This MCQ module is based on: Exercises and Summary – Sets
TOPIC 4 OF 45
Exercises and Summary – Sets
🎓 Class 11
Mathematics
CBSE
Theory
Ch 1 — Sets
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Exercises and Summary – Sets
Targeting Class 11 level in Sets, with Advanced difficulty.
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Miscellaneous Examples
Example 23
Show that the set of letters needed to spell "CATARACT" and the set of letters needed to spell "TRACT" are equal.
Solution
Let X be the set of letters in "CATARACT". Then\(X = \{C, A, T, R\}\)
Let Y be the set of letters in "TRACT". Then
\(Y = \{T, R, A, C\} = \{C, A, T, R\}\)
Since every element in X is in Y and every element in Y is in X, it follows that \(X = Y\).
Example 24
List all the subsets of the set \(\{-1, 0, 1\}\).
Solution
Let \(A = \{-1, 0, 1\}\). The subset of A having no element is the empty set \(\emptyset\).The subsets of A having one element are: \(\{-1\}\), \(\{0\}\), \(\{1\}\).
The subsets of A having two elements are: \(\{-1, 0\}\), \(\{-1, 1\}\), \(\{0, 1\}\).
The subset of A having three elements is A itself: \(\{-1, 0, 1\}\).
So, all the subsets of A are: \(\emptyset, \{-1\}, \{0\}, \{1\}, \{-1, 0\}, \{-1, 1\}, \{0, 1\}, \{-1, 0, 1\}\).
Total = \(2^3 = 8\) subsets.
Example 25
Show that \(A \cup B = A \cap B\) implies \(A = B\).
Solution
Let \(a \in A\). Then \(a \in A \cup B\). Since \(A \cup B = A \cap B\), \(a \in A \cap B\). So \(a \in B\). Therefore, \(A \subset B\).Similarly, if \(b \in B\), then \(b \in A \cup B = A \cap B\). So \(b \in A\). Therefore, \(B \subset A\).
Since \(A \subset B\) and \(B \subset A\), we have \(A = B\).
Miscellaneous Exercise on Chapter 1
Q1. Decide, among the following sets, which sets are subsets of one another:
\(A = \{x : x \in \mathbb{R} \text{ and } x \text{ satisfy } x^2 - 8x + 12 = 0\}\)
\(B = \{2, 4, 6\}\), \(C = \{2, 4, 6, 8, \ldots\}\), \(D = \{6\}\)
\(A = \{x : x \in \mathbb{R} \text{ and } x \text{ satisfy } x^2 - 8x + 12 = 0\}\)
\(B = \{2, 4, 6\}\), \(C = \{2, 4, 6, 8, \ldots\}\), \(D = \{6\}\)
\(x^2 - 8x + 12 = 0 \Rightarrow (x-2)(x-6) = 0 \Rightarrow x = 2, 6\). So \(A = \{2, 6\}\).
\(D = \{6\} \subset A = \{2, 6\} \subset B = \{2, 4, 6\} \subset C = \{2, 4, 6, 8, \ldots\}\).
So: \(D \subset A \subset B \subset C\).
\(D = \{6\} \subset A = \{2, 6\} \subset B = \{2, 4, 6\} \subset C = \{2, 4, 6, 8, \ldots\}\).
So: \(D \subset A \subset B \subset C\).
Q2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If \(x \in A\) and \(A \in B\), then \(x \in B\).
(ii) If \(A \subset B\) and \(B \in C\), then \(A \in C\).
(iii) If \(A \subset B\) and \(B \subset C\), then \(A \subset C\).
(iv) If \(A \not\subset B\) and \(B \not\subset C\), then \(A \not\subset C\).
(v) If \(x \in A\) and \(A \not\subset B\), then \(x \in B\).
(vi) If \(A \subset B\) and \(x \notin B\), then \(x \notin A\).
(i) If \(x \in A\) and \(A \in B\), then \(x \in B\).
(ii) If \(A \subset B\) and \(B \in C\), then \(A \in C\).
(iii) If \(A \subset B\) and \(B \subset C\), then \(A \subset C\).
(iv) If \(A \not\subset B\) and \(B \not\subset C\), then \(A \not\subset C\).
(v) If \(x \in A\) and \(A \not\subset B\), then \(x \in B\).
(vi) If \(A \subset B\) and \(x \notin B\), then \(x \notin A\).
(i) False. Let \(A = \{1\}\), \(B = \{\{1\}, 2\}\). Then \(1 \in A\) and \(A \in B\), but \(1 \notin B\) (elements of B are \(\{1\}\) and 2, not 1).
(ii) False. Let \(A = \{1\}\), \(B = \{1, 2\}\), \(C = \{\{1, 2\}, 3\}\). Then \(A \subset B\) and \(B \in C\), but \(A \notin C\).
(iii) True. Let \(x \in A\). Since \(A \subset B\), \(x \in B\). Since \(B \subset C\), \(x \in C\). Hence \(A \subset C\). (Transitivity of subsets.)
(iv) False. Let \(A = \{1, 2\}\), \(B = \{2, 3\}\), \(C = \{1, 2, 3\}\). Then \(A \not\subset B\) (since \(1 \notin B\)) and \(B \not\subset C\) is actually false here; let us try \(C = \{1, 3\}\). Then \(B = \{2,3\} \not\subset C = \{1,3\}\) (since \(2 \notin C\)). But \(A = \{1,2\} \not\subset C = \{1,3\}\) (since \(2 \notin C\)). This gives \(A \not\subset C\). Try: \(A = \{1\}\), \(B = \{2\}\), \(C = \{1, 3\}\). \(A \not\subset B\), \(B \not\subset C\), but \(A \subset C\). So the statement is false.
(v) False. Let \(A = \{1, 2\}\), \(B = \{2, 3\}\). Then \(1 \in A\) and \(A \not\subset B\), but \(1 \notin B\).
(vi) True. By contrapositive: if \(x \in A\), then since \(A \subset B\), \(x \in B\). Equivalently, \(x \notin B \Rightarrow x \notin A\).
(ii) False. Let \(A = \{1\}\), \(B = \{1, 2\}\), \(C = \{\{1, 2\}, 3\}\). Then \(A \subset B\) and \(B \in C\), but \(A \notin C\).
(iii) True. Let \(x \in A\). Since \(A \subset B\), \(x \in B\). Since \(B \subset C\), \(x \in C\). Hence \(A \subset C\). (Transitivity of subsets.)
(iv) False. Let \(A = \{1, 2\}\), \(B = \{2, 3\}\), \(C = \{1, 2, 3\}\). Then \(A \not\subset B\) (since \(1 \notin B\)) and \(B \not\subset C\) is actually false here; let us try \(C = \{1, 3\}\). Then \(B = \{2,3\} \not\subset C = \{1,3\}\) (since \(2 \notin C\)). But \(A = \{1,2\} \not\subset C = \{1,3\}\) (since \(2 \notin C\)). This gives \(A \not\subset C\). Try: \(A = \{1\}\), \(B = \{2\}\), \(C = \{1, 3\}\). \(A \not\subset B\), \(B \not\subset C\), but \(A \subset C\). So the statement is false.
(v) False. Let \(A = \{1, 2\}\), \(B = \{2, 3\}\). Then \(1 \in A\) and \(A \not\subset B\), but \(1 \notin B\).
(vi) True. By contrapositive: if \(x \in A\), then since \(A \subset B\), \(x \in B\). Equivalently, \(x \notin B \Rightarrow x \notin A\).
Q3. Let A, B, and C be the sets such that \(A \cup B = A \cup C\) and \(A \cap B = A \cap C\). Show that \(B = C\).
Let \(b \in B\). Then either \(b \in A\) or \(b \notin A\).
Case 1: If \(b \in A\), then \(b \in A \cap B = A \cap C\), so \(b \in C\).
Case 2: If \(b \notin A\), then since \(b \in B\), we have \(b \in A \cup B = A \cup C\). Since \(b \notin A\), we must have \(b \in C\).
In both cases, \(b \in C\). So \(B \subset C\).
By a symmetric argument (swapping B and C), \(C \subset B\).
Therefore, \(B = C\).
Case 1: If \(b \in A\), then \(b \in A \cap B = A \cap C\), so \(b \in C\).
Case 2: If \(b \notin A\), then since \(b \in B\), we have \(b \in A \cup B = A \cup C\). Since \(b \notin A\), we must have \(b \in C\).
In both cases, \(b \in C\). So \(B \subset C\).
By a symmetric argument (swapping B and C), \(C \subset B\).
Therefore, \(B = C\).
Q4. Show that the following four conditions are equivalent:
(i) \(A \subset B\) (ii) \(A - B = \emptyset\) (iii) \(A \cup B = B\) (iv) \(A \cap B = A\)
(i) \(A \subset B\) (ii) \(A - B = \emptyset\) (iii) \(A \cup B = B\) (iv) \(A \cap B = A\)
(i) \(\Rightarrow\) (ii): If \(A \subset B\), then every element of A is in B, so no element of A is outside B. Hence \(A - B = \emptyset\).
(ii) \(\Rightarrow\) (iii): If \(A - B = \emptyset\), every element of A is in B. So \(A \cup B = B\) (adding A contributes nothing new).
(iii) \(\Rightarrow\) (iv): If \(A \cup B = B\), then every element of A is in B (otherwise \(A \cup B\) would have elements not in B). So \(A \cap B = A\).
(iv) \(\Rightarrow\) (i): If \(A \cap B = A\), then every element of A is in both A and B, meaning every element of A is in B. So \(A \subset B\).
(ii) \(\Rightarrow\) (iii): If \(A - B = \emptyset\), every element of A is in B. So \(A \cup B = B\) (adding A contributes nothing new).
(iii) \(\Rightarrow\) (iv): If \(A \cup B = B\), then every element of A is in B (otherwise \(A \cup B\) would have elements not in B). So \(A \cap B = A\).
(iv) \(\Rightarrow\) (i): If \(A \cap B = A\), then every element of A is in both A and B, meaning every element of A is in B. So \(A \subset B\).
Q5. Show that if \(A \subset B\), then \(C - B \subset C - A\).
Let \(x \in C - B\). Then \(x \in C\) and \(x \notin B\).
Since \(A \subset B\), if \(x\) were in A, then \(x\) would be in B (contradiction). So \(x \notin A\).
Therefore \(x \in C\) and \(x \notin A\), which means \(x \in C - A\).
Hence \(C - B \subset C - A\).
Since \(A \subset B\), if \(x\) were in A, then \(x\) would be in B (contradiction). So \(x \notin A\).
Therefore \(x \in C\) and \(x \notin A\), which means \(x \in C - A\).
Hence \(C - B \subset C - A\).
Q6. Show that for any sets A and B:
\(A = (A \cap B) \cup (A - B)\) and \(A \cup (B - A) = A \cup B\).
\(A = (A \cap B) \cup (A - B)\) and \(A \cup (B - A) = A \cup B\).
Part 1: \((A \cap B) \cup (A - B)\). Let \(x \in A\). Either \(x \in B\) or \(x \notin B\). If \(x \in B\), then \(x \in A \cap B\). If \(x \notin B\), then \(x \in A - B\). Either way, \(x \in (A \cap B) \cup (A - B)\). Conversely, any element of \((A \cap B) \cup (A - B)\) must be in A. So \(A = (A \cap B) \cup (A - B)\).
Part 2: \(A \cup (B - A)\). \(B - A\) is the set of elements in B but not in A. So \(A \cup (B - A) = A \cup \{x \in B : x \notin A\}\). This gives all elements of A plus all elements of B that are not already in A, which is exactly \(A \cup B\).
Part 2: \(A \cup (B - A)\). \(B - A\) is the set of elements in B but not in A. So \(A \cup (B - A) = A \cup \{x \in B : x \notin A\}\). This gives all elements of A plus all elements of B that are not already in A, which is exactly \(A \cup B\).
Q7. Using properties of sets, show that:
(i) \(A \cup (A \cap B) = A\) (ii) \(A \cap (A \cup B) = A\)
(i) \(A \cup (A \cap B) = A\) (ii) \(A \cap (A \cup B) = A\)
(i) Since \(A \cap B \subset A\), every element of \(A \cap B\) is already in A. Therefore \(A \cup (A \cap B) = A\).
Alternatively: \(A \cup (A \cap B) = (A \cup A) \cap (A \cup B) = A \cap (A \cup B) = A\) (since \(A \subset A \cup B\)).
(ii) Since \(A \subset A \cup B\), \(A \cap (A \cup B) = A\).
Alternatively: \(A \cap (A \cup B) = (A \cap A) \cup (A \cap B) = A \cup (A \cap B) = A\).
Alternatively: \(A \cup (A \cap B) = (A \cup A) \cap (A \cup B) = A \cap (A \cup B) = A\) (since \(A \subset A \cup B\)).
(ii) Since \(A \subset A \cup B\), \(A \cap (A \cup B) = A\).
Alternatively: \(A \cap (A \cup B) = (A \cap A) \cup (A \cap B) = A \cup (A \cap B) = A\).
Q8. Show that \(A \cap B = A \cap C\) need not imply \(B = C\).
Counterexample: Let \(A = \{1, 2\}\), \(B = \{1, 3\}\), \(C = \{1, 4\}\).
Then \(A \cap B = \{1\}\) and \(A \cap C = \{1\}\). So \(A \cap B = A \cap C\), but \(B = \{1, 3\} \neq \{1, 4\} = C\).
Then \(A \cap B = \{1\}\) and \(A \cap C = \{1\}\). So \(A \cap B = A \cap C\), but \(B = \{1, 3\} \neq \{1, 4\} = C\).
Q9. Let A and B be sets. If \(A \cap X = B \cap X = \emptyset\) and \(A \cup X = B \cup X\) for some set X, show that \(A = B\).
We use: \(A = A \cap (A \cup X)\) (absorption law).
\(A = A \cap (A \cup X) = A \cap (B \cup X)\) (since \(A \cup X = B \cup X\)).
\(= (A \cap B) \cup (A \cap X) = (A \cap B) \cup \emptyset = A \cap B\).
Similarly, \(B = B \cap (B \cup X) = B \cap (A \cup X) = (B \cap A) \cup (B \cap X) = (B \cap A) \cup \emptyset = A \cap B\).
Therefore \(A = A \cap B = B\), so \(A = B\).
\(A = A \cap (A \cup X) = A \cap (B \cup X)\) (since \(A \cup X = B \cup X\)).
\(= (A \cap B) \cup (A \cap X) = (A \cap B) \cup \emptyset = A \cap B\).
Similarly, \(B = B \cap (B \cup X) = B \cap (A \cup X) = (B \cap A) \cup (B \cap X) = (B \cap A) \cup \emptyset = A \cap B\).
Therefore \(A = A \cap B = B\), so \(A = B\).
Q10. Find sets A, B and C such that \(A \cap B\), \(B \cap C\) and \(A \cap C\) are non-empty sets and \(A \cap B \cap C = \emptyset\).
Let \(A = \{1, 2\}\), \(B = \{2, 3\}\), \(C = \{1, 3\}\).
\(A \cap B = \{2\} \neq \emptyset\),
\(B \cap C = \{3\} \neq \emptyset\),
\(A \cap C = \{1\} \neq \emptyset\),
\(A \cap B \cap C = \emptyset\) (no element is common to all three).
\(A \cap B = \{2\} \neq \emptyset\),
\(B \cap C = \{3\} \neq \emptyset\),
\(A \cap C = \{1\} \neq \emptyset\),
\(A \cap B \cap C = \emptyset\) (no element is common to all three).
Q11. If R is the set of real numbers and Q is the set of rational numbers, then what is \(\mathbb{R} - \mathbb{Q}\)?
\(\mathbb{R} - \mathbb{Q}\) is the set of all real numbers that are NOT rational, i.e., the set of irrational numbers. This set includes numbers like \(\sqrt{2}\), \(\sqrt{3}\), \(\pi\), \(e\), etc.
Q12. State whether each of the following statement 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.
(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.
(i) False. 3 is common to both sets.
(ii) False. 'a' is common to both sets.
(iii) True. No element is common (even vs odd numbers with different offsets).
(iv) True. No element is common (all elements differ).
(ii) False. 'a' is common to both sets.
(iii) True. No element is common (even vs odd numbers with different offsets).
(iv) True. No element is common (all elements differ).
Summary
- A set is a well-defined collection of objects.
- A set which does not contain any element is called empty set.
- A set which consists of a definite number of elements is called finite set, otherwise it is called infinite set.
- Two sets A and B are said to be equal if they have exactly the same elements.
- A set A is said to be subset of a set B if every element of A is also an element of B. Intervals are subsets of \(\mathbb{R}\).
- The union of two sets A and B is the set of all those elements which are either in A or in B.
- The intersection of two sets A and B is the set of all elements which are common. The difference of two sets A and B in this order is the set of elements which belong to A but not to B.
- The complement of a subset A of universal set U is the set of all elements of U which are not the elements of A.
- For any two sets A and B, \((A \cup B)' = A' \cap B'\) and \((A \cap B)' = A' \cup B'\) — these are De Morgan's laws.
Key Formulae at a Glance
Union
\(A \cup B = \{x : x \in A \text{ or } x \in B\}\)
Commutative, Associative
Commutative, Associative
Intersection
\(A \cap B = \{x : x \in A \text{ and } x \in B\}\)
Commutative, Associative
Commutative, Associative
Complement
\(A' = U - A\)
\((A')' = A\), \(A \cup A' = U\)
\((A')' = A\), \(A \cup A' = U\)
De Morgan's Laws
\((A \cup B)' = A' \cap B'\)
\((A \cap B)' = A' \cup B'\)
\((A \cap B)' = A' \cup B'\)
Power Set
If \(|A| = n\), then \(|P(A)| = 2^n\)
Distributive Laws
\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
\(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\)
\(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\)
Historical Note
The modern theory of sets is considered to have been originated largely by Georg Cantor (1845–1918). His papers on set theory appeared between 1874 and 1897. His work was initially controversial — Richard Dedekind (1831–1916) welcomed it, while Leopold Kronecker (1810–1893) strongly opposed treating infinite sets on equal footing with finite ones.Gottlob Frege presented set theory as part of the principles of logic. However, Bertrand Russell (1872–1970) showed in 1902 that the assumption of the existence of a "set of all sets" leads to a paradox (Russell's Paradox). The first axiomatisation of set theory was published by Ernst Zermelo in 1908, followed by Abraham Fraenkel in 1922. John Von Neumann contributed the axiom of regularity in 1925, and later Paul Bernays gave a more satisfactory axiomatisation. This was known as Von Neumann–Bernays (VNB) or Gödel–Bernays (GB) set theory. Despite all these challenges, Cantor's set theory remains foundational to present-day mathematics.
Activity: Chapter Revision Challenge
Predict: Can you solve all 8 types of set problems without looking at notes? Test yourself below.
- Type 1 — Representation: Write \(\{x : x \text{ is a prime less than 20}\}\) in roster form.
- Type 2 — Empty set: Is \(\{x : x^2 = -1, x \in \mathbb{R}\}\) empty?
- Type 3 — Subsets: List all subsets of \(\{p, q\}\).
- Type 4 — Power set: How many elements in \(P(\{1,2,3,4\})\)?
- Type 5 — Union: Find \(\{1,2,3\} \cup \{3,4,5\}\).
- Type 6 — Intersection: Find \(\{1,2,3\} \cap \{3,4,5\}\).
- Type 7 — Difference: Find \(\{1,2,3,4\} - \{2,4,6\}\).
- Type 8 — Complement: If \(U = \{1,...,10\}\) and \(A = \{2,4,6,8\}\), find \(A'\).
1. \(\{2, 3, 5, 7, 11, 13, 17, 19\}\) 2. Yes, empty (no real number squares to −1) 3. \(\emptyset, \{p\}, \{q\}, \{p,q\}\) 4. \(2^4 = 16\) 5. \(\{1,2,3,4,5\}\) 6. \(\{3\}\) 7. \(\{1,3\}\) 8. \(\{1,3,5,7,9,10\}\)
Competency-Based Questions
A data analyst is working with customer databases for an e-commerce platform. She defines the following sets from 10,000 registered users: P = users who purchased in Jan (3,200), Q = users who purchased in Feb (2,800), and she finds that 1,500 users purchased in both months. The universal set U is all 10,000 registered users.
Q1. The marketing team wants to send a "Thank You" email to everyone who made at least one purchase in Jan or Feb. Using set operations, determine how many emails need to be sent. Then determine how many users bought in Jan only.
L3 Apply\(n(P \cup Q) = n(P) + n(Q) - n(P \cap Q) = 3200 + 2800 - 1500 = 4500\).
4,500 emails need to be sent.
Jan only: \(n(P - Q) = n(P) - n(P \cap Q) = 3200 - 1500 = 1700\) users.
4,500 emails need to be sent.
Jan only: \(n(P - Q) = n(P) - n(P \cap Q) = 3200 - 1500 = 1700\) users.
Q2. Analyse: The analyst notices that \(n(P \cup Q) = 4500\) but the total user base is 10,000. What does \((P \cup Q)'\) represent in business terms? If the company targets \((P \cup Q)'\) for a re-engagement campaign, how does this relate to the complement concept? Could \(P - Q\) and \(Q - P\) overlap? Justify using set properties.
L4 Analyse\((P \cup Q)'\) represents inactive users — those who purchased in neither Jan nor Feb. Count: \(10000 - 4500 = 5500\) users. This is exactly the complement of "purchasers" relative to the full user base.
\(P - Q\) and \(Q - P\) cannot overlap. If \(x \in P - Q\), then \(x \in P\) and \(x \notin Q\). If \(x \in Q - P\), then \(x \in Q\) and \(x \notin P\). These are contradictory, so \((P - Q) \cap (Q - P) = \emptyset\). This follows from the property that \(A - B\), \(A \cap B\), and \(B - A\) are always mutually disjoint.
\(P - Q\) and \(Q - P\) cannot overlap. If \(x \in P - Q\), then \(x \in P\) and \(x \notin Q\). If \(x \in Q - P\), then \(x \in Q\) and \(x \notin P\). These are contradictory, so \((P - Q) \cap (Q - P) = \emptyset\). This follows from the property that \(A - B\), \(A \cap B\), and \(B - A\) are always mutually disjoint.
Q3. A junior analyst claims: "If we simply add the Jan purchasers and Feb purchasers, we get \(3200 + 2800 = 6000\), which means 6000 unique users purchased something." Evaluate this reasoning. What mathematical error has been made, and how does the inclusion-exclusion principle correct it?
L5 EvaluateThe reasoning is flawed. Simply adding \(n(P) + n(Q)\) double-counts the 1,500 users who purchased in BOTH months. The inclusion-exclusion principle states: \(n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)\). The correct count is \(3200 + 2800 - 1500 = 4500\), not 6000. The error overstates unique customers by \(6000 - 4500 = 1500\), which is exactly \(n(P \cap Q)\) — the double-counted group.
Q4. Design a segmentation strategy using set operations that divides all 10,000 users into exactly 4 non-overlapping groups for targeted marketing. Define each group using set notation, calculate its size, and propose a different marketing action for each group.
L6 CreateFour segments using set operations:
S1 = \(P - Q\) (Jan only): 1,700 users. Action: Send Feb product recommendations.
S2 = \(P \cap Q\) (Both months): 1,500 users. Action: Offer loyalty rewards / VIP discount.
S3 = \(Q - P\) (Feb only): 1,300 users. Action: Recommend Jan best-sellers they missed.
S4 = \((P \cup Q)'\) (Neither): 5,500 users. Action: Re-engagement campaign with special first-purchase offer.
Verification: \(1700 + 1500 + 1300 + 5500 = 10000 = n(U)\). The four sets are mutually disjoint and their union is U, forming a partition of the universal set.
S1 = \(P - Q\) (Jan only): 1,700 users. Action: Send Feb product recommendations.
S2 = \(P \cap Q\) (Both months): 1,500 users. Action: Offer loyalty rewards / VIP discount.
S3 = \(Q - P\) (Feb only): 1,300 users. Action: Recommend Jan best-sellers they missed.
S4 = \((P \cup Q)'\) (Neither): 5,500 users. Action: Re-engagement campaign with special first-purchase offer.
Verification: \(1700 + 1500 + 1300 + 5500 = 10000 = n(U)\). The four sets are mutually disjoint and their union is U, forming a partition of the universal set.
Assertion–Reason Questions
Assertion (A): If \(A \cup B = A \cap B\), then \(A = B\).
Reason (R): Two sets are equal if and only if each is a subset of the other.
Reason (R): Two sets are equal if and only if each is a subset of the other.
Answer: (a) — Both true. If \(A \cup B = A \cap B\), then for any \(a \in A\), \(a \in A \cup B = A \cap B\), so \(a \in B\). Thus \(A \subset B\). Similarly \(B \subset A\). By R, \(A = B\). R is the underlying principle used in the proof.
Assertion (A): \(\mathbb{R} - \mathbb{Q}\) is the set of irrational numbers.
Reason (R): The set difference \(A - B\) contains those elements that are in A but not in B.
Reason (R): The set difference \(A - B\) contains those elements that are in A but not in B.
Answer: (a) — \(\mathbb{R} - \mathbb{Q}\) is the set of real numbers that are not rational, i.e., irrational numbers. R is the definition of set difference, which directly explains why A is true.
Assertion (A): For any set A, \(P(A)\) always contains at least 2 elements.
Reason (R): \(P(A)\) always contains \(\emptyset\) and A itself as elements.
Reason (R): \(P(A)\) always contains \(\emptyset\) and A itself as elements.
Answer: (d) — A is false. If \(A = \emptyset\), then \(P(\emptyset) = \{\emptyset\}\), which has only 1 element. R is true in general: both \(\emptyset\) and A are always subsets of A. But when \(A = \emptyset\), these two are the same element (\(\emptyset = A\)), so P(A) has only 1 element.
Frequently Asked Questions
How to solve NCERT Sets exercises for Class 11?
Identify the type of problem: representation, subset identification, set operations, or Venn diagram questions. Apply the relevant definitions and formulas. Practice each section systematically before miscellaneous exercises.
What are the important formulas for Sets in Class 11?
Key formulas include: n(A union B) = n(A) + n(B) - n(A intersection B), number of subsets = 2^n, De Morgan Laws, and the three-set union formula.
What topics are covered in NCERT Sets exercises?
NCERT Sets exercises cover set representations, types of sets, subsets and power sets, Venn diagrams, union and intersection operations, complement and difference of sets, and application problems.
How many exercises are in NCERT Class 11 Sets chapter?
NCERT Class 11 Sets chapter contains Exercise 1.1 through Exercise 1.6, plus a Miscellaneous Exercise. Each focuses on specific concepts building from basic to complex.
What are common mistakes in Sets problems?
Common mistakes include confusing subset with element-of notation, forgetting the empty set when listing subsets, errors in De Morgan Laws, and not accounting for intersection when computing union.
Frequently Asked Questions — Sets
What is Exercises and Summary - Sets in NCERT Class 11 Mathematics?
Exercises and Summary - Sets is a key concept covered in NCERT Class 11 Mathematics, Chapter 1: Sets. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Exercises and Summary - Sets step by step?
To solve problems on Exercises and Summary - Sets, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Sets?
The essential formulas of Chapter 1 (Sets) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Exercises and Summary - Sets important for the Class 11 board exam?
Exercises and Summary - Sets is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Exercises and Summary - Sets?
Common mistakes in Exercises and Summary - Sets include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Exercises and Summary - Sets?
End-of-chapter NCERT exercises for Exercises and Summary - Sets cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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