Courses
Courses for Kids
Free study material
Offline Centres
More
Store

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 - Sets

Last updated date: 12th Aug 2024
Total views: 11.1k
Views today: 0.11k

NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Miscellaneous Exercise Class 11 Chapter 1 Sets includes solutions to all Miscellaneous Exercise problems. The Miscellaneous Exercises NCERT Solutions for Class 11 Maths are based on the concepts presented in Maths Chapter 1 Sets. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the latest CBSE Class 11 Maths Syllabus in PDF format and practice it offline.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 1 Sets
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 1: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 1 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs
Competitive Exams after 12th Science

Access NCERT Solutions for Class 11 Maths Chapter 1 Sets

Miscellaneous Exercise

1. Decide among the following sets, which sets are the subsets of one and another:

$A=\left\{ x:x\in R\text{ satisfy }{{\text{x}}^{2}}-8x+12=0 \right\}$

$B=\left\{ 2,4,6 \right\}$

$C=\left\{ 2,4,6,8,... \right\},D=\left\{ 6 \right\}$

Ans: Given that,

$A=\left\{ x:x\in R\text{ satisfy }{{\text{x}}^{2}}-8x+12=0 \right\},B=\left\{ 2,4,6 \right\}$

$C=\left\{ 2,4,6,8,... \right\},D=\left\{ 6 \right\}$

$A=\left\{ x:x\in R\text{ satisfy }{{\text{x}}^{2}}-8x+12=0 \right\}$

${{x}^{2}}-8x+12=0$

$\left( x-2 \right)\left( x-6 \right)=0$

$x=2,6$

$A=\left\{ 2,6 \right\},B=\left\{ 2,4,6 \right\},$  $C=\left\{ 2,4,6,8,... \right\},D=\left\{ 6 \right\}$

A set A is said to be a subset of B if every element of A is also an element of B

$A\subset B$ if $a\in A,a\in B$

We can observe that,

$D\subset A\subset B\subset C$

$\therefore D\subset A,B,C\text{ and A}\subset B,C\text{ and B}\subset C$

2. In each of the following statements, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

i. If $x\in A\text{ and A}\in \text{B, then x}\in \text{B}$

Ans: $x\in A\text{ and A}\in \text{B, then x}\in \text{B}$

To determine whether the statement is true

The given statement is false

For example,

Let $A=\left\{ 1,2 \right\},B=\left\{ 1,\left\{ 1,2 \right\},\left\{ 3 \right\} \right\}$

Now $2\in \left\{ 1,2 \right\},\left\{ 1,2 \right\}\in \left\{ 1,\left\{ 1,2 \right\},\left\{ 3 \right\} \right\}$

But $2\notin \left\{ 1,\left\{ 1,2 \right\},\left\{ 3 \right\} \right\}$

Hence the given statement is false.

ii. If $A\subset B\text{ and }B\in C,thenA\in C$

Ans: $A\subset B\text{ and }B\in C,thenA\in C$

To determine whether the statement is true

The statement is false

For example,

Let $A=\left\{ 2 \right\},B=\left\{ 0,2 \right\},C=\left\{ 1,\left\{ 0,2 \right\},3 \right\}$

As $A\subset B,B\in C$ but $A\notin C$

iii. If $A\subset B\text{ and }B\subset C,thenA\subset C$

Ans:

$A\subset B\text{ and }B\subset C,thenA\subset C$

To determine whether the statement is true

The given statement is true.

Let $A\subset B,B\subset C$

Let $x\in A$

$x\in B$so $A\subset B$

$x\in C$so $B\subset C$

iv. If $A\not\subset B\text{ and }B\not\subset C,thenA\not\subset C$

Ans: $A\not\subset B\text{ and }B\not\subset C,thenA\not\subset C$

To determine whether the statement is true

The given statement is false

Let $A=\left\{ 1,2 \right\},B=\left\{ 0,6,8 \right\},C=\left\{ 0,1,2,6,9 \right\}$

Now by the statement,

$A\not\subset B\text{ and }B\not\subset C$

But $A\subset C$

v. If $x\in A\text{ and A}\not\subset \text{B, then x}\in \text{B}$

Ans: $x\in A\text{ and A}\not\subset \text{B, then x}\in \text{B}$

To determine whether the statement is true

The given statement is false

Let $A=\left\{ 3,5,7 \right\},B=\left\{ 3,4,6 \right\}$

Now, $5\in A$ and $A\not\subset B$

But $5\notin B$

vi. If $\text{A}\subset \text{B and x}\notin \text{B, then x}\notin \text{A}$

Ans: $\text{A}\subset \text{B and x}\notin \text{B, then x}\notin \text{A}$

To determine whether the statement is true

The given statement is true

Let $A\subset B$ and $x\notin B$

To show that $x\notin A$

Suppose $x\in A$

Then $x\in B$ which is a contradiction

$\therefore x\notin A$

3. Let $A,B$ and $C$ be the set such that $A\cup B=A\cup C$ and $A\cap B=A\cap C$. Show that $B=C$

Ans: To show that $B=C$

Let $x\in B$

$x\in A\cup B$ since $\left[ B\subset A\cup B \right]$

$x\in A\cup C$ since $\left[ A\cup B=A\cup C \right]$

$x\in A\text{ or }\in \text{C}$

Case I:

Also $x\in B$

$x\in A\cap B$ since $\left[ B\subset A\cap B \right]$

$x\in A\cap C$ since $\left[ A\cap B=A\cap C \right]$

$x\in A\text{ or x}\in \text{C}$

$\therefore x\in C$

$\therefore B\subset C$

Similarly, we can show that $C\subset B$

$\therefore$It has been proved that $B=C$

4. Show that the following four conditions are equivalent:

i. $A\subset B$

ii. $A-B=\varnothing$

iii. $A\cup B=B$

iv. $A\cap B=A$

Ans: To show that the above four conditions are equivalent

First showing that,

(i)⬄(ii),

Let $A\subset B$

To show,

$A-B=\varnothing$

If possible, suppose,

$A-B\ne \varnothing$

This means that there exist $x\in A,x\notin B$, which is impossible as $A\subset B$

$\therefore A-B=\varnothing$

$A\subset B$=> $A-B=\varnothing$

Let $A-B=\varnothing$

To show that,

$A\subset B$

Let $x\in B$ since if $x\notin B,$ then $A-B\ne \varnothing$

$\therefore$  $A-B=\varnothing$ => $A\subset B$

$\therefore$(i)⬄ (iii)

Let $A\subset B$

To show $A\cup B=B$

Clearly$B\subset A\cup B$

Let $x\in A\cup B$

$x\in A$ or $x\in B$

Case I:

$x\in A$

So that $x\in B$

$\therefore A\cup B\subset B$

Case II:

$x\in B$

Then $A\cup B=B$

Conversely let $A\cup B=B$

Let $x\in A$

$x\in A\cup B$ since $\left[ A\subset A\cup B \right]$

$x\in B$ since $A\cup B=B$

$\therefore A\subset B$

Hence (i) ⬄ (iii)

Now we have to show that (i) ⬄ (iv)

Let $A\subset B$

$A\cap B\subset A$

Let $x\in A$

We have to show that $x\in A\cap B$

As $A\subset B$, $x\in B$

$x\in A\cap B$

$A\subset A\cap B$

Hence $A=A\cap B$

Let $x\in A$

$x\in A\cap B$

$x\in A$ and $x\in B$

So that $x\in B$

$\therefore A\subset B$

Hence (i) ⬄ (iv)

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

Ans: Given that,

$A\subset B$

To show that,

$C-B\subset C-A$

Let $x\in C-B$

$x\in C$ and $x\in B$

$x\notin A\left[ A\subset B \right]$ and $x\in C$

$x\in C-A$

$\therefore C-B\subset C-A$

Hence it has been showed that $C-B\subset C-A$

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

$A=\left( A\cap B \right)\cup \left( A-B \right)$ and $A\cup \left( B-A \right)=A\cup B$

Ans: To show that,

$A=\left( A\cap B \right)\cup \left( A-B \right)$

Let $x\in A$

We have to show that $A=\left( A\cap B \right)\cup \left( A-B \right)$

Case I:

$x\in A\cap B$

Then $x\in A\cap B\subset \left( A\cup B \right)\cup \left( A-B \right)$

Case II:

$x\notin A\cap B$

$x\notin A$ or $x\notin B$

$x\notin B\left[ x\notin A \right]$

$x\notin A-B\subset \left( A\cup B \right)\cup \left( A-B \right)$

$\therefore A\subset \left( A\cap B \right)\cup \left( A-B \right)$                                          ……(1)

It is clear that,

$A\cap B\subset A$ and $A-B\subset A$

$\therefore \left( A\cap B \right)\cup \left( A-B \right)\subset A$                                          ……(2)

From (1) and (2) we obtain that,

$A=\left( A\cap B \right)\cup \left( A-B \right)$

To prove that,

$A\cup \left( B-A \right)\subset A\cup B$

Let $x\in A\cup \left( B-A \right)$

$x\in A$ or ($x\in B$ and $x\notin A$)

($x\in A$ or $x\in B$) and ($x\in A$ or $x\notin A$)

$x\in A\cup B$

$\therefore A\cup \left( B-A \right)\subset A\cup B$                                             ……(3)

Next we show that $A\cup \left( B-A \right)\subset A\cup B$

Let $y\in A\cup B$

$y\in A$ or $y\in B$

($y\in A$ or $y\in B$) and ($y\in A$or $y\notin A$)

$y\in A\cup \left( B-A \right)$

$\therefore A\cup \left( B-A \right)\subset A\cup B$

Hence from (3) and (4) we obtain that$A\cup \left( B-A \right)=A\cup B$

Hence proved the statement

7. Using properties of sets, show that

i. $A\cup \left( A\cap B \right)=A$

Ans: To show that, $A\cup \left( A\cap B \right)=A$ by using the property offsets

We know that,

$A\subset A$

$A\cap B\subset A$

$\therefore A\cup \left( A\cap B \right)\subset A$                             ……(1)

And $A\subset A\cup \left( A\cap B \right)$                         ……(2)

From (1) and (2) we get that

$A\cup \left( A\cap B \right)=A$

Hence it has been showed that $A\cup \left( A\cap B \right)=A$

ii. $A\cap \left( A\cup B \right)=A$

Ans: To show that $A\cap \left( A\cup B \right)=A$ by using the property of sets

$A\cap \left( A\cup B \right)=\left( A\cap A \right)\cup \left( A\cap B \right)$

$=A\cup \left( A\cap B \right)$

$=A$(from (1))

Hence it has been shown that $A\cap \left( A\cup B \right)=A$

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

Ans: Given that,

$A\cap B=A\cap C$

To show that the above does not imply $B=C$

Let $A=\left\{ 0,1 \right\},B=\left\{ 0,2,3 \right\},C=\left\{ 0,4,5 \right\}$

Now,

$A\cap B=\left\{ 0 \right\}=A\cap C$

But $B\ne C$ since $2\in B$ and not to $C$

$\therefore$It has been shown that $A\cap B=A\cap C$ does not necessarily imply $B=C$

9. Let $A$ and $B$ be sets. If $A\cap X=B\cap X=\varnothing$ and $A\cup X=B\cup X$ for some text $X$, show that $A=B$

(Hints: $A=A\cap \left( A\cup X \right),B=B\cap \left( B\cup X \right)$ and use distributive law)

Ans: Given that,

$A\cap X=B\cap X=\varnothing$

$A\cup X=B\cup X$

To show that,

$A=B$

We know that,

$A=A\cap \left( A\cup X \right)$

$=A\cap \left( B\cup X \right)$ since $A\cup X=B\cup X$

$=\left( A\cap B \right)\cup \left( A\cap X \right)$ By distributive law

$=\left( A\cap B \right)\cup \varnothing$ since $A\cap X=\varnothing$

$=A\cap B$                                              ……(1)

Now consider$B=B\cap \left( B\cup X \right)$

$=B\cap \left( A\cup X \right)$ since $B\cup X=A\cup X$

$=\left( B\cap A \right)\cup \left( B\cap X \right)$ By distributive law

$=\left( B\cap A \right)\cup \varnothing$ since $B\cap X=\varnothing$

$=B\cap A$

$=A\cap B$                                               ……(2)

It is possible only when $A=B$

So from (1) and (2), we get,

$A=B$

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

Ans: Given that,

$A\cap B,B\cap C,A\cap C$ are non-empty subsets and

$A\cap B\cap C=\varnothing$

Let $A=\left\{ 0,2 \right\},B=\left\{ 1,2 \right\},C=\left\{ 2,0 \right\}$

Now,

$A\cap B=\left\{ 1 \right\}$

$B\cap C=\left\{ 1,2 \right\}$

$A\cap C=\left\{ 0 \right\}$

And $A\cap B\cap C=\varnothing$

$\therefore$It has been showed that $A\cap B,B\cap C$ and $A\cap C$ are non-empty subsets but however $A\cap B\cap C=\varnothing$

Conclusion

NCERT Miscellaneous Exercise Class 11 Chapter 1 Maths is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

Class 11 Maths Chapter 1: Exercises Breakdown

 Exercise Number of Questions Exercise 1.1 6 Questions & Solutions Exercise 1.2 6 Questions & Solutions Exercise 1.3 8 Questions & Solutions Exercise 1.4 12 Questions & Solutions Exercise 1.5 7 Questions & Solutions

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 - Sets

1. What is the best way to use Class 11 Maths Ch 1 Miscellaneous Exercise Solutions for this exercise?

First, try to solve the problems on your own. Then, check your answers with the NCERT solutions to see if you did them correctly. This helps you learn and correct any mistakes.

2. What tips can help with solving the Class 11 Maths Ch 1 Miscellaneous Exercise Solutions?

Some tips include using Venn diagrams to understand set operations, practising different types of functions, and breaking down complex problems into smaller steps. Regular practice and reviewing solutions can make solving these problems easier.

3. What is the difference between a subset and a proper subset in Class 11 Miscellaneous Exercise Chapter 1?

A subset includes all elements of another set, which means every element of set A is also an element of set B. A proper subset, however, includes all elements but is not equal to the set itself, implying that set A has fewer elements than set B. This distinction is essential for solving problems related to set theory and operations in Class 11 Miscellaneous Exercise Chapter 1.

4. What are the operations on sets in NCERT Class 11 Maths Chapter 1 Miscellaneous Solutions?

Operations on sets include union, intersection, and difference. The union of sets combines all elements from the sets, the intersection includes only the common elements, and the difference subtracts elements of one set from another. Practising Class 11 Maths Chapter 1 Miscellaneous Solutions help in understanding their application in various mathematical problems.

5. What is the use of Venn diagrams in sets in NCERT Class 11 Maths Chapter 1 Miscellaneous Solutions?

Venn diagrams visually represent the relationships between different sets. They are used to illustrate operations such as union, intersection, and difference. Mastering this tool is crucial for solving complex problems as it provides a clear and intuitive way to understand set operations and their results.

6. How to solve problems involving the complement of a set in class 11 maths chapter 1 miscellaneous exercise?

The complement of a set includes all elements not in the given set, usually within a universal set context. To solve these problems, identify the universal set and subtract the elements of the given set. Understanding this concept helps in tackling various questions related to set operations and logical deductions.

7. What topics are in the Miscellaneous Exercise Chapter 1 Class 11?

The Miscellaneous Exercise Chapter 1 Class 11 covers many topics like sets, relations, and functions. It includes problems with Venn diagrams, different types of sets, subsets, power sets, and Cartesian products.

8. How should I solve problems in the Miscellaneous Exercise Chapter 1 Class 11?

To solve these problems, make sure you understand the basic ideas of sets and functions. Practice each type of problem and use NCERT solutions to see step-by-step answers.

9. Are these questions important for exams from Class 11 Maths Ch 1 Miscellaneous Exercise Solutions?

Yes, these questions are important for exams. They help you practice and understand the different concepts taught in the chapter, which can appear in your exams.

10. How do Class 11 Maths Ch 1 Miscellaneous Exercise Solutions help with the Miscellaneous Exercise?

NCERT Solutions provides detailed answers for each question. They show you the steps to solve problems, making it easier to understand and learn the methods.