OP Malhotra Class 11 Maths Solutions Chapter 1 Sets Exercise 1 (C)

Question 1. Suggest a universal set for each of the following :
(i) {Jaipur, Chennai, Bangalore, Itanagar}
(ii) {Narmada, Cauvery, Mahanadi, Jhelum}
(iii) {Asia, Europe, Antarctica}
(iv) {Earth, Mars, Venus}
(v) given set containing some multiples of 5.
Answer:
(i) The given set contains elements that are state capitals of India. So, a universal set can be the set of all state capitals of India. This set would include all such cities.
(ii) The elements in this set are rivers of India. Therefore, a suitable universal set is the set of all rivers of India.
(iii) The elements in this set are continents. Thus, the universal set can be considered as the set of all continents.
(iv) Each element in the given set is a planet in our solar system. Hence, the universal set can be taken as the set of all planets in our solar system.
(v) The given set contains multiples of 5. A specific universal set could be \( \{x : x = 5p, p \in W, 0 \le p \le 10\} \). This set defines all multiples of 5 from 0 up to 50, which covers many scenarios.
In simple words: For each group, we need to find a larger group that includes all of them. For example, cities are state capitals, rivers are rivers of India, Asia is a continent, Earth is a planet, and 5 is a multiple of 5.

🎯 Exam Tip: When suggesting a universal set, ensure it is broad enough to include all elements of the given sets and is clearly defined by a common property.

Question 2. What universal set may be proposed for the following sets?
(i) The set of parallelograms
(ii) The set of irrational numbers
(iii) The set of positive even numbers
Answer:
(i) The given set is the set of all parallelograms. Since every parallelogram is a type of quadrilateral, the universal set can be the set of all quadrilaterals. Quadrilaterals are four-sided shapes, and parallelograms fit this description perfectly.
(ii) Real numbers include both rational and irrational numbers. So, if we are talking about irrational numbers, the set of all real numbers can be a universal set for them. Real numbers cover all numbers on the number line.
(iii) The given set is the set of all positive even numbers. Since all positive even numbers are natural numbers, the set of all natural numbers can be taken as the universal set. Natural numbers start from 1 and go up endlessly.
In simple words: For parallelograms, quadrilaterals are the bigger group. For irrational numbers, real numbers are the bigger group. For positive even numbers, natural numbers are the bigger group.

🎯 Exam Tip: Always choose the smallest possible universal set that contains all the elements of the given set, without being overly specific or too general.

Question 3. Solve the following equations :
(i) \( \{x | 2x + 6 = 0, x \in Z\} \)
(ii) \( \{x | 5x + 16 = 1, x \in N\} \)
(iii) \( \{x | 2x - 3 < 7, x \in W\} \)
(iv) \( \{x | 4x - 25 > 13, x \in W\} \)
(v) \( \{y | \frac { 5y }{ 3 } - 7 \le 13, y \text{ is a prime number}\} \)
Answer:
(i) Given the equation \( 2x + 6 = 0 \), where \( x \in Z \) (integers).
\( 2x = -6 \)
\( x = -3 \)
Since \( -3 \) is an integer, the solution set is \( \{-3\} \).

(ii) Given the equation \( 5x + 16 = 1 \), where \( x \in N \) (natural numbers).
\( 5x = 1 - 16 \)
\( 5x = -15 \)
\( x = -3 \)
Since \( -3 \) is not a natural number, the solution set is an empty set \( \Phi \). This means there are no natural numbers that satisfy the equation.

(iii) Given the inequality \( 2x - 3 < 7 \), where \( x \in W \) (whole numbers).
\( 2x < 7 + 3 \)
\( 2x < 10 \)
\( x < 5 \)
The whole numbers less than 5 are 0, 1, 2, 3, 4. So, the solution set is \( \{0, 1, 2, 3, 4\} \). Whole numbers start from zero and include all positive integers.

(iv) Given the inequality \( 4x - 25 > 13 \), where \( x \in W \) (whole numbers).
\( 4x > 13 + 25 \)
\( 4x > 38 \)
\( x > \frac{38}{4} \)
\( x > \frac{19}{2} \)
\( x > 9.5 \)
The whole numbers greater than 9.5 are 10, 11, 12, and so on. So, the solution set is \( \{10, 11, 12, \dots\} \).

(v) Given the inequality \( \frac{5y}{3} - 7 \le 13 \), where \( y \) is a prime number.
\( \frac{5y}{3} \le 13 + 7 \)
\( \frac{5y}{3} \le 20 \)
\( 5y \le 20 \times 3 \)
\( 5y \le 60 \)
\( y \le 12 \)
The prime numbers less than or equal to 12 are 2, 3, 5, 7, 11. Prime numbers are numbers greater than 1 that are only divisible by 1 and themselves. So, the solution set is \( \{2, 3, 5, 7, 11\} \).
In simple words: Solve each problem step-by-step. Make sure your answer fits the type of number allowed for 'x' or 'y' (like integers, natural numbers, whole numbers, or prime numbers).

🎯 Exam Tip: Always pay close attention to the specified domain (e.g., integers, natural numbers, whole numbers, prime numbers) for the variable, as this greatly impacts the final solution set.

Question 4. \( \xi = \left\{-2 \frac{1}{2},-1, \sqrt{2}, 3.5, \sqrt{30}, \sqrt{36}\right\} \)
X = {integers}, Y = {irrational numbers}
List the members of (i) X (ii) Y
Answer:
Given universal set \( \xi = \left\{-2 \frac{1}{2},-1, \sqrt{2}, 3.5, \sqrt{30}, \sqrt{36}\right\} \).
Let's simplify the elements in \( \xi \):
\( -2 \frac{1}{2} = -2.5 \)
\( \sqrt{36} = 6 \)
So, \( \xi = \{-2.5, -1, \sqrt{2}, 3.5, \sqrt{30}, 6\} \).

(i) X is the set of integers.
Integers are whole numbers, including negative whole numbers and zero, but not fractions or decimals unless they simplify to a whole number. From \( \xi \), the integers are \( -1 \) and \( 6 \).
So, \( X = \{-1, 6\} \).

(ii) Y is the set of irrational numbers.
Irrational numbers are numbers that cannot be expressed as a simple fraction, meaning their decimal representation is non-terminating and non-repeating. From \( \xi \), the irrational numbers are \( \sqrt{2} \) (approximately 1.414...) and \( \sqrt{30} \) (approximately 5.477...). These numbers go on forever without a pattern.
So, \( Y = \{\sqrt{2}, \sqrt{30}\} \).
In simple words: First, look at all the numbers given. Then, pick out the whole numbers for X, and pick out the numbers that are unending, non-repeating decimals for Y.

🎯 Exam Tip: Remember that square roots of non-perfect squares (like \( \sqrt{2} \), \( \sqrt{30} \)) are irrational, while square roots of perfect squares (like \( \sqrt{36} \)) are rational and often integers.

Question 5. \( \xi = \{40, 41, 42, 43, 44, 45, 46, 47, 48, 49\} \)
A = {prime numbers}, B = {odd numbers}
(i) Place the ten numbers in the correct places on the diagram.
(ii) Write the set B ∩ A.
Answer:
Given \( \xi = \{40, 41, 42, 43, 44, 45, 46, 47, 48, 49\} \).
First, let's identify the members of sets A and B from \( \xi \):
Prime numbers in \( \xi \) are numbers greater than 1 divisible only by 1 and themselves. So, \( A = \{41, 43, 47\} \).
Odd numbers in \( \xi \) are numbers not divisible by 2. So, \( B = \{41, 43, 45, 47, 49\} \).

(i) Placing the numbers in the Venn diagram:
The intersection \( A \cap B \) (numbers that are both prime and odd) is \( \{41, 43, 47\} \).
Numbers in A only (prime but not odd) - None (all primes here are odd except 2, which is not in \( \xi \)).
Numbers in B only (odd but not prime) are \( \{45, 49\} \).
Numbers in \( \xi \) but not in A or B (even numbers) are \( \{40, 42, 44, 46, 48\} \).

(ii) The set \( B \cap A \) represents the elements that are common to both B (odd numbers) and A (prime numbers). These are the numbers that are both odd and prime from the given universal set.
\( B \cap A = \{41, 43, 47\} \).
In simple words: We find which numbers are prime and which are odd from the given list. Then we place them in the correct sections of the Venn diagram: numbers that are both prime and odd go in the middle, numbers that are only odd go in the 'odd' circle but not middle, and other numbers go outside the circles. Finally, we list the numbers that are both prime and odd.

🎯 Exam Tip: When filling Venn diagrams, always start by identifying the intersection (common elements) first, then fill the unique parts of each set, and finally place any remaining elements from the universal set outside the circles.

Question 6. Shade the regions as directed?
(i) A \( \cap \) B
(ii) (A \( \cup \) B)'
(iii) Complementary set B
Answer:
(i) Shading A \( \cap \) B: This region represents the intersection of set A and set B, which means the elements common to both A and B. It is the overlapping area of the two circles.

(ii) Shading (A \( \cup \) B)': This region represents the complement of the union of A and B. It includes all elements in the universal set \( \xi \) that are neither in A nor in B. This means the area outside both circles. The union means "A or B or both".

(iii) Shading Complementary set B (B'): This region includes all elements in the universal set \( \xi \) that are not in set B. It means the area outside circle B but within the universal set rectangle. The complement of a set includes everything not in that set.

In simple words: For A intersect B, shade only where the circles overlap. For (A union B) complement, shade everything outside both circles. For B complement, shade everything outside circle B, including circle A's non-overlapping part.

🎯 Exam Tip: Clearly understand the definitions of intersection (\( \cap \)), union (\( \cup \)), and complement (') to accurately shade Venn diagrams. Complement means 'not in the set'.

Question 7. On the Venn diagrams shade the regions :
(i) A' \( \cap \) C'
(ii) (A \( \cup \) C) \( \cap \) B
Answer:
(i) Shading A' \( \cap \) C': This represents the intersection of the complement of A and the complement of C. In other words, it is the region outside both A and C. This means everything in the universal set except for A and C. The region is the complement of (A union C).

(ii) Shading (A \( \cup \) C) \( \cap \) B: This represents the intersection of the union of A and C with set B. First, we consider the area covered by A or C (or both). Then, we find the part of this area that also overlaps with B. This means shading the parts of B that intersect with A, the parts of B that intersect with C, and the part where all three intersect. This is basically any part of B that is also in A or C.

In simple words: For A' intersect C', shade everything outside both A and C circles. For (A union C) intersect B, shade only the parts of B that are inside A or inside C (or both).

🎯 Exam Tip: When dealing with multiple operations (like union, intersection, complement), perform them in the correct order, usually working from inside parentheses outwards.

Question 8.
(i) On a copy of the Venn diagram, shade the set A \( \cup \) (B \( \cap \) C).
(ii) Express in set notation the subset shaded in the Venn diagram.
(iii) Express in set notation, as simply as possible, the subset shaded in the Venn diagram.
Answer:
(i) Shading A \( \cup \) (B \( \cap \) C): This region includes all elements in set A, plus any elements that are in both B and C. It covers the entire circle A and the overlapping area between B and C. The union means 'A or the intersection of B and C'.

(ii) For the diagram with sets P and Q, if the shaded portion consists of P but does not belong to Q, this is the set difference \( P - Q \). This can also be written as \( P \cap Q' \), where \( Q' \) is the complement of Q. This shows elements that are unique to P.

(iii) For the diagram with sets A, B, C where the intersection of B and C, but outside A, is shaded, this represents \( B \cap C \cap A' \). This means elements that are in both B and C, but are not in A. It is the common part of B and C, excluding any part that overlaps with A.
In simple words: For (i), color all of circle A, and then only the part where B and C overlap. For (ii), if P is shaded but not where it overlaps with Q, it's P minus Q. For (iii), if the overlap of B and C is shaded, but not where A is, it's the intersection of B and C, and outside A.

🎯 Exam Tip: When expressing shaded regions in set notation, mentally break down the shaded area into simpler set operations (union, intersection, complement) and combine them logically.

Question 9. Answer true or false. Refer to the given below figure
(i) A \( \subset \) D
(ii) B \( \subset \) C
(iii) A \( \notin \) B
(iv) C \( \subset \) D
Answer:
Let's analyze the Venn diagram provided:

(i) A \( \subset \) D (A is a subset of D): Looking at the diagram, set A is completely inside set D. This statement is true.

(ii) B \( \subset \) C (B is a subset of C): Looking at the diagram, C is a small circle, and B is a larger circle that contains C. So C is a subset of B, not B of C. This statement is false. All elements of C are in B, but not all elements of B are in C.

(iii) A \( \notin \) B (A is not an element of B): This notation typically refers to an element not being in a set. If it means "A is not a subset of B", it's true because A is not fully contained in B. If it implies that set A itself is not an element within set B (which it isn't, as sets contain elements, not other sets as elements in this context), it's also true. Assuming the common interpretation in Venn diagrams, A is not entirely within B. So, the statement is true.

(iv) C \( \subset \) D (C is a subset of D): Looking at the diagram, set C is completely inside set D. This statement is true. Every element of C is also an element of D.
In simple words: Look at the circles in the picture. If one circle is entirely inside another, it's a subset. If not, it's not a subset. Check each statement to see if it matches the picture.

🎯 Exam Tip: Carefully distinguish between "is a subset of" (\( \subset \)) and "is an element of" (\( \in \)) symbols, and interpret the visual representation of containment correctly in Venn diagrams.

Question 10. Describe the shaded areas in the following Venn diagrams.
(i)
(ii)
(iii)
(iv)
Answer:
(i) The shaded portion consists of elements of A that do not belong to set B. This can be described as \( A \setminus B \) or \( A \cap B' \). This region shows what is in A but not in B. This is also known as "A only".

(ii) The shaded portion consists of all elements that do not belong to set A, set B, and set C, which means they are outside their union. This region can be described as \( (A \cup B \cup C)' \). It represents elements in the universal set that are not in any of the three sets.

(iii) The shaded portion consists of common elements of A and B', and C. This means elements that are in A, not in B, and also in C. This region can be described as \( A \cap B' \cap C \). It is the area where A and C overlap, but specifically excluding any part that also overlaps with B.

(iv) The shaded portion consists of common elements of C' and A, where A is a subset of B. This means elements that are in A, but not in C. Since A is inside B, it's the part of A that does not overlap with C. This can be described as \( A \cap C' \). It is the section of A that is outside C.
In simple words: For (i), it's what's in A but not B. For (ii), it's everything outside all three circles. For (iii), it's where A and C cross, but outside B. For (iv), it's what's in A but not C.

🎯 Exam Tip: When describing shaded regions, identify the sets involved and the logical operations (intersection, union, complement, difference) that define the shaded area. Start from the most specific parts and expand.

Question 11. Use the given diagram to shade the following regions.
(i) A' \( \cap \) B \( \cap \) C
(ii) A' \( \cap \) B \( \cap \) C'
(iii) (A \( \cap \) B \( \cap \) C)'
Answer:
(i) Shading A' \( \cap \) B \( \cap \) C: This region represents elements that are not in A, but are in both B and C. It is the part where B and C overlap, excluding any portion that is also part of A.

(ii) Shading A' \( \cap \) B \( \cap \) C': This region represents elements that are not in A, are in B, and are not in C. It is the part of B that is outside both A and C. It is the section of B unique from A and C's overlap.

(iii) Shading (A \( \cap \) B \( \cap \) C)': This represents the complement of the intersection of A, B, and C. This means it includes all elements in the universal set except for the region where all three sets A, B, and C overlap. It covers almost the entire diagram except for the very center.

In simple words: For (i), shade where B and C cross, but not A. For (ii), shade only the part of B that is outside both A and C. For (iii), shade everything except the very center where A, B, and C all cross paths.

🎯 Exam Tip: Always analyze the innermost operations first (like \( \cap \) or \( \cup \)), then apply complements ('), and then proceed to outer operations, just like in algebra.

Question 12. X = {all triangles}, P = {isosceles triangles}, Z = {equilateral triangles}, Draw a Venn diagram to illustrate the relationship between these sets.
Answer:
We are given three sets:
X = {all triangles}
P = {isosceles triangles}
Z = {equilateral triangles}
We know the relationships between these types of triangles:
\( \implies \) Every equilateral triangle is also an isosceles triangle (since it has at least two equal sides). So, \( Z \subset P \).
\( \implies \) Every isosceles triangle is a type of triangle. So, \( P \subset X \).
\( \implies \) Therefore, \( Z \subset P \subset X \). This means equilateral triangles are a subset of isosceles triangles, which are in turn a subset of all triangles. This nested relationship is clearly shown in the diagram.

In simple words: Equilateral triangles are a special kind of isosceles triangle, and isosceles triangles are a special kind of all triangles. So, the smallest circle for equilateral triangles goes inside the circle for isosceles triangles, which then goes inside the biggest circle for all triangles.

🎯 Exam Tip: When dealing with geometric shapes in Venn diagrams, always recall their definitions and how they relate to each other (e.g., all squares are rectangles, but not all rectangles are squares).

Question 13. Draw a Venn diagram to show the relationship between the following sets :
\( \xi \) = {quadrilaterals}, A = {parallelograms}, B = {rectangles}, C = rhombuses
Show in your diagram the region that represents the set of squares.
Answer:
Given sets:
\( \xi \) = {quadrilaterals}
A = {parallelograms}
B = {rectangles}
C = {rhombuses}
The relationships are:

\( \implies \) Rectangles are parallelograms with all angles equal (90 degrees). So, \( B \subset A \).
\( \implies \) Rhombuses are parallelograms with all sides equal. So, \( C \subset A \).
\( \implies \) Squares are both rectangles and rhombuses (all sides equal AND all angles 90 degrees). So, the set of squares is \( B \cap C \).
\( \implies \) All these shapes are quadrilaterals.
The region representing the set of squares is the intersection of B and C. This is also the smallest common area of B and C, showing that squares inherit properties from both. Squares are perfect examples of these two shapes combined.

In simple words: The big box is for all quadrilaterals. Inside that, draw a large circle for parallelograms (A). Then, draw one circle for rectangles (B) and another for rhombuses (C) inside the parallelogram circle, making sure they overlap. The part where B and C overlap is the region for squares.

🎯 Exam Tip: Always start with the most general set (universal set) and then nest subsets within it. The intersection of overlapping specific sets (like rectangles and rhombuses) often represents a more specialized shape (like squares).

Question 14. Using \( \xi \) = {books}, N = {novels} and D = {detective novels}, represent the following statement as a Venn diagram, 'Some novels are not detective novels.
Answer:
Given:
\( \xi \) = {books} (Universal set)
N = {novels}
D = {detective novels}
The statement 'Some novels are not detective novels' means that there are some novels that exist which do not fall into the category of detective novels. This implies that there is a part of the 'novels' set (N) that is outside the 'detective novels' set (D). It also implies that there are some detective novels that are, by definition, also novels. Therefore, D is a subset of N.
The shaded portion represents the non-detective novels, showing the part of N that is outside D.

In simple words: Draw a big rectangle for all books. Inside it, draw a circle for novels (N). Inside the novel circle, draw a smaller circle for detective novels (D). To show "some novels are not detective novels," shade the part of the novels circle that is outside the detective novels circle.

🎯 Exam Tip: When representing statements like 'Some X are not Y', ensure the shading correctly indicates the part of X that is exclusively X, without including any part of Y.

Question 15. Illustrate by a Venn diagram the rela- tionship between the sets A, B and C given that B \( \subset \) A, C \( \subset \) B' and A \( \cap \) C = \( \Phi \).
Answer:
We need to draw a Venn diagram for sets A, B, and C based on the following conditions:

\( \implies \) B \( \subset \) A: Set B is a subset of set A, meaning all elements of B are also in A. Visually, circle B must be completely inside circle A.

\( \implies \) C \( \subset \) B': Set C is a subset of the complement of B, meaning all elements of C are outside set B. Visually, circle C must not overlap with circle B at all.

\( \implies \) A \( \cap \) C = \( \Phi \): The intersection of A and C is an empty set, meaning there are no common elements between A and C. Visually, circle A and circle C must not overlap. This condition ensures that even though B is inside A, C remains entirely separate from A, consistent with C being outside B.

To satisfy all conditions, B is inside A, C is completely separate from B, and C is also completely separate from A.

In simple words: Draw a big circle A. Draw a smaller circle B completely inside A. Then, draw circle C completely outside circle B and also completely outside circle A. This shows B is inside A, C is outside B, and A and C do not touch.

🎯 Exam Tip: When drawing Venn diagrams from given conditions, address each condition sequentially, building the diagram step-by-step to ensure all rules are satisfied simultaneously. The empty set condition ( \( \Phi \) ) is critical for showing no overlap.

Question 16. If \( \xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \), A = \( \{1, 3, 6, 10\} \), B = \( \{1, 3, 5, 7, 9\} \) and C = \( \{1, 4, 9\} \) then list the elements of the following sets.
(i) A \( \cap \) B
(ii) B \( \cup \) C
(iii) C'
(iv) n(B)
(v) B \( \cap \) C
(vi) A \( \cap \) C
(vii) A \( \cup \) C
(ix) (A \( \cap \) B)'
(x) A – B
(xi) B – A
(xii) (A \( \cup \) B \( \cup \) C)
(xiii) (A \( \cap \) B \( \cap \) C)'
(xiv) A \( \Delta \) B
Answer:
Given: \( \xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \)
A = \( \{1, 3, 6, 10\} \)
B = \( \{1, 3, 5, 7, 9\} \)
C = \( \{1, 4, 9\} \)

(i) A \( \cap \) B (Intersection of A and B): These are the common elements in both A and B.
\( A \cap B = \{1, 3, 6, 10\} \cap \{1, 3, 5, 7, 9\} = \{1, 3\} \).

(ii) B \( \cup \) C (Union of B and C): These are all distinct elements in B or C or both.
\( B \cup C = \{1, 3, 5, 7, 9\} \cup \{1, 4, 9\} = \{1, 3, 4, 5, 7, 9\} \).

(iii) C' (Complement of C): These are all elements in \( \xi \) that are not in C.
\( C' = \xi \setminus C = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \setminus \{1, 4, 9\} = \{2, 3, 5, 6, 7, 8, 10\} \).

(iv) n(B) (Number of elements in B): This is the count of distinct elements in set B.
\( n(B) = 5 \).

(v) B \( \cap \) C (Intersection of B and C): These are the common elements in both B and C.
\( B \cap C = \{1, 3, 5, 7, 9\} \cap \{1, 4, 9\} = \{1, 9\} \).

(vi) A \( \cap \) C (Intersection of A and C): These are the common elements in both A and C.
\( A \cap C = \{1, 3, 6, 10\} \cap \{1, 4, 9\} = \{1\} \).

(vii) A \( \cup \) C (Union of A and C): These are all distinct elements in A or C or both.
\( A \cup C = \{1, 3, 6, 10\} \cup \{1, 4, 9\} = \{1, 3, 4, 6, 9, 10\} \).

(ix) (A \( \cap \) B)' (Complement of A \( \cap \) B): These are all elements in \( \xi \) that are not in \( A \cap B \). First, find \( A \cap B = \{1, 3\} \).
\( (A \cap B)' = \xi \setminus \{1, 3\} = \{2, 4, 5, 6, 7, 8, 9, 10\} \).

(x) A – B (Set difference A minus B): These are elements in A but not in B.
\( A \setminus B = \{1, 3, 6, 10\} \setminus \{1, 3, 5, 7, 9\} = \{6, 10\} \).

(xi) B – A (Set difference B minus A): These are elements in B but not in A.
\( B \setminus A = \{1, 3, 5, 7, 9\} \setminus \{1, 3, 6, 10\} = \{5, 7, 9\} \).

(xii) (A \( \cup \) B \( \cup \) C) (Union of A, B, and C): These are all distinct elements in A, B, or C.
\( A \cup B \cup C = \{1, 3, 6, 10\} \cup \{1, 3, 5, 7, 9\} \cup \{1, 4, 9\} = \{1, 3, 4, 5, 6, 7, 9, 10\} \).

(xiii) (A \( \cap \) B \( \cap \) C)' (Complement of A \( \cap \) B \( \cap \) C): These are all elements in \( \xi \) that are not in \( A \cap B \cap C \). First, find \( A \cap B \cap C = \{1, 3, 6, 10\} \cap \{1, 3, 5, 7, 9\} \cap \{1, 4, 9\} = \{1\} \).
\( (A \cap B \cap C)' = \xi \setminus \{1\} = \{2, 3, 4, 5, 6, 7, 8, 9, 10\} \).

(xiv) A \( \Delta \) B (Symmetric difference of A and B): These are elements that are in A or B but not in their intersection. It can also be found as \( (A \setminus B) \cup (B \setminus A) \).
From (x), \( A \setminus B = \{6, 10\} \).
From (xi), \( B \setminus A = \{5, 7, 9\} \).
\( A \Delta B = \{6, 10\} \cup \{5, 7, 9\} = \{5, 6, 7, 9, 10\} \).
In simple words: For each item, we combine or separate the numbers from sets A, B, and C based on the rules. Intersection means common numbers. Union means all numbers together, without repeats. Complement means numbers in the main universal set but not in the specific set. Difference means numbers in the first set but not the second. Symmetric difference means numbers that are in one set or the other, but not in both.

🎯 Exam Tip: Systematically list all elements of each set first. Then, apply the set operations step-by-step. For complements, always refer back to the universal set \( \xi \).

Question 17. Let L = {Setters of CRICKET}, M = {Setters of CATERPILLAR} and N = {Setters of CARE TAKER} find : (i) L \( \cup \) M (ii) M \( \cup \) N (iii) L \( \cup \) N
Answer:
First, let's list the distinct letters in each word to form the sets:
L = {letters of CRICKET} = {C, R, I, K, E, T}
M = {letters of CATERPILLAR} = {C, A, T, E, R, P, I, L}
N = {letters of CARE TAKER} = {C, A, R, E, T, K}

(i) L \( \cup \) M (Union of L and M): This set contains all distinct letters that are in L or M or both.
\( L \cup M = \{C, R, I, K, E, T\} \cup \{C, A, T, E, R, P, I, L\} = \{C, A, T, E, R, P, I, L, K\} \).

(ii) M \( \cup \) N (Union of M and N): This set contains all distinct letters that are in M or N or both.
\( M \cup N = \{C, A, T, E, R, P, I, L\} \cup \{C, A, R, E, T, K\} = \{C, A, T, E, R, P, I, L, K\} \).

(iii) L \( \cup \) N (Union of L and N): This set contains all distinct letters that are in L or N or both.
\( L \cup N = \{C, R, I, K, E, T\} \cup \{C, A, R, E, T, K\} = \{C, R, I, K, E, T, A\} \).
In simple words: For each part, we combine the letters from the given words. We only list each unique letter once, even if it appears in both words.

🎯 Exam Tip: When forming sets from words, make sure to list only the *distinct* letters. For union operations, list all unique elements present in any of the combined sets.

Question 18. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) \( \{x : x \text{ is a positive multiple of } 3\} \)
(ii) \( \{x: x \text{ is a prime number}\} \)
(iii) \( \{x : x \text{ is a natural number divisible by } 3 \text{ and } 5\} \)
(iv) \( \{x: x + 5 = 8\} \)
(v) \( \{x : 2x + 5 = 9\} \)
(vi) \( \{x: x \ge 7\} \)
(vii) \( \{x: x \in N \text{ and } 2x + 1 > 10\} \)
Answer:
The universal set \( \xi \) is N = {1, 2, 3, ...} (the set of natural numbers).

(i) Let A = \( \{x: x \text{ is a positive multiple of } 3\} = \{3, 6, 9, 12, \dots\} \).
The complement A' is \( \{x: x \text{ is not a positive multiple of } 3\} \). This means it includes all natural numbers except those that can be divided evenly by 3.

(ii) Let B = \( \{x: x \text{ is a prime number}\} = \{2, 3, 5, 7, 11, \dots\} \).
The complement B' is \( \{x: x \text{ is not a prime number}\} \). This includes 1 and all composite numbers (numbers with more than two factors) from the natural numbers. For example, 1, 4, 6, 8, 9, 10, etc.

(iii) Let C = \( \{x: x \text{ is a natural number divisible by } 3 \text{ and } 5\} \).
A number divisible by both 3 and 5 is divisible by their least common multiple, which is 15. So, C = \( \{x: x \text{ is a natural number divisible by } 15\} = \{15, 30, 45, \dots\} \).
The complement C' is \( \{x: x \text{ is a natural number which is not divisible by } 15\} \). This includes all natural numbers that are not a multiple of 15.

(iv) Let D = \( \{x: x + 5 = 8\} \).
Solving the equation \( x + 5 = 8 \implies x = 3 \). So, D = \( \{3\} \).
The complement D' is \( \{x: x \in N \text{ and } x \ne 3\} \). This includes all natural numbers except for the number 3.

(v) Let E = \( \{x: 2x + 5 = 9\} \).
Solving the equation \( 2x + 5 = 9 \implies 2x = 4 \implies x = 2 \). So, E = \( \{2\} \).
The complement E' is \( \{x: x \in N \text{ and } x \ne 2\} \). This includes all natural numbers except for the number 2.

(vi) Let F = \( \{x: x \ge 7\} \).
This set includes all natural numbers greater than or equal to 7: F = \( \{7, 8, 9, 10, \dots\} \).
The complement F' is \( \{x: x \in N \text{ and } x < 7\} \). This includes all natural numbers less than 7: F' = \( \{1, 2, 3, 4, 5, 6\} \).

(vii) Let G = \( \{x: x \in N \text{ and } 2x + 1 > 10\} \).
Solving the inequality \( 2x + 1 > 10 \implies 2x > 9 \implies x > \frac{9}{2} \implies x > 4.5 \).
Since \( x \) must be a natural number, G = \( \{5, 6, 7, 8, \dots\} \).
The complement G' is \( \{x: x \in N \text{ and } 2x + 1 \le 10\} \). This means \( x \le 4.5 \).
So, G' = \( \{1, 2, 3, 4\} \). This includes natural numbers up to 4.
In simple words: For each set, we find all the numbers that are *not* in that set, but are still natural numbers. For example, if a set has multiples of 3, its complement has all natural numbers that are *not* multiples of 3.

🎯 Exam Tip: When finding complements, always remember the universal set you are working within (here, natural numbers). Clearly state the condition for the complement in set-builder notation if asked.

Question 19. Refer to the Venn diagram. List the elements of the following sets.
(i) \( (A \cap B) \cup C \)
(ii) \( A \cap (B \cup C) \)
(iii) \( (A \cap C) \cap (B \cap C) \)
(iv) \( (A \cup B) \cap (B \cap C) \)
(v) \( B - C \)
(vi) \( C - B \)
(vii) \( B \Delta C \)
Answer: From the given Venn diagram, we can identify the elements of each set:
\( \xi = \{-4, -3, -2, -1, 0, 3, 4, 5, 6\} \)
\( A = \{-4, -3, -2, -1, 3, 4\} \)
\( B = \{0, -2, -1, 3\} \)
\( C = \{3, 4, 5, 6\} \)

(i) To find \( (A \cap B) \cup C \):
First, find the intersection of A and B: \( A \cap B = \{-2, -1, 3\} \)
Next, find the union of \( (A \cap B) \) and C: \( (A \cap B) \cup C = \{-2, -1, 3\} \cup \{3, 4, 5, 6\} = \{-2, -1, 3, 4, 5, 6\} \)

(ii) To find \( A \cap (B \cup C) \):
First, find the union of B and C: \( B \cup C = \{0, -2, -1, 3\} \cup \{3, 4, 5, 6\} = \{0, -2, -1, 3, 4, 5, 6\} \)
Next, find the intersection of A and \( (B \cup C) \): \( A \cap (B \cup C) = \{-4, -3, -2, -1, 3, 4\} \cap \{0, -2, -1, 3, 4, 5, 6\} = \{-2, -1, 3, 4\} \)

(iii) To find \( (A \cap C) \cap (B \cap C) \):
First, find the intersection of A and C: \( A \cap C = \{-4, -3, -2, -1, 3, 4\} \cap \{3, 4, 5, 6\} = \{3, 4\} \)
Next, find the intersection of B and C: \( B \cap C = \{0, -2, -1, 3\} \cap \{3, 4, 5, 6\} = \{3\} \)
Finally, find the intersection of \( (A \cap C) \) and \( (B \cap C) \): \( (A \cap C) \cap (B \cap C) = \{3, 4\} \cap \{3\} = \{3\} \)

(iv) To find \( (A \cup B) \cap (B \cap C) \):
First, find the union of A and B: \( A \cup B = \{-4, -3, -2, -1, 3, 4\} \cup \{0, -2, -1, 3\} = \{-4, -3, -2, -1, 0, 3, 4\} \)
Next, find the intersection of B and C: \( B \cap C = \{3\} \)
Finally, find the intersection of \( (A \cup B) \) and \( (B \cap C) \): \( (A \cup B) \cap (B \cap C) = \{-4, -3, -2, -1, 0, 3, 4\} \cap \{3\} = \{3\} \)

(v) To find \( B - C \):
\( B - C = \{x \mid x \in B \text{ and } x \notin C\} = \{0, -2, -1, 3\} - \{3, 4, 5, 6\} = \{0, -2, -1\} \)

(vi) To find \( C - B \):
\( C - B = \{x \mid x \in C \text{ and } x \notin B\} = \{3, 4, 5, 6\} - \{0, -2, -1, 3\} = \{4, 5, 6\} \)

(vii) To find \( B \Delta C \) (symmetric difference):
\( B \Delta C = (B - C) \cup (C - B) \)
Using results from (v) and (vi): \( B \Delta C = \{0, -2, -1\} \cup \{4, 5, 6\} = \{0, -2, -1, 4, 5, 6\} \)
In simple words: Look at the diagram and list out all the numbers in each set. Then, follow the rules for intersection (common numbers), union (all numbers), and difference (numbers in one set but not the other) to find the elements for each new set. The symmetric difference means all elements that are in either set, but not in their common part.

🎯 Exam Tip: Always list out the elements of each set clearly from the Venn diagram before performing any set operations. This helps avoid errors, especially with negative numbers.

Question 20. If \( \xi = \{1, 2, 3, 4\} \), \( A = \{1, 4\} \), \( B = \{1, 3\} \), then list the elements of
(i) \( A' \)
(ii) \( B' \)
(iii) \( (A \cap B)' \)
(iv) \( (A \cup B)' \)
(v) \( A' \cap B' \)
(vi) \( A' \cup B' \)
Also show that \( (A \cap B)' = A' \cup B' \) and \( (A \cup B)' = A' \cap B' \)
Answer: Given the universal set \( \xi = \{1, 2, 3, 4\} \), set \( A = \{1, 4\} \), and set \( B = \{1, 3\} \).

(i) To find \( A' \) (complement of A):
\( A' = \xi - A = \{1, 2, 3, 4\} - \{1, 4\} = \{2, 3\} \)

(ii) To find \( B' \) (complement of B):
\( B' = \xi - B = \{1, 2, 3, 4\} - \{1, 3\} = \{2, 4\} \)

(iii) To find \( (A \cap B)' \):
First, find the intersection of A and B: \( A \cap B = \{1, 4\} \cap \{1, 3\} = \{1\} \)
Next, find the complement of \( (A \cap B) \): \( (A \cap B)' = \xi - \{1\} = \{2, 3, 4\} \)

(iv) To find \( (A \cup B)' \):
First, find the union of A and B: \( A \cup B = \{1, 4\} \cup \{1, 3\} = \{1, 3, 4\} \)
Next, find the complement of \( (A \cup B) \): \( (A \cup B)' = \xi - \{1, 3, 4\} = \{2\} \)

(v) To find \( A' \cap B' \):
Using results from (i) and (ii): \( A' \cap B' = \{2, 3\} \cap \{2, 4\} = \{2\} \)

(vi) To find \( A' \cup B' \):
Using results from (i) and (ii): \( A' \cup B' = \{2, 3\} \cup \{2, 4\} = \{2, 3, 4\} \)

Verification of De Morgan's Laws:
1. Show that \( (A \cap B)' = A' \cup B' \):
From (iii), we have \( (A \cap B)' = \{2, 3, 4\} \).
From (vi), we have \( A' \cup B' = \{2, 3, 4\} \).
Since both sides are equal, \( (A \cap B)' = A' \cup B' \) is shown to be true.

2. Show that \( (A \cup B)' = A' \cap B' \):
From (iv), we have \( (A \cup B)' = \{2\} \).
From (v), we have \( A' \cap B' = \{2\} \).
Since both sides are equal, \( (A \cup B)' = A' \cap B' \) is shown to be true.
In simple words: The complement of a set means all the items in the universal set that are not in that specific set. When you combine operations, remember that the complement of an intersection is the union of the complements, and the complement of a union is the intersection of the complements. These are known as De Morgan's Laws.

🎯 Exam Tip: De Morgan's Laws are fundamental in set theory. Always write down the universal set, the given sets, and their complements clearly before applying the rules. This step-by-step approach ensures accuracy in calculations.