Rbse Solutions for Class 11 maths Chapter 1 Miscellaneous Exercise | Sets

Last Updated on November 24, 2025 by Aman Singh

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 1 Miscellaneous Exercise (Sets). Learn to prove fundamental set properties (Absorption Law, Distributive Law), demonstrate set equivalences (A⊂B⟺A∪B=B), and disprove statements using effective counterexamples. Essential practice for mastering subset relations and set algebra proofs.

This exercise covers fundamental concepts and properties of Sets, including subsets, set operations (union, intersection, difference), and proofs involving equality and equivalence.

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1. Deciding Subset Relationships

First, we define the elements of each set clearly:

• Set A: $A = \{x : x \in R \text{ and } x^2 – 8x + 12 = 0 \}$.We solve the quadratic equation: $x^2 – 8x + 12 = 0 \implies (x-2)(x-6) = 0$.Thus, $\mathbf{A = \{2, 6\}}$.
• Set B: $\mathbf{B = \{2, 4, 6\}}$.
• Set C: $\mathbf{C = \{2, 4, 6, 8, \dots \}}$ (The set of positive even integers).
• Set D: $\mathbf{D = \{6\}}$.

Now, we check the subset relationships ($\subset$):

1. A and B: Since every element of $A$ (2 and 6) is in $B$, $A \subset B$ is true.
2. A and C: Since every element of $A$ (2 and 6) is in $C$, $A \subset C$ is true.
3. A and D: Since $2 \in A$ but $2 \notin D$, $A \not\subset D$.
4. B and C: Since every element of $B$ (2, 4, 6) is in $C$, $B \subset C$ is true.
5. C and D: Since $2 \in C$ but $2 \notin D$, $C \not\subset D$.
6. D and A: Since $6 \in D$ and $6 \in A$, $D \subset A$ is true.
7. D and B: Since $6 \in D$ and $6 \in B$, $D \subset B$ is true.
8. D and C: Since $6 \in D$ and $6 \in C$, $D \subset C$ is true.

Conclusion: The subset relations are $D \subset A \subset B \subset C$ (and $D \subset B$, $D \subset C$).

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2. True or False Statements (with Proof/Counterexample)

(i) If $x \in A$ and $A \in B$, then $x \in B$.False. This confuses element membership with set membership.Counterexample: Let $A = \{1\}$ and $B = \{\{1\}, 2\}$.

Here, $x=1$ is an element of $A$ ($1 \in A$). $A$ is an element of $B$ ($A \in B$).

But $x=1$ is not an element of $B$ (the elements of $B$ are $\{1\}$ and $2$). $1 \notin B$.

(ii) If $A \subset B$ and $B \in C$, then $A \in C$.False. This again confuses subset with element membership.Counterexample: Let $A = \{1\}$, $B = \{1, 2\}$, and $C = \{\{1, 2\}, 3\}$.

Here, $A \subset B$ is true. $B$ is an element of $C$ ($B \in C$).

But $A$ is not an element of $C$ (the elements of $C$ are $\{1, 2\}$ and $3$). $A \notin C$.

(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$.True. This is the transitive property of subsets.Proof: Let $x$ be an arbitrary element such that $x \in A$.

Since $A \subset B$, $x \in B$.

Since $B \subset C$, and $x \in B$, it follows that $x \in C$.

Since $x \in A$ implies $x \in C$, we conclude that $A \subset C$.

(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$.False. The non-subset relationship is not transitive.Counterexample: Let $A = \{1, 2\}$, $B = \{2, 3\}$, and $C = \{1, 2, 4\}$.

• $A \not\subset B$ (since $1 \in A$ but $1 \notin B$).
• $B \not\subset C$ (since $3 \in B$ but $3 \notin C$).
• However, $A \subset C$ is True (since $1 \in C$ and $2 \in C$).

(v) If $x \in A$ and $A \not\subset B$, then $x \in B$.False. If $A \not\subset B$, it means there is at least one element in $A$ that is not in $B$. The element $x$ could be that element.Counterexample: Let $A = \{1, 2\}$ and $B = \{2, 3\}$.

Here, $A \not\subset B$. Let $x=1$. $x \in A$ is true, but $x \notin B$ is true.

(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$.True. This is the contrapositive of the definition of $A \subset B$.Proof (by contradiction):

Assume $A \subset B$ and $x \notin B$ are true, but the conclusion $x \notin A$ is false.

If $x \notin A$ is false, then $x \in A$ must be true.

Since $x \in A$ and we are given $A \subset B$, it must follow that $x \in B$.

But this contradicts our initial premise that $x \notin B$.

Therefore, our initial assumption that the conclusion is false must be wrong, so $x \notin A$ is true.

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3. Show that if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.

We need to show that $B \subset C$ and $C \subset B$.

Part 1: Show $B \subset C$

Let $x$ be an arbitrary element such that $x \in B$. We consider two cases for $x$:

• Case 1: $x \in A$If $x \in B$ and $x \in A$, then $x \in A \cap B$.Since $A \cap B = A \cap C$, we have $x \in A \cap C$.If $x \in A \cap C$, then $x \in C$.
• Case 2: $x \notin A$If $x \in B$, then $x \in A \cup B$.Since $A \cup B = A \cup C$, we have $x \in A \cup C$.$x \in A \cup C$ means $x \in A$ or $x \in C$.Since we are in the case $x \notin A$, it must be that $x \in C$.

In both cases, $x \in C$. Therefore, $B \subset C$.

Part 2: Show $C \subset B$

Similarly, let $y$ be an arbitrary element such that $y \in C$.

• Case 1: $y \in A$If $y \in C$ and $y \in A$, then $y \in A \cap C$.Since $A \cap C = A \cap B$, we have $y \in A \cap B$.If $y \in A \cap B$, then $y \in B$.
• Case 2: $y \notin A$If $y \in C$, then $y \in A \cup C$.Since $A \cup C = A \cup B$, we have $y \in A \cup B$.$y \in A \cup B$ means $y \in A$ or $y \in B$.Since we are in the case $y \notin A$, it must be that $y \in B$.

In both cases, $y \in B$. Therefore, $C \subset B$.

Since $B \subset C$ and $C \subset B$, we conclude that $B = C$.

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4. Show that the following four conditions are equivalent: (i) $A \subset B$, (ii) $A – B = \phi$, (iii) $A \cup B = B$, (iv) $A \cap B = A$.

We show the chain of equivalence: (i) $\iff$ (ii) $\iff$ (iii) $\iff$ (iv).

A. (i) $A \subset B \iff$ (ii) $A – B = \phi$

• ($\implies$): Assume $A \subset B$. $A – B$ contains elements $x$ such that $x \in A$ and $x \notin B$. If $A \subset B$, every element of $A$ is in $B$, so there are no elements $x$ such that $x \in A$ and $x \notin B$. Hence, $A – B = \phi$.
• ($\impliedby$): Assume $A – B = \phi$. This means there are no elements $x$ such that $x \in A$ and $x \notin B$. This is the definition of $A \subset B$.

B. (i) $A \subset B \iff$ (iii) $A \cup B = B$

• ($\implies$): Assume $A \subset B$. The union $A \cup B$ contains all elements in $A$ or in $B$. Since $A$ is entirely contained in $B$, adding $A$ to $B$ yields no new elements beyond those in $B$. Hence, $A \cup B = B$.
• ($\impliedby$): Assume $A \cup B = B$. If $A \cup B = B$, it implies that all elements of $A$ must be present in $B$ for the union to equal $B$. Hence, $A \subset B$.

C. (i) $A \subset B \iff$ (iv) $A \cap B = A$

• ($\implies$): Assume $A \subset B$. The intersection $A \cap B$ contains all elements common to $A$ and $B$. Since $A$ is entirely contained in $B$, all elements of $A$ are common to both sets. Hence, $A \cap B = A$.
• ($\impliedby$): Assume $A \cap B = A$. If the intersection of $A$ and $B$ results in $A$, it means that every element of $A$ must also be an element of $B$. Hence, $A \subset B$.

Since all four conditions are equivalent to $A \subset B$, they are equivalent to one another.

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5. Show that if $A \subset B$, then $C – B \subset C – A$.

We need to show that every element in $C – B$ is also an element in $C – A$.

Let $x$ be an arbitrary element such that $x \in C – B$.

By the definition of set difference, this means:

1. $x \in C$
2. $x \notin B$

We are given the premise $A \subset B$. By the contrapositive property (from Q.2(vi)), if $A \subset B$ and $x \notin B$, then $x \notin A$.

Since $x \notin B$ (from step 2) and $A \subset B$ (given), we must have:

3. $x \notin A$

Combining (1) $x \in C$ and (3) $x \notin A$, by the definition of set difference, we get:

$$x \in C – A$$

Since $x \in C – B$ implies $x \in C – A$, we conclude that $C – B \subset C – A$.

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6. Show that for any sets $A$ and $B$,

(i) $A = (A \cap B) \cup (A – B)$

We show both sides are subsets of each other.

• Proof $A \subset (A \cap B) \cup (A – B)$:Let $x \in A$. We consider two possibilities: either $x \in B$ or $x \notin B$.
  • If $x \in B$, then $x \in A$ and $x \in B$, so $x \in A \cap B$.
  • If $x \notin B$, then $x \in A$ and $x \notin B$, so $x \in A – B$.In either case, $x$ is in $A \cap B$ or $A – B$, so $x \in (A \cap B) \cup (A – B)$.
• Proof $(A \cap B) \cup (A – B) \subset A$:Let $x \in (A \cap B) \cup (A – B)$. This means $x \in (A \cap B)$ or $x \in (A – B)$.
  • If $x \in A \cap B$, then $x \in A$ and $x \in B$. Hence, $x \in A$.
  • If $x \in A – B$, then $x \in A$ and $x \notin B$. Hence, $x \in A$.In both cases, $x \in A$.

Since both conditions hold, $A = (A \cap B) \cup (A – B)$ is proven.

(ii) $A \cup (B – A) = A \cup B$

We use set algebra properties.

$$A \cup (B – A) = A \cup (B \cap A’)$$

Using the Distributive Law of union over intersection:

$$A \cup (B \cap A’) = (A \cup B) \cap (A \cup A’)$$

Since $A \cup A’$ is the universal set $U$:

$$(A \cup B) \cap (A \cup A’) = (A \cup B) \cap U$$

The intersection of any set with the universal set is the set itself:

$$(A \cup B) \cap U = \mathbf{A \cup B}$$

Thus, $A \cup (B – A) = A \cup B$ is proven.

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7. Using properties of sets, show that

(i) $A \cup (A \cap B) = A$ (Absorption Law)

Using the Identity Law ($A = A \cap U$) and the Distributive Law of union over intersection:

$$A \cup (A \cap B) = (A \cap U) \cup (A \cap B)$$

$$A \cup (A \cap B) = A \cap (U \cup B)$$

Since $U \cup B = U$ (Universal set property):

$$A \cap (U \cup B) = A \cap U$$

Since $A \cap U = A$ (Identity Law):

$$A \cap U = \mathbf{A}$$

(ii) $A \cap (A \cup B) = A$ (Absorption Law)

Using the Identity Law ($A = A \cup \phi$) and the Distributive Law of intersection over union:

$$A \cap (A \cup B) = (A \cup \phi) \cap (A \cup B)$$

$$A \cap (A \cup B) = A \cup (\phi \cap B)$$

Since $\phi \cap B = \phi$ (Null set property):

$$A \cup (\phi \cap B) = A \cup \phi$$

Since $A \cup \phi = A$ (Identity Law):

$$A \cup \phi = \mathbf{A}$$

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8. Show that $A \cap B = A \cap C$ need not imply $B = C$.

We need to provide a counterexample where the intersection equality holds, but the sets $B$ and $C$ are different.

Counterexample:

Let the universal set $U = \{1, 2, 3, 4\}$.

Let $A = \{1\}$

Let $B = \{1, 2\}$

Let $C = \{1, 3\}$

Now, check the intersections:

$$A \cap B = \{1\} \cap \{1, 2\} = \{1\}$$

$$A \cap C = \{1\} \cap \{1, 3\} = \{1\}$$

We have $A \cap B = A \cap C$ (since both equal $\{1\}$).

However, $B \ne C$ (since $2 \in B$ but $2 \notin C$).

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9. Show that if $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$, then $A = B$.

We are given:

1. $A \cap X = \phi$
2. $B \cap X = \phi$
3. $A \cup X = B \cup X$

We start by trying to express $A$ and $B$ using the given equations and set properties (as hinted).

Step 1: Express $A$

We know $A$ is a subset of $A \cup X$, so $A = A \cap (A \cup X)$.

Using the given $A \cup X = B \cup X$:

$$A = A \cap (B \cup X)$$

Now, apply the Distributive Law of intersection over union:

$$A = (A \cap B) \cup (A \cap X)$$

Using (1), $A \cap X = \phi$:

$$A = (A \cap B) \cup \phi$$

By the Identity Law, $A \cup \phi = A$:(I) $A = A \cap B$

From (I), by equivalence in Q.4(iv), we conclude $\mathbf{A \subset B}$.

Step 2: Express $B$

Similarly, $B = B \cap (B \cup X)$.

Using the given $B \cup X = A \cup X$:

$$B = B \cap (A \cup X)$$

Apply the Distributive Law:

$$B = (B \cap A) \cup (B \cap X)$$

Using (2), $B \cap X = \phi$:

$$B = (B \cap A) \cup \phi$$

(II) $B = B \cap A$

From (II), since $A \cap B = B \cap A$, we have $B = A \cap B$.

Step 3: Conclusion

From (I), $A = A \cap B$.

From (II), $B = A \cap B$.

Since $A$ and $B$ are both equal to the same set $A \cap B$, we conclude that $A = B$.

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10. Find sets $A, B$ and $C$ such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C = \phi$.

We need three sets where every pair has at least one common element, but there is no element common to all three sets.

Example Solution:

Let $A, B, C$ contain the following unique elements:

• $A = \{1, 2\}$
• $B = \{2, 3\}$
• $C = \{1, 3\}$

Now, check the conditions:

1. $A \cap B$: $\{1, 2\} \cap \{2, 3\} = \{2\}$. This is non-empty.
2. $B \cap C$: $\{2, 3\} \cap \{1, 3\} = \{3\}$. This is non-empty.
3. $A \cap C$: $\{1, 2\} \cap \{1, 3\} = \{1\}$. This is non-empty.
4. $A \cap B \cap C$: $(\{1, 2\} \cap \{2, 3\}) \cap \{1, 3\} = \{2\} \cap \{1, 3\} = \mathbf{\phi}$.

All conditions are satisfied.