NCERT Class 11 Maths Sets Miscellaneous Exercise Guide

Deep-Dive into Set Identities & Proof Techniques

To solve problems in the Sets Miscellaneous Exercise, you must first master the art of writing algebraic set proofs. Many questions ask you to show that two sets are equal, such as proving $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

In set theory, asserting that two sets $A$ and $B$ are equal ($A = B$) means they contain exactly the same elements. To prove this rigorously, we cannot simply draw a Venn diagram (Venn diagrams are great for intuition, but CBSE examiners demand analytical proofs). Instead, we use the method of Double Inclusion. This means we must prove two conditions:

1. $A \subseteq B$ (Every element of $A$ is also in $B$)
2. $B \subseteq A$ (Every element of $B$ is also in $A$)

To show $A \subseteq B$, we select an arbitrary, general element $x$ from $A$ and write logical deduction steps to show that $x$ must also belong to $B$. This is written mathematically as: Let $x \in A \implies \dots \implies x \in B$. Similarly, to show $B \subseteq A$, we select an arbitrary element $y \in B$ and show that $y \in A$. Once both directions are proven, we can confidently conclude that $A = B$. Understanding this fundamental tool of mathematical logic is the key to conquering the theoretical problems in the miscellaneous exercise.

The Step-by-Step Blueprint to Prove Set Equality (A = B)

1. Step 1: Choose an Arbitrary Element from the Left-Hand Side (LHS) — Start by assuming there is an element $x$ such that $x \in LHS$. Use set definition rules to break down the expression. For example, if $x \in (P \cup Q)$, write $x \in P$ or $x \in Q$. If $x \in (P \cap Q)$, write $x \in P$ and $x \in Q$.
2. Step 2: Logically Manipulate to Reach the Right-Hand Side (RHS) — Apply distributive, associative, or identity laws to transform the logical statements about $x$. Show that these conditions inevitably mean $x \in RHS$. This establishes that $LHS \subseteq RHS$.
3. Step 3: Reverse the Process for RHS — Choose an arbitrary element $y$ such that $y \in RHS$. Work backward through the logical rules to show that $y \in LHS$. This establishes that $RHS \subseteq LHS$.
4. Step 4: Conclude Equality — State that since $LHS \subseteq RHS$ and $RHS \subseteq LHS$, the two sets are equal ($LHS = RHS$).

Fully Worked Solutions to Core Miscellaneous Problems

• Example 1 (Algebraic Proof): Show that for any sets $A$ and $B$, $A = (A \cap B) \cup (A - B)$. Step 1 (Proving $A \subseteq (A \cap B) \cup (A - B)$): Let $x$ be an arbitrary element of $A$. Case I: Let $x \in B$. Since we assumed $x \in A$, we have $x \in A$ and $x \in B$, which means $x \in (A \cap B)$. Case II: Let $x \notin B$. Since $x \in A$ and $x \notin B$, we have $x \in (A - B)$. In both cases, $x \in (A \cap B)$ or $x \in (A - B)$, which means $x \in (A \cap B) \cup (A - B)$. Thus, $A \subseteq (A \cap B) \cup (A - B)$ — (Equation 1) Step 2 (Proving $(A \cap B) \cup (A - B) \subseteq A$): Let $y \in (A \cap B) \cup (A - B)$. This implies $y \in (A \cap B)$ or $y \in (A - B)$. If $y \in (A \cap B)$, then $y \in A$ and $y \in B$. Thus, $y \in A$. If $y \in (A - B)$, then $y \in A$ and $y \notin B$. Thus, $y \in A$. In either case, $y \in A$. Thus, $(A \cap B) \cup (A - B) \subseteq A$ — (Equation 2) Step 3 (Conclusion): From Equations 1 and 2, we conclude that $A = (A \cap B) \cup (A - B)$.
• Example 2 (Three-Set Word Problem): In a survey of 60 people, 25 read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, and 3 read all three newspapers. Find the number of people who read at least one of the newspapers. Step 1 (Identify given values): Total surveyed, $n(U) = 60$ $n(H) = 25$, $n(T) = 26$, $n(I) = 26$ $n(H \cap I) = 9$, $n(H \cap T) = 11$, $n(T \cap I) = 8$ $n(H \cap T \cap I) = 3$ Step 2 (Apply the Three-Set Union Formula): We know that the number of people who read at least one newspaper is given by $n(H \cup T \cup I)$: $n(H \cup T \cup I) = n(H) + n(T) + n(I) - n(H \cap T) - n(H \cap I) - n(T \cap I) + n(H \cap T \cap I)$ Step 3 (Substitute values and calculate): $n(H \cup T \cup I) = 25 + 26 + 26 - 11 - 9 - 8 + 3$ $n(H \cup T \cup I) = 77 - 28 + 3 = 52$ Final Answer: 52 people read at least one newspaper.

Exam Traps & Crucial Notational Tips

1. Don't Confuse Element ($\in$) and Subset ($\subseteq$): This is the most common point of failure. If $x \in A$ and $A \subseteq B$, then $x \in B$. However, if $A \in B$ (where $A$ is treated as an element of power set $B$), $A$ is not necessarily a subset of $B$.
2. Always State the Method: In CBSE exams, if you use a Venn diagram to solve a proving question without writing algebraic statements, you will lose 50% of the marks. Always back up your diagrams with mathematical logic.
3. De Morgan's Signs: Remember that taking a complement flips the operator: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$. Getting this wrong breaks the entire logical flow of your proof.

Practice Questions with Solutions

• Q: Prove that if $A \subseteq B$, then $C - B \subseteq C - A$. A: Step 1: Let $x$ be an arbitrary element in $C - B$. Step 2: By definition of set difference, $x \in C$ and $x \notin B$. Step 3: We are given that $A \subseteq B$. This means if any element is in $A$, it must be in $B$. Logically, its contrapositive is also true: if an element is not in $B$, it cannot be in $A$ ($x \notin B \implies x \notin A$). Step 4: Since $x \in C$ and $x \notin A$, by definition of set difference, we have $x \in C - A$. Step 5: Thus, every element of $C - B$ is also in $C - A$. Final answer: Therefore, $C - B \subseteq C - A$ is proven.
• Q: For any two sets $A$ and $B$, is it true that $P(A) \cup P(B) = P(A \cup B)$? Justify your answer with a counterexample. A: Step 1: Let us test this statement with small sets. Let $A = \{1\}$ and $B = \{2\}$. Step 2: Find $A \cup B = \{1, 2\}$. Step 3: Write down the power sets. $P(A) = \{\emptyset, \{1\}\}$ $P(B) = \{\emptyset, \{2\}\}$ $P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$ Step 4: Find the union of individual power sets: $P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\}$ Step 5: Compare the two sets. We see that $P(A) \cup P(B) \neq P(A \cup B)$ because the set $\{1, 2\}$ is in $P(A \cup B)$ but not in $P(A) \cup P(B)$. Final answer: No, the statement is false. The counterexample shows $P(A) \cup P(B) \neq P(A \cup B)$.
• Q: Out of 100 students in a school, 50 play cricket, 30 play football, and 20 play hockey. 10 play cricket and football, 5 play football and hockey, and 10 play cricket and hockey. If 3 play all three games, find the number of students who do not play any of the three games. A: Step 1: Define the sets. Let $C$, $F$, and $H$ represent students playing cricket, football, and hockey respectively. Step 2: Write the given cardinalities: $n(C) = 50$, $n(F) = 30$, $n(H) = 20$ $n(C \cap F) = 10$, $n(F \cap H) = 5$, $n(C \cap H) = 10$ $n(C \cap F \cap H) = 3$ Total students, $n(U) = 100$. Step 3: Calculate the union $n(C \cup F \cup H)$: $n(C \cup F \cup H) = 50 + 30 + 20 - (10 + 5 + 10) + 3$ $= 100 - 25 + 3 = 78$ Step 4: Find the students who do not play any game by subtracting from the universal set: Students playing none $= n(U) - n(C \cup F \cup H) = 100 - 78 = 22$. Final answer: 22 students do not play any of the three games.
• Q: Show that if $P(A) = P(B)$, then $A = B$. A: Step 1: We will show $A \subseteq B$ and $B \subseteq A$. Step 2: To show $A \subseteq B$, let $x \in A$. This means $\{x\} \subseteq A$. Step 3: By definition of power set, $\{x\} \in P(A)$. Step 4: Since $P(A) = P(B)$, we have $\{x\} \in P(B)$. This implies $\{x\} \subseteq B$, which means $x \in B$. Thus, $A \subseteq B$. Step 5: Similarly, let $y \in B$. Then $\{y\} \subseteq B \implies \{y\} \in P(B)$. Step 6: Since $P(B) = P(A)$, $\{y\} \in P(A) \implies \{y\} \subseteq A \implies y \in A$. Thus, $B \subseteq A$. Final answer: Since $A \subseteq B$ and $B \subseteq A$, we have $A = B$ proved.

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