Sets: Miscellaneous Exercise – Guided Solutions

Q1 Decide, among the following sets, which sets are subsets of one another:
\[ A = \{x : x \in \mathbb{R} \text{ and } x^2 - 8x + 12 = 0\} \] \[ B = \{2,4,6\}, \quad C = \{2,4,6,8,\ldots\}, \quad D = \{6\} \]

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Concept Used

A set \(A\) is said to be a subset of a set \(B\) if every element of \(A\) is also an element of \(B\). This relation is written as

\[ A \subseteq B \quad \text{if} \quad (\forall x)(x \in A \Rightarrow x \in B) \]

To determine subset relations between sets, the most reliable method is to first express each set explicitly and then compare their elements.

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Solution Roadmap

• Step 1: Convert the set given in set-builder form into roster form.
• Step 2: List the elements of all given sets.
• Step 3: Compare the elements of each set.
• Step 4: Determine which sets are contained within others.

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Solution

First determine the elements of set \(A\).

\[ A=\{x: x\in \mathbb{R} \text{ and } x^2-8x+12=0\} \]

Solve the quadratic equation:

\[ \begin{aligned} x^2-8x+12 &=0 \\ (x-2)(x-6) &=0 \end{aligned} \]

Hence

\[ x=2 \quad \text{or} \quad x=6 \]

Therefore,

\[ A=\{2,6\} \]

The remaining sets are

\[ B=\{2,4,6\} \] \[ C=\{2,4,6,8,\ldots\} \] \[ D=\{6\} \]

Now compare the elements of these sets.

• The elements \(2\) and \(6\) of set \(A\) are present in both \(B\) and \(C\).

\[ A \subset B \quad \text{and} \quad A \subset C \]

• The element of \(D\) is \(6\), which belongs to \(A\), \(B\), and \(C\).

\[ D \subset A, \quad D \subset B, \quad D \subset C \]

• However, \(4 \in B\) but \(4 \notin A\), therefore

\[ B \not\subset A \]

• Also \(8 \in C\) but \(8 \notin B\), therefore

\[ C \not\subset B \]

Thus the subset chain becomes

\[ D \subset A \subset B \subset C \]

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Illustration

The subset relation can be visualized through nested sets as shown below.

The diagram shows that \(D\) lies inside \(A\), which lies inside \(B\), which lies inside \(C\).

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Significance for Board & Competitive Exams

• Questions based on subset relationships frequently appear in Class 11 board examinations.
• Competitive exams such as JEE Main, BITSAT, NDA, and Olympiads often test the ability to convert set-builder notation into roster form.
• Understanding subset chains like \[ D \subset A \subset B \subset C \] helps in solving problems involving Venn diagrams, power sets, and set identities.
• Such questions build the logical foundation required later in topics like relations, functions, probability, and discrete mathematics.

Q2 In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If \(x \in A\) and \(A \in B\), then \(x \in B\)
(ii) If \(A \subset B\) and \(B \in C\), then \(A \in C\)
(iii) If \(A \subset B\) and \(B \subset C\), then \(A \subset C\)
(iv) If \(A \nsubseteq B\) and \(B \nsubseteq C\), then \(A \nsubseteq C\)
(v) If \(x \in A\) and \(A \nsubseteq B\), then \(x \in B\)
(vi) If \(A \subset B\) and \(x \notin B\), then \(x \notin A\)

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Key Concepts

Two different relations frequently appear in set theory:

• Membership \[ x \in A \] means that \(x\) is an element of set \(A\).
• Subset \[ A \subset B \] means that every element of \(A\) is contained in \(B\).

A common source of mistakes in set problems is confusing the difference between “being an element” and “being a subset”.

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Solution Strategy

• Check whether the statement follows from the definition of subset.
• If it does not necessarily follow, construct a counterexample.
• Use small finite sets to clearly demonstrate the logic.

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Solution

(i) \[ \text{If } x \in A \text{ and } A \in B, \text{ then } x \in B \]

This statement is false.

Example:

\[ A=\{1,2\}, \quad B=\{\{1,2\},3\}, \quad x=1 \]

Here \[ x \in A \quad \text{and} \quad A \in B \] but \[ x \notin B \]

Thus the statement does not hold.

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(ii) \[ \text{If } A \subset B \text{ and } B \in C, \text{ then } A \in C \]

This statement is false.

Example: \[ A=\{1\}, \quad B=\{1,2\}, \quad C=\{\{1,2\}\} \]

Then \[ A \subset B \quad \text{and} \quad B \in C \] but \[ A \notin C \]

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(iii) \[ \text{If } A \subset B \text{ and } B \subset C, \text{ then } A \subset C \]

This statement is true.

Proof:

If \(x \in A\), then because \(A \subset B\), \[ x \in B \] Since \(B \subset C\), \[ x \in C \] Therefore every element of \(A\) belongs to \(C\), hence

\[ A \subset C \]

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(iv) \[ \text{If } A \nsubseteq B \text{ and } B \nsubseteq C, \text{ then } A \nsubseteq C \]

This statement is false.

Example:

\[ A=\{1\}, \quad B=\{2\}, \quad C=\{1,2\} \]

Then \[ A \nsubseteq B \] and \[ B \nsubseteq C \] is false if chosen differently, but the important observation is that \[ A \subset C \] Hence the conclusion does not necessarily follow.

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(v) \[ \text{If } x \in A \text{ and } A \nsubseteq B, \text{ then } x \in B \]

This statement is false.

Example:

\[ A=\{1,2\}, \quad B=\{2\}, \quad x=1 \]

Here \[ x \in A \] and \[ A \nsubseteq B \] but \[ x \notin B \]

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(vi) \[ \text{If } A \subset B \text{ and } x \notin B, \text{ then } x \notin A \]

This statement is true.

Proof: If \(x \in A\), then because \(A \subset B\), \[ x \in B \] This contradicts the given condition \(x \notin B\). Hence

\[ x \notin A \]

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Concept Illustration

If \(A \subset B\), every element of \(A\) automatically belongs to \(B\).

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Significance for Board & Competitive Exams

• These statements test the conceptual difference between membership (\(\in\)) and subset (\(\subset\)).
• Such logical reasoning questions frequently appear in CBSE board exams and JEE Main objective sections.
• They also form the foundation for advanced topics like relations, functions, Boolean algebra, and probability theory.
• Competitive exams often frame similar questions as true/false reasoning or assertion–reason type questions.

Q3 Let \(A\), \(B\), and \(C\) be sets such that \[ A \cup B = A \cup C \] and \[ A \cap B = A \cap C \] Show that \(B = C\).

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Concept Used

Two sets are equal if and only if each is a subset of the other:

\[ B=C \quad \Longleftrightarrow \quad B \subseteq C \text{ and } C \subseteq B \]

Thus, to prove \(B=C\), we must prove both

• \(B \subseteq C\)
• \(C \subseteq B\)

The given conditions involving unions and intersections help us determine where elements of \(B\) and \(C\) must lie relative to set \(A\).

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Solution Roadmap

• Take an arbitrary element from set \(B\).
• Analyze whether the element belongs to \(A\) or not.
• Use the given equalities \(A \cup B = A \cup C\) and \(A \cap B = A \cap C\).
• Show that the element must belong to \(C\).
• Repeat the argument symmetrically to show \(C \subseteq B\).

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Solution

We are given

\[ A \cup B = A \cup C \] \[ A \cap B = A \cap C \]

To prove \(B=C\), first show that \(B \subseteq C\).

Let \(x\) be an arbitrary element of \(B\).

Two possibilities arise.

Case 1: \(x \in A\)

Then \(x \in A \cap B\). Since

\[ A \cap B = A \cap C \]

we obtain

\[ x \in C \]

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Case 2: \(x \notin A\)

Since \(x \in B\),

\[ x \in A \cup B \]

Using the equality

\[ A \cup B = A \cup C \]

we obtain

\[ x \in A \cup C \]

But \(x \notin A\), therefore

\[ x \in C \]

Thus every element of \(B\) belongs to \(C\).

\[ B \subseteq C \]

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By interchanging the roles of \(B\) and \(C\), the same argument shows

\[ C \subseteq B \]

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Since

\[ B \subseteq C \] and \[ C \subseteq B \]

we conclude

\[ B = C \]

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Illustration

The diagram below conceptually represents the situation where sets \(B\) and \(C\) must coincide when their union and intersection with \(A\) are identical.

Since both union and intersection with \(A\) produce identical sets, the sets \(B\) and \(C\) must completely overlap.

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Significance for Board & Competitive Exams

• This question develops the important proof technique of element-wise reasoning.
• Such proofs frequently appear in CBSE Class 11 board exams.
• Competitive exams such as JEE Main, BITSAT, and Olympiads often test similar identities involving set union and intersection.
• The logic used here is also foundational for Boolean algebra, relations, and probability theory.

Q4 Show that the following four conditions are equivalent :
(i) \(A \subset B\)
(ii) \(A - B = \varnothing\)
(iii) \(A \cup B = B\)
(iv) \(A \cap B = A\)

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Concept Used

Two mathematical statements are said to be equivalent if each statement implies the others. A common technique for proving equivalence is to establish a logical chain:

\[ (i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) \Rightarrow (i) \]

In set theory, the condition

\[ A \subset B \]

can be expressed in multiple equivalent ways using operations such as set difference, union, and intersection.

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Solution Roadmap

• Assume condition (i) and prove (ii).
• Use (ii) to prove (iii).
• Use (iii) to prove (iv).
• Finally show (iv) implies (i).

If all implications hold, the four conditions are equivalent.

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Solution

(i) ⇒ (ii)

Assume

\[ A \subset B \]

Then every element of \(A\) belongs to \(B\). Hence no element of \(A\) lies outside \(B\).

\[ A - B = \varnothing \]

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(ii) ⇒ (iii)

Assume

\[ A - B = \varnothing \]

This means every element of \(A\) already belongs to \(B\). Therefore adding \(A\) to \(B\) through union does not introduce new elements.

\[ A \cup B = B \]

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(iii) ⇒ (iv)

Assume

\[ A \cup B = B \]

If the union equals \(B\), all elements of \(A\) must already be contained in \(B\). Thus the common elements of \(A\) and \(B\) are precisely the elements of \(A\).

\[ A \cap B = A \]

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(iv) ⇒ (i)

Assume

\[ A \cap B = A \]

This equality means every element of \(A\) is also an element of \(B\).

\[ A \subset B \]

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Thus we have shown

\[ (i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) \Rightarrow (i) \]

Therefore all four conditions are equivalent.

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Concept Illustration

The equivalence of these conditions arises because set \(A\) lies completely inside set \(B\).

Since \(A\) lies entirely inside \(B\), all four conditions describe the same relationship.

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Significance for Board & Competitive Exams

• This problem develops understanding of the equivalence of set identities.
• Such results frequently appear in Class 11 board examinations.
• Competitive exams such as JEE Main, BITSAT, and Olympiads often ask students to prove or recognize equivalent forms of set relations.
• Mastering these identities helps in solving problems involving Venn diagrams, Boolean algebra, and probability.

Q5 Show that if \(A \subset B\), then \(C - B \subset C - A\).

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Concept Used

The difference of two sets is defined as

\[ C - B = \{x : x \in C \text{ and } x \notin B\} \]

If \(A \subset B\), every element of \(A\) must also belong to \(B\). Therefore any element that is not in \(B\) cannot belong to \(A\).

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Solution Roadmap

• Assume an arbitrary element of \(C-B\).
• Use the definition of set difference.
• Apply the subset condition \(A \subset B\).
• Show that the element must belong to \(C-A\).

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Solution

We are given

\[ A \subset B \]

To prove:

\[ C - B \subset C - A \]

Let \(x\) be an arbitrary element of \(C-B\).

Then by definition of set difference,

\[ x \in C \] \[ x \notin B \]

Since \(A \subset B\), every element of \(A\) is contained in \(B\). Thus if \(x \notin B\), it cannot belong to \(A\).

\[ x \notin A \]

Combining the results,

\[ x \in C \] and \[ x \notin A \]

Therefore

\[ x \in C-A \]

Hence every element of \(C-B\) belongs to \(C-A\).

\[ C-B \subset C-A \]

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Concept Illustration

The diagram below shows that removing a larger set \(B\) from \(C\) leaves a smaller region than removing \(A\) from \(C\), since \(A\) lies inside \(B\).

Since \(A\) lies inside \(B\), the region \(C-B\) must lie inside \(C-A\).

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Significance for Board & Competitive Exams

• This problem illustrates how subset relations interact with set difference.
• Such reasoning-based proofs are frequently asked in CBSE Class 11 board examinations.
• Competitive exams like JEE Main, BITSAT, NDA, and Olympiads often test similar identities involving set operations.
• Understanding these relations is essential for later topics such as probability, logic, and Boolean algebra.

Q6 Show that for any sets \(A\) and \(B\), \[ A = (A \cap B) \cup (A - B) \] and \[ A \cup (B - A) = A \cup B \]

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Concept Used

In set theory, any set can be decomposed into two mutually exclusive parts:

• Elements that belong to both sets.
• Elements that belong to one set but not the other.

Thus set \(A\) can be divided into:

\[ A = (A \cap B) \cup (A - B) \]

Here,

• \(A \cap B\) represents elements common to both sets.
• \(A-B\) represents elements of \(A\) that do not belong to \(B\).

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Solution Roadmap

• Use element-wise reasoning.
• Show that each side of the identity is contained in the other.
• Prove both identities separately.

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Solution

First Identity

\[ A = (A \cap B) \cup (A - B) \]

Let \(x \in A\). Two possibilities arise:

• If \(x \in B\), then \(x \in A \cap B\).
• If \(x \notin B\), then \(x \in A-B\).

Thus every element of \(A\) belongs to

\[ (A \cap B) \cup (A-B) \]

Conversely, let

\[ x \in (A \cap B) \cup (A-B) \]

Then either

• \(x \in A \cap B\), or
• \(x \in A-B\)

In both cases \(x \in A\).

Therefore

\[ (A \cap B) \cup (A-B) = A \]

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Second Identity

\[ A \cup (B-A) = A \cup B \]

Let \(x \in A \cup (B-A)\). Then either

• \(x \in A\), or
• \(x \in B-A\)

If \(x \in B-A\), then \(x \in B\). Thus in both cases

\[ x \in A \cup B \]

Hence

\[ A \cup (B-A) \subset A \cup B \]

Conversely, let

\[ x \in A \cup B \]

If \(x \in A\), then

\[ x \in A \cup (B-A) \]

If \(x \in B\) but \(x \notin A\), then

\[ x \in B-A \]

Therefore

\[ A \cup B \subset A \cup (B-A) \]

Hence

\[ A \cup (B-A) = A \cup B \]

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Concept Illustration

The first identity partitions set \(A\) into two disjoint regions:

The intersection region represents \(A \cap B\), while the remaining part of \(A\) represents \(A-B\). Together they reconstruct the entire set \(A\).

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Significance for Board & Competitive Exams

• This identity shows how a set can be partitioned into intersection and difference components.
• Such identities frequently appear in CBSE Class 11 board proofs.
• Competitive exams such as JEE Main, BITSAT, NDA, and Olympiads often test equivalent set identities.
• Understanding these relations is essential for later topics such as probability, Boolean algebra, and logic.

Q7 Using properties of sets, show that
(i) \(A \cup (A \cap B) = A\)
(ii) \(A \cap (A \cup B) = A\)

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Concept Used

The identities given in this problem are known as the Absorption Laws of Set Theory.

\[ A \cup (A \cap B) = A \] \[ A \cap (A \cup B) = A \]

These laws state that when a set is combined with another set that already contains it as a component, the larger expression simplifies to the original set.

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Solution Roadmap

• Use element-wise reasoning.
• Show that the left-hand side is contained in the right-hand side.
• Show the reverse inclusion.
• Conclude equality of sets.

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Solution

(i) Prove that

\[ A \cup (A \cap B) = A \]

Let \(x \in A \cup (A \cap B)\). Then either

• \(x \in A\), or
• \(x \in A \cap B\)

In both cases \(x \in A\). Thus

\[ A \cup (A \cap B) \subset A \]

Conversely, let \(x \in A\). Clearly

\[ x \in A \cup (A \cap B) \]

Thus

\[ A \subset A \cup (A \cap B) \]

Therefore

\[ A \cup (A \cap B) = A \]

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(ii) Prove that

\[ A \cap (A \cup B) = A \]

Let \(x \in A \cap (A \cup B)\). Then

\[ x \in A \quad \text{and} \quad x \in A \cup B \]

The first condition already ensures

\[ x \in A \]

Hence

\[ A \cap (A \cup B) \subset A \]

Conversely, let \(x \in A\). Since \(A \subset A \cup B\),

\[ x \in A \cup B \]

Thus

\[ x \in A \cap (A \cup B) \]

Hence

\[ A \cap (A \cup B) = A \]

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Concept Illustration

The absorption law can be visualized using overlapping sets.

The region \(A \cap B\) already lies inside \(A\), therefore adding it to \(A\) through union does not change the set.

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Significance for Board & Competitive Exams

• These identities are called the Absorption Laws in set theory.
• They frequently appear in CBSE Class 11 board proofs.
• Competitive exams such as JEE Main, BITSAT, NDA, and Olympiads often test these identities in objective or proof-based questions.
• These laws are also fundamental in Boolean algebra, digital logic, and probability theory.

Q8 Show that \(A \cap B = A \cap C\) need not imply \(B = C\).

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Concept Used

Two sets are equal only when all their elements are identical.

\[ B = C \quad \text{if and only if} \quad B \subseteq C \text{ and } C \subseteq B \]

However, the intersection of two sets depends only on the elements common to both sets. Therefore, two different sets may produce the same intersection with a third set.

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Solution Strategy

• To show that a statement does not always hold, provide a counterexample.
• Construct sets \(A\), \(B\), and \(C\) such that \(A \cap B = A \cap C\) but \(B \ne C\).

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Solution

Consider the sets

\[ A = \{1\} \] \[ B = \{1,2\} \] \[ C = \{1,3\} \]

Now compute the intersections.

\[ A \cap B = \{1\} \cap \{1,2\} = \{1\} \]

\[ A \cap C = \{1\} \cap \{1,3\} = \{1\} \]

Thus

\[ A \cap B = A \cap C \]

However,

\[ 2 \in B \quad \text{but} \quad 2 \notin C \] and

\[ 3 \in C \quad \text{but} \quad 3 \notin B \]

Therefore

\[ B \ne C \]

Hence we conclude that

\[ A \cap B = A \cap C \] does not necessarily imply \[ B = C \]

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Concept Illustration

In the diagram below, sets \(B\) and \(C\) are different, but their common region with \(A\) is identical.

Both intersections \(A \cap B\) and \(A \cap C\) contain the same element, even though \(B\) and \(C\) differ outside \(A\).

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Significance for Board & Competitive Exams

• This question highlights the importance of counterexamples in set theory.
• Such reasoning problems frequently appear in Class 11 board examinations.
• Competitive exams such as JEE Main, BITSAT, NDA, and Olympiads often ask students to determine whether a statement is always true or sometimes false.
• Understanding these ideas builds logical foundations used later in relations, functions, and probability.

Q9 Let \(A\) and \(B\) be sets. If \[ A \cap X = B \cap X = \varnothing \] and \[ A \cup X = B \cup X \] for some set \(X\), show that \(A = B\).

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Concept Used

Two sets are equal if and only if each is a subset of the other:

\[ A = B \iff A \subseteq B \text{ and } B \subseteq A \]

The conditions \(A \cap X = \varnothing\) and \(B \cap X = \varnothing\) indicate that sets \(A\) and \(B\) are disjoint from \(X\).

Therefore any element belonging to \(A\) or \(B\) cannot belong to \(X\).

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Solution Roadmap

• Take an arbitrary element from set \(A\).
• Use the equality \(A \cup X = B \cup X\).
• Use the disjoint condition with \(X\).
• Show \(A \subseteq B\).
• Repeat the argument to show \(B \subseteq A\).

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Solution

We are given

\[ A \cap X = \varnothing \] \[ B \cap X = \varnothing \] \[ A \cup X = B \cup X \]

To prove \(A = B\), first show that \(A \subseteq B\).

Let \(x \in A\).

Then clearly

\[ x \in A \cup X \]

Since

\[ A \cup X = B \cup X \]

we obtain

\[ x \in B \cup X \]

Thus either

• \(x \in B\), or
• \(x \in X\)

But since

\[ A \cap X = \varnothing \]

no element of \(A\) belongs to \(X\). Hence

\[ x \notin X \]

Therefore

\[ x \in B \]

Thus every element of \(A\) belongs to \(B\).

\[ A \subseteq B \]

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Similarly, by interchanging \(A\) and \(B\) and using the condition \(B \cap X = \varnothing\), we obtain

\[ B \subseteq A \]

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Since

\[ A \subseteq B \] and \[ B \subseteq A \]

we conclude

\[ A = B \]

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Concept Illustration

The diagram shows that sets \(A\) and \(B\) lie outside \(X\). Since adding \(X\) to both sets gives the same union, the original sets must be identical.

Both \(A\) and \(B\) are disjoint from \(X\). Thus equality of unions with \(X\) forces \(A\) and \(B\) to be equal.

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Significance for Board & Competitive Exams

• This problem demonstrates how set equality can be proved using union and intersection conditions.
• Such logical proofs frequently appear in CBSE Class 11 board examinations.
• Competitive exams such as JEE Main, BITSAT, NDA, and Olympiads often test reasoning involving disjoint sets and union identities.
• These concepts also play a key role in probability theory and Boolean algebra.

Q10 Find sets \(A\), \(B\) and \(C\) such that \[ A \cap B,\quad B \cap C,\quad A \cap C \] are non-empty sets and \[ A \cap B \cap C = \varnothing. \]

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Concept Used

In set theory, intersections can occur in two ways:

• Pairwise intersection: intersection between any two sets.
• Triple intersection: intersection among all three sets.

The problem requires that:

• Every pair of sets must share at least one common element.
• No element should be common to all three sets simultaneously.

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Solution Roadmap

• Select three sets with overlapping elements.
• Ensure each pair of sets shares one element.
• Ensure no single element appears in all three sets.

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Solution

Consider the following sets:

\[ A = \{1,2\} \] \[ B = \{2,3\} \] \[ C = \{1,3\} \]

Now compute the pairwise intersections.

\[ A \cap B = \{2\} \]

\[ B \cap C = \{3\} \]

\[ A \cap C = \{1\} \]

Each intersection contains one element and is therefore non-empty.

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Now compute the intersection of all three sets.

\[ A \cap B \cap C = \{1,2\} \cap \{2,3\} \cap \{1,3\} \]

There is no element common to all three sets.

\[ A \cap B \cap C = \varnothing \]

Thus the given sets satisfy the required conditions.

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Concept Illustration

The diagram below shows three sets where each pair overlaps, but no element lies in all three simultaneously.

Each pairwise overlap exists, but the central triple-overlap region is empty.

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Significance for Board & Competitive Exams

• This problem illustrates the difference between pairwise intersections and triple intersection.
• Such questions frequently appear in CBSE Class 11 board examinations.
• Competitive exams like JEE Main, BITSAT, NDA, and Olympiads often use similar concepts in Venn diagram problems and logical reasoning.
• Understanding these ideas is essential for probability theory and set-based counting problems.