a + b$, where $a \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $b \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The constraint given is $a + b = 8$ [Constraint based on problem statement].
To find all possible values for $x$, we iterate through all possible values of the tens digit ($a$) and determine the corresponding units digit ($b$) such that $b = 8 - a$. Since $x$ is a natural number, $a$ must be at least 1.
| Tens digit ($a$) | Units digit ($b = 8 - a$) | Resulting Number ($x$) |
|---|---|---|
| 1 | 7 | 17 |
| 2 | 6 | 26 |
| 3 | 5 | 35 |
| 4 | 4 | 44 |
| 5 | 3 | 53 |
| 6 | 2 | 62 |
| 7 | 1 | 71 |
| 8 | 0 | 80 |
We check if any other values exist. If $a=9$, $b = 8 - 9 = -1$, which is not a natural digit [By definition of decimal numeral system]. Thus, the set is complete.
The roster form requires listing all elements of the set separated by commas and enclosed within curly braces [Per the definition of Roster/Tabular form].
Final Solution: $C = \{17, 26, 35, 44, 53, 62, 71, 80\}$
Solution:
The set $A$ is defined using the set-builder notation: $A = \{x : x \text{ is an odd natural number}\}$. To list the elements of this set, we invoke the following definitions:
We generate the elements of the set by substituting consecutive values of $n$ into the expression for odd numbers:
| $n$ | Expression ($2n - 1$) | Resulting Element |
|---|---|---|
| 1 | $2(1) - 1$ | 1 |
| 2 | $2(2) - 1$ | 3 |
| 3 | $2(3) - 1$ | 5 |
| 4 | $2(4) - 1$ | 7 |
The set $A$ constitutes an infinite sequence along the number line representing discrete odd points.
By the Roster Method (or Tabular Form), we list all elements within curly braces, separated by commas. [Per the Axiom of Extensionality, a set is defined solely by its elements].
$A = \{1, 3, 5, 7, 9, \dots\}$
Final Solution: The set of odd natural numbers is represented in roster form as $A = \{1, 3, 5, 7, 9, \dots\}$.
Solution:
The set $B$ is defined by the set-builder form $B = \{x : x \in \mathbb{N} \text{ and } x < 6\}$. This notation requires the identification of two specific constraints:
We evaluate the natural numbers against the condition $x < 6$:
| Candidate ($x$) | Constraint Check ($x < 6$) | Status |
|---|---|---|
| 1 | $1 < 6$ | Valid |
| 2 | $2 < 6$ | Valid |
| 3 | $3 < 6$ | Valid |
| 4 | $4 < 6$ | Valid |
| 5 | $5 < 6$ | Valid |
| 6 | $6 < 6$ | Invalid |
In roster form, elements of a set are listed explicitly within curly braces, separated by commas. [Per the definition of the Roster/Tabular method]. Including all verified elements from Step 2:
$B = \{1, 2, 3, 4, 5\}$
Final Solution: $B = \{1, 2, 3, 4, 5\}$
Solution:
The set $E$ is defined by the property of containing all unique letters found in the word "TRIGONOMETRY". In set theory, a set is defined as a collection of well-defined, distinct objects [Per the Zermelo-Fraenkel set theory axiom of extensionality].
We extract each character from the word T-R-I-G-O-N-O-M-E-T-R-Y sequentially:
According to the standard definition of a set, each element must be unique; repeated elements are counted only once [Per the Fundamental Principle of Set Representation]. We identify and remove duplicates:
| Character | Frequency | Status |
|---|---|---|
| T | 2 | Include once |
| R | 2 | Include once |
| I | 1 | Include once |
| G | 1 | Include once |
| O | 2 | Include once |
| N | 1 | Include once |
| M | 1 | Include once |
| E | 1 | Include once |
| Y | 1 | Include once |
By collecting the unique letters identified in Step 3, we arrange them into the roster format, where elements are enclosed in curly braces $\{ \}$ and separated by commas:
$E = \{T, R, I, G, O, N, M, E, Y\}$
Final Solution: $E = \{T, R, I, G, O, N, M, E, Y\}$
Solution:
The universal set of concern is the English alphabet, $U = \{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z\}$. [Per the definition of the English alphabet].
The set $F$ is defined by two conditions:
To determine the elements of $F$, we filter the set $S$ to exclude vowels.
| Candidate Letter | Constraint Status |
|---|---|
| a | Vowel (Exclude) |
| b | Consonant (Include) |
| c | Consonant (Include) |
| d | Consonant (Include) |
| e | Vowel (Exclude) |
| f | Consonant (Include) |
| g | Consonant (Include) |
| h | Consonant (Include) |
| i | Vowel (Exclude) |
| j | Consonant (Include) |
By collecting the included elements identified in Step 3, we construct the set $F$. [By the Roster Method of set representation].
Final Solution: $F = \{b, c, d, f, g, h, j\}$
Solution:
The set is defined as $A = \{x : x \in \mathbb{Z} \text{ and } -3 \le x < 7\}$. To convert this into roster form, we must identify all elements that satisfy the given conditions simultaneously [Per the definition of Set-Builder Notation].
Based on the defined interval $[-3, 7)$, we list the integers sequentially starting from the lower bound ($-3$) and ending at the largest integer less than $7$ (which is $6$):
| Lower Bound | Upper Bound | Valid Integers |
|---|---|---|
| $-3$ | $7$ | $-3, -2, -1, 0, 1, 2, 3, 4, 5, 6$ |
By the Roster Method, elements of a set are listed within curly braces $\{ \}$, separated by commas [Per Set Theory conventions].
$A = \{-3, -2, -1, 0, 1, 2, 3, 4, 5, 6\}$
Final Solution: A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
Solution:
A set is defined as a well-defined collection of objects. In mathematics, "well-defined" implies that there is an objective, universal criterion that allows one to decide, without ambiguity, whether a specific object belongs to the collection or not [Per Cantor’s Set Theory].
The collection in question is: "A collection of most dangerous animals of the world."
To determine if this collection constitutes a set, we must examine the term "most dangerous."
Since the membership of this collection cannot be determined with absolute certainty due to the lack of a universal, objective, and well-defined rule, the collection fails the fundamental requirement of set theory.
Because the collection depends on the opinion of the observer rather than an objective mathematical or scientific standard, the rule defining the collection is not "well-defined."
Final Solution: The collection of the "most dangerous animals of the world" is not a set because the term "most dangerous" is subjective and lacks a universally accepted, precise definition.
Solution:
The set $A$ is given by the roster method as $A = \{1, 2, 3, 4, 5, 6\}$. This implies that the elements belonging to set $A$ are exclusively the integers from 1 to 6 inclusive [Definition of a Set].
To determine the correct relation for the element $10$ with respect to set $A$, we perform an exhaustive comparison:
| Candidate Element ($x$) | Membership Status in $A$ |
|---|---|
| 1 | $1 \in A$ |
| ... | ... |
| 6 | $6 \in A$ |
| 10 | $10 \notin A$ |
Since $10$ is not found in the collection $\{1, 2, 3, 4, 5, 6\}$, it fails the membership criteria [Set Membership Axiom].
Final Solution: 10 $\notin$ A
Solution:
In set theory, a set is defined as a well-defined collection of distinct objects. The term "well-defined" implies that for any given object, there must be an unambiguous criterion to determine whether that object belongs to the collection or not [Per Cantor’s Definition of a Set].
The collection in question is: "The collection of questions in this Chapter."
| Criterion | Application to "Questions in this Chapter" |
|---|---|
| Well-defined | Yes; the boundary of the set is defined by the specific text provided by the publisher. |
| Objectivity | Yes; any observer checking the chapter would identify the same set of questions. |
| Existence | Yes; the set is a finite subset of the Universal Set of all mathematical problems. |
Because the collection of questions in the chapter is fixed and possesses a clear, objective boundary that allows for definitive membership testing, it satisfies all axiomatic requirements of a mathematical set.
Final Solution: Yes, the collection of questions in this Chapter is a set.
Solution:
The set $\{1, 2, 3, 6\}$ contains positive integers that divide the number 6 without leaving a remainder. In set-builder notation, this is expressed as: $x \in \mathbb{N}$ such that $x$ is a divisor of 6.
Matching: Matches with (c) $\{x : x \text{ is a natural number and divisor of 6}\}$. [Per the definition of divisors in natural numbers].
The elements 2 and 3 are both prime numbers. Furthermore, both 2 and 3 are divisors of 6 (since $6 \div 2 = 3$ and $6 \div 3 = 2$).
Matching: Matches with (a) $\{x : x \text{ is a prime number and a divisor of 6}\}$. [Per the Fundamental Theorem of Arithmetic regarding prime factorization].
This set represents the unique letters found in the word "MATHEMATICS". When writing a set in roster form, duplicate elements are listed only once. The letters are M, A, T, H, E, M (repeat), A (repeat), T (repeat), I, C, S.
Matching: Matches with (d) $\{x : x \text{ is a letter of the word MATHEMATICS}\}$. [Per the property of sets that elements are distinct and order is irrelevant].
The elements are odd numbers ($2n-1$ for $n=1,2,3,4,5$) and all values are strictly less than 10.
Matching: Matches with (b) $\{x : x \text{ is an odd natural number less than 10}\}$. [Per the definition of odd natural numbers in the interval $(0, 10)$].
| Roster Form | Set-Builder Form |
|---|---|
| (i) $\{1, 2, 3, 6\}$ | (c) $\{x : x \text{ is natural number and divisor of 6}\}$ |
| (ii) $\{2, 3\}$ | (a) $\{x : x \text{ is a prime number and a divisor of 6}\}$ |
| (iii) $\{M, A, T, H, E, I, C, S\}$ | (d) $\{x : x \text{ is a letter of the word MATHEMATICS}\}$ |
| (iv) $\{1, 3, 5, 7, 9\}$ | (b) $\{x : x \text{ is an odd natural number less than 10}\}$ |
Final Solution: The correct matches are (i)-(c), (ii)-(a), (iii)-(d), and (iv)-(b).
Solution:
Given the set $A = \{5, 25, 125, 625\}$. We observe that each element is a positive integral power of the base number $5$:
Generalizing this, every element $x$ in the set follows the rule $x = 5^n$, where $n$ is a natural number [Definition of Natural Numbers, $\mathbb{N}$].
The set consists of exactly four elements, corresponding to the values of $n = 1, 2, 3, 4$. Therefore, the variable $n$ must satisfy the following inequality constraint:
$1 \le n \le 4$ where $n \in \mathbb{N}$
The set-builder notation requires defining a variable, assigning the rule for the variable, and stating the domain constraint. We define the set as follows:
$A = \{x : x = 5^n, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}$
Final Solution: \{x : x = 5^n, n \in \mathbb{N}, 1 \le n \le 4\}
Solution:
Given the set $A = \{2, 4, 6, . . .\}$, we analyze the constituent elements:
Let $x$ represent an arbitrary element of the set. Each element $x$ can be expressed in the form $x = 2n$, where $n$ is a natural number. The set of all natural numbers is denoted by $\mathbb{N} = \{1, 2, 3, . . .\}$. Since the sequence $\{2, 4, 6, . . .\}$ is infinite, $n$ must span the entire set of natural numbers $\mathbb{N}$. [By Definition of Natural Numbers and Even Integers]
The set-builder form is structured as $\{x : P(x)\}$, where $P(x)$ is the property that defines the membership of the set.
The property here is that $x$ is an even natural number, or algebraically, $x = 2n$ such that $n \in \mathbb{N}$.
Combining the variable definition and the property, the set-builder form is written as:
$A = \{x : x = 2n, n \in \mathbb{N}\}$
Alternatively, this can also be expressed as:
$A = \{x \in \mathbb{N} : x \text{ is an even number}\}$
Final Solution: \{x : x = 2n, n \in \mathbb{N}\}
Solution:
According to Cantor’s set theory, a set is defined as a well-defined collection of objects [Definition per Set Theory Axioms]. The term "well-defined" implies that for any given object $x$ and a collection $A$, there exists an unambiguous rule to determine whether $x \in A$ or $x \notin A$.
The collection in question is: "The collection of all boys in your class."
To determine if the collection is a set, we test the criterion of objectivity:
| Criterion | Application |
|---|---|
| Objectivity | The roster of a specific class is fixed and verifiable by school administration records. |
| Membership | For any individual person, one can definitively categorize them as either being in the class or not, and as being a boy or not. |
Since the criteria for membership are based on factual attributes (enrollment status and gender identification as recorded in institutional documentation), the collection is not subject to individual opinion, ambiguity, or variable interpretation [Ref: Axiom of Specification].
Because the collection is well-defined and the boundary for membership is clear, fixed, and objective, it satisfies the fundamental criteria of a mathematical set.
Final Solution: The collection of all boys in your class is a set, because it is a well-defined collection of objects where membership is objective and verifiable.
Solution:
The set $B$ is defined by the condition $x \in \mathbb{Z}$ (where $\mathbb{Z}$ is the set of integers) subject to the interval constraints:
$-\frac{1}{2} < x < \frac{9}{2}$
[By the property of real numbers, we evaluate the decimal equivalents to identify the boundaries of the interval].
We must select all integers $x$ such that $-0.5 < x < 4.5$. We test consecutive integers:
| Integer ($x$) | Comparison to Interval | Status |
|---|---|---|
| $-1$ | $-1 < -0.5$ | False |
| $0$ | $-0.5 < 0 < 4.5$ | True |
| $1$ | $-0.5 < 1 < 4.5$ | True |
| $2$ | $-0.5 < 2 < 4.5$ | True |
| $3$ | $-0.5 < 3 < 4.5$ | True |
| $4$ | $-0.5 < 4 < 4.5$ | True |
| $5$ | $5 > 4.5$ | False |
[Per the definition of the set of integers $\mathbb{Z} = \{..., -2, -1, 0, 1, 2, ...\}$, we include only discrete values falling strictly between the calculated bounds].
Collecting the valid integers identified in Step 2, we list the elements within the curly braces according to the roster method [Definition of Roster/Tabular Form of Sets].
Final Solution: $B = \{0, 1, 2, 3, 4\}$
Solution:
The symbol $\in$ denotes that an element belongs to a set (is a member of the set), while the symbol $\notin$ denotes that an element does not belong to a set (is not a member of the set). [Per the definition of Set Membership in Zermelo-Fraenkel Set Theory].
The set $A$ is defined by the roster method as:
$A = \{1, 2, 3, 4, 5, 6\}$
The elements contained within set $A$ are exclusively the integers $\{1, 2, 3, 4, 5, 6\}$.
To determine the correct symbol for $0 \dots A$, we examine if the element $0$ exists within the list of elements defining set $A$:
Since the integer $0$ is not listed among the defined members of set $A$, it fails the condition of membership [By the Principle of Extensionality, a set is determined by its elements].
Final Solution: 0 $\notin$ A
Solution:
The set $C$ is defined by the set-builder notation: $C = \{x : x \in \mathbb{Z}, x^2 \le 4\}$.
To find the elements of $C$, we must solve for $x$ in the inequality $x^2 - 4 \le 0$.
Factoring the difference of squares [By Algebraic Identity $a^2 - b^2 = (a-b)(a+b)$]:
$(x - 2)(x + 2) \le 0$
The roots of the corresponding equation $(x - 2)(x + 2) = 0$ are $x = 2$ and $x = -2$. These values divide the number line into three intervals: $(-\infty, -2]$, $[-2, 2]$, and $[2, \infty)$.
Since $x$ must be an integer, we identify all integers $x$ such that $x^2 \le 4$ within the closed interval $[-2, 2]$.
| $x$ | $x^2$ | $x^2 \le 4$ |
|---|---|---|
| $-2$ | $4$ | True |
| $-1$ | $1$ | True |
| $0$ | $0$ | True |
| $1$ | $1$ | True |
| $2$ | $4$ | True |
Final Solution: $C = \{-2, -1, 0, 1, 2\}$
Solution:
To determine the correct relation between an element $x$ and a set $A$, we apply the fundamental definition of set membership [Per Zermelo-Fraenkel Set Theory]:
The set is defined by the roster method as:
$A = \{1, 2, 3, 4, 5, 6\}$
The elements contained within set $A$ are explicitly listed as the natural numbers from $1$ to $6$. Any integer $n$ such that $1 \le n \le 6$ satisfies the condition $n \in A$.
We evaluate the given value $4$ against the contents of set $A$:
Based on the analysis that the digit $4$ is explicitly present within the set $A$, we use the membership symbol $\in$.
Final Solution: 4 $\in$ A
Solution:
The set $A$ is defined by the roster method as $A = \{1, 2, 3, 4, 5, 6\}$. This implies that the set contains exactly six distinct elements: $1, 2, 3, 4, 5, \text{and } 6$. Any value not present within these curly braces is strictly excluded from the set [Per the definition of set membership].
We test the candidate element $x = 8$ against the set $A$:
The symbol $\in$ denotes "is an element of," implying a truth value of $True$ if the element is contained within the set. The symbol $\notin$ denotes "is not an element of," implying a truth value of $True$ if the element is not found within the set [By Axiom of Extensionality].
Since the number $8$ is not included in the listing of elements for set $A$, the correct relation is expressed as $8 \notin A$.
Final Solution: 8 $\notin$ A
Solution:
In set theory, a set is defined as a well-defined collection of distinct objects. When representing a set in Roster (or Tabular) form, every unique element of the set is listed exactly once, separated by commas, and enclosed within curly brackets $\{ \}$. [Per the principles of Zermelo-Fraenkel set theory regarding the extensionality of sets].
The given condition defines set $D$ as containing the letters of the word "LOYAL". We must perform a lexicographical breakdown of the word to identify individual letters:
A fundamental property of sets is that elements must be distinct. If an element is repeated in the descriptive criteria, it appears only once in the roster representation [By the Axiom of Specification].
| Character | Status |
|---|---|
| L | Included |
| O | Included |
| Y | Included |
| A | Included |
| L | Excluded (Duplicate) |
By collecting the unique identifiers identified in Step 3, we construct the set $D$:
$D = \{L, O, Y, A\}$
Final Solution: $D = \{L, O, Y, A\}$
Solution:
The symbol $\in$ denotes that an object is an element of a specific set (membership), while the symbol $\notin$ denotes that an object is not an element of a specific set (non-membership) [Per Zermelo-Fraenkel set theory definition of the element relation].
The set $A$ is defined by the roster method as:
$A = \{1, 2, 3, 4, 5, 6\}$
This implies the collection of elements contained within $A$ is limited to the integers $1$ through $6$ inclusive.
We examine the given element $x = 2$ against the set $A$.
| Element | Relationship | Set |
|---|---|---|
| $2$ | $\in$ | $\{1, 2, 3, 4, 5, 6\}$ |
Final Solution: 2 $\in$ A