Sets: Exercise 1.5 – 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
Let
\(U = \{1,2,3,4,5,6,7,8,9\}\),
\(A = \{1,2,3,4\}\),
\(B = \{2,4,6,8\}\), and
\(C = \{3,4,5,6\}\).
Find:
(i) \(A'\)
(ii) \(B'\)
(iii) \((A \cup C)'\)
(iv) \((A \cup B)'\)
(v) \((A')'\)
(vi) \((B - C)'\)
Concept Used
In set theory, the complement of a set contains all elements of the universal set that are not present in the given set.
If \(U\) is the universal set and \(A \subseteq U\), then
\[ A' = U - A \]
Important related concepts used in this problem:
- Union: \(A \cup B\) contains elements belonging to either \(A\) or \(B\).
- Set Difference: \(A - B\) contains elements of \(A\) that are not in \(B\).
- Double Complement Law: \((A')' = A\).
Solution Roadmap
To solve the problem efficiently:
- Use the definition \(A' = U - A\).
- Compute unions like \(A \cup C\) before finding complements.
- Compute set differences such as \(B - C\).
- Finally subtract each result from the universal set \(U\).
Illustration (Venn Concept)
This Venn representation helps visualize union, complement and set difference operations.
Solution
\( \begin{aligned} U &= \{1,2,3,4,5,6,7,8,9\} \\ A &= \{1,2,3,4\} \\ B &= \{2,4,6,8\} \\ C &= \{3,4,5,6\} \end{aligned} \)
Since complements are taken with respect to the universal set \(U\), we compute each result accordingly.
\[ (i)\quad A' = U - A = \{5,6,7,8,9\} \]
\[ (ii)\quad B' = U - B = \{1,3,5,7,9\} \]
First compute the union:
\[ A \cup C = \{1,2,3,4,5,6\} \]
\[ (iii)\quad (A \cup C)' = U - (A \cup C) = \{7,8,9\} \]
Next,
\[ A \cup B = \{1,2,3,4,6,8\} \]
\[ (iv)\quad (A \cup B)' = U - (A \cup B) = \{5,7,9\} \]
\[ (v)\quad (A')' = A = \{1,2,3,4\} \]
Now compute the difference:
\[ B - C = \{2,8\} \]
\[ (vi)\quad (B - C)' = U - (B - C) = \{1,3,4,5,6,7,9\} \]
Final Results
\( \begin{aligned} A' &= \{5,6,7,8,9\} \\ B' &= \{1,3,5,7,9\} \\ (A \cup C)' &= \{7,8,9\} \\ (A \cup B)' &= \{5,7,9\} \\ (A')' &= \{1,2,3,4\} \\ (B - C)' &= \{1,3,4,5,6,7,9\} \end{aligned} \)
Why This Problem Is Important
- Board Exams: Questions on complements, union and set difference frequently appear in CBSE Class 11 and Class 12 board examinations.
- Competitive Exams: Understanding complement operations is essential for solving problems in JEE, NEET, NDA, CUET and Olympiads, especially when dealing with Venn diagrams and logical set identities.
- Foundation for Probability: Many probability formulas rely on complements, such as \(P(A') = 1 - P(A)\).
- Logical Thinking: These operations train students to analyze sets systematically using algebraic rules.
Q2
If \(U = \{a,b,c,d,e,f,g,h\}\), find the complements of the following sets:
(i) \(A = \{a,b,c\}\)
(ii) \(B = \{d,e,f,g\}\)
(iii) \(C = \{a,c,e,g\}\)
(iv) \(D = \{f,g,h,a\}\)
Concept Used
The complement of a set consists of all elements of the universal set that are not contained in the given set.
If \(U\) is the universal set and \(A \subseteq U\), then
\[ A' = U - A \]
Thus, to find the complement of a set, we simply remove the elements of that set from the universal set.
Solution Roadmap
To determine each complement:
- Write the universal set \(U\).
- Identify elements present in the given set.
- Remove those elements from \(U\).
- The remaining elements form the complement.
Concept Illustration
The shaded circle represents set \(A\). All elements in the universal set but outside the circle form the complement \(A'\).
Solution
\( \begin{aligned} U &= \{a,b,c,d,e,f,g,h\} \end{aligned} \)
The complement of a set is obtained by removing all its elements from the universal set \(U\).
\[ (i)\quad A = \{a,b,c\} \] \[ A' = U - A = \{d,e,f,g,h\} \]
\[ (ii)\quad B = \{d,e,f,g\} \] \[ B' = U - B = \{a,b,c,h\} \]
\[ (iii)\quad C = \{a,c,e,g\} \] \[ C' = U - C = \{b,d,f,h\} \]
\[ (iv)\quad D = \{f,g,h,a\} \] \[ D' = U - D = \{b,c,d,e\} \]
Final Results
\( \begin{aligned} A' &= \{d,e,f,g,h\} \\ B' &= \{a,b,c,h\} \\ C' &= \{b,d,f,h\} \\ D' &= \{b,c,d,e\} \end{aligned} \)
Why This Problem Is Important
- Board Exams: Complement problems are frequently asked in CBSE Class 11 Mathematics examinations.
- Competitive Exams: Complement operations are widely used in JEE, NEET, NDA, CUET and Olympiad questions involving Venn diagrams.
- Logical Reasoning: Understanding complements helps students solve problems related to set identities and Boolean logic.
- Probability Foundation: Complement sets form the basis of the probability rule \(P(A') = 1 - P(A)\).
Q3
Taking the set of natural numbers as the universal set, write down the complements of the following
sets:
(i) {x : x is an even natural number}
(ii) {x : x is an odd natural number}
(iii) {x : x is a positive multiple of 3}
(iv) {x : x is a prime number}
(v) {x : x is a natural number divisible by 3 and 5}
(vi) {x : x is a perfect square}
(vii) {x : x is a perfect cube}
(viii) {x : x + 5 = 8}
(ix) {x : 2x + 5 = 9}
(x) {x : x ≥ 7}
(xi) {x : x ∈ N and 2x + 1 > 10}
Concept Used
In set theory, the complement of a set consists of all elements of the universal set that do not belong to the given set.
\[ A' = U - A \]
In this problem, the universal set is the set of all natural numbers:
\[ U = \mathbb{N} = \{1,2,3,4,5,\ldots\} \]
Therefore, the complement of each set will contain all natural numbers that do not satisfy the given condition.
Solution Roadmap
- Identify the property defining each set.
- Determine which natural numbers satisfy that condition.
- The complement contains natural numbers that do not satisfy that property.
- For algebraic expressions, solve the equation or inequality first.
Concept Illustration
The circle represents set \(A\). All natural numbers outside the circle form the complement \(A'\).
Solution
\[ U = \{1,2,3,4,5,\ldots\} \]
The complement of each set consists of natural numbers that do not satisfy the given condition.
\[ (i)\quad A = \{x : x \text{ is an even natural number}\} \] \[ A' = \{x : x \text{ is an odd natural number}\} \]
\[ (ii)\quad B = \{x : x \text{ is an odd natural number}\} \] \[ B' = \{x : x \text{ is an even natural number}\} \]
\[ (iii)\quad C = \{x : x \text{ is a positive multiple of }3\} \] \[ C' = \{x : x \text{ is not a multiple of }3\} \]
\[ (iv)\quad D = \{x : x \text{ is a prime number}\} \] \[ D' = \{x : x \text{ is a non-prime natural number}\} \]
\[ (v)\quad E = \{x : x \text{ is divisible by both }3 \text{ and }5\} \]
Numbers divisible by both 3 and 5 are multiples of \(15\).
\[ E' = \{x : x \text{ is not divisible by both }3 \text{ and }5\} \]\[ (vi)\quad F = \{x : x \text{ is a perfect square}\} \] \[ F' = \{x : x \text{ is not a perfect square}\} \]
\[ (vii)\quad G = \{x : x \text{ is a perfect cube}\} \] \[ G' = \{x : x \text{ is not a perfect cube}\} \]
\[ (viii)\quad H = \{x : x + 5 = 8\} \]
Solving the equation:
\[ x + 5 = 8 \Rightarrow x = 3 \] \[ H = \{3\} \] \[ H' = \mathbb{N} - \{3\} \]\[ (ix)\quad I = \{x : 2x + 5 = 9\} \]
Solving:
\[ 2x + 5 = 9 \] \[ 2x = 4 \] \[ x = 2 \] \[ I = \{2\} \] \[ I' = \mathbb{N} - \{2\} \]\[ (x)\quad J = \{x : x \ge 7\} \] \[ J' = \{1,2,3,4,5,6\} \]
\[ (xi)\quad K = \{x : x \in \mathbb{N} \text{ and } 2x + 1 > 10\} \]
Solve the inequality:
\[ 2x + 1 > 10 \] \[ 2x > 9 \] \[ x > 4.5 \]Thus,
\[ K = \{5,6,7,8,\ldots\} \] \[ K' = \{1,2,3,4\} \]Why This Problem Is Important
- Board Exams: Complement problems using set-builder notation frequently appear in CBSE Class 11 examinations.
- Competitive Exams: These concepts are important for JEE, NEET, NDA, CUET and Olympiads, particularly in Venn diagram and logical reasoning questions.
- Mathematical Thinking: Students learn to translate real conditions into mathematical set-builder notation.
- Foundation for Advanced Topics: Complement operations are later used in probability theory and Boolean algebra.
Q4
If \(U = \{1,2,3,4,5,6,7,8,9\}\), \(A = \{2,4,6,8\}\) and \(B = \{2,3,5,7\}\), verify that:
(i) \((A \cup B)' = A' \cap B'\)
(ii) \((A \cap B)' = A' \cup B'\)
Concept Used
These identities are known as De Morgan’s Laws in set theory. They describe how complements interact with union and intersection.
\[ (A \cup B)' = A' \cap B' \]
\[ (A \cap B)' = A' \cup B' \]
These laws state that:
- The complement of a union equals the intersection of complements.
- The complement of an intersection equals the union of complements.
Solution Roadmap
To verify each identity:
- Compute union or intersection first.
- Find its complement using the universal set \(U\).
- Compute complements of the individual sets.
- Evaluate the right-hand side.
- Compare both results.
Concept Illustration
The overlapping circles represent sets \(A\) and \(B\). De Morgan’s laws describe relationships between complements, unions, and intersections.
Solution
\( \begin{aligned} U &= \{1,2,3,4,5,6,7,8,9\} \\ A &= \{2,4,6,8\} \\ B &= \{2,3,5,7\} \end{aligned} \)
We verify each identity by evaluating both sides.
(i) Verify \( (A \cup B)' = A' \cap B' \)
\[ A \cup B = \{2,3,4,5,6,7,8\} \]
\[ (A \cup B)' = U - (A \cup B) \]
\[ (A \cup B)' = \{1,9\} \]
Now compute complements of individual sets:
\[ A' = U - A = \{1,3,5,7,9\} \]
\[ B' = U - B = \{1,4,6,8,9\} \]
\[ A' \cap B' = \{1,9\} \]
Thus,
\[ (A \cup B)' = A' \cap B' \]
Hence, the first identity is verified.
(ii) Verify \( (A \cap B)' = A' \cup B' \)
\[ A \cap B = \{2\} \]
\[ (A \cap B)' = U - (A \cap B) \]
\[ (A \cap B)' = \{1,3,4,5,6,7,8,9\} \]
Now compute the right-hand side.
\[ A' = \{1,3,5,7,9\} \]
\[ B' = \{1,4,6,8,9\} \]
\[ A' \cup B' = \{1,3,4,5,6,7,8,9\} \]
Thus,
\[ (A \cap B)' = A' \cup B' \]
Hence, the second identity is also verified.
Conclusion
Both identities satisfy the conditions of De Morgan’s Laws:
\[ (A \cup B)' = A' \cap B' \]
\[ (A \cap B)' = A' \cup B' \]
Why This Problem Is Important
- Board Exams: Verification of De Morgan’s laws is a common question in CBSE Class 11 Mathematics examinations.
- Competitive Exams: These identities are widely used in JEE, NEET, NDA, CUET, Olympiads for solving complex Venn diagram and logical reasoning problems.
- Boolean Logic: De Morgan’s laws are also fundamental in computer science, digital logic, and Boolean algebra.
- Problem Simplification: They help simplify complicated set expressions efficiently.
Q5
Draw appropriate Venn diagram for each of the following:
(i) \((A \cup B)'\)
(ii) \(A' \cap B'\)
(iii) \((A \cap B)'\)
(iv) \(A' \cup B'\)
Concept Used
Venn diagrams visually represent relationships between sets. The shaded region represents the elements belonging to the required set.
These diagrams illustrate the complement operations and also help verify De Morgan’s Laws.
Solution
(i) \((A \cup B)'\)
Region outside both sets is shaded.
(ii) \(A' \cap B'\)
Same region as (i). Outside both sets.
(iii) \((A \cap B)'\)
Everything except the intersection.
(iv) \(A' \cup B'\)
Same region as (iii).
Observation
From the diagrams we observe that:
\[ (A \cup B)' = A' \cap B' \]
\[ (A \cap B)' = A' \cup B' \]
Thus the diagrams visually confirm De Morgan’s Laws.
Why This Problem Is Important
- Board Exams: Venn diagram representation of set identities is commonly asked in CBSE Class 11 exams.
- Competitive Exams: These diagrams are frequently used in JEE, NEET, NDA, CUET and Olympiads to solve logical set problems.
- Visual Understanding: Venn diagrams help students understand relationships between union, intersection and complements.
- Foundation for Probability: Venn diagrams later become essential in probability and statistics problems.
Q6 Let \(U\) be the set of all triangles in a plane. If \(A\) is the set of all triangles with at least one angle different from \(60^\circ\), what is \(A'\)?
Concept Used
The complement of a set contains all elements of the universal set that do not belong to the given set.
\[ A' = U - A \]
In geometry, an important property of triangles is:
\[ \text{Sum of interior angles of a triangle} = 180^\circ \]
If all three angles of a triangle are equal, each angle must be
\[ 60^\circ \]
Such triangles are called equilateral triangles.
Solution Roadmap
- Identify the universal set \(U\).
- Understand the condition defining set \(A\).
- Determine triangles that do not satisfy this condition.
- Those triangles form the complement \(A'\).
Concept Illustration
An equilateral triangle has all three angles equal to \(60^\circ\).
Solution
\[ U = \{x : x \text{ is a triangle in a plane}\} \]
\[ A = \{x : x \text{ is a triangle with at least one angle different from } 60^\circ\} \]
Thus, set \(A\) includes all triangles that are not equiangular.
The complement \(A'\) contains triangles that do not satisfy this condition.
\[ A' = \{x : x \text{ is a triangle in which all angles are } 60^\circ\} \]
A triangle with all angles equal to \(60^\circ\) is an equilateral triangle.
\[ A' = \{\text{all equilateral triangles}\} \]
Conclusion
The complement of set \(A\) consists of all triangles whose three angles are \(60^\circ\). Therefore,
\[ A' = \text{set of all equilateral triangles} \]
Why This Problem Is Important
- Board Exams: Conceptual questions connecting geometry with set theory frequently appear in CBSE Class 11 Mathematics exams.
- Competitive Exams: Logical interpretation of complements is useful in JEE, NDA, CUET and Olympiad problems.
- Conceptual Understanding: The problem reinforces both triangle angle properties and complement operations.
- Interdisciplinary Thinking: It connects set theory with geometry, an important mathematical skill.
Q7
Fill in the blanks to make each of the following a true statement:
(i) \(A \cup A' = \ldots\)
(ii) \(\varnothing' \cap A = \ldots\)
(iii) \(A \cap A' = \ldots\)
(iv) \(U' \cap A = \ldots\)
Concept Used
In set theory, complements follow some important identities. If \(U\) is the universal set and \(A\subseteq U\), then:
\[ A' = U - A \]
Some fundamental complement laws are:
- \(A \cup A' = U\)
- \(A \cap A' = \varnothing\)
- \(\varnothing' = U\)
- \(U' = \varnothing\)
These identities are frequently used to simplify expressions involving sets.
Solution Roadmap
- Use complement identities.
- Replace complements of special sets (\(\varnothing\) and \(U\)).
- Apply union and intersection rules.
Concept Illustration
A set \(A\) and its complement \(A'\) together form the universal set \(U\).
Solution
Let \(U\) denote the universal set and \(A\) be any subset of \(U\).
\[ (i)\quad A \cup A' = U \]
A set together with its complement contains every element of the universal set.
\[ (ii)\quad \varnothing' \cap A \]
Since the complement of the empty set is the universal set,
\[ \varnothing' = U \]
\[ U \cap A = A \]
\[ \therefore \quad \varnothing' \cap A = A \]
\[ (iii)\quad A \cap A' = \varnothing \]
A set and its complement have no common elements.
\[ (iv)\quad U' \cap A \]
The complement of the universal set is the empty set:
\[ U' = \varnothing \]
\[ \varnothing \cap A = \varnothing \]
Final Answers
\[ \begin{aligned} (i) &\quad A \cup A' = U \\ (ii) &\quad \varnothing' \cap A = A \\ (iii) &\quad A \cap A' = \varnothing \\ (iv) &\quad U' \cap A = \varnothing \end{aligned} \]
Why This Problem Is Important
- Board Exams: Complement identities frequently appear in CBSE Class 11 Mathematics examinations.
- Competitive Exams: These laws are essential for solving problems in JEE, NEET, NDA, CUET and Olympiads.
- Boolean Logic: These identities form the foundation of Boolean algebra used in computer science and digital electronics.
- Problem Simplification: Complement laws help simplify complex set expressions quickly.
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