NCERT Class 11 Maths Chapter 1 Notes Sets - Download PDF

NCERT Class 11 Maths Chapter 1 Notes Sets - Download PDF

Edited By Ramraj Saini | Updated on Jan 30, 2024 02:57 PM IST

Class 11 Math chapter 1 sets notes are provided here. These notes are created by expert team at Careers360 considering the need of students. These Notes covers all topics, sub topics, concepts, formulae, exercises, and solved problems that are very useful to score well in the boards and competitive exams. In chapter 1 we will be discussing topics such as sets, representation of sets, types of sets, subset, power set, Venn diagrams, operations of sets, Laws of algebra.

Sets are collections of objects with specific characteristics. An empty set contains no elements. A finite set holds a definite number of elements, while an infinite set has an indefinite number. If two sets, P and Q, contain the exact same elements, they are considered equal. Set P is a subset of set Q if all elements in P also exist in Q. The power set [A(P)] of a set P consists of all possible subsets of P. The union of sets P and Q contains all elements found in either set. The intersection of sets P and Q comprises elements common to both. Similarly, the difference between sets P and Q, when in the same order, contains elements from P but not Q.

Also, students can refer,

Sets:

Set is a well-defined collection of elements. These elements may include numbers, symbols, variables, etc.

We generally represent them using capital letter(A, B, C, D, Z, N etc)

Elements in the sets are represented by small letters(a, b, c, d, e)

Elements of sets are represented using closed brackets { } .

Examples of sets:

  • i) set of first n natural numbers: N = 1, 2, 3, 4, 5,.....,n
  • ii) set of rational numbers: R.
  • iii) set of whole numbers: W.
  • iv) set of positive integers: Z+
  • v) set of negative integers: Z-

Representation of Sets:

We have 2 methods to represent a set:

Roaster form or tabular form:

In this form, elements are separated with commas and within braces { }.

Eg: set of natural numbers below 10: { 1, 2, 3, 4, 5, 6, 7, 8, 9}

Set-builder form: In this form, each element contains a common character that any other element outside the set does not contain.
Eg: F = {x: x is a natural number that divides 24}

Types of Sets:

Empty set: A set that does not contain any element and is empty such sets are called empty (or) null (or) void sets.

Represented as: { } or ∅

Eg: A={x:1<x,x is a natural number}

So A is an empty set as we don't have any natural element that is less than 1.

Finite set: Sets that contain a fixed or finite number of elements are called finite sets.

Represented by: A= { , , , , , }

Eg: A={natural numbers <20}

A= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

NOTE: Empty set is also a finite set.

Infinite sets: Sets that are not finite or the sets that are infinite are called infinite sets.

Represented by: {, , , , , , , , ……………………………}

Eg: A={n natural numbers}

A = {1, 2, 3, 4, 5, 6, 7, 8, 9,...............................}

NOTE: All finite sets can be represented in set-builder form but infinite sets cannot be represented in set-builder form because they have no fixed pattern.

Equal sets: Sets where two sets have exactly the same elements.

Represented with the symbol: “ = “

Examples:

i) Let A, B be two sets.

A= set of natural even elements less than 10

A = {2, 4, 6, 8}

B = set of multiples of 2 less than 10 that are natural

B = {2, 4, 6, 8}

Here A = B. Since every element of A is also an element of B and vice versa.

ii) A = {1, 2, 3, 4}

B = {1, 3, 2, 1, 4, 1}

Then also A = B, the set elements do not change even if repeated.

The order of the elements also doesn’t matter.

Subset:

Any Set that has every element of another set is called a subset.

Representation: A⊂B if a ∈A ⇒ a ∈B

The above representation means that A should have every element of set B, but B need not have every element of A.

NOTE: Every set is a subset of itself

Eg: A = {1, 3, 5, 7, 9}

B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Here A⊂B because every element of A is an element of B and vice versa doesn’t matter

Venn diagram of subset:

1646647931489

Proper set: If A⊂B and A≠B then such set A is called a proper subset of B.

B is called the superset of A.

Eg: A = {1, 2, 3, 4, 5, 6}

B = {1, 2, 3, 4, 5, 6, 7, 8}

Here A⊂B and A≠B.

So A is a proper subset of B.

B is a superset of A.

Intervals of subsets:

When a,b ∈ real numbers, a<b then,

Set of real numbers: {x: a<x<b} is open interval, represented as (a, b).

The set that includes end points also then such intervals are closed intervals.

{y: a ≤ y ≤ b}, closed interval represented by [a, b].

Eg: (p, q] ⇒ {x: p < x ≤ q}

Here in the set elements includes q and exclude p.

Pictorial representation of intervals:

1646647930985

Power set: It is a set that is a collection of all subsets including null or empty sets and itself denotes a power set.

Represented by p(A).

Eg: A = {1, 2, 3, 4}

Subsets: ∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1,4}, {2,3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}

Total subsets:16

Points :

  1. First is a null set

  2. Next singleton sets(sets with one element}

  3. Next with two elements

  4. The set itself

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Remember: {1, 2} = {2, 1}

(order doesn’t matter so it should be written only once)

Formula to find a number of subsets:

m= number of elements in set A

\\ Formula: 2^m \\ Eg: A=\left \{ 1,2,3,4 \right \}\ t = 2^4= 16. \ \text{We found the same above.} \\ A=\left \{ 1,2,3,4 \right \}

Subsets: ∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}

Total subsets:16

Universal Set:

The set that includes all the elements is called a universal set.

Represented using U.

Venn Diagram:

1646647931234

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} all belong to the universal set.

Operation of Sets:

Union of sets: The set contains all the elements of A and B and repeating elements of both sets A, B is taken only once.

Represented by: ∪

Read as A union B.

Set builder form of union sets:

{y: y∈A or y∈B}

Venn diagram of union:

1646647930704

Eg: A = {1, 2, 3, 4, 5} B = {2, 4, 5, 6, 7, 8}

A∪B = {1, 2, 3, 4, 5, 6, 7, 8}

Properties of union:

  1. A∪B = B∪A is called commutative property.

  2. A∪(B∪C) = (A∪B)∪C is associative property.

  3. A∪∅ = A is identity law

  4. A∪A is an idempotent law

  5. U∪A is union law

The intersection of Sets:

This set contains the elements that are common in the sets.

Represented as: ∩

Read as A intersection B.

Set builder form of intersection:

{y: y ∈ A and y ∈ B}

Venn diagram of AB:

1646647932075

Eg: A = {1, 2, 3, 4, 5, 6} B = (2, 4, 6, 8, 10}

A∩B = {2, 4, 6}

Properties of intersection:

a)A∩B = B∩A is called commutative property.

b) A∩(B∩C) = (A∩B)∩C is an associative property.

c) A∩∅ = ∅ is identity law

d) A∩A is idempotent law

e) U∩A is union law

f) A∩(B∪C) = (A∩B)∪(A∩C) is distribution property.

Difference of Sets:

In the set the elements belong to A but do not belong to B, such sets are called Differences of sets.

Represented as: A-B

Set builder form: {y: y ∈ A and y ∉ B}

Venn diagram of A - B:

1646647930427

Eg: A = {1, 2, 3, 4, 5}

B = {2, 3, 4, 8, 9}

A - B = {1, 5}

Here even though we have {2, 3, 4} in A as they belong to B also do not include them in A - B.

Complement of set:

The elements other than A that is in the universe come under complement elements.

Represented by: A’

Set builder form: {y: y ∈ U and y ∉ A}

Venn diagram:

1646647932568

Eg: A = {y: y ∈ U and not divisor of 24}

A’ = {1, 2, 3, 4, 6, 8, 12, 24}

Properties of complement sets:

  1. A∪A’ = U (complement laws)

  2. A∩ A’ = ∅ (complement laws)

  3. (A∪B)’ = A’∩B’ (Demorgan’s law)

  4. (A∩B)’ = A’∪ B’ (Demorgan’s law)

  5. (A’)’ = A (double complement)

  6. U’ = ∅ and ∅’= U

Theorems and Proofs:

Theorems 1:

n(A∪B) = n(A) + n (B) - n(A∩B)

Proof:

n ( A – B) + n ( A ∩ B ) + n ( B – A )

n(A) - n ( A ∩ B ) + n(B) - n( A ∩ B ) + n( A ∩ B )

n(A) + n(B) - n( A ∩ B )

Hence proved;

Theorem 2: n(A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )

Proof:

n ( A ∪ B ∪ C ) = n (A) + n ( B ∪ C ) – n [ A ∩ ( B ∪ C ) ]

n (A) + n ( B ) + n ( C ) – n ( B ∩ C ) – n [ A ∩ ( B ∪ C ) ]

n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)] n ( A ∩ B ) + n ( A ∩ C ) – n (A ∩ B ∩ C)

Similarly for the other two:

Finally proved;

n ( A ∪ B ∪ C ) = n (A) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )

With this topic we conclude NCERT Class 11 chapter 1 notes.

The link for NCERT textbook pdf is given below:

URL: ncert.nic.in/ncerts/l/kemh101.pdf

Significance of NCERT Class 11 Maths Chapter 1 Notes:

NCERT Class 11 Maths chapter 1 notes will be very much helpful for students to score maximum marks in their 11 board exams. NCERT Class 11 Mathematics chapter 1 is also very useful to cover major topics of Class 11 CBSE Mathematics Syllabus. The CBSE Class 11 Maths chapter 1 will help students to understand the formulas, statements, and rules in detail. This pdf also contains previous year’s questions and NCERT textbook pdf. The next part includes FAQ’s or frequently asked questions along with topic-wise explanations. These topics can also be downloaded from Class 11 chapter 1 Sets pdf download.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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