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CBSE - Class 11 Mathematics Sets Worksheet
EXERCISE 1.4
Worksheet Answers
Solution:
The difference of two sets $Y$ and $X$, denoted by $Y - X$, is defined as the set of all elements that belong to $Y$ but do not belong to $X$. Mathematically, this is expressed as:
$Y - X = \{ y \mid y \in Y \text{ and } y \notin X \}$
[Per the Set-Theoretic Definition of Relative Complement]
We evaluate each element $y \in Y$ to determine if it is present in $X$:
| Element of $Y$ | Presence in $X$ | Include in $Y - X$? |
|---|---|---|
| $f$ | Not in $X$ | Yes |
| $b$ | In $X$ | No |
| $d$ | In $X$ | No |
| $g$ | Not in $X$ | Yes |
By collecting the elements identified in Step 3 that satisfy the condition ($y \in Y$ and $y \notin X$):
$Y - X = \{ f, g \}$
Final Solution: $Y - X = \{ f, g \}$
Solution:
The problem provides four distinct sets defined by the roster method:
The intersection of two sets $A$ and $D$, denoted by $A \cap D$, is defined as the set containing all elements $x$ such that $x \in A$ AND $x \in D$. [By the Set-Theoretic Definition of Intersection].
We perform a comparative analysis to identify common elements between the two sets:
| Set A Elements | Set D Elements | Commonality Check |
|---|---|---|
| $3$ | $15, 17$ | No |
| $5$ | $15, 17$ | No |
| $7$ | $15, 17$ | No |
| $9$ | $15, 17$ | No |
| $11$ | $15, 17$ | No |
Since there are no elements that satisfy the condition $x \in A \cap D$, the resulting set is the empty set (or null set). [By the definition of Disjoint Sets].
Final Solution: $A \cap D = \emptyset$ or $\phi$
Solution:
The set difference of two sets $B$ and $A$, denoted as $B - A$, is defined as the set of all elements that belong to set $B$ but do not belong to set $A$. Mathematically, this is expressed as:
$B - A = \{x : x \in B \text{ and } x \notin A\}$
[Per the Definition of Relative Complement in Set Theory]
We are provided with the following sets:
To determine $B - A$, we evaluate each element of $B$ to check for its presence in $A$:
| Element ($x \in B$) | Is $x \in A$? | Retain in $B - A$? |
|---|---|---|
| 4 | No | Yes |
| 8 | No | Yes |
| 12 | Yes | No |
| 16 | No | Yes |
| 20 | No | Yes |
By removing the intersection $\{12\}$ from set $B$, we obtain the set containing the remaining elements.
$B - A = \{4, 8, 16, 20\}$
Final Solution: $B - A = \{4, 8, 16, 20\}$
Solution:
The problem provides four distinct sets containing discrete elements as follows:
The union of two sets, denoted as $A \cup C$, is defined as the set of all elements that are members of set $A$, set $C$, or both. Mathematically, this is expressed as:
$A \cup C = \{x : x \in A \text{ or } x \in C\}$ [By the definition of Set Union]
To find the union, we combine the elements of set $A$ and set $C$, ensuring that duplicate elements are listed only once [Per the Property of Idempotency and Set Representation Standards].
List elements of $A$: $1, 2, 3, 4$
List elements of $C$: $5, 6, 7, 8$
Combining these sets: $\{1, 2, 3, 4, 5, 6, 7, 8\}$
| Source Set | Elements Contained |
|---|---|
| Set A | 1, 2, 3, 4 |
| Set C | 5, 6, 7, 8 |
| Union Result | 1, 2, 3, 4, 5, 6, 7, 8 |
Final Solution: $A \cup C = \{1, 2, 3, 4, 5, 6, 7, 8\}$
Solution:
The universal sets provided for the calculation are defined as follows:
The union of two sets, $B$ and $C$, is defined as the set of all elements that are members of $B$, or $C$, or both. [Per the definition of Set Union: $B \cup C = \{x : x \in B \text{ or } x \in C\}$].
$B \cup C = \{7, 9, 11, 13\} \cup \{11, 13, 15\}$
By listing all unique elements from both sets and removing duplicates:
$B \cup C = \{7, 9, 11, 13, 15\}$
The intersection of set $A$ and the resulting set $(B \cup C)$ is defined as the set containing elements common to both. [Per the definition of Set Intersection: $A \cap (B \cup C) = \{x : x \in A \text{ and } x \in (B \cup C)\}$].
$A \cap (B \cup C) = \{3, 5, 7, 9, 11\} \cap \{7, 9, 11, 13, 15\}$
Analyzing the common elements:
Final Solution: $A \cap (B \cup C) = \{7, 9, 11\}$
Solution:
The problem requires the determination of the set difference, denoted as $A - D$. By definition, the difference of two sets $A$ and $D$ is the set of elements which belong to $A$ but do not belong to $D$.
Symbolically, $A - D = \{x : x \in A \text{ and } x \notin D\}$. [Per the Set Theoretic Definition of Relative Complement].
We evaluate each element of set $A$ to determine if it is present in set $D$:
| Element in $A$ | Presence in $D$ | Status |
|---|---|---|
| $3$ | No | Retain |
| $6$ | No | Retain |
| $9$ | No | Retain |
| $12$ | No | Retain |
| $15$ | Yes | Exclude |
| $18$ | No | Retain |
| $21$ | No | Retain |
By collecting all elements of $A$ that are not in $D$, we obtain:
$A - D = \{3, 6, 9, 12, 18, 21\}$
Final Solution: $A - D = \{3, 6, 9, 12, 18, 21\}$
Solution:
The intersection of two sets $X$ and $Y$, denoted by $X \cap Y$, is defined as the set containing all elements that are members of both set $X$ AND set $Y$ simultaneously [Per the Definition of Intersection in Set Theory]. Formally:
$X \cap Y = \{ x : x \in X \text{ and } x \in Y \}$
| Element | Exists in $X$? | Exists in $Y$? | Included in $X \cap Y$? |
|---|---|---|---|
| $a$ | Yes | No | No |
| $b$ | Yes | Yes | Yes |
| $c$ | Yes | No | No |
| $d$ | Yes | Yes | Yes |
| $f$ | No | Yes | No |
| $g$ | No | Yes | No |
Extracting the common elements identified in Step 3, we observe that elements $b$ and $d$ satisfy the condition of belonging to both sets $X$ and $Y$.
$X \cap Y = \{ b, d \}$
Final Solution: $X \cap Y = \{ b, d \}$
Solution:
To analyze the intersection, we must first express the given sets in their explicit form based on the properties provided:
The intersection of two sets $B$ and $C$, denoted by $B \cap C$, is defined as the set containing all elements that are common to both set $B$ and set $C$.
Mathematically: $B \cap C = \{x : x \in B \text{ and } x \in C\}$
We observe the nature of elements in sets $B$ and $C$:
| Set | Property | Parity |
|---|---|---|
| $B$ | Even Natural Numbers | Divisible by 2 |
| $C$ | Odd Natural Numbers | Not divisible by 2 |
By the Fundamental Theorem of Arithmetic and the definition of parity, an integer cannot be both even and odd simultaneously. Since $B$ contains only even numbers and $C$ contains only odd numbers, there exists no natural number $x$ such that $x \in B$ and $x \in C$ [Law of Contradiction].
Since the sets $B$ and $C$ share no common elements, their intersection is the empty set (or null set), denoted by the symbol $\phi$ or $\{\}$.
Final Solution: $B \cap C = \emptyset$
Solution:
Two sets $A$ and $B$ are defined as disjoint if and only if their intersection is an empty set (or null set). Mathematically, this is expressed as:
$A \cap B = \phi$
[Definition: A set $A$ and a set $B$ are disjoint if there exists no element $x$ such that $x \in A$ and $x \in B$.]
Let the given sets be defined as follows:
To determine if the sets are disjoint, we evaluate the intersection of set $A$ and set $B$. The intersection consists of all elements that are common to both sets.
$A \cap B = \{x \mid x \in A \text{ and } x \in B\}$
Comparing the elements:
| Set A Elements | Set B Elements | Common Elements |
|---|---|---|
| 2, 6, 10 | 3, 7, 11 | None |
$A \cap B = \phi$
Since there are no elements shared between Set $A$ and Set $B$, the intersection is indeed the empty set, $\phi$. According to the definition established in Step 1, this confirms that the sets satisfy the condition for being disjoint.
Final Solution: The statement is True. Since the intersection of the two sets $\{2, 6, 10\}$ and $\{3, 7, 11\}$ is the empty set ($\phi$), they are disjoint sets.
Solution:
The set difference of two sets, denoted as $A - B$ or $A \setminus B$, is defined as the set containing elements that are members of set $A$ but are not members of set $B$. Mathematically, this is expressed as:
$A - B = \{x : x \in A \text{ and } x \notin B\}$
[By Definition of Set Difference]
To find $A - B$, we iterate through every element of set $A$ and verify its existence in set $B$. If an element exists in both sets, it is excluded from the resulting set.
| Element of A ($x$) | Presence in B ($x \in B$) | Action |
|---|---|---|
| 3 | No | Retain |
| 6 | No | Retain |
| 9 | No | Retain |
| 12 | Yes | Exclude |
| 15 | No | Retain |
| 18 | No | Retain |
| 21 | No | Retain |
Gathering the retained elements from Step 3:
$A - B = \{3, 6, 9, 15, 18, 21\}$
Final Solution: $A - B = \{3, 6, 9, 15, 18, 21\}$
Solution:
To evaluate the intersection $B \cap D$, we must first express the sets $B$ and $D$ in their Roster (tabular) form based on the provided definitions:
The intersection of two sets $B$ and $D$, denoted $B \cap D$, is the set containing all elements $x$ such that $x \in B$ AND $x \in D$ [Per the definition of Set Intersection].
We examine the elements of set $B$ and determine which are present in set $D$:
| Element Type | Analysis |
|---|---|
| $2$ | $2$ is an even natural number ($2 \in B$) and $2$ is the only even prime number ($2 \in D$). Thus, $2 \in B \cap D$. |
| Other even numbers ($4, 6, 8, \dots$) | Any even number $x > 2$ has at least three divisors: 1, 2, and $x$. Therefore, $x$ is composite, not prime. Thus, $x \notin D$. |
Since the only element common to both sets is the number 2, we define the resultant set as:
$B \cap D = \{2\}$
Final Solution: $B \cap D = \{2\}$
Solution:
As per the definition of sets in set theory [Ref: Zermelo-Fraenkel Set Theory], the elements are uniquely defined as follows:
The intersection of three sets $A, C,$ and $D$, denoted as $A \cap C \cap D$, is defined as the set of elements that are common to all three sets. Mathematically, $x \in (A \cap C \cap D) \iff x \in A \land x \in C \land x \in D$ [Per the Associative Law of Intersection: $(A \cap C) \cap D = A \cap (C \cap D)$].
First, we isolate the common elements between set $A$ and set $C$:
$A = \{3, 5, 7, 9, \mathbf{11}\}$
$C = \{\mathbf{11}, 13, 15\}$
Comparing the two, the only element present in both sets is $11$.
$A \cap C = \{11\}$
Next, we find the intersection of the resulting set from Step 3 with set $D$:
$(A \cap C) = \{11\}$
$D = \{15, 17\}$
We observe the set of elements common to $\{11\}$ and $\{15, 17\}$. Since there are no elements shared between these two sets, the intersection is the empty set (or null set) [Per the definition of Disjoint Sets].
Final Solution: $A \cap C \cap D = \phi$ or $\emptyset$ (The Empty Set)
Solution:
Two sets $A$ and $B$ are said to be disjoint if and only if they have no elements in common. Mathematically, this is expressed as:
$A \cap B = \phi$
[Definition: Two sets $A$ and $B$ are disjoint if their intersection is the empty set $\phi$, implying there exists no element $x$ such that $x \in A$ and $x \in B$.]
Let the given sets be represented as follows:
To determine if the sets are disjoint, we must calculate the intersection $A \cap B$, which contains elements present in both set $A$ and set $B$.
$A \cap B = \{ x : x \in A \text{ and } x \in B \}$
Comparing the elements:
| Set $A$ Elements | Set $B$ Elements | Commonality |
|---|---|---|
| $2$ | $3$ | None |
| $6$ | $7$ | None |
| $10$ | $11$ | None |
| $14$ | $15$ | None |
Upon evaluation, it is evident that no element of $A$ is present in $B$, and no element of $B$ is present in $A$.
Therefore, $A \cap B = \phi$.
Since the intersection of the two sets is empty, the sets possess no common elements.
Final Solution: The statement is True. Because $\{ 2, 6, 10, 14 \} \cap \{ 3, 7, 11, 15 \} = \phi$, the sets are disjoint.
Solution:
The problem provides two sets, $X$ and $Y$, defined by their elements:
The union of two sets $A$ and $B$, denoted by $A \cup B$, is defined as the set of all elements which are members of $A$, or members of $B$, or members of both $A$ and $B$. [Per Set Theory Definition: $A \cup B = \{x : x \in A \text{ or } x \in B\}$].
To compute $X \cup Y$, we combine all unique elements present in either set while adhering to the principle that set notation does not recognize redundant elements. [By the Principle of Idempotency/Set Definition].
| Source | Elements |
|---|---|
| Set $X$ | 1, 3, 5 |
| Set $Y$ | 1, 2, 3 |
We list all elements present in at least one of the sets:
$X \cup Y = \{1, 3, 5\} \cup \{1, 2, 3\}$
$X \cup Y = \{1, 1, 2, 3, 3, 5\}$
Removing duplicate entries to satisfy the definition of a set: [By the Property of Distinctness in Set Construction].
$X \cup Y = \{1, 2, 3, 5\}$
Final Solution: $X \cup Y = \{1, 2, 3, 5\}$
Solution:
The problem provides four distinct sets as follows:
The difference of two sets $A$ and $C$, denoted as $A - C$, is defined as the set of all elements that belong to set $A$ but do not belong to set $C$ [Per the definition of Set Difference: $A - C = \{x : x \in A \text{ and } x \notin C\}$].
We evaluate each element of set $A$ to determine its presence in set $C$:
| Element ($x \in A$) | Is $x \in C$? | Include in $A - C$? |
|---|---|---|
| 3 | No | Yes |
| 6 | Yes | No |
| 9 | No | Yes |
| 12 | Yes | No |
| 15 | No | Yes |
| 18 | No | Yes |
| 21 | No | Yes |
Collecting the elements from $A$ that are not members of $C$, we obtain the resultant set:
$A - C = \{3, 9, 15, 18, 21\}$
Final Solution: $A - C = \{3, 9, 15, 18, 21\}$
Solution:
The union of two sets $A$ and $B$, denoted by $A \cup B$, is defined as the set of all elements which are members of set $A$, or members of set $B$, or members of both. Symbolically, this is expressed as:
$A \cup B = \{x : x \in A \text{ or } x \in B\}$
[By the Axiom of Specification in Zermelo-Fraenkel Set Theory]
We are given the following sets:
The cardinality of set $A$ is $n(A) = 5$, and the cardinality of set $B$ is $n(B) = 3$.
To find $A \cup B$, we list all unique elements found in either $A$ or $B$, ensuring that duplicate elements are represented only once in accordance with the property of sets [Per the Principle of Extensionality].
Combining the identified unique elements into a single set:
$A \cup B = \{a, e, i, o, u, b, c\}$
Standardizing the notation for clarity (alphabetical/set order):
$A \cup B = \{a, b, c, e, i, o, u\}$
Final Solution: $A \cup B = \{a, b, c, e, i, o, u\}$
Solution:
The problem provides four distinct sets as follows:
The intersection of two sets $B$ and $C$, denoted by $B \cap C$, is defined as the set containing all elements $x$ such that $x \in B$ AND $x \in C$. [Per the definition of Set Intersection in Zermelo-Fraenkel Set Theory].
To find $B \cap C$, we compare the elements of set $B$ and set $C$ to extract commonalities:
| Set $B$ | Set $C$ | Common Elements |
|---|---|---|
| $\{7, 9, 11, 13\}$ | $\{11, 13, 15\}$ | $\{11, 13\}$ |
Mathematically, the evaluation is as follows:
$B \cap C = \{ x : x \in \{7, 9, 11, 13\} \text{ and } x \in \{11, 13, 15\} \}$The elements $11$ and $13$ are the only members present in both input sets.
Final Solution: $B \cap C = \{11, 13\}$
Solution:
The problem requires the operation of set difference $B - D$. The given sets are:
The difference of two sets $B$ and $D$, denoted as $B - D$, is defined as the set of all elements that are in $B$ but are not in $D$. Formally, this is expressed as:
$B - D = \{x : x \in B \text{ and } x \notin D\}$
[Per the definition of set difference in set theory]
To find the elements of $B - D$, we systematically compare each element of $B$ against the elements of $D$:
By collecting the elements that satisfy the condition $\{x : x \in B, x \notin D\}$, we obtain:
$B - D = \{4, 8, 12, 16\}$
Final Solution: $B - D = \{4, 8, 12, 16\}$
Solution:
To determine the intersection $A \cap C$, we must first express the sets in roster form or clear descriptive notation based on the provided definitions:
The intersection of two sets $A$ and $C$, denoted as $A \cap C$, is defined as the set of all elements that are common to both $A$ and $C$ [By the Definition of Set Intersection]:
$A \cap C = \{x : x \in A \text{ and } x \in C\}$
We analyze the membership criteria:
Since every element $x$ in set $C$ belongs to the set of natural numbers ($A$), the common elements between $A$ and $C$ are simply the elements of $C$ itself.
$A \cap C = \{x : x \text{ is an odd natural number}\}$
Final Solution: A \cap C = \{x : x \text{ is an odd natural number}\}
Solution:
The set difference of two sets $X$ and $Y$, denoted as $X - Y$ (or $X \setminus Y$), is defined as the set of all elements that are members of $X$ but are not members of $Y$. Mathematically, this is expressed as:
$X - Y = \{ x : x \in X \text{ and } x \notin Y \}$
[Per the Axiom of Specification in Zermelo-Fraenkel set theory]
The sets provided are:
We evaluate each element $x \in X$ to determine if it exists in $Y$:
| Element ($x \in X$) | Status in $Y$ | Action |
|---|---|---|
| $a$ | $a \notin Y$ | Keep in result |
| $b$ | $b \in Y$ | Discard |
| $c$ | $c \notin Y$ | Keep in result |
| $d$ | $d \in Y$ | Discard |
By retaining only those elements found in $X$ that are absent in $Y$, we collect the remaining elements $\{a, c\}$.
Final Solution: $X - Y = \{ a, c \}$
