RBSE Solutions Class 11 Maths Chapter 1 Sets More Questions

 

Question 1. Solution of equation \(x^2 + x - 2 = 0\) in Roster form is:
(A) {1, 2}
(B) {-1, 2}
(C) {-1, -2}
(D) {1, -2}
Answer: (D) {1, -2}
In simple words: To solve the equation, we factor it to find the values of x. These values are the elements that form the set in roster form. The solutions here are x = -2 and x = 1.

🎯 Exam Tip: Always double-check your factorization by expanding the factors to ensure they match the original quadratic equation.

 

Question 2. Which of the following sets represents the vowels of the English alphabet?
(A) {a, b, c, d, e}
(B) {e, i, o, u}
(C) {a, e, i, o}
(D) {a, e, i, o, u}
Answer: (D) {a, e, i, o, u}
In simple words: The vowels are specific letters of the alphabet that make an open sound without the tongue touching the roof of the mouth. There are five such letters in English.

🎯 Exam Tip: Remember the five main English vowels (a, e, i, o, u) for quick set identification tasks.

 

Question 3. Set builder form of set A = {1, 4, 9, 16, 25,.......} will be:
(A) {x: x is an odd natural no.)
(B) {x: x is an even natural no.}
(C) {x: x is square of natural no.}
(D) {x: x is a prime natural no.}
Answer: (C) {x: x is square of natural no.}
In simple words: Look at the numbers in the set: 1 is \(1^2\), 4 is \(2^2\), 9 is \(3^2\), and so on. This pattern shows that each number is the square of a natural number.

🎯 Exam Tip: To convert to set builder form, identify the common property or rule that all elements in the set follow.

 

Question 4. Which set is infinite of the following:
(A) {x: x \( \in \) N and (x - 1)(x - 2) = 0}
(B) {x: x \( \in \) N and \(x^2 = 4\)}
(C) {x: x \( \in \) N and 2x - 1 = 0}
(D) {x: x \( \in \) N and x is a prime no.}
Answer: (D) {x: x \( \in \) N and x is a prime no.}
In simple words: An infinite set has endless elements. Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves, and there is no largest prime number. Options A, B, and C result in finite sets with specific solutions.

🎯 Exam Tip: Remember that natural numbers are positive integers (1, 2, 3...), and prime numbers continue indefinitely, making their set infinite.

 

Question 5. If A = {0}, B = {x : x > 15 and x < 5}, C = {x : x - 5 = 0}, D = {x : \(x^2 = 25\)}, E = {x : x is a positive integer root of equation \(x^2 - 2x - 15 = 0\)}, then pair of equal sets is:
(A) А, В
(B) B, C
(C) B, C
(D) C, E
Answer: (D) C, E
In simple words: First, find the elements for each set. Set B is empty because no number is both greater than 15 and less than 5. Set C contains only 5. Set E also contains only 5 as its positive integer root. Since C and E have the exact same elements, they are equal.

🎯 Exam Tip: Convert all sets from set-builder form to roster form first to easily compare their elements for equality.

 

Question 6. For sets \( \Phi \), A = {1, 3}, B = {1, 5, 9}, C = {1, 5, 7, 9} True option is:
(A) A \( \subset \) B
(B) B \( \subset \) C
(C) C \( \subset \) B
(D) A \( \subset \) C
Answer: (B) B \( \subset \) C
In simple words: A set is a subset of another if all its elements are also present in the other set. All the numbers in set B ({1, 5, 9}) are also found in set C ({1, 5, 7, 9}), making B a subset of C.

🎯 Exam Tip: To check if a set is a subset of another, ensure every single element of the first set is present in the second set.

 

Question 7. If A = {2, 4, 6, 8} and B = {1, 4, 7, 8} then A – B and B – A will be respectively:
(A) {2, 6}; {1, 7}
(B) {1, 7}; {4, 8}
(C) {1, 7}; {2, 6}
(D) {4, 8}; {1, 7}
Answer: (A) {2, 6}; {1, 7}
In simple words: A - B means the elements that are in A but not in B. For set B - A, it means the elements that are in B but not in A. After finding these unique elements for both operations, we get {2, 6} and {1, 7}.

🎯 Exam Tip: Set difference (A - B) means elements exclusively in A. Do not include elements common to both sets.

 

Question 8. Which of the following statement is true?
(A) A \( \cap \) B = \( \Phi \implies \) A = \( \Phi \) and B = \( \Phi \)
(B) A – B = \( \Phi \implies \) A \( \subset \) B
(C) A \( \cup \) B = \( \Phi \implies \) A \( \subset \) B
(D) None of these
Answer: (B) A – B = \( \Phi \implies \) A \( \subset \) B
In simple words: If A - B is an empty set, it means there are no elements in A that are not also in B. This condition directly implies that every element of A must be in B, which is the definition of A being a subset of B.

🎯 Exam Tip: Understand the definitions: A - B = \( \Phi \) means A has no unique elements, so A must be contained within B.

 

Question 9. If A \( \cap \) B = \( \Phi \) then:
(A) A – B = \( \Phi \)
(B) A - B = A
Answer: (B) A - B = A
In simple words: If the intersection of sets A and B is empty, it means A and B have no common elements. So, when you remove elements of B from A (A - B), you are not removing anything from A, because there's nothing common to remove. Therefore, A - B simply remains A.

🎯 Exam Tip: When two sets are disjoint (their intersection is empty), subtracting one from the other leaves the original set unchanged.

 

Question 10. Shaded portion of the following Venn diagram represents:

(A) A \( \cup \) B
(B) A \( \cap \) B
(C) A - B
(D) B - A
Answer: (C) A - B
In simple words: The shaded part includes all the elements that belong only to set A and none of the elements that are also in set B. This clearly shows the set difference A - B.

🎯 Exam Tip: For Venn diagrams, A - B represents the region of circle A that does not overlap with circle B.

 

Question 11. If A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, then A – B will be:
(A) {1, 3, 5, 8}
(B) {1, 3, 5}
(C) {1, 2, 3, 4, 5, 6, 8}
(D) { }
Answer: (B) {1, 3, 5}
In simple words: To find A - B, we take all the numbers in set A and remove any numbers that are also in set B. The numbers {2, 4, 6} are in both, so removing them from A leaves us with {1, 3, 5}.

🎯 Exam Tip: When performing set difference, carefully compare each element of the first set against the second set to identify unique elements.

 

Question 13. If U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5} then:
(A) (A \( \cup \) B)' = {2, 3, 4, 5}
(B) B - A = {4, 5}
(C) A – B = {2, 4, 5}
(D) (A \( \cup \) B) = {3}
Answer: (B) B - A = {4, 5}
In simple words: We want elements present in B but not in A. Set B is {3, 4, 5} and set A is {2, 3}. The element 3 is common. So, removing 3 from B leaves {4, 5}.

🎯 Exam Tip: Always list the elements of each set clearly before performing any set operations to avoid errors.

 

Question 14. The shaded portion of the following Venn diagram represents:

(A) (A \( \cap \) B) \( \cap \) C
(B) (A \( \cup \) B) \( \cap \) C
(C) (A \( \cap \) B) \( \cup \) C
(D) (A \( \cap \) B) \( \cup \) (A \( \cap \) C)
Answer: (D) (A \( \cap \) B) \( \cup \) (A \( \cap \) C)
In simple words: The diagram shows two distinct shaded regions: the area where set A and set B overlap (A \( \cap \) B), and the area where set A and set C overlap (A \( \cap \) C). The total shaded area is the combination (union) of these two overlapping parts.

🎯 Exam Tip: When a Venn diagram shows separate, non-contiguous shaded regions, it usually represents a union of those individual regions.

 

Question 15. Which of Venn diagram represents set A – (B \( \cap \) C) is:

Answer: The diagram (B) represents set A – (B \( \cap \) C).
In simple words: This diagram should show all parts of set A shaded, except for the tiny region where A, B, and C all overlap. This is because B \( \cap \) C is the common area of B and C, and we are removing any part of this common area that falls within A.

🎯 Exam Tip: When visualizing A - (B \( \cap \) C), first identify the (B \( \cap \) C) region, then shade all of A while making sure to exclude any portion of (B \( \cap \) C) that overlaps with A.

 

Question 16. Which of the following are sets? Justify your answer.
(i) Collection of even natural less than 8.
(ii) The collection of big cities in India.
(iii) The collection of various geometrical figures.
(iv) The collection of all the integers which divides 46.
(v) The collection of the best 20 cricket batsmen of the world.
(vi) The collection of all even integers.
(vii) The collection of literature written by the poet Kalidas.
(viii) The collection of Greatmen contributed to Indian culture.
Answer:
(i) Even natural numbers less than 8 are {2, 4, 6}. This is a set because its elements are clearly defined and can be easily identified. For example, 10 is not in this collection.
(ii) The collection of big cities in India is not a set because the term "big" is subjective. What one person considers a big city, another might not. There is no clear rule to include or exclude cities.
(iii) The collection of various geometrical figures is not a set because "various" is not a specific criterion. It's unclear which figures should be included and which should be left out. A collection must be well-defined.
(iv) The numbers which divide 46 are { -46, -23, -2, -1, 1, 2, 23, 46}. This is a set because the divisors of 46 are precisely defined, making it a well-defined collection. Divisors include both positive and negative integers.
(v) The collection of the best 20 cricket batsmen of the world is not a set because "best" is subjective. Different people will have different opinions on who the best batsmen are, so the collection cannot be clearly defined.
(vi) The collection of all even integers is a set. Even integers are numbers that can be divided by 2 with no remainder. This rule is very clear, so we can definitively say whether any integer is part of this collection or not. For example, {..., -4, -2, 0, 2, 4, ...}.
(vii) The collection of literature written by the poet Kalidas is a set. The author's works are specific and verifiable, so this collection is well-defined and unambiguous. We can definitively identify which pieces of literature belong to this collection.
(viii) The collection of Greatmen contributed to Indian culture is not a set because "Greatmen" is a subjective term. There is no universal agreement on who qualifies as a "Greatman," making the collection ill-defined.
In simple words: A collection is a set only if its members are very clearly defined, with no room for different opinions. If you can easily tell what belongs and what doesn't, it's a set. If it depends on someone's feeling or opinion, it's not a set.

🎯 Exam Tip: For a collection to be a set, its elements must be well-defined, meaning there is no ambiguity about whether an object belongs to the collection or not.

 

Question 17. Write the following sets in roster form.
(i) A = {x : x \( \in \) N, 2 \( \le \) x \( \le \) 9}
(ii) B = {x : x is a two digit natural number sum of whose digits is 6}
(iii) C = {set of all the letters of word MATHEMATICS}
(iv) D = {x : x is a prime number less than 50}
Answer:
(i) For set A, we need natural numbers (1, 2, 3...) that are between 2 and 9, including 2 and 9.
\( \implies \) A = {2, 3, 4, 5, 6, 7, 8, 9}.
(ii) For set B, we need two-digit natural numbers where the sum of their digits is 6.
\( \implies \) The numbers are 15 (1+5=6), 24 (2+4=6), 33 (3+3=6), 42 (4+2=6), and 51 (5+1=6).
\( \implies \) B = {15, 24, 33, 42, 51}.
(iii) For set C, we list all the unique letters from the word "MATHEMATICS". Repeating letters are only listed once.
\( \implies \) C = {M, A, T, H, E, I, C, S}.
(iv) For set D, we list prime numbers (numbers only divisible by 1 and themselves) that are less than 50.
\( \implies \) D = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}. Prime numbers are the building blocks of other numbers through multiplication.
In simple words: Roster form means listing all the items of a set inside curly brackets { } with commas in between. For each part, we just need to find all the numbers or letters that fit the rule given.

🎯 Exam Tip: Always check for unique elements when listing letters from a word and ensure all conditions (e.g., "two-digit", "prime") are met for numerical sets.

 

Question 18. If A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4} and C = {4, 6, 8, 10} then insert appropriate symbols:
(i) 4...A, 5.....B
(ii) 2..A, 4...C
(iii) B...A, A....C
(iv) A - B...C
(v) A....B = B
(vi) B - C,...{2}
(vii) B \( \cap \) C = {....}
(viii) B \( \cup \) C - A = {....}
Answer:
(i) For 4...A, we check if 4 is an element of A. Yes, 4 \( \in \) A. For 5.....B, we check if 5 is an element of B. No, 5 \( \notin \) B.
\( \implies \) 4 \( \in \) A, 5 \( \notin \) B.
(ii) For 2..A, we check if 2 is an element of A. Yes, 2 \( \in \) A. For 4...C, we check if 4 is an element of C. Yes, 4 \( \in \) C.
\( \implies \) 2 \( \in \) A, 4 \( \in \) C.
(iii) For B...A, we check if B is a subset of A. All elements of B ({2, 3, 4}) are in A. Yes, B \( \subset \) A. For A....C, we check if A is a subset of C. Not all elements of A are in C. No, A \( \not\subset \) C.
\( \implies \) B \( \subset \) A, A \( \not\subset \) C.
(iv) First, find A - B. A - B = {1, 2, 3, 4, 5, 6} - {2, 3, 4} = {1, 5, 6}. Now, compare {1, 5, 6} with C = {4, 6, 8, 10}. Are all elements of {1, 5, 6} in C? No. So {1, 5, 6} \( \not\subset \) C.
\( \implies \) A - B \( \not\subset \) C.
(v) To make A....B = B true, the operation must be intersection. A \( \cap \) B = {1, 2, 3, 4, 5, 6} \( \cap \) {2, 3, 4} = {2, 3, 4}. Since {2, 3, 4} is equal to B, then A \( \cap \) B = B is correct. This shows that B is a subset of A.
\( \implies \) A \( \cap \) B = B.
(vi) First, find B - C. B - C = {2, 3, 4} - {4, 6, 8, 10} = {2, 3}. Now, compare {2, 3} with {2}. They are not equal.
\( \implies \) B - C \( \neq \) {2}.
(vii) Find B \( \cap \) C. B \( \cap \) C = {2, 3, 4} \( \cap \) {4, 6, 8, 10} = {4}.
\( \implies \) B \( \cap \) C = {4}.
(viii) First, find B \( \cup \) C. B \( \cup \) C = {2, 3, 4} \( \cup \) {4, 6, 8, 10} = {2, 3, 4, 6, 8, 10}. Next, find (B \( \cup \) C) - A. (B \( \cup \) C) - A = {2, 3, 4, 6, 8, 10} - {1, 2, 3, 4, 5, 6} = {8, 10}.
\( \implies \) B \( \cup \) C - A = {8, 10}.
In simple words: We need to use symbols like 'is an element of' (\( \in \)), 'is not an element of' (\( \notin \)), 'is a subset of' (\( \subset \)), 'is not a subset of' (\( \not\subset \)), 'equals' (=), or 'not equals' (\( \neq \)). For each blank, we check the relationship between the set or element and the other set.

🎯 Exam Tip: Clearly write out the roster form of any derived sets (like A - B or B \( \cup \) C) before comparing them, to avoid mistakes.

 

Question 19. Give two examples of each of the following:
(i) Null set
(ii) Finite set
(iii) Infinite set
(iv) Universal set
Answer:
(i) Null set (or empty set): A set with no elements. It is represented by \( \Phi \) or { }.
(a) The set of natural numbers less than 1. (There are no natural numbers less than 1).
(b) The set of positive natural numbers that lie between 2 and 3. (There are no natural numbers between 2 and 3).
(ii) Finite set: A set with a definite, countable number of elements.
(a) The set of prime numbers less than 10. ({2, 3, 5, 7}).
(b) The set of all months in a year. ({January, February, ..., December}).
(iii) Infinite set: A set with an unlimited, uncountable number of elements.
(a) The set of all even natural numbers. ({2, 4, 6, 8, ...}).
(b) The set of all rational numbers. (Any number that can be written as a fraction p/q, where p and q are integers and q \( \neq \) 0).
(iv) Universal set: A set that contains all elements relevant to a particular context or discussion. It is denoted by U.
(a) If we are talking about students in a specific school, the universal set would be all the students in that school. This set contains every possible element in a particular context.
(b) If we are discussing different types of numbers like integers, natural numbers, and rational numbers, the universal set could be the set of all real numbers.
In simple words: A null set is empty. A finite set has a countable number of items. An infinite set has endless items. A universal set is the big set that holds all the possible items we are interested in for a particular problem.

🎯 Exam Tip: Always use well-defined criteria for your examples to ensure they clearly illustrate the properties of each set type.

 

Question 20. If A = {a, b, c, d}, B = {p, q, r} and C = {a, b, p, q}, then examine the validity of the following:
(i) A - B = C
(ii) B - C \( \neq \) A
(iii) B - A = \( \Phi \)
(iv) (A \( \cup \) B) – C = {c, d, r}
(v) (A \( \cup \) B) \( \cap \) C = C
Answer:
Given sets: A = {a, b, c, d}, B = {p, q, r}, C = {a, b, p, q}.
(i) Examine A - B = C:
First, calculate A - B: A - B = {a, b, c, d} - {p, q, r} = {a, b, c, d}.
Now, compare with C: C = {a, b, p, q}.
Since {a, b, c, d} \( \neq \) {a, b, p, q}, the statement A - B = C is False.
(ii) Examine B - C \( \neq \) A:
First, calculate B - C: B - C = {p, q, r} - {a, b, p, q} = {r}.
Now, compare with A: A = {a, b, c, d}.
Since {r} \( \neq \) {a, b, c, d}, the statement B - C \( \neq \) A is True.
(iii) Examine B - A = \( \Phi \):
First, calculate B - A: B - A = {p, q, r} - {a, b, c, d} = {p, q, r}.
Now, compare with \( \Phi \): \( \Phi \) is the empty set. {p, q, r} is not empty.
Since {p, q, r} \( \neq \Phi \), the statement B - A = \( \Phi \) is False.
(iv) Examine (A \( \cup \) B) – C = {c, d, r}:
First, calculate A \( \cup \) B: A \( \cup \) B = {a, b, c, d} \( \cup \) {p, q, r} = {a, b, c, d, p, q, r}.
Next, calculate (A \( \cup \) B) - C: {a, b, c, d, p, q, r} - {a, b, p, q} = {c, d, r}.
Now, compare with {c, d, r}. Since {c, d, r} = {c, d, r}, the statement (A \( \cup \) B) – C = {c, d, r} is True.
(v) Examine (A \( \cup \) B) \( \cap \) C = C:
From (iv), A \( \cup \) B = {a, b, c, d, p, q, r}.
Next, calculate (A \( \cup \) B) \( \cap \) C: {a, b, c, d, p, q, r} \( \cap \) {a, b, p, q} = {a, b, p, q}.
Now, compare with C. Since {a, b, p, q} = C, the statement (A \( \cup \) B) \( \cap \) C = C is True.
In simple words: To check if each statement is correct, we first perform the set operations (like union, intersection, or difference) on the left side. Then, we compare the result with the set or condition given on the right side. This tells us if the statement is true or false.

🎯 Exam Tip: Work through each side of the equality or inequality separately, simplifying expressions to their roster form, then compare the final sets for validity.

 

Question 21. State whether the following statement is true or false: If A = \( \Phi \), then P(A) contains only one element.
Answer: True
In simple words: The power set P(A) is the set of all possible subsets of A. If A is an empty set, its only subset is the empty set itself. So, P(A) will be { \( \Phi \) }, which contains only one element.

🎯 Exam Tip: Remember that the power set of an empty set is a set containing only the empty set, not an empty set itself.

 

Question 22. Write the following sets in interval form.
(i) {x: x \( \in \) R, a < x < b}
(ii) {x: x \( \in \) R, 3 < x \( \le \) 5}
(iii) {x: x \( \in \) R, 0 \( \le \) x < 8}
(iv) {x: x \( \in \) R, -1 \( \le \) x \( \le \) 5}
Answer:
(i) For {x: x \( \in \) R, a < x < b}, the numbers are strictly between a and b, not including a or b.
\( \implies \) (a, b).
(ii) For {x : x \( \in \) R, 3 < x \( \le \) 5}, the numbers are strictly greater than 3 but less than or equal to 5.
\( \implies \) (3, 5].
(iii) For {x : x \( \in \) R, 0 \( \le \) x < 8}, the numbers are greater than or equal to 0 but strictly less than 8.
\( \implies \) [0, 8).
(iv) For {x: x \( \in \) R, -1 \( \le \) x \( \le \) 5}, the numbers are greater than or equal to -1 and less than or equal to 5.
\( \implies \) [-1, 5].
In simple words: Interval form is a shorthand way to write a range of numbers. Round brackets () mean "not including the number," while square brackets [] mean "including the number." We use these to show the start and end of the range.

🎯 Exam Tip: Pay close attention to the inequality signs (<, \( \le \)) to correctly use round or square brackets in interval notation.

 

Question 23. Write the following intervals in set builder form.
(i) (2, 5)
(ii) [0, 7)
(iii) [2, 10]
(iv) (-5, 0]
Answer:
(i) For (2, 5), the numbers are strictly between 2 and 5.
\( \implies \) {x : x \( \in \) R, 2 < x < 5}.
(ii) For [0, 7), the numbers are greater than or equal to 0 but strictly less than 7.
\( \implies \) {x : x \( \in \) R, 0 \( \le \) x < 7}.
(iii) For [2, 10], the numbers are greater than or equal to 2 and less than or equal to 10.
\( \implies \) {x : x \( \in \) R, 2 \( \le \) x \( \le \) 10}.
(iv) For (-5, 0], the numbers are strictly greater than -5 but less than or equal to 0.
\( \implies \) {x : x \( \in \) R, -5 < x \( \le \) 0}. Real numbers are often used when defining intervals, as they can represent all values in a continuous range.
In simple words: Set builder form describes the rule that all numbers in an interval follow. We say "x is a real number (x \( \in \) R)" and then add the conditions for x (like "x is greater than 2 but less than 5").

🎯 Exam Tip: Always specify that 'x' is a real number (x \( \in \) R) when converting intervals to set-builder form, unless otherwise specified.

 

Question 24. If A = {x : x \( \in \) N, 2 \( \le \) x \( \le \) 9} and B = {x : x is two digits natural number, sum of which digits is 8}, then find the following sets:
(i) A \( \cup \) B
(ii) A \( \cap \) B
(iii) A - B
(iv) (A - B) \( \cup \) (B - A)
Answer:
First, write sets A and B in roster form:
A = {x : x \( \in \) N, 2 \( \le \) x \( \le \) 9}
\( \implies \) A = {2, 3, 4, 5, 6, 7, 8, 9}.
B = {x : x is two digits natural number, sum of whose digits is 8}
\( \implies \) B = {17 (1+7=8), 26 (2+6=8), 35 (3+5=8), 44 (4+4=8), 53 (5+3=8), 62 (6+2=8), 71 (7+1=8)}.
So, B = {17, 26, 35, 44, 53, 62, 71}.

(i) Find A \( \cup \) B (union of A and B):
A \( \cup \) B includes all unique elements from both A and B.
\( \implies \) A \( \cup \) B = {2, 3, 4, 5, 6, 7, 8, 9} \( \cup \) {17, 26, 35, 44, 53, 62, 71}
\( \implies \) A \( \cup \) B = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71}.

(ii) Find A \( \cap \) B (intersection of A and B):
A \( \cap \) B includes elements common to both A and B. Looking at A and B, there are no common elements.
\( \implies \) A \( \cap \) B = \( \Phi \) (empty set). The intersection of disjoint sets is always empty.

(iii) Find A - B (elements in A but not in B):
Since A and B have no common elements (A \( \cap \) B = \( \Phi \)), removing elements of B from A leaves A unchanged.
\( \implies \) A - B = {2, 3, 4, 5, 6, 7, 8, 9}.

(iv) Find (A - B) \( \cup \) (B - A):
From (iii), A - B = {2, 3, 4, 5, 6, 7, 8, 9}.
Next, find B - A (elements in B but not in A). Since A and B are disjoint, B - A = B.
\( \implies \) B - A = {17, 26, 35, 44, 53, 62, 71}.
Now, find the union of these two results: (A - B) \( \cup \) (B - A) = {2, 3, 4, 5, 6, 7, 8, 9} \( \cup \) {17, 26, 35, 44, 53, 62, 71}.
This is the same as A \( \cup \) B.
\( \implies \) (A - B) \( \cup \) (B - A) = {2, 3, 4, 5, 6, 7, 8, 9, 17, 26, 35, 44, 53, 62, 71}.
In simple words: We first list out all the numbers for set A and set B. Then, we find their union (all unique numbers from both), their intersection (numbers they share), the numbers only in A, and finally, the union of numbers unique to A and numbers unique to B.

🎯 Exam Tip: Always convert set-builder forms to roster forms first. This makes performing union, intersection, and difference operations much clearer and reduces errors.

 

Question 25. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4} and C = {4, 6, 8, 10} then find the value of following sets:
(i) (A \( \cup \) B) \( \cap \) B
(ii) (A \( \cap \) B) \( \cup \) C
(iii) A' \( \cup \) B'
(iv) (A \( \cup \) B)'
Answer:
Given Universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Given sets A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4}, C = {4, 6, 8, 10}.

(i) Find (A \( \cup \) B) \( \cap \) B:
First, find A \( \cup \) B: A \( \cup \) B = {1, 2, 3, 4, 5, 6} \( \cup \) {2, 3, 4} = {1, 2, 3, 4, 5, 6}.
Next, find the intersection of (A \( \cup \) B) with B: (A \( \cup \) B) \( \cap \) B = {1, 2, 3, 4, 5, 6} \( \cap \) {2, 3, 4} = {2, 3, 4}.
Notice that {2, 3, 4} is exactly set B. This result makes sense because B is a subset of A \( \cup \) B, so their intersection is B itself.
\( \implies \) (A \( \cup \) B) \( \cap \) B = {2, 3, 4}.

(ii) Find (A \( \cap \) B) \( \cup \) C:
First, find A \( \cap \) B: A \( \cap \) B = {1, 2, 3, 4, 5, 6} \( \cap \) {2, 3, 4} = {2, 3, 4}.
Next, find the union of (A \( \cap \) B) with C: (A \( \cap \) B) \( \cup \) C = {2, 3, 4} \( \cup \) {4, 6, 8, 10} = {2, 3, 4, 6, 8, 10}.
\( \implies \) (A \( \cap \) B) \( \cup \) C = {2, 3, 4, 6, 8, 10}.

(iii) Find A' \( \cup \) B':
First, find A' (complement of A): A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}.
Next, find B' (complement of B): B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10}.
Now, find the union of A' and B': A' \( \cup \) B' = {7, 8, 9, 10} \( \cup \) {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10}.
\( \implies \) A' \( \cup \) B' = {1, 5, 6, 7, 8, 9, 10}.

(iv) Find (A \( \cup \) B)':
First, find A \( \cup \) B: A \( \cup \) B = {1, 2, 3, 4, 5, 6} \( \cup \) {2, 3, 4} = {1, 2, 3, 4, 5, 6}.
Next, find the complement of (A \( \cup \) B): (A \( \cup \) B)' = U - (A \( \cup \) B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}.
\( \implies \) (A \( \cup \) B)' = {7, 8, 9, 10}. (This also aligns with De Morgan's law: (A \( \cup \) B)' = A' \( \cap \) B').
In simple words: We perform each set operation step-by-step. Union combines all unique elements. Intersection finds common elements. Complement means elements in the universal set but not in the given set. Follow the order of operations carefully.

🎯 Exam Tip: When dealing with complements, always remember to define and list the elements of the universal set (U) first, as complements are relative to U.

 

Question 26. Verify De Morgan's laws using the sets from Question 25.
Answer:
Given Universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
Given sets A = {1, 2, 3, 4, 5, 6}, B = {2, 3, 4}.

**De Morgan's First Law: (A \( \cup \) B)' = A' \( \cap \) B'**
**Left Hand Side (LHS): (A \( \cup \) B)'**
First, find A \( \cup \) B: A \( \cup \) B = {1, 2, 3, 4, 5, 6} \( \cup \) {2, 3, 4} = {1, 2, 3, 4, 5, 6}.
Next, find the complement of (A \( \cup \) B): (A \( \cup \) B)' = U - (A \( \cup \) B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}.

**Right Hand Side (RHS): A' \( \cap \) B'**
First, find A': A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {1, 2, 3, 4, 5, 6} = {7, 8, 9, 10}.
Next, find B': B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10}.
Now, find the intersection of A' and B': A' \( \cap \) B' = {7, 8, 9, 10} \( \cap \) {1, 5, 6, 7, 8, 9, 10} = {7, 8, 9, 10}.
Since LHS = {7, 8, 9, 10} and RHS = {7, 8, 9, 10}, we have LHS = RHS.
\( \implies \) **(A \( \cup \) B)' = A' \( \cap \) B' is verified.**

**De Morgan's Second Law: (A \( \cap \) B)' = A' \( \cup \) B'**
**Left Hand Side (LHS): (A \( \cap \) B)'**
First, find A \( \cap \) B: A \( \cap \) B = {1, 2, 3, 4, 5, 6} \( \cap \) {2, 3, 4} = {2, 3, 4}.
Next, find the complement of (A \( \cap \) B): (A \( \cap \) B)' = U - (A \( \cap \) B) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {2, 3, 4} = {1, 5, 6, 7, 8, 9, 10}.

**Right Hand Side (RHS): A' \( \cup \) B'**
From the first law, we already found A' = {7, 8, 9, 10} and B' = {1, 5, 6, 7, 8, 9, 10}.
Now, find the union of A' and B': A' \( \cup \) B' = {7, 8, 9, 10} \( \cup \) {1, 5, 6, 7, 8, 9, 10} = {1, 5, 6, 7, 8, 9, 10}.
Since LHS = {1, 5, 6, 7, 8, 9, 10} and RHS = {1, 5, 6, 7, 8, 9, 10}, we have LHS = RHS.
\( \implies \) **(A \( \cap \) B)' = A' \( \cup \) B' is verified.**
In simple words: De Morgan's laws show how set complements, unions, and intersections are related. We check each side of the equation separately by listing the elements. If both sides result in the same set of elements, the law is proven true for these specific sets.

🎯 Exam Tip: Clearly show each step of calculating sets like A \( \cup \) B, (A \( \cup \) B)', A', and B' before comparing the final results of LHS and RHS for De Morgan's laws.

 

Question 27. Using Venn diagrams, represent the sets given below:
(i) (A \( \cup \) B) \( \cap \) C
(ii) (A \( \cap \) B) \( \cup \) C
(iii) A' \( \cup \) B'
(iv) (A \( \cup \) B)'
Answer:
(i) To represent (A \( \cup \) B) \( \cap \) C:
This set includes all elements that are in C AND either in A or in B (or both). It's the region where the union of A and B overlaps with C.

(ii) To represent (A \( \cap \) B) \( \cup \) C:
This set includes all elements that are in C, along with any elements that are common to both A and B. It's the union of the central lens of A and B with the entire circle of C.

(iii) To represent A' \( \cup \) B':
This set includes all elements that are outside A, or outside B (or both). It's everything in the universal set except the intersection of A and B.

(iv) To represent (A \( \cup \) B)':
This set includes all elements in the universal set that are not in A and not in B. It's the region outside both A and B circles.

In simple words: For each set definition, we draw a Venn diagram and shade the area that includes the elements described by the set operation. For three sets, we use three overlapping circles. For two sets, we use two circles. Complements mean shading outside the specified sets.

🎯 Exam Tip: Carefully identify the order of operations in complex set expressions (like union before intersection, or complement last) to shade the Venn diagram correctly.

 

Question 28. With the help of Venn diagram prove that:
(i) (A \( \cup \) B)' = A' \( \cap \) B'
(ii) (A \( \cap \) B)' = A' \( \cup \) B'
Answer:
(i) To prove (A \( \cup \) B)' = A' \( \cap \) B' using Venn diagrams (De Morgan's First Law):
**Left Hand Side (LHS): (A \( \cup \) B)'**
First, consider A \( \cup \) B, which is the entire area covered by circles A and B. Then, (A \( \cup \) B)' represents the area *outside* both circles A and B, but within the universal set U.

Shaded region represents (A \( \cup \) B)'.

**Right Hand Side (RHS): A' \( \cap \) B'**
A' represents the area outside circle A. B' represents the area outside circle B. The intersection A' \( \cap \) B' is the area that is *both* outside A *and* outside B. This means it's the area outside both A and B, which is the same region as (A \( \cup \) B)'.

Shaded region represents A' \( \cap \) B'. Both Venn diagrams show the same shaded region, thus (A \( \cup \) B)' = A' \( \cap \) B' is proven.

(ii) To prove (A \( \cap \) B)' = A' \( \cup \) B' using Venn diagrams (De Morgan's Second Law):
**Left Hand Side (LHS): (A \( \cap \) B)'**
First, consider A \( \cap \) B, which is the overlapping area of circles A and B. Then, (A \( \cap \) B)' represents all the area *outside* this overlapping region, but within the universal set U.

Shaded region represents (A \( \cap \) B)'.

**Right Hand Side (RHS): A' \( \cup \) B'**
A' represents the area outside circle A (shaded horizontally). B' represents the area outside circle B (shaded vertically). The union A' \( \cup \) B' means all the areas that are either outside A or outside B (or both). This covers everything in U *except* the common region A \( \cap \) B. This is the same region as (A \( \cap \) B)'.

Shaded region represents A' \( \cup \) B'. Both Venn diagrams show the same shaded region, thus (A \( \cap \) B)' = A' \( \cup \) B' is proven.
In simple words: De Morgan's laws tell us how complements work with unions and intersections. We draw a picture (Venn diagram) for each side of the law. If the shaded parts of both pictures are exactly the same, then the law is proven to be true with Venn diagrams.

🎯 Exam Tip: For De Morgan's laws, always draw separate diagrams for the Left Hand Side and the Right Hand Side, and show that their shaded regions match exactly to prove the identity.

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