Class 11 Mathematics Chapter 1 Sets

Key Concept: Definition Understanding

c) Empty Set

[Solution Description]

The definition of an empty set is a set that does not contain any element. This matches with option C.

Key Concept: Equal Sets Identification

a) Yes, they are equal

[Solution Description]

A set is considered equal to another if both have exactly the same elements with no regard to order. Here, $A$ contains $\{a, b, c\}$ and $B$ also contains these same elements $\{c, b, a\}$. Since they have identical elements, the sets are equal.

Key Concept: Definition Recall

b) Set of even prime numbers other than 2

[Solution Description] The empty set is defined as a set with no elements. It can be represented by conditions that cannot be satisfied.

Key Concept: Interval Types

c) Half-Open Interval

[Solution Description]

In interval notation, the square bracket $[$ or $]$ indicates that the endpoint is included (closed), while the parenthesis $($or $)$ indicates that the endpoint is not included (open). Therefore, the interval $[a, b)$ includes $a$ but not $b$, making it a half-open interval.

Key Concept: Symbol Usage

a) $\in$

[Solution Description] The symbol $Q^+$ represents the set of positive rational numbers. Since 7 is a positive whole number, it can be expressed as $\frac{7}{1}$, which is a positive rational number. Therefore, 7 belongs to the set $Q^+$.

Key Concept: Real-world Union

c) All employees and managers in the company

[Solution Description] The union $E \cup M$ consists of elements that are either in set $E$ or set $M$ or both. Hence, it represents all individuals who are either employees, managers, or both within the company context.

Key Concept: Complement Identification

a) $\{1, 2, 7, 8, 9\}$

[Solution Description] The complement of a set $A$, denoted as $A’$, includes all elements in the universal set $U$ that are not in $A$. Here, $A = \{3, 4, 5, 6\}$, so $A’$ should include the elements $\{1, 2, 7, 8, 9\}$.

Key Concept: Set Examples

d) $\{x : x \text{ is an integer} \}$

[Solution Description] The set of integers is infinite because it includes all whole numbers both negative and positive, extending indefinitely in both directions.

Key Concept: Element Order

a) {2, 3, 5, 7}

[Solution Description]

The prime numbers less than 10 are 2, 3, 5, and 7. Since the order does not matter in a set, both {2, 3, 5, 7} and {7, 5, 3, 2} are correct representations.

Key Concept: Complex Subset Relations, Infinite Set Analysis

a) $A \subset B \subset C$

[Solution Description]

The set $A$ includes natural numbers whose squares are less than 100: i.e., $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. $B$ includes integers between -10 and 10: i.e., $B = \{-9, -8, …, 8, 9\}$. $C$ contains real numbers such that $x^3 \leq 27$: i.e., $C = (-\infty, 3]$.

Notice that $A \subset B$ since all elements of $A$ are also in $B$, but not vice versa as $B$ contains negative integers. Then, every element of $B$ is in $C$ because for integer $x$ within given range, $x^3 \leq 27$. Hence, $A \subset B \subset C$ holds true.

Key Concept: Finite vs Infinite

b) Infinite

[Solution Description]

The set in question can be described as $\{5, 10, 15, 20, \ldots\}$. Since there is no upper limit to how many multiples of 5 exist beyond 0, this set continues indefinitely without end. Therefore, it is an infinite set.

Key Concept: Basic Interval Length

b) 6

[Solution Description]

To find the length of an interval $[a, b)$, we calculate $b – a$. Here, $a = 2$ and $b = 8$.

So, the length will be:

$8 – 2 = 6$

Thus, the length of the interval $[2, 8)$ is $6$.

Key Concept: Difference of Sets

a) $\{a, b, d\}$

[Solution Description]

To determine the difference $M – N$, we remove elements from $M$ that are also in $N$. So, $M – N = \{a, b, d\}$ as these elements are not present in set $N$.

Key Concept: Real-world Application

c) Students who do not play guitar

[Solution Description]

In this context, $G’$ represents the complement of set $G$. By definition, it includes all elements in the universal set $U$ that are not in set $G$. Therefore, $G’$ represents students who do not play guitar.

Key Concept: Associative Union

a) $\{1, 3, 5, 7, 9\}$

[Solution Description]

To find $(X \cup Y) \cup Z$, we first find $X \cup Y$. The union of sets $X = \{1, 3, 5\}$ and $Y = \{5, 7\}$ gives $X \cup Y = \{1, 3, 5, 7\}$.

Next, we take this result and find the union with set $Z = \{7, 9\}$. Therefore, $(X \cup Y) \cup Z = \{1, 3, 5, 7, 9\}$.

Key Concept: Subset Identification

c) $\{2, 4\}$

[Solution Description] A subset includes elements that are entirely present in another set. Here, $\{2, 4\$ is a subset as both 2 and 4 are elements in $\{1, 2, 3, 4\}$.

Key Concept: Application in Context

c) $\{3\}$

[Solution Description]

We begin by calculating $Y \cap Z$:

$$Y \cap Z = \{1, 2, 3, 4\} \cap \{3, 5, 7\} = \{3\}$$

Next, we calculate $X \cap (Y \cap Z)$:

$X \cap \{3\} = \{3, 6, 9\} \cap \{3\} = \{3\}$

Thus, the answer is $\{3\}$.

Key Concept: Basic Set Comparison

b) No

[Solution Description]

For two sets to be equal, they must contain the exact same elements. Set $E$ contains \{a, b, c\}, while set $F$ contains {c, b, d}. The element ‘a’ in set $E$ is not present in set $F$, and the element ‘d’ in set $F$ is not present in set $E$. Therefore, $E \neq F$.

Key Concept: Subset Identification

b) $\{1, 3\}$

[Solution Description] To determine whether a given set is a subset of another set, check if all elements of the first set are contained within the second set. The set $\{1, 3\}$ contains only elements that are present in $\{1, 2, 3, 4\}$.

Key Concept: Intersection Analysis

b) $\{6\}$

[Solution Description] The intersection of two sets $P$ and $Q$, denoted by $P \cap Q$, consists of all elements common to both sets. First, we identify each set: $P = \{2, 4, 6, 8\}$ and $Q = \{3, 6, 9\}$. The common element between these sets is 6, so $P \cap Q = \{6\}$.

Key Concept: Law of Double Complementation

a) $\{1, 3, 5\}$

[Solution Description] By the law of double complementation, $(C’)’ = C$. Let’s identify $C$:

$C = \{1, 3, 5\}$ since these are the odd numbers in $U$.

Applying the law of double complementation, $(C’)’ = C = \{1, 3, 5\}$.

Key Concept: Complement of a Set

a) $\{a, b, f\}$

[Solution Description]

The complement of a set $D$, denoted by $D’$ or $U – D$, consists of all elements in the universal set $U$ that are not in $D$. Here, $U – D = \{a, b, f\}$ because these are the elements present in $U$ but not in $D$.

Key Concept: Power Set Calculation

c) 16

[Solution Description] The number of elements in a power set is calculated using the formula $2^n$, where $n$ is the number of elements in the original set. Here, set $B$ has 4 elements. Thus, the power set will have $2^4 = 16$ elements.

Key Concept: Set Operations

b) $\{e, f\}$

[Solution Description]

First, find the union $C \cup D$:

– $C \cup D = \{a, b, c, d\}$ (combine all elements from both $C$ and $D$ without repetition).

Now, find the complement of this union with respect to $U$.

– Complement of $C \cup D$: elements in $U$ but not in $C \cup D$ are $\{e, f\}$.

Therefore, $(C \cup D)’ = \{e, f\}$.

Key Concept: Set Operations

a) $\{a, c\}$

[Solution Description] The difference between two sets $A$ and $B$, denoted by $A – B$, is the set of elements that are in $A$ but not in $B$. Therefore, we need to find the elements present in $A$ that are not in $B$. In this case, elements $b$ and $d$ are common in both sets, so they do not appear in $A – B$. Thus, the difference is $\{a, c\}$.

Key Concept: Set Relationships

b) $Z \subset R$

[Solution Description] According to the subset relationship $N \subset Z \subset Q \subset R$, natural numbers ($N$) are a subset of integers ($Z$), which in turn are a subset of rational numbers ($Q$), and all these are subsets of real numbers ($R$). Hence, the correct statement that follows this relationship is option b: $Z \subset R$.

Key Concept: Roster Form

b) $\{2, 4, 6\}$

[Solution Description] To express a set in roster form, list each element of the set within curly braces. The even natural numbers less than or equal to 6 are 2, 4, and 6.

Key Concept: Definition Recognition

b) Infinite

[Solution Description]

According to the definition, an infinite set consists of elements that do not have a specific end point or limit in terms of quantity. A finite set, by contrast, does have a clear boundary or total element count.

Hence, a set with an endlessly expanding list of components is termed “Infinite.”

Correct answer: b.

Key Concept: Difference of Sets

d) $\{m, o\}$

[Solution Description] The difference of sets $P – Q$ consists of elements that belong to $P$ but not to $Q$. Given $P = \{m, n, o, p\}$ and $Q = \{n, p, q\}$, the difference is $P – Q = \{m, o\}$.

Key Concept: Interval Conversion

a) $\{ x : 2 < x \leq 6 \}$ [Solution Description] The interval $(2, 6]$ includes all real numbers $x$ such that $x$ is greater than 2 and less than or equal to 6. This can be represented in set-builder notation as $\{ x : 2 < x \leq 6 \}$.

Key Concept: Interval Comparison

a) $(-3, 4]$ is a subset of $[-3, 5)$

[Solution Description]

The interval $(-3, 4] = \{ x : -3 < x \leq 4 \}$ includes all numbers greater than -3 and up to 4 including 4 itself. The interval $[-3, 5) = \{ x : -3 \leq x < 5 \}$ includes all numbers from -3 and up to but not including 5. Since -3 is included in $[-3, 5)$ but not in $(-3, 4]$, and 4 is included in both intervals, $(-3, 4]$ is a subset of $[-3, 5)$.

Key Concept: Set Relations

c) $T \subset R$

[Solution Description] Natural numbers ($N$) are a subset of rational numbers, not irrational ones, so $N \not\subset T$. However, irrational numbers ($T$) are a subset of real numbers since all irrational numbers are real. Thus, $T \subset R$ is true.

Key Concept: Union with Universal Set

d) $U$

[Solution Description] By the Law of U, the union of the universal set $U$ with any set $A$ is always the universal set $U$. Thus, $U \cup A = U$.

Key Concept: Rational vs Irrational

c) $\sqrt{7}$

[Solution Description] Among the given options, $\sqrt{7}$ cannot be expressed as a fraction of two integers and hence it is an irrational number.

Key Concept: Identify Law

c) Idempotent Law

[Solution Description] The equation $B \cup B = B$ uses the Idempotent Law for union, which states that the union of a set with itself is the set itself.

Key Concept: Commutative Law

c) $A \cap B = B \cap A$

[Solution Description] According to the commutative law of intersection, for any two sets $A$ and $B$, it holds that $A \cap B = B \cap A$. This means the order in which you intersect the sets does not affect the result.

Key Concept: Element Belonging

b) No

[Solution Description] To determine if ‘z’ belongs to set $A$, we check if it is listed within the set’s elements. The set $A$ includes all lowercase letters from ‘a’ to ‘y’, but it does not include ‘z’.

Key Concept: Basic Difference

b) $\{3, 5\}$

[Solution Description]

To find $A – B$, we need to identify elements that are in set $A$ but not in set $B$. The elements in $A$ are $\{3, 4, 5, 6\}$ and those in $B$ are $\{4, 6\}$. Removing the elements common to both sets (i.e., $4$ and $6$), we have:

$$A – B = \{3, 5\}$$

Therefore, the answer is $\{3, 5\}$.

Key Concept: Double Complement

c) $\{1, 2, 6, 7\}$

[Solution Description] By the law of double complementation, $(C’)’ = C$. So to find $C$, we need to determine $(C’)’$ which is $U \setminus C’$.

Calculate $C$:

$$C = U \setminus C’ = \{1, 2, 3, 4, 5, 6, 7\} \setminus \{3, 4, 5\} = \{1, 2, 6, 7\}$$

Thus, the original set $C = \{1, 2, 6, 7\}$.

Key Concept: Definition Recall

a) A = {x : 1 < x < 2, x is a natural number} [Solution Description] An empty set is a set with no elements in it. The set A = {x : 1 < x < 2, x is a natural number} is an empty set because there are no natural numbers between 1 and 2.

Key Concept: Subset Identification

b) $\{1, 2\} \subseteq \{1, 2, 3\}$

[Solution Description]

Let’s consider two sets $A = \{1, 2\}$ and $B = \{1, 2, 3\}$. A set $A$ is a subset of $B$ if every element of $A$ is also an element of $B$. Here, both elements 1 and 2 in set $A$ are present in set $B$. Thus, $A \subset B$.

Key Concept: Interval Identification

c) $[-3, 5)$

[Solution Description]

To identify a half-open interval, we need one endpoint included and one excluded. The interval $[-3, 5)$ includes $-3$ but excludes $5$. This is the representation of a half-open interval starting from $-3$ and extending up to, but not including, $5$.

Key Concept: Set Difference

b) $\{1, 5, 7\}$

[Solution Description] The set difference $X – Y$ includes elements that are in $X$ but not in $Y$. Here, $1$, $5$, and $7$ are in $X$ but not in $Y$, so $X – Y = \{1, 5, 7\}$.

Key Concept: Historical Developments

d) Leopold Kronecker

[Solution Description] Leopold Kronecker was one of the main critics of Georg Cantor’s work on set theory. He did not accept Cantor’s ideas about infinite sets and believed that mathematics should be based only on finite numbers.

Key Concept: Set Equality

c) Yes, because they have the same elements regardless of order.

[Solution Description] Sets are considered equal if they contain exactly the same elements, irrespective of order. Here, $P$ and $Q$ contain the same elements $\{1, 3, 5\}$.

Key Concept: Definition Recall

c) $\phi$

[Solution Description] An empty set is denoted by either the Greek letter $\phi$ or curly brackets $\{\}$.

Key Concept: Venn Diagram Interpretation

a) Elements are only in $G$

[Solution Description] When interpreting a Venn diagram for $G – H$, the shaded area should represent elements present only in $G$ and not in $H$. This means the shaded area contains elements exclusive to set $G$.

Key Concept: Set Operations

b) $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$

[Solution Description] The union of two sets $A \cup B$ consists of all elements in either set $A$ or set $B$. So, we combine all distinct elements from both $A$ and $B$:

$A \cup B = \{2, 4, 6, 8\} \cup \{1, 3, 5, 7, 9\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$

Key Concept: Set Equation Solving

a) $\{1, 3, 4, 5, 6\}$

[Solution Description]

To solve this, apply the commutative law for union, which states that $A \cup B = B \cup A$. Therefore, we can compute $A \cup B$ as follows:

First list all unique elements from both sets $A$ and $B$:

– From $A$: 1, 3, 5

– From $B$: 3, 4, 6

Combine them to get: $\{1, 3, 5, 4, 6\}$. Now remove duplicate elements:

– Resulting set is $\{1, 3, 4, 5, 6\}$.

Therefore, $A \cup B = \{1, 3, 4, 5, 6\}$.

Key Concept: Set-Builder Form

a) $A = \{x : x = 4n, n \in \mathbb{N}, 1 \leq n \leq 5\}$

[Solution Description] The given set is a sequence of numbers with a common difference of 4. We can describe this set using set-builder notation as follows: Let $A = \{x : x = 4n, n \in \mathbb{N}, 1 \leq n \leq 5\}$. This means $x$ is a multiple of 4 where $n$ ranges from 1 to 5.

Key Concept: Property Identification

c) $x$ is an odd number under 10

[Solution Description] The set $\{x : x \text{ is an odd number under 10}\}$ consists of numbers that are both odd and less than 10. Therefore, the common property is that each member of the set is an odd number below 10.

Key Concept: Subset Definition

b) $A \subseteq B$

[Solution Description] Set $A$ is a subset of set $B$ if every element in $A$ is also in $B$. In this case, both elements 1 and 2 from set $A$ are present in set $B$, hence $A \subseteq B$.

Key Concept: Roster Form Recognition

b) $\{1, 3, 5, 7\}$

[Solution Description] The set-builder notation describes all odd integers greater than 0 and less than 8, which are $1, 3, 5, 7$. Hence, the roster form is $\{1, 3, 5, 7\}$.

Key Concept: Basic Definition

a) If every element of A is in B, but A is not equal to B

[Solution Description]

For a set $A$ to be a proper subset of set $B$, every element of $A$ must also be an element of $B$, and $A$ should not be equal to $B$. Hence, if $A \subset B$, then $A$ has fewer elements than $B$.

A proper subset implies that $A = B$ is not possible, hence $a$ is correct.

Key Concept: Set Operations

d) $\phi$

[Solution Description]

The intersection of two sets $A \cap B$ contains elements that are present in both $A$ and $B$. Here, $A = \{1, 3, 5, 7\}$ and $B = \{2, 4, 6, 8\}$. None of the elements in $A$ appear in $B$, so $A \cap B = \phi$ (the empty set).

Key Concept: Interval Subset Analysis, Real World Subset Application

c) $A \cap B = (2, 4)$

[Solution Description]

To analyze if $A$ is a subset or intersects with $B$, we look for overlapping regions. The interval $A = [1, 4)$ includes numbers from $1$ up to but not including $4$. The interval $B = (2, 6]$ includes numbers greater than $2$ up to and including $6$. Taking the intersection $A \cap B$, any number $x$ in this set must satisfy both $1 \leq x < 4$ (from $A$) and $2 < x \leq 6$ (from $B$). Therefore, the intersection $A \cap B = (2, 4)$. Since $(2, 4)$ is neither equal to $A$ nor $B$, $A$ is not a subset of $B$ and vice versa. However, they do intersect.

Key Concept: Definition Recall

b) Two sets are equal if every element of one set is in the other, and vice versa.

[Solution Description] Two sets are considered equal if they contain exactly the same elements, regardless of order or repetition.

Key Concept: Subset Identification

a) Yes

[Solution Description] A set $A$ is a subset of another set $B$ if every element of $A$ is also an element of $B$. Here, both elements 2 and 4 are part of the set $B$, hence $A \subset B$.

Key Concept: Roster Form

b) $\{2, 4, 6, 8\}$

[Solution Description] The set-builder notation describes $x$ as an even number less than 10. Even numbers less than 10 are 2, 4, 6, and 8. Therefore, the roster form of set $D$ is $\{2, 4, 6, 8\}$.

Key Concept: Operations on Sets

c) $\{1, 2, 3, 4, 5\}$

[Solution Description] The union of two sets $A$ and $B$, denoted $A \cup B$, contains all elements that are in $A$, or in $B$, or in both. For $A = \{1, 2, 3\}$ and $B = \{3, 4, 5\}$, the union would include each distinct element from both sets: $\{1, 2, 3, 4, 5\}.$

Key Concept: Element Count

c) 3

[Solution Description] To find the number of elements in the set $C$, we simply count each unique item within the set. Here, the set $C$ contains three days: Monday, Tuesday, and Wednesday. Therefore, $n(C) = 3$.

Key Concept: Set Equality Verification

b) $C = D$

[Solution Description]

Two sets are considered equal if they contain exactly the same elements. Here, both sets $C$ and $D$ have the same elements $\{a, b, c, d\}$ even though their orders are different. Hence, $C = D$.