SETS NCERT Solutions Miscellaneous Exercise for Class 11 Maths provides a complete set of mixed questions covering all concepts of the chapter. This exercise is designed to test your understanding of sets, subsets, set operations, Venn diagrams, and laws of sets in one place. It is highly useful for revision and competitive exam preparation for engineering entrance tests.
Chapter 1 SETS Miscellaneous Exercise NCERT Solutions for Class 11 Maths
NCERT NCERT Question 1 : Decide, among the following sets, which sets are subsets of one another:
$A = \{x : x \in \mathbb{R} \text{ and } x \text{ satisfy } x^2 – 8x + 12 = 0\}$
$B = \{2, 4, 6\}$
$C = \{2, 4, 6, 8, \dots\}$
$D = \{6\}$
Solution :
First simplify set $A$:
Solve the equation
$$x^2 – 8x + 12 = 0$$
Factorizing,
$$(x – 6)(x – 2) = 0$$
So,
$$x = 6 \quad \text{or} \quad x = 2$$
Thus,
$$A = \{2, 6\}$$
A set $X$ is a subset of a set $Y$ if every element of $X$ is also an element of $Y$.
Therefore:
- $\{2,6\} \subset \{2,4,6\}$ ⇒ $A \subset B$
- $\{2,6\} \subset \{2,4,6,8,\dots\}$ ⇒ $A \subset C$
- $\{2,4,6\} \subset \{2,4,6,8,\dots\}$ ⇒ $B \subset C$
- $\{6\} \subset \{2,6\}$ ⇒ $D \subset A$
- $\{6\} \subset \{2,4,6\}$ ⇒ $D \subset B$
- $\{6\} \subset \{2,4,6,8,\dots\}$ ⇒ $D \subset C$
Final Answer
$$A \subset B,\quad A \subset C,\quad B \subset C,\quad D \subset A,\quad D \subset B,\quad D \subset C$$
NCERT Question 2: 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 $x \notin A$.
Solution :
(i) If $x \in A$ and $A \in B$, then $x \in B$.
Answer: False.
Counterexample: Let $x=1$, $A=\{1,2,3\}$ and $B=\{\{1,2,3\},\{4,5,6\}\}$. Then $x\in A$ and $A\in B$, but $x\notin B$ because elements of $B$ are sets, not the number $1$; specifically $1\notin B$.
(ii) If $A \subset B$ and $B \in C$, then $A \in C$.
Answer: False.
Counterexample: Let $A=\{1\}$, $B=\{1,3\}$ and $C=\{\{1,3\},\{5,7\}\}$. Then $A\subset B$ and $B\in C$, but $A\notin C$ since $C$ contains $\{1,3\}$ and $\{5,7\}$, not $\{1\}$.
(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$.
Answer: True.
Proof: $A\subset B$ means every element of $A$ is in $B$. $B\subset C$ means every element of $B$ is in $C$. Therefore every element of $A$ (being in $B$) is also in $C$. Hence $A\subset C$.
(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$.
Answer: False.
Counterexample: Let $A=\{1,2\}$, $B=\{3,4,5\}$ and $C=\{1,2,6,7\}$. Here $A\not\subset B$ (because $1\notin B$) and $B\not\subset C$ (because $3\notin C$), yet $A\subset C$ (every element $1,2$ of $A$ is in $C$). Thus the given implication is false.
(v) If $x \in A$ and $A \not\subset B$, then $x \in B$.
Answer: False.
Counterexample: Let $x=1$, $A=\{1,2\}$ and $B=\{3,4,5\}$. Then $x\in A$ and $A\not\subset B$ (since $1\notin B$), but $x\notin B$. So the statement is false.
(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$.
Answer: True.
Proof : $A\subset B$ means every element of $A$ is also an element of $B$. If $x\notin B$, then $x$ cannot be an element of any subset $A$ of $B$. Hence $x\notin A$.
NCERT Question 3 : Let $A, B,$ and $C$ be the sets such that
$$A \cup B = A \cup C \quad \text{and} \quad A \cap B = A \cap C.$$
Show that $B = C$.
Solution :
Given:
$$A \cup B = A \cup C \quad …(1)$$
$$A \cap B = A \cap C \quad …(2)$$
We must show:
$$B = C$$
Proof
Let $x \in B$.
We consider two cases:
Case 1: $x \in A$
Since $x \in B$ and $x \in A$, then
$$x \in A \cap B$$
Using (2):
$$A \cap B = A \cap C \Rightarrow x \in A \cap C$$
Thus,
$$x \in C$$
Case 2: $x \notin A$
Since $x \in B$, then
$$x \in A \cup B$$
Using (1):
$$A \cup B = A \cup C \Rightarrow x \in A \cup C$$
As $x \notin A$, it must be that:
$$x \in C$$
From both cases, we have
$$x \in B \Rightarrow x \in C \quad \Rightarrow \quad B \subset C$$
Similarly, reversing the role of $B$ and $C$, we get:
$$C \subset B$$
Thus,
$$B \subset C \quad \text{and} \quad C \subset B$$
Therefore,
$$\boxed{B = C}$$
NCERT Question 4 : Show that the following four conditions are equivalent :
(i) $A \subset B$
(ii) $A – B = \varnothing$
(iii) $A \cup B = B$
(iv) $A \cap B = A$
Solution :
We will show each condition implies the other.
(i) ⇒ (ii)
If $A \subset B$, then every element of $A$ is also in $B$.
So, $A$ has no element outside $B$.
Therefore,
$$A – B = \varnothing$$
(i) ⇒ (iii)
Since all elements of $A$ are in $B$:
$$A \cup B = B$$
(i) ⇒ (iv)
Since $A \subset B$, every element of $A$ lies in $B$, so:
$$A \cap B = A$$
Now, reverse implications:
(ii) ⇒ (i)
Given $A – B = \varnothing$, there is no element of $A$ outside $B$.
Hence,
$$A \subset B$$
(iii) ⇒ (i)
From $A \cup B = B$, every element of $A$ is already in $B$.
Thus,
$$A \subset B$$
(iv) ⇒ (i)
From $A \cap B = A$, every element of $A$ also belongs to $B$.
Thus,
$$A \subset B$$
Final Conclusion
All four statements are logically equivalent:
$$A \subset B \iff A – B = \varnothing \iff A \cup B = B \iff A \cap B = A$$
NCERT Question 5 : Show that if $A \subset B$, then $C – B \subset C – A$
Solution :
Let
$$A = \{1,2\}, \quad B = \{1,2,3,4,5\} \quad \text{and} \quad C = \{2,5,6,7,8\}$$
Since every element of $A$ is in $B$:
$$A \subset B$$
Now find the difference sets:
Elements in $C$ but not in $B$:
$$C – B = \{6,7,8\}$$
Elements in $C$ but not in $A$:
$$C – A = \{5,6,7,8\}$$
Checking subset relation
$\{6,7,8\}$ — these elements all appear in $\{5,6,7,8\}$
So we can clearly see:
$$C – B \subset C – A$$
Final Answer
$$\boxed{C – B \subset C – A}$$
NCERT Question 6 : Assume that $P(A) = P(B)$. Show that $A = B$
Solution :
$P(X)$ represents the power set of a set $X$.
Power set means the set of all subsets of a given set.
Example:
If $A = \{1,2\}$, then
$$P(A) = \{\varnothing, \{1\}, \{2\}, \{1,2\}\}$$
To prove $A = B$, we must prove:
- $A \subset B$
- $B \subset A$
Since $A$ is a set,
$$A \in P(A)$$
Given:
$$P(A) = P(B)$$
So:
$$A \in P(B)$$
That means $A$ is a subset of $B$:
$$A \subset B \quad …(1)$$
Similarly,
$$B \in P(B)$$
Since $P(A) = P(B)$,
$$B \in P(A)$$
So $B$ is a subset of $A$:
$$B \subset A \quad …(2)$$
From (1) and (2):
$$A \subset B \quad \text{and} \quad B \subset A \Rightarrow A = B$$
Final Answer
$$\boxed{A = B}$$
Question 7: Is it true that for any sets $A$ and $B$, $P(A) \cup P(B) = P(A \cup B)$? Justify your answer
Solution:
It is False.
Let:
$$A = \{1, 2\}$$
$$B = \{2, 3\}$$
Then:
$$A \cup B = \{1, 2, 3\}$$
Power sets:
$$P(A) = \{\varnothing, \{1\}, \{2\},\{1, 2\}\}$$
$$P(B) = \{\varnothing, \{2\}, \{3\}, \{2, 3\}\}$$
Their union:
$$P(A) \cup P(B) = \{\varnothing, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}\} \quad \text{…(1)}$$
Now:
$$P(A \cup B) = \{\varnothing, \{1\}, \{2\}, \{3\},\{1, 2\}, \{2, 3\}, \{1, 3\}, \{1, 2, 3\}\} \quad \text{…(2)}$$
From (1) and (2):
$$P(A) \cup P(B) \ne P(A \cup B)$$
Hence, the statement is False.
NCERT Question 8: Show that for any sets $A$ and $B$,
$A = (A \cap B) \cup (A – B)$ and $A \cup (B – A) = (A \cup B)$
Solution:
Proof 1:
$$A = (A \cap B) \cup (A – B)$$
$$A =(A \cap B) \cup (A – B)$$
$$A =(A \cap B) \cup (A \cap B’) $$
since $(A – B) = (A \cap B’)$
$$A = A \cap (B \cup B’)$$
$$A = A \cap (U) = A $$
Since, $ B \cup B’ = U \text{ (Universal set)}$
Hence, it is proved that:
$$A = (A \cap B) \cup (A – B)$$
Proof 2:
$$ A \cup (B – A) = (A \cup B) $$
$$A \cup (B – A) = A \cup (B \cap A’) $$
since $(B – A) = (B \cap A’)$
$$(A \cup B) \cap (A \cup A’) \quad \text{// distributive law}$$
$$(A \cup B) \cap U \quad \text{// } A \cup A’ = U \text{ (Universal set)}$$
$$(A \cup B)$$
Hence, it is proved that:
$$A \cup (B – A) = (A \cup B)$$
NCERT Question 9: Using properties of sets, show that
(i) $A \cup (A \cap B) = A$
(ii) $A \cap (A \cup B) = A$
Solution:
(i) $A \cup (A \cap B)$
$$A \cup (A \cap B)$$
$$(A \cup A) \cap (A \cup B) \quad \text{// distributive law}$$
$$A \cap (A \cup B) \quad \text{// } A \cup A = A$$
$$A \quad \text{// absorption law}$$
Hence,
$$A \cup (A \cap B) = A$$
(ii) $A \cap (A \cup B)$
$$A \cap (A \cup B)$$
$$(A \cap A) \cup (A \cap B) \quad \text{// distributive law}$$
$$A \cup (A \cap B) \quad \text{// } A \cap A = A$$
$$A \quad \text{// absorption law}$$
Hence,
$$A \cap (A \cup B) = A$$
NCERT Question 10: Show that $A \cap B = A \cap C$
need not imply $B = C$
Solution:
Let us take a counterexample.
Let
$$A = \{1,2\}, \quad B = \{2,3\}, \quad C = \{2,4\}$$
Here, clearly $B \ne C$.
Now,
$$A \cap B = \{2\}$$
$$A \cap C = \{2\}$$
So,
$$A \cap B = A \cap C \quad \text{but} \quad B \ne C$$
Therefore, it is not necessary that $B = C$ whenever $A \cap B = A \cap C$.
Important Chapter Links
Sets NCERT Solutions Miscellaneous Exercise for Class 11 Maths covers all major topics of the chapter including representation of sets, types of sets, subsets, power sets, set operations, Venn diagrams, and laws of sets. This section includes a variety of mixed questions that help students revise the entire chapter and test their conceptual understanding. With detailed step-by-step solutions by Maths Anand Classes, students can improve accuracy and problem-solving skills. It is highly useful for CBSE exams as well as competitive exams of India.
FAQs of Chapter-1 Sets Miscellaneous Exercise NCERT Solutions for Class 11 Maths
What is the Miscellaneous Exercise in Sets?
It is a mixed set of questions covering all topics of the chapter for revision.
Why is this exercise important?
It helps in testing overall understanding and improves problem-solving skills.
Does it include all concepts of Sets?
Yes, it covers all major concepts of the chapter.
Are Venn diagrams required in this exercise?
Yes, many problems can be solved using Venn diagrams.
Is this useful for competitive exams?
Yes, it is very useful for exams like JEE, NDA, IMUCET, and Merchant Navy.
Are solutions by Maths Anand Classes reliable?
Yes, they are designed to be clear, accurate, and exam-oriented.
How should I practice this exercise?
Attempt questions first, then check solutions to improve accuracy.
Is this exercise difficult?
It may be slightly challenging as it combines all concepts.
Exercise-wise NCERT Solutions
- Basic definition of sets
- Writing sets in roster and set-builder form
- Types of sets
- Finite and infinite sets
- Subsets and proper subsets
- Number of subsets
- Set operations (union, intersection, complement)
- Advanced problems on set operations and Venn diagrams
Miscellaneous Exercise
- Mixed problems covering all concepts of the chapter