Sets: Exercise 1.4 – Guided Solutions
Work through every question with crystal‑clear steps, stress–strain visuals and exam‑oriented hints – all in a sleek deep‑space layout.
Q1
Find the union of each of the following pairs of sets :
(i) X = {1, 3, 5}, Y = {1, 2, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6}
B = {x : x is a natural number and 6 < x <
10}
(v) A = {1, 2, 3}, B = φ
Concept Theory : Union of Sets
In set theory, the union of two sets contains all elements that belong to either of the sets or to both.
\[ A \cup B = \{x : x \in A \text{ or } x \in B\} \]
Important properties of union:
- \(A \cup B = B \cup A\) (Commutative law)
- \((A \cup B) \cup C = A \cup (B \cup C)\) (Associative law)
- \(A \cup \varnothing = A\)
- \(A \cup A = A\)
Solution Roadmap (How to Solve Union Problems)
- Write both sets clearly.
- Collect all distinct elements appearing in either set.
- Remove repetitions.
- Arrange elements logically (usually ascending order).
Solution
(i)
\[ \begin{aligned} X &= \{1,3,5\} \\ Y &= \{1,2,3\} \end{aligned} \] Union contains every distinct element: \[ X \cup Y = \{1,2,3,5\} \]
(ii)
\[ \begin{aligned} A &= \{a,e,i,o,u\} \\ B &= \{a,b,c\} \end{aligned} \] \[ A \cup B = \{a,b,c,e,i,o,u\} \]
(iii)
\[ \begin{aligned} A &= \{3,6,9,12,\ldots\} \\ B &= \{1,2,3,4,5\} \end{aligned} \] \[ A \cup B = \{1,2,3,4,5,6,9,12,\ldots\} \]
(iv)
\[ \begin{aligned} A &= \{2,3,4,5,6\} \\ B &= \{7,8,9\} \end{aligned} \] \[ A \cup B = \{2,3,4,5,6,7,8,9\} \]
(v)
\[ \begin{aligned} A &= \{1,2,3\} \\ B &= \varnothing \end{aligned} \] Using the property \(A \cup \varnothing = A\) \[ A \cup B = \{1,2,3\} \]
Why This Question is Important
Understanding union of sets is foundational in mathematics and is frequently tested in board examinations and competitive entrance exams.
- Helps in solving Venn diagram problems
- Used in probability and statistics
- Essential for logic, relations and functions
- Frequently appears in JEE, NDA, CUET, Olympiad and other entrance tests
Mastering union operations builds the base for more advanced topics like set identities, inclusion–exclusion principle and probability theory.
Q2 Let A = { a, b }, B = { a, b, c }. Is \(A \subset B\) ? What is \(A \cup B\) ?
Concept Used : Subset and Union Relationship
A set \(A\) is said to be a subset of a set \(B\) if every element of \(A\) is also an element of \(B\).
\[ A \subset B \quad \text{if} \quad x \in A \Rightarrow x \in B \]
A key property connecting subset and union is:
\[ \text{If } A \subset B \text{ then } A \cup B = B \]
Solution Roadmap
- Compare elements of set \(A\) with set \(B\).
- Check whether every element of \(A\) appears in \(B\).
- If yes, then \(A\) is a subset of \(B\).
- The union will simply be the larger set.
Solution
\[ \begin{aligned} A &= \{a,b\} \\ B &= \{a,b,c\} \end{aligned} \] Both elements \(a\) and \(b\) of set \(A\) belong to set \(B\). \[ A \subset B \] Therefore, \[ A \cup B = \{a,b,c\} \]
Thus, when one set is completely contained in another set, their union equals the larger set.
Illustration (Subset Relationship)
Exam Importance
- Subset–union relationships are frequently asked in CBSE board examinations.
- Important for solving Venn diagram and set identity problems.
- Forms the conceptual base for probability questions in JEE and other entrance exams.
Q3 If \(A \subset B\), then what is \(A \cup B\) ?
Concept Used
When a set \(A\) is a subset of set \(B\), it means that every element of \(A\) is already present in \(B\).
Therefore, combining both sets does not introduce any new element.
\[ A \subset B \Rightarrow A \cup B = B \]
Logical Explanation
Since all elements of \(A\) are already contained in \(B\), the union simply remains the larger set \(B\).
Thus,
\[ A \cup B = B \]
Conceptual Visualization
Why This Concept Matters
- This property simplifies many problems involving unions.
- Useful in solving set algebra identities.
- Appears frequently in JEE, NDA, Olympiad and aptitude exams.
Q4 If \(A=\{1,2,3,4\},\; B=\{3,4,5,6\},\; C=\{5,6,7,8\},\; D=\{7,8,9,10\}\), find:
(i) \(A \cup B\)
(ii) \(A \cup C\)
(iii) \(B \cup C\)
(iv) \(B \cup D\)
(v) \(A \cup B \cup C\)
(vi) \(A \cup B \cup D\)
(vii) \(B \cup C \cup D\)
Concept Theory : Union of Multiple Sets
The union of sets contains all elements that belong to at least one of the sets.
\[ A \cup B = \{x : x \in A \text{ or } x \in B\} \]
For more than two sets, union follows the associative property.
\[ A \cup B \cup C = (A \cup B) \cup C \]
This means we can combine sets step by step while removing repeated elements.
Solution Roadmap
- Write the sets clearly.
- Combine all elements from the given sets.
- Remove duplicate elements.
- Arrange elements in increasing order.
Given Sets
\[ \begin{aligned} A &= \{1,2,3,4\} \\ B &= \{3,4,5,6\} \\ C &= \{5,6,7,8\} \\ D &= \{7,8,9,10\} \end{aligned} \]
Solution
\[ \begin{aligned} \text{(i)}\quad A \cup B &= \{1,2,3,4,5,6\} \\ \\ \text{(ii)}\quad A \cup C &= \{1,2,3,4,5,6,7,8\} \\ \\ \text{(iii)}\quad B \cup C &= \{3,4,5,6,7,8\} \\ \\ \text{(iv)}\quad B \cup D &= \{3,4,5,6,7,8,9,10\} \end{aligned} \]
\[ \begin{aligned} \text{(v)}\quad A \cup B \cup C &= (A \cup B) \cup C \\ &= \{1,2,3,4,5,6\} \cup \{5,6,7,8\} \\ &= \{1,2,3,4,5,6,7,8\} \end{aligned} \]
\[ \begin{aligned} \text{(vi)}\quad A \cup B \cup D &= (A \cup B) \cup D \\ &= \{1,2,3,4,5,6\} \cup \{7,8,9,10\} \\ &= \{1,2,3,4,5,6,7,8,9,10\} \end{aligned} \]
\[ \begin{aligned} \text{(vii)}\quad B \cup C \cup D &= (B \cup C) \cup D \\ &= \{3,4,5,6,7,8\} \cup \{7,8,9,10\} \\ &= \{3,4,5,6,7,8,9,10\} \end{aligned} \]
Illustration : Overlapping Sets
The diagram shows how elements spread across multiple sets while union collects all unique elements.
Why This Question Is Important
- Tests understanding of union of multiple sets.
- Builds foundation for Venn diagram reasoning.
- Important for solving probability problems in higher mathematics.
- Frequently appears in CBSE board exams, JEE, CUET, NDA and Olympiads.
Mastering multi-set unions prepares students for advanced topics such as inclusion–exclusion principle and probability theory.
Q5 Find the intersection of each pair of sets of Question 1 above.
Concept Theory : Intersection of Sets
The intersection of two sets consists of all elements that are common to both sets.
\[ A \cap B = \{x : x \in A \text{ and } x \in B\} \]
Important properties of intersection:
- \(A \cap B = B \cap A\) (Commutative law)
- \((A \cap B) \cap C = A \cap (B \cap C)\) (Associative law)
- \(A \cap \varnothing = \varnothing\)
- If two sets have no common element, they are called disjoint sets.
Solution Roadmap
- Write both sets clearly.
- Identify elements common to both sets.
- List only the common elements.
- If no common element exists, the intersection is the empty set \( \varnothing \).
Solutions
\[ \begin{aligned} \text{(i)}\quad X &= \{1,3,5\} \\ Y &= \{1,2,3\} \\ X \cap Y &= \{1,3\} \end{aligned} \]
\[ \begin{aligned} \text{(ii)}\quad A &= \{a,e,i,o,u\} \\ B &= \{a,b,c\} \\ A \cap B &= \{a\} \end{aligned} \]
\[ \begin{aligned} \text{(iii)}\quad A &= \{3,6,9,12,\ldots\} \\ B &= \{1,2,3,4,5\} \\ A \cap B &= \{3\} \end{aligned} \]
\[ \begin{aligned} \text{(iv)}\quad A &= \{2,3,4,5,6\} \\ B &= \{7,8,9\} \\ A \cap B &= \varnothing \end{aligned} \]
Since the two sets have no common element, they are called disjoint sets.
\[ \begin{aligned} \text{(v)}\quad A &= \{1,2,3\} \\ B &= \varnothing \\ A \cap B &= \varnothing \end{aligned} \]
Why This Question Is Important
- Intersection is a fundamental concept used in Venn diagram problems.
- Essential for solving problems in probability and statistics.
- Frequently tested in CBSE board exams and competitive entrance tests such as JEE, CUET, NDA and Olympiads.
Understanding intersections also helps students learn advanced concepts such as set identities, inclusion–exclusion principle and probability theory.
Q6 If \(A = \{3,5,7,9,11\}\), \(B = \{7,9,11,13\}\), \(C = \{11,13,15\}\) and \(D = \{15,17\}\), find:
(i) \(A \cap B\)
(ii) \(B \cap C\)
(iii) \(A \cap C \cap D\)
(iv) \(A \cap C\)
(v) \(B \cap D\)
(vi) \(A \cap (B \cup C)\)
(vii) \(A \cap D\)
(viii) \(A \cap (B \cup D)\)
(ix) \( (A \cap B) \cap (B \cup C)\)
(x) \( (A \cup D) \cap (B \cup C)\)
Concept Theory : Intersection of Sets
The intersection of sets consists of elements that are common to all sets involved in the operation.
\[ A \cap B = \{x : x \in A \text{ and } x \in B\} \]
Important identities used in this problem:
- \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
- \((A \cap B) \cap C = A \cap B \cap C\)
- If no common element exists → result is \( \varnothing \)
Solution Roadmap
- Write all sets clearly.
- Find common elements between the required sets.
- Apply union first if brackets contain \( \cup \).
- Apply intersection to obtain elements common to all involved sets.
Given Sets
\[ \begin{aligned} A &= \{3,5,7,9,11\} \\ B &= \{7,9,11,13\} \\ C &= \{11,13,15\} \\ D &= \{15,17\} \end{aligned} \]
Solutions
\[ \begin{aligned} (i)\quad A \cap B &= \{7,9,11\} \\ \\ (ii)\quad B \cap C &= \{11,13\} \end{aligned} \]
\[ \begin{aligned} (iii)\quad A \cap C \cap D &= (A \cap C) \cap D \\ &= \{11\} \cap \{15,17\} \\ &= \varnothing \end{aligned} \]
\[ \begin{aligned} (iv)\quad A \cap C &= \{11\} \\ \\ (v)\quad B \cap D &= \varnothing \end{aligned} \]
\[ \begin{aligned} (vi)\quad A \cap (B \cup C) \end{aligned} \] First find union: \[ B \cup C = \{7,9,11,13,15\} \] Then intersection: \[ A \cap (B \cup C) = \{3,5,7,9,11\} \cap \{7,9,11,13,15\} \] \[ = \{7,9,11\} \]
\[ (vii)\quad A \cap D = \varnothing \]
\[ \begin{aligned} (viii)\quad A \cap (B \cup D) \end{aligned} \] First union: \[ B \cup D = \{7,9,11,13,15,17\} \] Intersection: \[ A \cap (B \cup D) = \{3,5,7,9,11\} \cap \{7,9,11,13,15,17\} \] \[ = \{7,9,11\} \]
\[ \begin{aligned} (ix)\quad (A \cap B) \cap (B \cup C) \end{aligned} \] \[ A \cap B = \{7,9,11\} \] \[ B \cup C = \{7,9,11,13,15\} \] \[ (A \cap B) \cap (B \cup C) = \{7,9,11\} \]
\[ \begin{aligned} (x)\quad (A \cup D) \cap (B \cup C) \end{aligned} \] First unions: \[ A \cup D = \{3,5,7,9,11,15,17\} \] \[ B \cup C = \{7,9,11,13,15\} \] Intersection: \[ (A \cup D) \cap (B \cup C) = \{7,9,11,15\} \]
Illustration : Set Relationships
Intersection identifies the elements lying in the overlapping region of sets.
Why This Question Is Important
- Tests understanding of intersection operations involving multiple sets.
- Helps students practice combining union and intersection operations.
- Builds foundation for Venn diagram problems and probability.
- Frequently appears in CBSE Board exams, JEE, CUET, NDA and Olympiads.
Mastery of these operations prepares students for advanced concepts like set identities, Boolean algebra and probability theory.
Q7 If \(A = \{x : x \text{ is a natural number}\}\), \(B = \{x : x \text{ is an even natural number}\}\), \(C = \{x : x \text{ is an odd natural number}\}\) and \(D = \{x : x \text{ is a prime number}\}\), find:
(i) \(A \cap B\)
(ii) \(A \cap C\)
(iii) \(A \cap D\)
(iv) \(B \cap C\)
(v) \(B \cap D\)
(vi) \(C \cap D\)
Concept Theory : Intersection of Number Sets
The intersection of two sets contains elements common to both sets.
\[ A \cap B = \{x : x \in A \text{ and } x \in B\} \]
In this question we use important number classifications:
- Natural numbers: \(1,2,3,4,\ldots\)
- Even numbers: multiples of 2
- Odd numbers: numbers not divisible by 2
- Prime numbers: numbers greater than 1 having exactly two factors
Solution Roadmap
- Write each set in roster form.
- Identify elements common to both sets.
- Use properties of number sets such as even, odd, and prime numbers.
Representation of Sets
\[ \begin{aligned} A &= \{1,2,3,4,\ldots\} \\ B &= \{2,4,6,8,\ldots\} \\ C &= \{1,3,5,7,\ldots\} \\ D &= \{2,3,5,7,11,\ldots\} \end{aligned} \]
Solutions
\[ (i)\quad A \cap B \] All even numbers are natural numbers. \[ A \cap B = \{2,4,6,8,\ldots\} = B \]
\[ (ii)\quad A \cap C \] All odd numbers are natural numbers. \[ A \cap C = \{1,3,5,7,\ldots\} = C \]
\[ (iii)\quad A \cap D \] All prime numbers are natural numbers. \[ A \cap D = \{2,3,5,7,11,\ldots\} = D \]
\[ (iv)\quad B \cap C \] Even and odd numbers cannot be the same. \[ B \cap C = \varnothing \]
\[ (v)\quad B \cap D \] Only one prime number is even. \[ B \cap D = \{2\} \]
\[ (vi)\quad C \cap D \] Odd prime numbers: \[ C \cap D = \{3,5,7,11,\ldots\} \]
Illustration : Even vs Odd Numbers
Even and odd numbers are disjoint sets, so their intersection is the empty set.
Illustration : Prime Numbers within Natural Numbers
Why This Question Is Important
- Builds understanding of relationships between number sets.
- Important for solving Venn diagram problems involving numbers.
- Strengthens conceptual clarity about even, odd and prime numbers.
- Commonly tested in CBSE Board exams, JEE, CUET, NDA and Olympiads.
This concept also forms the basis for problems in number theory, probability and mathematical logic.
Q8 Which of the following pairs of sets are disjoint?
(i) \( \{1,2,3,4\} \) and \( \{x : x \text{ is a natural number and } 4 \le x \le 6\} \)
(ii) \( \{a,e,i,o,u\} \) and \( \{c,d,e,f\} \)
(iii) \( \{x : x \text{ is an even integer}\} \) and \( \{x : x \text{ is an odd integer}\} \)
Concept Theory : Disjoint Sets
Two sets are said to be disjoint if they have no elements in common.
\[ A \cap B = \varnothing \]
If the intersection of two sets is the empty set, the sets are disjoint.
Solution Roadmap
- Convert the given sets into roster form wherever possible.
- Find the intersection of the two sets.
- If the intersection is \( \varnothing \), the sets are disjoint.
Solution
\[ (i)\quad \{1,2,3,4\} \text{ and } \{x : x \text{ is a natural number and } 4 \le x \le 6\} \] Second set: \[ \{4,5,6\} \] Intersection: \[ \{1,2,3,4\} \cap \{4,5,6\} = \{4\} \] Since element \(4\) is common, \[ \text{The sets are NOT disjoint.} \]
\[ (ii)\quad \{a,e,i,o,u\} \text{ and } \{c,d,e,f\} \] Intersection: \[ \{a,e,i,o,u\} \cap \{c,d,e,f\} = \{e\} \] Since element \(e\) is common, \[ \text{The sets are NOT disjoint.} \]
\[ (iii)\quad \{x : x \text{ is an even integer}\} \text{ and } \{x : x \text{ is an odd integer}\} \] No number can be both even and odd. \[ \text{Intersection} = \varnothing \] Therefore, \[ \text{These sets are disjoint.} \]
Illustration : Disjoint Sets
Even and odd integers do not overlap, so their intersection is the empty set.
Illustration : Non-Disjoint Sets
If sets share at least one element, they are not disjoint.
Final Conclusion
Only the pair given in (iii) consists of disjoint sets.
Why This Question Is Important
- Helps students understand the concept of disjoint sets.
- Important for solving Venn diagram problems.
- Used extensively in probability and statistics.
- Frequently asked in CBSE board exams, JEE, CUET and Olympiads.
Understanding disjoint sets is essential for learning advanced topics such as mutually exclusive events in probability.
Q9 If \(A=\{3,6,9,12,15,18,21\}\), \(B=\{4,8,12,16,20\}\), \(C=\{2,4,6,8,10,12,14,16\}\), \(D=\{5,10,15,20\}\), find:
(i) \(A-B\)
(ii) \(A-C\)
(iii) \(A-D\)
(iv) \(B-A\)
(v) \(C-A\)
(vi) \(D-A\)
(vii) \(B-C\)
(viii) \(B-D\)
(ix) \(C-B\)
(x) \(D-B\)
(xi) \(C-D\)
(xii) \(D-C\)
Concept Theory : Difference of Sets
The difference of two sets \(A-B\) consists of all elements of \(A\) that do not belong to \(B\).
\[ A-B=\{x:x\in A \text{ and } x\notin B\} \]
Important idea:
- \(A-B \neq B-A\)
- Difference operation removes common elements.
Solution Strategy
- Write the given sets clearly.
- Compare elements of the first set with the second set.
- Remove elements that appear in both sets.
- The remaining elements form the difference set.
Given Sets
\[ \begin{aligned} A &= \{3,6,9,12,15,18,21\}\\ B &= \{4,8,12,16,20\}\\ C &= \{2,4,6,8,10,12,14,16\}\\ D &= \{5,10,15,20\} \end{aligned} \]
Solutions
\[ \begin{aligned} (i)\quad A-B &= \{3,6,9,15,18,21\} \end{aligned} \]
\[ \begin{aligned} (ii)\quad A-C &= \{3,9,15,18,21\} \end{aligned} \]
\[ \begin{aligned} (iii)\quad A-D &= \{3,6,9,12,18,21\} \end{aligned} \]
\[ \begin{aligned} (iv)\quad B-A &= \{4,8,16,20\} \end{aligned} \]
\[ \begin{aligned} (v)\quad C-A &= \{2,4,8,10,14,16\} \end{aligned} \]
\[ \begin{aligned} (vi)\quad D-A &= \{5,10,20\} \end{aligned} \]
\[ \begin{aligned} (vii)\quad B-C &= \{20\} \end{aligned} \]
\[ \begin{aligned} (viii)\quad B-D &= \{4,8,12,16\} \end{aligned} \]
\[ \begin{aligned} (ix)\quad C-B &= \{2,6,10,14\} \end{aligned} \]
\[ \begin{aligned} (x)\quad D-B &= \{5,10,15\} \end{aligned} \]
\[ \begin{aligned} (xi)\quad C-D &= \{2,4,6,8,12,14,16\} \end{aligned} \]
\[ \begin{aligned} (xii)\quad D-C &= \{5,15,20\} \end{aligned} \]
Conceptual Illustration
The shaded region of set A excluding B represents the set difference \(A-B\).
Why This Question Is Important
- Helps understand set difference operations.
- Important for solving Venn diagram problems.
- Forms the basis for complement and probability problems.
- Frequently tested in CBSE Board exams, JEE, CUET, NDA and Olympiads.
Mastering set difference also helps students understand advanced topics such as set identities, complements and probability theory.
Q10 If \(X = \{a,b,c,d\}\) and \(Y = \{f,b,d,g\}\), find
(i) \(X - Y\)
(ii) \(Y - X\)
(iii) \(X \cap Y\)
Concept Theory
Two important set operations are used here:
- Difference of sets: \(A-B\) contains elements belonging to \(A\) but not to \(B\).
- Intersection of sets: \(A \cap B\) contains elements common to both sets.
\[ A-B=\{x:x\in A \text{ and } x\notin B\} \]
\[ A\cap B=\{x:x\in A \text{ and } x\in B\} \]
Solution Roadmap
- Write the sets clearly.
- Identify elements common to both sets.
- Remove common elements when computing set difference.
- Keep only common elements for intersection.
Given Sets
\[ \begin{aligned} X &= \{a,b,c,d\} \\ Y &= \{f,b,d,g\} \end{aligned} \]
Solution
\[ (i)\quad X-Y \] Elements of \(X\) not in \(Y\): \[ X-Y=\{a,c\} \]
\[ (ii)\quad Y-X \] Elements of \(Y\) not in \(X\): \[ Y-X=\{f,g\} \]
\[ (iii)\quad X\cap Y \] Common elements of both sets: \[ X\cap Y=\{b,d\} \]
Venn Diagram Illustration
The middle region represents \(X \cap Y\), while the non-overlapping parts represent the differences \(X-Y\) and \(Y-X\).
Why This Question Is Important
- Helps students clearly distinguish between intersection and difference of sets.
- Important for solving Venn diagram problems.
- Forms the conceptual base for set identities and probability.
- Frequently appears in CBSE Board exams, JEE, CUET, NDA and Olympiads.
Understanding these operations strengthens logical reasoning and prepares students for advanced topics like set algebra and probability theory.
Q11 If \(R\) is the set of real numbers and \(Q\) is the set of rational numbers, find \(R-Q\).
Concept Theory : Real and Rational Numbers
The set of real numbers contains all numbers on the number line. It consists of two main categories:
- Rational numbers (\(Q\)) – numbers that can be written as a fraction \( \frac{p}{q} \), where \(p,q\) are integers and \(q \ne 0\).
- Irrational numbers – numbers that cannot be expressed as a fraction.
\[ R = Q \cup (\text{irrational numbers}) \]
Thus, the real number system is composed of both rational and irrational numbers.
Solution Roadmap
- Identify the meaning of the sets \(R\) and \(Q\).
- Understand the set difference operation \(R-Q\).
- Remove all rational numbers from the real number set.
Solution
\[ \begin{aligned} R &= \{\text{all real numbers}\} \\ Q &= \{\text{all rational numbers}\} \end{aligned} \]
The difference \(R-Q\) means all elements of \(R\) that are not in \(Q\).
\[ R-Q=\{\text{real numbers that are not rational}\} \]
But real numbers that are not rational are called irrational numbers.
\[ R-Q=\{\text{all irrational numbers}\} \]
Illustration : Real Number System
The inner region represents rational numbers \(Q\), while the remaining part of \(R\) represents irrational numbers.
Examples of Irrational Numbers
\[ \sqrt{2},\ \sqrt{3},\ \pi,\ e,\ \sqrt{5} \]
These numbers cannot be written as fractions of integers.
Why This Question Is Important
- Helps students understand the structure of the real number system.
- Important for topics like number systems and algebra.
- Frequently appears in CBSE Board exams, JEE, CUET and Olympiads.
- Useful in advanced topics like limits, calculus and analysis.
Q12 State whether each of the following statements is true or false. Justify your answer.
(i) \( \{2,3,4,5\} \) and \( \{3,6\} \) are disjoint sets.
(ii) \( \{a,e,i,o,u\} \) and \( \{a,b,c,d\} \) are disjoint sets.
(iii) \( \{2,6,10,14\} \) and \( \{3,7,11,15\} \) are disjoint sets.
(iv) \( \{2,6,10\} \) and \( \{3,7,11\} \) are disjoint sets.
Concept Theory : Disjoint Sets
Two sets are said to be disjoint if they have no elements in common.
\[ A \cap B = \varnothing \]
If the intersection of two sets is the empty set, the sets are disjoint.
Solution Roadmap
- Write the sets clearly.
- Check whether any element appears in both sets.
- If a common element exists → sets are not disjoint.
- If no common element exists → sets are disjoint.
Solution
\[ (i)\quad \{2,3,4,5\} \text{ and } \{3,6\} \] Intersection: \[ \{2,3,4,5\} \cap \{3,6\} = \{3\} \] Since \(3\) is common, The statement is FALSE.
\[ (ii)\quad \{a,e,i,o,u\} \text{ and } \{a,b,c,d\} \] Intersection: \[ \{a,e,i,o,u\} \cap \{a,b,c,d\} = \{a\} \] Since element \(a\) is common, The statement is FALSE.
\[ (iii)\quad \{2,6,10,14\} \text{ and } \{3,7,11,15\} \] No element is common. \[ \text{Intersection} = \varnothing \] The statement is TRUE.
\[ (iv)\quad \{2,6,10\} \text{ and } \{3,7,11\} \] No element is common. \[ \text{Intersection} = \varnothing \] The statement is TRUE.
Illustration : Disjoint vs Non-Disjoint Sets
If two sets share elements, they are not disjoint. If they have no common elements, they are disjoint.
Final Result
- (i) False
- (ii) False
- (iii) True
- (iv) True
Why This Question Is Important
- Strengthens understanding of disjoint sets.
- Useful for solving Venn diagram problems.
- Important for probability concepts such as mutually exclusive events.
- Frequently asked in CBSE Board exams, JEE, CUET, NDA and Olympiads.
Understanding disjoint sets is essential for advanced topics like probability theory, set identities and logical reasoning.
Exercise 1.4 – Chapter Summary & Formula Sheet
This section provides a quick revision of the key concepts used in NCERT Class 11 Mathematics Chapter 1 – Sets (Exercise 1.4). Students can use it for rapid revision before exams.
1. Union of Sets
The union of two sets contains all elements that belong to either set.
\[ A \cup B = \{x : x \in A \text{ or } x \in B\} \]
- Combines elements from both sets
- Repeated elements appear only once
2. Intersection of Sets
The intersection of two sets contains elements common to both sets.
\[ A \cap B = \{x : x \in A \text{ and } x \in B\} \]
3. Difference of Sets
The difference \(A-B\) contains elements belonging to \(A\) but not to \(B\).
\[ A-B = \{x : x \in A \text{ and } x \notin B\} \]
4. Disjoint Sets
Two sets are called disjoint if they have no elements in common.
\[ A \cap B = \varnothing \]
5. Subset Relationship
If every element of set \(A\) belongs to set \(B\), then \(A\) is a subset of \(B\).
\[ A \subset B \]
Important identity:
\[ A \subset B \Rightarrow A \cup B = B \]
Important Set Identities
- \(A \cup B = B \cup A\)
- \(A \cap B = B \cap A\)
- \(A \cup \varnothing = A\)
- \(A \cap \varnothing = \varnothing\)
- \(A - A = \varnothing\)
Venn Diagram Summary
Venn diagrams help visualize relationships between sets such as union, intersection, and difference.
Exam Tips
- Always remove duplicate elements when writing unions.
- For intersection, look only for common elements.
- For difference \(A-B\), remove elements of \(B\) from \(A\).
- Check carefully whether sets are disjoint.
Competitive Exam Relevance
Set theory forms the foundation for many topics used in higher mathematics and competitive exams such as:
- JEE Main & Advanced
- CUET
- NDA
- Olympiad mathematics
Concepts of union, intersection, and disjoint sets are also used in probability, logic, and computer science.
🚀 Sets Visualization Engine
Experiment with set operations interactively. Enter elements separated by commas. Example: 1,2,3,4
Input Sets
Result
Visualization
🤖 AI Set Solver
Type a set problem using simple format. Examples:
A = {1,2,3,4}
B = {3,4,5}
Find A ∩ B
A = {x : x ≤ 10 and even}
B = {2,3,5,7}
Find A ∩ B
Result
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