Unit-I

Chapter 1

SETS

Who is the founder of set theory?

Georg Cantor, in full Georg Ferdinand Ludwig Philipp Cantor, (born March 3, 1845, St. Petersburg, Russia—died January 6, 1918, Halle, Germany), German mathematician who founded set theory and introduced the mathematically meaningful concept of transfinite numbers, indefinitely large but distinct from one another.

Between the years 1874 and 1897, the German mathematician and logician Georg Cantor created a theory of abstract sets of entities and made it into a mathematical discipline.

SET THEORY

Definition: Set is a collection of well defined or  well-determined objects or elements or numbers or members or functions .

Well defined means known characteristics and properties clearly expressed, explained, or described.

N.B.:-

(i) Sets, are generally denoted by capital English alphabets like A, B, C, P, Q, R etc.

(ii) Objects,elements,members of a set synonymous terms or similar or homogeneous terms.

(iii) The elements of a set are represented by small letters a,b,c,x,y,z, etc..

(iv) If a is an element of a set A, then a is belongs to A.   i.e., a ∈  A 

(v) If b is not an element of a set A, then b does not belong to A   i.e., b ∉   A

Examples:     

(i) The collection of three boys: Mohan, Ram, Prakash.

(ii) The collection N of all natural numbers.

(iii) The collection R of all real numbers.

(iv) The collection of all first year students of DAV

(v)  The collection of the roots of the equation x²⁷-1=0

But the following collections are not sets.

(i)  The collection of some natural numbers.

(ii) The collection of the politicians of India.

(iii) Collection of good cricket players of India.

These are not sets because the terms ‘some’ in (i) politicians in (ii) and good cricket players in (iii) are not well defined.

Example: Collection of tall students in your school.

It is not a set because tall is not well defined.

Similarly , 

Interesting, honesty, intelligence, better, tall ,etc… are not well defined.

Notations: -

• N: the set of natural numbers
• W: the set of Whole numbers
• I=Z: the set of Integer numbers
• Z⁺: the set of Positive Integer numbers
• Z^- : the set of negative Integer numbers
• Q : the set of Rational numbers
• Q’=T : the set of Irrational numbers
• C: the set of Complex numbers
• ∃ : there exists
• ∄ : there does not exists

Description or Representation of a set:  A set is generally described in two ways.

(1) Roster method / Tabular method / set by extension:

In this method the elements of the set are written within the curly brackets ({  }) and separated by a comma (,).

To write a set in Roster method elements are not to be repeated.

Example:        

(i) Set of all vowels of English alphabets.      =          {a, e, i, o u}

(ii) Set of all primes less than 8 =  {2, 3, 5, 7}

(2) Set Builder Method / Set selector method / method of specification:

It is represented by a symbol or variable that all the elements  of set possess a single common property and within the curly bracket which is not possessed by any elements outside the set.         

i.e., a set is described by a charactering properly of its elements.

Example: (i) The set of all even natural numbers is written by

A = {x|  x = 2n for n Î N}

(ii) The set of all prime numbers in between 5 and 15 is written by

B = { x|  x is a prime no. & 5 < x < 15}

Note:   If an element a  remains in a set A, we say etc. a belongs to A and can be written by x Î A. If a does not remain in A we say a does not belongs to A and is written by x Ï A.

Types of sets:

[1] Empty set / Null set / Void set / Phi set (ɸ):

A set is said to be empty set or null set or void set if it has no element. i.e., ɸ={ }

Example:

(i)  The set of all the months which are of 35 days.

(ii)   {x² | x² < 0}

[2] Singleton Set:

A set consisting of a single element is called a singleton set.

Example:

(i) A= { 5 }

(ii)  B= {Rama}

(iii) Present prime minister of India etc.

[3] Finite set:

A set is said to be finite if it is void set or its elements can be counted by natural numbers for any natural number. n Î N and the counting can be terminated for some n.

Example:

(i)  Set of even prime numbers.

(ii) {Arun, Mohan, Ramesh}

(iii) { 1, 2, 3, 4, -------- 100}

[4] Cardinality of a finite set:

The cardinality or cardinal number of a finite set is defined as the number of elements present in that set.

Cardinality of a set is denoted by n (A) on |A|

[5] Infinite sets:

A set whose elements cannot be listed by the natural numbers for any natural no. nÎN.

Example: A= set of points on a line.

N.B.:- All infinite sets cannot be described in the roster method.

[6] Equivalent sets:

Two finite sets A & B are said to be equivalent if their cardinal numbers are same i.e. n (A) = n (B)

Example: A = {1, 2, 3, 4, 5},  B = {a, b, c, d, e)

[7] Equal sets:

Two sets A and B are said to be equal if every elements of A is an element of B.

Example: {0, 1, 2, 3} = {2, 1, 0, 3}

[8] Subsets & super sets:

Let A and B be two sets such that every element of A is also an element of B, then A is said to be the subset of B and is denoted by AÌ B. and B is called the super set of A and denoted by B É A.

If n(A)=n, then Total No. of Subset of set A = 2ⁿ

Example:

            A = { a, e, i, o, u}

            B = {a, b, c, d, e, --------- x, y, z}

            Hence A Ì B & B É A

Theorem: The empty set (or void set) is a subset of every set. Empty set is represented by f.

Proof: Let A be any set and f must be a member of A .

            Þ  f Í A

Theorem:

The total number of subsets of a finite set containing n elements is 2ⁿ.

Proof: A be a finite set & |A| = n

Consider those subsets of A that have r elements. Since no. of selections or r elements from n different things is

Hence the total no. of subsets is

[9] Proper subset / proper set :

If A Ì B and B contain at least one element that is not in A, then A is called a proper subset of B.

Only itself is not Proper subset.

If n(A)=n, then Total No. of Proper Subset of set A = 2ⁿ-1

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

Properties : (i) f is subset of every set’

                   (ii) If AÍ B and BÍ A ,then A=B

                   (iii) If A Í B and B Í C then A Í C

                   (iv) A Í AUB , B Í AUB

                   (v) If AÍ B and xÎ A, then x Î B

                   (vi) A Í A

                   (vii) A Ç  B Í A  ,  AÇB Í B

[10] Power set:

            The collection of all the subsets of a set A is called the power set of A. It is denoted by P(A).

Example: A = {a, b, c}

                P(A) = {f, {a}, {b}, {c}, {a,b}, {b,c}, {c, a}, {a, b, c}}

Note:   The cardinality of the power set of a set A having n elements is 2ⁿ.

[11] Universal set  (U  / S  / x / w):

A set that contains all sets in a given context is called the universal set, denoted by E or U.

Example: When we study 2-dimensional co-ordinate geometry, then the set of all the points in xy-plane is the universal set.

 

Chapter 2

Relation and Function

Cartesian Product / Product Simply:

After French philosopher, mathematician and Scientists René Descartes (1596-1650),whose formulation of analytic geometry gave rise to the concept.

In 17^th century the invention of Cartesian coordinates system by Rene Descartes( Latin name: cartesius) revolutionized mathematics by providing the first systematic link between  Euclidean Geometry and Algebra

Who invented the Cartesian product?

René Descartes invented the Cartesian product. It derives the name from the same person. René formulated analytic geometry which helped in the origination of this concept which we further generalize in terms of direct product.

An ordered pair means that two elements are taken from each set.’

 For two non-empty sets (say A & B), the first element of the pair is from one set A and the second element is taken from the second set B.

If {a,b} is a set consisting of the elements of a,b, it is called pair.

If we specify a to be the first element and b to be the second element , then we call {a,b} an order pair and Write it as (a,b).

Cartesian Products of Sets:

Let A and B be the two sets such that A is a set of three colors of tables and B is a set of three colors of chairs objects, i.e.,

A = {brown, green, yellow}

B = {red, blue, purple},

Let’s find the number of pairs of colored objects that we can make from a set of tables and chairs in different combinations. They can be paired as given below:

(brown, red), (brown, blue), (brown, purple), (green, red), (green, blue), (green, purple), (yellow, red), (yellow, blue), (yellow, purple)

There are nine such pairs in the Cartesian product since three elements are there in each of the defined sets A and B. The above-ordered pairs represent the definition for the Cartesian product of sets given. This product is denoted by “A × B”.

Suppose there are two non-empty sets A and B.

So, the Cartesian product of A and B is the set of all ordered pairs of elements from A and B.

i.e., A × B = {(a , b) : a ∊ A, b ∊ B}

Example: Let A = {a_1,a_2,a₃,a₄} and B = {b₁,b₂}

Then, The Cartesian product of A and B will be;

A × B = {(a_1,b₁₎, (a_2,b₁),( a_3,b₁),( a_4,b₁).( a_1,b₂),( a_2,b₂),( a_3,b₂),( a_4,b₂ )}

Example: Let us say, A = {1,2} and B = { a,b,c}

Therefore, A × B = {(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)}.

This set has 8 ordered pairs. We can also represent it as in a tabular form.

Note: Two ordered pair X and Y are equal, if and only if the corresponding first elements and second elements are equal.

Example: Suppose, A = {cow, horse} B = {egg, juice}

 then, A×B = {(cow, egg), (horse, juice), (cow, juice), (horse, egg)}

If either of the two sets is a null set, i.e., either A = Φ or B = Φ, then, A × B = Φ i.e., A × B will also be a null set

Number of Ordered Pairs

For two non-empty sets, A and B. If the number of elements of A is p

 i.e., n(A) = p & that of B is q

 i.e., n(B) = q,

then the number of ordered pairs in Cartesian product will be n(A × B) = n(A) × n(B) = pq.

Properties:

1. The Cartesian Product is non-commutative: A × B ≠ B × A.
2. The cardinality of the Cartesian Product is defined as the number of elements in A × B and is equal to the product of cardinality of both sets:

|A × B| = |A| * |B|

3. A × B = ∅, if either A = ∅ or B = ∅
4. If (x,y) = (a,b) ,then x=a , y=b
5. A×B=B×A, if only A=B
6. The Cartesian product is associative:

(A×B)×C=A×(B×C). It means the Cartesian product of the three-set is the same, i.e., it doesn’t depend upon which bracket is multiplied first as the final result will be the same.

7. Distributive property over a set intersection:

A×(B∩C)=(A×B)∩(A×C)

8.  Distributive property over set union:

A×(B∪C)=(A×B)∪(A×C)

9.  If A⊆B, then A×C⊆B×C for any set C.
10. AxBxC  = {(a,b,c) : aÎA, bÎ B ,cÎ C}

Here (a,b,c) is called ordered triplet.

Solved Examples:

1. Let A = {–1, 2, 3} and B = {1, 3}. Determine
(i) A × B

(ii) B × A
(iii) B × B

(iv) A × A

Solution:

According to the question,

A = {–1, 2, 3} and B = {1, 3}

(i) A × B

{–1, 2, 3} × {1, 3}

So, A × B = {(–1, 1), (–1, 3), (2, 1), (2, 3), (3, 1), (3, 3)}

Hence, the Cartesian product = {(–1, 1), (–1, 3), (2, 1), (2, 3), (3, 1), (3, 3)}

(ii) B × A.

{1, 3} × {–1, 2, 3}

So, B × A = {(1, –1), (1, 2), (1, 3), (3, –1), (3, 2), (3, 3)}

Hence, the Cartesian product = {(1, –1), (1, 2), (1, 3), (3, –1), (3, 2), (3, 3)}

(iii) B × B

{1, 3} ×{1, 3}

So, B × B = {(1, 1), (1, 3), (3, 1), (3, 3)}

Hence, the Cartesian product = {(1, 1), (1, 3), (3, 1), (3, 3)}

(iv) A × A

{–1, 2, 3} × {–1, 2, 3}

So, A × A = {(–1, –1), (–1, 2), (–1, 3), (2, –1), (2, 2), (2, 3), (3, –1), (3, 2), (3, 3)}

Hence,

the Cartesian product ={(–1, –1), (–1, 2), (–1, 3), (2, –1), (2, 2), (2, 3), (3, –1), (3, 2), (3, 3)}

2. If P = {x : x < 3, x ∈ N}, Q = {x : x ≤ 2, x ∈ W}. Find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.

Solution:

According to the question,

P = {x: x < 3, x ∈N}, Q = {x : x ≤ 2, x ∈W} where W is the set of whole numbers

P = {1, 2}

Q = {0, 1, 2}

Now

(P∪Q) = {1, 2}∪{0, 1, 2} = {0, 1, 2}

And,

(P∩Q) = {1, 2}∩{0, 1, 2} = {1, 2}

We need to find the Cartesian product of (P∪Q) = {0, 1, 2} and (P∩Q) = {1, 2}

So,

(P∪Q) × (P∩Q) = {0, 1, 2} × {1, 2}

= {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

Hence, the Cartesian product = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

3. If A = {x : x ∈ W, x < 2}, B = {x : x ∈ N, 1 < x < 5}, C = {3, 5} find
(i) A × (B ∩ C)
(ii) A × (B ∪ C)

Solution:

According to the question,

A = {x: x ∈ W, x < 2}, B = {x : x ∈N, 1 < x < 5} C = {3, 5}; W is the set of whole numbers

A = {x: x ∈ W, x < 2} = {0, 1}

B = {x : x ∈N, 1 < x < 5} = {2, 3, 4}

(i)

(B∩C) = {2, 3, 4} ∩ {3, 5}

(B∩C) = {3}

A × (B∩C) = {0, 1} × {3} = {(0, 3), (1, 3)}

Hence, the Cartesian product = {(0, 3), (1, 3)}

(ii)

(B∪C) = {2, 3, 4} ∪ {3, 5}

(B∪C) = {2, 3, 4, 5}

A × (B∪C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Hence, the Cartesian product = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

 

4. : The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.

Solution:

We know that,

If n(A) = p and n(B) = q, then n(A × B) = pq

From the given,

n(A × A) = 9

n(A) × n(A) = 9,

n(A) = 3 ……(i)

The ordered pairs (-1, 0) and (0, 1) are two of the nine elements of A × A.

Therefore, A × A = {(a, a) : a ∈ A}

Hence, -1, 0, 1 are the elemets of A. …..(ii)

From (i) and (ii),

A = {-1, 0, 1}

The remaining elements of set A × A are (-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0) and (1, 1).

Chapter 3

Trigonometric Functions

Angles, Measurement of Anges (Degree, Radian, Gradian) and Notational Convention:

Tri = three  , gono= sides ,metric from metron(greek word) = study

The study of three sides is called trigonometry.

In other words , the trigonometry is that branch of mathematics which deals with measurement of sides and angles of a triangle.

Angles: a figure traced by rotating a given ray about its end points.

Otherwise An angle is formed when two straight lines or rays meet at a common endpoint. The common point of contact is called the vertex of an angle. The word angle comes from a Latin word named 'angulus,' meaning “corner.”

 

 For measurement of sides, we need

1. Units (cm,m,mm,hm,km,etc..)
2. Devices ( Scale, tape, thread, gauge  ,etc.,)

For measurement of angles , we need

1. Units (degree, radian , gradian , staradian,etc..)
2. Devices ( Protactor ,setsquare , angualar , trigonometric functions  ,etc.,)

Three systems for measurement of angles.

1. Sexagesimal system(FPS) or English System:
2. Centesimal System(CGS) or French System
3. Circular System

Sexagesimal system(FPS) or English System:

The sexagesimal system was an ancient system of counting, calculation, and numerical notation that used powers of 60 much as the decimal system uses powers of 10. Rudiments of the ancient system survive in vestigial form in our division of the hour into 60 minutes and the minute into 60 seconds.

In Sexagesimal System, an angle is measured in degrees, minutes and seconds.

A complete rotation describes 360°. In this system, a right angle is divided into 90 equal parts and each such part is called a Degree (1°); a degree is divided into 60 equal parts and each such part is called a Sexagesimal Minute (1’) and a minute is further sub-divided into 60 equal parts, each of which is called a Sexagesimal Second (1’’). In short, 

Centesimal System(CGS) or French System:

In Centesimal System, an angle is measured in grades, minutes and seconds. In this system, a right angle is divided into 100

equal parts and each such part is called a Grade (1^g); again, a grade is divided into 100 equal parts and each such part is called a Centesimal Minute (1‵) ; and a minute is further sub-divided into 100 equal parts, each of which is called a Centesimal Second (1‶). In short,
 

Note: (i) Clearly, minute and second in sexagesimal and centesimal systems are different.

For example,

The first relation is used to reduce an angle of sexagesimal system to centesimal system and the second is used to reduce an angle of centesimal system to sexagesimal system.

Circular System:

In this System, an angle is measured in radians. In higher mathematics angles are usually measured in circular system. In this system a radian is considered as the unit for the measurement of angles.

Definition of Radian: A radian is an angle subtended at the center of a circle by an arc whose length is equal to the radius.

The circular measure of an angle is the number of radians it contains.

Thus the circular (radian) measure of a right angle is 

If an angle is given without mentioning units, it is assumed to be in radians. The relation between degree measures and circular (radian) measures of some standard angles are given below:

Radians = Degrees × π / 180

In terms of degrees, one complete counterclockwise revolution is 360° and whereas in radians, one complete counterclockwise revolution is 2π. These statements can be equated as:

One complete counterclockwise revolution in degrees = 360°
One complete counterclockwise revolution in radians = 2π

Radians = Degrees × (π/180°). We can follow the steps given below to calculate the measure of an angle given in degrees to radians.

• Note down the measure of the angle given in degrees.
• We know, 1°= (π)/180 radians. So, to convert the angle given in degrees to radians we multiply it with π/180°.
  Angle in Radians = Angle in Degrees × π/180°.
• Simplify the values and express the answer in radians.

Example: Convert 90 degrees to radians.

Solution: 90° = 90° × (π/180°) = π/2.

• Angles are measured in degrees, radians and gradian
• One full revolution is equal to 2π rad (or) 360°,100^g.
• 1° = 0.017453 radians and 1 rad = 57.2958°.
• To convert an angle from degrees to radians, we multiply it by π/180°.
• To convert an angle from radians to degrees, we multiply it by 180°/π.

Example 1: In a circle with center O, points A and B are two points on the circle. ∠AOB = 60°. Convert ∠AOB's angle measure from degrees to radians.

Solution:

We have, ∠AOB = 60°.

Degrees to radians conversion formula is given as (Degrees × π)/180°

Thus converting the given angle we get,

∠AOB in Radians = ∠AOB in degrees × (π/180°)

∠AOB in Radians = 60° × (π/180°)

∠AOB in Radians = π/3

Answer: ∠AOB = π/3 rad.

Convert the following degree measure to radian measure.

a) 20 degrees b) 28 degrees

Solution: To convert the following degree measure to radians we will use the following steps:

a) 20° × (π/180°) = π/9 Radians.

b) 28° × (π/180°) = 7π/45 Radians.

 

VENN-DIAGRAMS & OPERATIONS ON SETS

Venn-diagrams:

The graphical representation of a set is called Venn-diagram.

A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while circles that do not overlap do not share those traits. Venn diagrams help to visually represent the similarities and differences between two concepts.

A diagram used to represent all possible relations of different sets. A Venn diagram can be represented by any closed figure, whether it be a Circle or a Polygon (square, hexagon, etc.). But usually, we use circles to represent each set. 

A large rectangle is used to represent the universal set and it is usually denoted by the symbol E or sometimes U.

All the other sets are represented by circles or closed figures within this larger rectangle.

Every set is the subset of the universal set U.

Operations on sets:

Union f sets:

Let A and B be two sets, then the union of A & B is the set of all those elements which belong to A or to B or to both A and B. It is represented as AÈB.

  

Example: 

Properties:

(1) A È B = B È A           (Commutative law)                      

(2) A Í A È B & B Í A È B

(3) A È A = A (Idempotent law)

(4) A È f = A (Identity law)

(5) A È U = U where U is the universal set.

The union of n sets i.e. A₁, A₂, A₃ ……… A_n is denoted by

A₁ È A₂ È A₃ ---------- È A_n =

(6) AÍ B Þ AUCÍ BUC  then A Í AUC for any C

(7) AUB = B  iff  A Í B

 Intersection of Two sets:

Let A & B are two sets. The intersection of A & B, denoted by A Ç B is a set of all elements those belongs to both A & B.

 Properties of Intersection:

 Facts:

 (1) A Ç B = B Ç A (Commutative law)

 (2) A Ç A = A (Idempotent law)

 (3) A Ç f = f (Identity law)

 (4) A Ç B Í A & A Ç B Í B

 (5) A Ç U = A

Distributive for intersection over union

A È (BÇ C) = (A È B) Ç (A È C)

Distributive property for union over

A Ç (B È C) = (A Ç B) È (A Ç C)

The intersection of n different sets is written by

A₁ Ç A ₂ Ç A₃ ------- Ç A_n = 

Disjoint sets:

If the intersection of two sets A & B is a null set, then we say A & B are two disjoint sets.

Example:

A = {a, b, c}

B = {Rama, Shyama, Gopal}

 Difference of sets:

Difference of sets:

The difference of two sets A & B, denoted by A – B, is the set of all element of A which do not belongs to B.  i.e.  A – B = {x| x Î A and x Ï B}

Example:

A = {1, 2, 3, 4, 5, 6}  B = {2, 5, 7, 11, 12, 13}

A – B = {1, 3, 4, 6}

 

Symmetric Difference of sets:

Let A and B are two sets. Then the symmetric difference of A & B, written by A D B is defined as

i.e. the set of all elements which are not common to both A & B.

Example: A = {1, 2, 3, 4, 5, 6, 7, 8},  B = {5, 7, 8, 10, 11, 12}

A D B = {1, 2, 3, 4, 6, 10, 11, 12}

Note:   A D B = B D A

 Complement of a set (w.r.t. universal set U)

Let U be the universal set & A be any set  st. A Ì C. Then the complement of A with respect to U, denoter by A^c (or A¹) or U –A, is defined as the set of all elements of U which donot belongs to A.

 

Example:

Z = { -----------  -3, -2, -1, 0, 1, 2, 3, ------------}  be the universal set.

A = { ----------  -4, -2, 0, 2, 4, --------------}

Then    A^c = { ------------   -3, -1, 1, 3, -------------}

Properties of complement:

Facts:

1. A Ç A^c = f
2. U^c = f & f^c = U
3. A È A^c = U
4. (A^c) = A
5. Demorgan’s laws

•  (i)  (AÈB)^| = A^| Ç B^|
•  (ii) (A Ç B)^| = A^| È B^|

Practical Problems on Union and Intersection(Applications of Sets)

Daily Life Examples Of Sets

1. In Kitchen. Kitchen is the most relevant example of sets. ...
2. School Bags. ...
3. Shopping Malls. ...
4. Universe. ...
5. Playlist. ...
6. Rules. ...
7. Representative House.   etc…

Cardinality of Finite sets:

If A & B are two finite sets, then the cardinality of A È B is given by

|AÈB| = |A| + |B| - |A Ç B|

Note:  

(1) n(AÈB )= |AÈB| = |A| + |B|  if A Ç B = f, i.e. A & B are two finite disjoint sets.

 (2) n(AÈBÈC)=|AÈBÈC| = |A| + |B| + |C| - |AÇB| - |BÇC| - |CÇA| + |AÇBÇC|

  (3)  n (A – B) = n (A ∪ B) – n (B)

n (B – A) = n (B) – n (A ∩ B)

 (4) n (A ∪ B)′= n (U) – n (A ∪ B) = n (U) – n (A) – n (B) + n (A ∩ B)

 (5)  n ( A ∪ B )' = n ( U) – n ( A ∪ B)

Order pair:

Let A be a set and a, b Î A, then the order pair of a & b in A denoted by (a, b). Here a is called first co-ordinate & b is called the second co-ordinate.

N.B.    In a order pair (a, b), the respective order of the entries is always constant, we can not alter their respective orders. If we do so, it will be an different element.

Cartesian product of two sets:

Definition:

If A and B are two non-empty sets, then their Cartesian product, denoted by A X B is defined to be the set of order pairs as

A X B = {(a, b)  |  x ÎA, b Î B}

Example:  A = {a, b, c}

                  B = {1, 2, 3}

                  A X B = {(a, 1), (a, 2), (a,3), (b,1), (b, 2), (b,3), (b,1), (c, 2), (c, 3)}

Note:  

(1) A X B ¹ B X A is A ¹ B i.e. Cartesian product is not commutative.

(2) If cardinality of A is n and cardinality of B is n then cardinality of     (A X B) is mn.

 (3) A X (B Ç C) = (A X B) Ç (A X C)

 (4) A X (B ÈC) = (A X B) È (A X C)

Solved examples:

Q.1: Prove that A – B = A Ç B^/, where A & B are any two sets.1

Proof:  Let x Î A –B be an arbitrary element.

Û x Î& x Ï B

Û x Î A & x Î B^/

Û x Î AÇB^/

\ A – B Ì A Ç B^/                              ----------- (1)

And A Ç B^/ Ì A – B                          ----------- (2)

From 1 & 2 we have,

A – B = A Ç B^/

Q.2: Prove that A È B = B È A, where A & B are any two sets.

Proof: Let x Î A È B be an arbitrary element.

Now x Î A È B

Û x Î A or x Î B

Û x Î B or x Î A

Û x Î B È A

\ A È B Í B È A                             -------------- (i)

And    B È A Í A È B                       --------------- (ii)

From (i) & (ii), we have

A È B = B È A

Q.3: Prove that A È (B Ç C) =(AÈB) Ç (A È C), for any three sets A, B & C.

Proof. Let x Î A È (B ÇC) be an arbitrary element.

Now x Î A È (B Ç C)

Û x Î A  or x Î (B Ç C)

Û x Î A or (x Î B and x Î C)

Û (x Î A or x Î B) and x Î A or x Î C)

Û x Î A È B and x Î A È C

 Û x Î (A È B) Ç (A È C)

\  A È (BÇC) Í  (A È B) Ç (A È c)                       -------------- (i)

And (A È B) Ç (A È C) = A È (B Ç C)                   -------------- (ii)

From (i) and (ii) we have

A È (BÇC) =  (A È B) Ç (A È c)

Q.4: Prove that A – (A – B) Í B for any two set A & B.

Proof:  Let x Î A – (A –B) be arbitrary.

Þ  x Î A and x Ï (A – B)

Þ  x Î A and (x Ï A and x Î B)

Þ x Î B

Þ A – (A – B) Í B

Q.5: For any three sets A, B and C, prove that  A Ç (B Ç C) = (A Ç B) Ç C

Proof: Let x Î A Ç (B Ç C) be an arbitrary element.

Now x Î A Ç (B Ç C)

Û x Î A and x Î (B Ç C)

Û x Î A and (x Î B and x Î C)

Û (x Î A and x Î B) and x Î C

Û x Î A Ç B) Ç C

\A Ç (B Ç C) Í (A Ç B) Ç C                      ------------ (1)

And (A Ç B) Ç C Í A Ç (B Ç C)                 ------------ (2)

From 1 & 2, we have

A Ç (B Ç C) = (A Ç B) Ç C

Q.6: If A D B = B D C, then prove that A = C for any three set A, B and C.

Proof:  Given that A D B = B D C

Þ (A È B) – (A Ç B) = (B È C) – (B Ç C)

Þ A ÈB = BÈC   and A Ç B = B Ç C

Þ (A ÈB) Ç B^c = (C È B) Ç B^c and A Ç B = B Ç C

Þ (A Ç B^c) È f = (C Ç B^c) È f and A Ç B = B Ç C

Þ (A Ç B^c) = C Ç B^c and A Ç B = B Ç C

Þ A – B = C – B and A Ç B = B Ç C

Þ (A – B) È A Ç B = (C – B) È (B Ç C)

Þ A = C

Q.7: Prove that for any three sets A, B, C

|AÈBÈC| = |A| + |B| + |C| - |A ÇB| - |BÇC| - |CÇA| + |AÇBÇC|

Proof: Let B È C = D

Now    |AÈBÈC|        = |AÈD|

= |A| + |D| - |A Ç D|

= |A| + |BÈC| - |AÇD|

= |A| + |B| + |C| - |BÇC| - |AÇ(BÈC)|

= |A| + |B| + |C| - |BÇC| - |(AÇB) È (AÇC)|

= |A| + |B| + |C| - |BÇC| - {|AÇB| + |AÇC| - |(AÇB) Ç (AÇC)|}

= |A| + |B| + |C| - |BÇC| - |AÇB|-|AÇC| + |AÇBÇC|

Þ |AÈBÈC| = |A| + |B| + |C| - |AÇB| - |BÇC| - |CÇA| + |AÇBÇC|

Q.8: A company must hire 25 programmers to handle systems programming jobs and 40 programmers for applications programming of those hired, 10 will be expected to perform jobs for both types. How many programmers must be hired?

Ans.     A = Set of system programmers hired

            B = Set of application programmers hired.

            According to the question

            |A|  = 25, |B| = 40 ,     |A Ç B| = 10

            The number of programmers that must be hired is

            |A È B| = |A| + |B| - |A Ç B|

            = 20 + 40 – 10 = 55

            Hence 55 programmers must be hired.

Q.9: In a class of 35 students, 17 have taken mathematics, 10 have taken mathematics but not economics. Find the number of students who have taken both mathematics and economics and the number of students who have taken economics but not mathematics. If it is given that each student can have either mathematics or economic or both.

Ans.     Let      

A = Set of students who have taken mathematics.

B = Set of students who have taken economics.

Hence as per question,

|A È B| = 35

|A| = 17

|A – B| = 10

Now    |(A)| = |A – B| + |A Ç B|

Þ 17 = 10 + |A Ç B|

Þ |A Ç B| = 7

Hence 7 students have taken both mathematics and economics.

Now    |A ÈB| = |A| + |B| - |A Ç B|

Þ 35 = 17 + |B| - 7

 

Þ |B| = 35 – 10

= 25

Now   |B| = |B – A| + |A Ç B|

Þ 25 = |B – A| + 7

Þ |B – A| = 25 – 7

= 18

Hence 18 students have taken economics but not mathematics.

Q.10: In a class, 38 students play football. 15 play basketball, 20 play cricket.

In a survey of 60 readers, it was found that 25 read Times of India, 26 read The Hindu, 26 read Indian Express, 9 read both Times of India and The Hindu, 11 both Times of India and Indian Express, 8 read The Hindu and Indian Express and 3 read all the three News Papers. Find the number of readers who read at least one of the three News Papers and also find the no. of people who read only Times of India.

 

Ans:    Let       T = Set of readers who read Times of India.

                        H = Set of readers who read The Hindu.

                        E = Set of people who read Indian Express.

 

According to the question

                        |T| = 25                                          

                        |H| = 26                                                                                  

                        |E| = 26                                                                       

                        |TÇ H| = 9

                        |TÇE| = 11,  |H Ç E| = 8                                            

            &         |TÇHÇE| = 3 

            Now |TÈ H È E| = |T| + |H| + |E| - |TÇH| - |HÇE| - |EÇT| + |TÇHÇE|

                                        = 25 + 26 + 26 – 9 -11 – 8 + 3 = 52

            Thus the number of readers who read at least one, of the three news paper is 52.

            Now the number of readers who read only Times of India

                        = |T| - |TÇH| - |TÇS| + |TÇHÇE|

                        = 25 – 9 -11 + 3 = 8

            Hence 8 readers read only Times of India.

 

Q.11: If A = {1, 2, 3}      B = {4, 5, 6}. Find A X B & B X A

Ans:    A = {1, 2, 3}

            B = {4, 5, 6}

            A X B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

            B X A = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}

 

Q.12: Prove that for any three sets A, B, C A X (BÇC) = (A X B) Ç (A X C)

Proof: Let (x, y) Î A x (B Ç c) be an arbitrary order pair.

                  (x, y) Î A X (B Ç C)

            Û  x ÎA  and y Î (B Ç C)

            Û  x ÎA and (y Î B, y Î C)

            Û  (x ÎA ,y ÎB) and (x Î A, y Î C)

            Û  (x, y) Î (A X B) and (x, y) Î (A X C)

            Û  (x, y) Î (A X B) and (x, y) Î (A X C)

            Û  (x, y) Î (A X B) Ç (A X C)

            \  A X (BÇC) Í (A X B) Ç (A X C)                       ------------------- (i)

            And (A X B) Ç (A X C) Í A X (B Ç C)                   ------------------- (ii)

From (i) and (ii) we have

A X (BÇC) = (A X B) Ç (A X C)

Q.13: Prove that A X (B – C) = (A X B) – (A X C) for any three sets A, B & C

Proof:  Let (x, y) Î A X (B – C) be arbitrary (x, y) Î A X (B – C)

            Û  x Î A , y Î (B – C)

            Û  x Î A ,(y ÎB and y Ï C)

            Û  x Î A and y Î B) and (x Î A but y Ï C)

            Û  (x, y) Î A X B and (x, y) Ï A X C

            Û  (x, y) Î (A X B) – (A X C)

            \   A X (B – C) Í (A X B) – (A X C)                      --------------- (1)

            And (A X B) – (A X C) Í A X (B – C)                     --------------- (2)

            From 1 and 2, we have,

            A X (B – C) = (A X B) – (A X C)

Relations and Types of Relations

Relation:

Relation is association between two well-defined  objects.

Relation  in real life give us the link between any two objects or entities. In our daily life, we come across many patterns and links that characterize relations such as a relation of a father and a son, brother and sister, etc.

Relations can be represented in different forms such as arrow representation, algebraic form, set-builder form, graphically, roster form, and tabular form. 

Definition:

Let A and B be two non empty sets.

Then R  :  A ® B  is said to be a relation if R Í AxB .

The element of A (first element) of AxB in the relation is called Domain or Pre-image of relation R.

The element of B (second element) of AxB in the relation is called Range or image of relation R.

The whole B set of AxB in the relation is called Codomain of relation R.

Range Í Codomain

Example : Define a relation R from A to A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y = x + 1}. Determine the domain, codomain and range of R.

Solution: We can see that A = {1, 2, 3, 4, 5, 6} is the domain and codomain of R.

To determine the range, we determine the values of y for each value of x, that is, when x = 1, 2, 3, 4, 5, 6

• • x = 1, y = 1 + 1 = 2;
  • x = 2, y = 2 + 1 = 3;
  • x = 3, y = 3 + 1 = 4;
  • x = 4, y = 4 + 1 = 5;
  • x = 5, y = 5 + 1 = 6;
  • x = 6, y = 6 + 1 = 7.

Since 7 does not belong to A and the relation R is defined on A, hence, x = 6 has no image in A.

Therefore range of R = {2, 3, 4, 5, 6}

Answer: Domain = Codomain = {1, 2, 3, 4, 5, 6}, Range = {2, 3, 4, 5, 6}

Types of Relations

Given below is a list of different types of relations:

1. Empty Relation
2. Universal Relation
3. Identity Relation
4. Inverse Relation
5. Reflexive Relation
6. Symmetric Relation
7. Transitive Relation
8. Equivalence Relation

1) Empty Relation - A relation is an empty relation if it has no elements, that is, no element of set A is mapped or linked to any element of A. It is denoted by R = ∅.

For example, if set A = {1, 2, 3} then, one of the void relations can be R = {x, y} where, |x – y| = 8. For empty relation,

R = φ ⊂ A × A

2) Universal Relation - A relation R in a set A is a universal relation if each element of A is related to every element of A, i.e., R = A × A. It is called the full relation.

Consider set A = {a, b, c}. Now one of the universal relations will be R = {x, y} where, |x – y| ≥ 0. For universal relation,

R = A × A

3) Identity Relation - A relation R on A is said to be an identity relation if each element of A is related to itself, that is, R = {(a, a) : for all a ∈ A}

 For example, in a set A = {a, b, c}, the identity relation will be I = {a, a}, {b, b}, {c, c}. For identity relation,

I = {(a, a), a ∈ A}

4) Inverse Relation - Define R to be a relation from set P to set Q i.e., R ∈ P × Q. The relation R-1 is said to be an Inverse relation if R-1 from set Q to P is denoted by R^-1 = {(q, p): (p, q) ∈ R}.

For example if set A = {(a, b), (c, d)}, then inverse relation will be R^-1 = {(b, a), (d, c)}. So, for an inverse relation,

R^-1 = {(b, a): (a, b) ∈ R}

5) Reflexive Relation - A binary relation R defined on a set A is said to be reflexive if, for every element a ∈ A, we have aRa, that is, (a, a) ∈ R.

 For example, consider a set A = {1, 2,}. Now an example of reflexive relation will be R = {(1, 1), (2, 2), (1, 2), (2, 1)}. The reflexive relation is given by-

(a, a) ∈ R

6) Symmetric Relation - A binary relation R defined on a set A is said to be symmetric if and only if, for elements a, b ∈ A, we have aRb, that is, (a, b) ∈ R, then we must have bRa, that is, (b, a) ∈ R.

 An example of symmetric relation will be R = {(1, 2), (2, 1)} for a set A = {1, 2}. So, for a symmetric relation,

aRb ⇒ bRa, ∀ a, b ∈ A

Example: For the set P={a,b}, the relation R={(a,b),(b,a)} is called symmetric relation, where a,b∈P.

7) Transitive Relation - A relation R is transitive if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for a, b, c ∈ A

Example: For the set A ={a,b,c}, the relation R={(a,b),(b,c),(a,c)} is called transitive relation, where a,b,c∈ A.

8) Equivalence Relation - A relation R defined on a set A is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive.

Conditions:

1. If the relation (R) is reflexive, then all the elements of set A are mapped with itself, such that for every x∈ A , then (x,x)∈R.

2. The relation (R) is symmetric on set A, if (x,y)∈R, then (y,x)∈R, such that a,b∈ A.

3. The relation R on set A, if (x,y)∈R and (y,z)∈R, then (x,z)∈R, for all a,b,c∈ A is called transitive relation

Other Types of Relations:

The other types of relations based on mapping of two sets are given as follows:

1) One to One Relations

A relation is said to be a One to One relation if all the distinct elements of one set are related or mapped to distinct elements of another set.

2) One to Many Relations

A relation is said to be One to Many relations if the one element of first set is mapped to more than one element second set.

3) Many to One Relation

A relation is said to be Many to One relation if  the more than one element of first set is mapped to one element second set.

4) Many to Many Relation

A relation is said to be Many to Many relation if  the more than one element of first set is mapped to more than one element second set.

Solved Examples:

Q.1. Let A be the set of two male persons in a family. R be a relation defined onset A is “is a brother of”, check whether R is symmetric or not?

Ans:

Let a,b are two persons in a family, then a,b∈A.

The given relation on set “is a brother of”.

If a is the brother of b, then b is also the brother of a.

R={(a,b),(b,a)}

Hence, the relation R is symmetric.

Q.2. Find the inverse relation of R={(1,2),(3,4),(5,6)}.

Ans:

Given relation is R={(1,2),(3,4),(5,6)}.

Consider R={(a,b):a∈P,b∈Q}. be the relation from set P to set Q, then the relation from set Q to set P is known as inverse relation, such that R–1:Q→P={(a,b):(a,b)∈R}.

So, inverse relation is obtained by taking the reverse of the given ordered pairs.

R–1:{(2,1),(4,3),(6,5)}

Q.3. Find the identity relation on the set P={x,y,z}.

Ans:

We know that the relation (I) is identity, then all the elements of set P are related with itself, such that for every a∈P, then (a,a)∈I.

Given set is P={x,y,z}

Then the identity relation is given by I={(x,x),(y,y),(z,z)}.

Q.4. Let P={1,2,3},R be a relation defined on set P as “is greater than” and R={(2,1),(3,2),(3,1)}. Verify R is transitive.

Ans:

Given set P={1,2,3},

Let P={1,2,3}, then we have

(b,a)=(2,1)→2 is greater than 1.

(c,b)=(3,2)→3 is greater than 2.

(c,a)=(3,1)→3 is greater than 1.

Thus in a transitive relation, if (x,y)∈R,(y,z)∈R, then (x,z)∈R.

So, the relation R={(2,1),(3,2),(3,1)} is transitive.

Q.5. Let set A contains the bag of red balls. Then the relation R of getting white balls from the bag of red balls is what type of relation?

Ans:

Given set A contains the bag contains the red balls. From the bag of red balls, relation R of getting the white ball is not possible. So, the given relation is an empty relation.

Q6. Given R = {(x, y) : x, y ∈ W, x² + y² = 25}. Find the domain and Range of R.

Ans:

According to the question,

R = {(x, y) : x, y ∈W, x² + y² = 25}

R = {(0,5), (3,4), (4, 3), (5,0)}

The domain of R consists of all the first elements of all the ordered pairs of R.

Domain of R = {0, 3, 4, 5}

The range of R contains all the second elements of all the ordered pairs of R.

Range of R = {5, 4, 3, 0}

Q7. If R₁ = {(x, y) | y = 2x + 7, where x ∈ R and – 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R₁.

Ans:

According to the question,

R₁ = {(x, y) | y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5} is a relation

The domain of R₁ consists of all the first elements of all the ordered pairs of R₁, i.e., x,

It is also given – 5 ≤ x ≤ 5.

Therefore,

Domain of R₁ = [–5, 5]

The range of R contains all the second elements of all the ordered pairs of R₁, i.e., y

It is also given y = 2x + 7

Now x ∈ [–5,5]

Multiply LHS and RHS by 2,

We get,

2x ∈ [–10, 10]

Adding LHS and RHS with 7,

We get,

2x + 7 ∈ [–3, 17]

Or, y ∈ [–3, 17]

So,

Range of R₁ = [–3, 17]

Q.8. If R₂ = {(x, y) | x and y are integers and x² + y² = 64} is a relation. Then find R₂.

Ans:

We have,

R₂ = {(x, y) | x and y are integers and x² + y² – 64}

So, we get,

x² = 0 and y² = 64 or x² = 64 and y² = 0

x = 0 and y = ±8 or x = ±8 and y = 0

Therefore, R₂ = {(0, 8), (0, –8), (8,0), (–8,0)}

Q.9. If R₃ = {(x, |x| ) |x is a real number} is a relation. Then find domain and range of R₃.

Ans:

According to the question,

R₃ = {(x, |x|) |x is a real number} is a relation

Domain of R₃ consists of all the first elements of all the ordered pairs of R₃, i.e., x,

It is also given that x is a real number,

So, Domain of R₃ = R

Range of R contains all the second elements of all the ordered pairs of R₃, i.e., |x|

It is also given that x is a real number,

So, |x| = |R|

⇒ |x|≥0,

i.e., |x| has all positive real numbers including 0

Hence,

Range of R₃ = [0, ∞)

Q.10: Assume that A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, such that x, y ∈ A}. Determine and write down its range, domain, and codomain.

Solution:

It is given that the relation R from A to A is given by R = {(x, y): 3x – y = 0, where x, y ∈ A}.

It means that R = {(x, y) : 3x = y, where x, y ∈ A}

Hence, R = {(1, 3), (2, 6), (3, 9), (4, 12)}

We know that the domain of R is defined as the set of all first elements of the ordered pairs in the given relation.

Hence, the domain of R = {1, 2, 3, 4}

To determine the codomain, we know that the entire set A is the codomain of the relation R.

Therefore, the codomain of R = A = {1, 2, 3,…,14}

As it is known that, the range of R is defined as the set of all second elements in the relation ordered pair.

Hence, the Range of R is given by = {3, 6, 9, 12}

Functions and Types of Functions

Functions

Function is a special type of relation.

Definition:-

A relation 'f' is said to be a function, if every element of a non-empty set X, has only one image or range to a non-empty set Y. Or. If 'f' is the function from X to Y and (x,y) ∊ f, then f(x) = y, where y is the image of x, under function f and x is the preimage of y, under 'f'.

Or

A relation ‘f’ is said to be a function, if every element of a non-empty set X, has only one image or range to a non-empty set Y.

Or

If ‘f’ is the function from X to Y and (x,y) ∊ f, then f(x) = y, where y is the image of x, under function f and x is the preimage of y, under ‘f’. It is denoted as;

f: X → Y.

Example: N be the set of Natural numbers and the relation R be defined as;

R = {(a,b) : b=a², a,b ∈ N}. State whether R is a relation function or not.

Solution: From the relation R = {(a,b) : b=a², a,b ∈ N}, we can see for every value of natural number, their is only one image. For example, if a=1 then b =1, if a=2 then b=4 and so on.

Therefore, R is a relation function here.

Types of Functions

There are various types of functions in mathematics which are explained below in detail. The different function types covered here are:

• One – one function (Injective function)
• Many – one function
• Onto – function (Surjective Function)
• Into – function
• Bijective – function
• Inverse -- function

One – one function (Injective function)

If each element in the domain of a function has a distinct image in the co-domain, the function is said to be one – one function.

For examples f; R ->R given by f(x) = 3x + 5 is one – one.

Many – one function

On the other hand, if there are at least two elements in the domain whose images are same, the function is known as many to one.

For example f : R-> R given by f(x) = x² + 1 is many one.

Onto – function (Surjective Function)

A function is called an onto function if each element in the co-domain has at least one pre – image in the domain.

Into – function

If there exists at least one element in the co-domain which is not an image of any element in the domain then the function will be Into function.

(Q) Let A = {x : 1 < x < 1} = B be a mapping f : A ->B, find the nature of the given function (f).

f(x) = |x|

 f (x) = |1|

Solution for x = 1 & -1

Hence it is many one the Range of f(x) from [-1, 1] is

[0,1] which is not equal to co-domain. Hence it is into function.

Lets say we have function,

f(x)={x²;x≥0−x²;x<0

For different values of Input, we have different output hence it is one – one function also it manage is equal to its co-domain hence it is onto also.

Special Types of Functions

• Constant Function
• Identity Function
• Polynomial function
• Linear Function
• Identical Function
• Quadratic Function
• Rational Function
• Algebraic Functions
• Cubic Function
• Modulus Function
• Signum Function
• Greatest Integer Function
• Smallest Integer Function
• Step Function
• Characteristics Function
• Indicator Function
• Fractional Part Function
• Exponential Function
• Logarithmic Function
•  Sinusoidal. Function
• Even and Odd Function
• Periodic Function
• Composite Function

Polynomial function

A real valued function f : P → P defined by y = f (a)=h₀+h₁ a+…..+h_n a_n, where n ϵ N, and h₀+h₁+…..+h_n ϵ P, for each a ϵ P, is called polynomial function.

• N = a Natural Number.
• The degree of Polynomial function is the highest power in the expression.
• If the degree is zero, it’s called a constant function.
• If the degree is one, it’s called a linear function. Example: b = a+1.
• Graph type: Always a straight line.

So, a polynomial function can be expressed as :

f(x)=a_nx_n+a_n−1x_n−1+…..+a₁x₁+a₀

The highest power in the expression is known as the degree of the polynomial function. The different types of polynomial functions based on the degree are:

1. The polynomial function is called a Constant function if the degree is zero.
2. The polynomial function is called a Linear if the degree is one.
3. The polynomial function is Quadratic if the degree is two.
4. The polynomial function is Cubic if the degree is three.

Linear Function

All functions in the form of ax + b where a, b∈R & a ≠ 0 are called as linear functions. The graph will be a straight line. In other words, a linear polynomial function is a first-degree polynomial where the input needs to be multiplied by m and added to c. It can be expressed by f(x) = mx + c.

For example, f(x) = 2x + 1 at x = 1

f(1) = 2.1 + 1 = 3

f(1) = 3

Another example of linear function is y = x + 3

Identical Function

Two functions f and g are said to be identical if

(a) The domain of f = domain of g

(b) The range of f = the Range of g

(c) f(x) = g(x)∀  x∈D_f & D_g

For example f(x) = x & g(x) =11/x

Solution: f(x) = x is defined for all x

But g(x) = 11/x is not defined of x = 0

Hence it is identical for x ∈R -{0}

Quadratic Function

All functions in the form of y = ax² + bx + c where a, b, c∈R, a ≠ 0 will be known as Quadratic function. The graph will be parabolic.

At x=−b±D2, we will get its maximum on minimum value depends on the leading coefficient and that value will be −D4a (where D = Discriminant)

In simpler terms,

A Quadratic polynomial function is a second degree polynomial and it can be expressed as;

F(x) = ax² + bx + c, and a is not equal to zero.

Where a, b, c are constant and x is a variable.

Example, f(x) = 2x² + x – 1 at x = 2

If x = 2, f(2) = 2.2² + 2 – 1 = 9

For Example: y = x² + 1

Read More: Quadratic Function Formula

Rational Function

These are the real functions of the type f(a)g(a) where f (a) and g (a) are polynomial functions of a defined in a domain, where g(a) ≠ 0.

• For example f : P – {– 6} → P defined by f (a) = f(a+1)g(a+2), ∀aϵP – {–6 }is a rational function.
• Graph type: Asymptotes (the curves touching the axes lines).

Algebraic Functions

A function that consists of a finite number of terms involving powers and roots of independent variable x and fundamental operations such as addition, subtraction, multiplication, and division is known as an algebraic equation.

For Example,

f(x)=5x3−2x2+3x+6, g(x)=3x+4(x−1)2.

Cubic Function

A cubic polynomial function is a polynomial of degree three and can be expressed as;

F(x) = ax³ + bx² + cx + d and a is not equal to zero.

In other words, any function in the form of f(x) = ax³ + bx² + cx + d, where a, b, c, d∈R & a ≠ 0

For example: y = x³

MODULUS FUNCTION:

A Function f(x) : R à R is said to be Modulus function if y=f(x) = |x| 

Domain of f = R

Range of f = R⁺U {0}

Codomain=R

 

If x = -5, then y = f(x) = – (-5) = 5, since x is less than zero

If x = 10, then y = f(x) = 10, since x is greater than zero

If x = 0, then y = f(x) = 0, since x is equal to zero

When x = -3 then y = |-3| = 3

When x = -2 then y = |-2| = 2

When x = -1 then y = |-1| = 0

When x = 0 then y = |0| = 0

When x = 1 then y = |1| = 1

When x = 2 then y = |2| = 2

When x = 3 then y = |3| = 3

Signum Function:

A Function f(x) : R à R is said to be signum function if  

Domain of f = R

Range of f = {-1,0,1}

Codomain=R

If x = -5, then y = f(x) = – 1 , since x is less than zero

If x = 5, then y = f(x) = 1 , since x is greater than zero

If x = 0, then y = f(x) = 0 , since x is equal to zero

Greatest Integer Function:

A  Function f(x) : R à R is said to be greatest integer function if y=f(x)=[x]

or

The function f: R , R defined by f(x) = [x], x R assumes the value of the greatest integer, less

than or equal to x. Such a function is called the greatest integer function.

Domain of f = R

Range of f = Z

Codomain=R

[1.15] = 1  , [1.9] =1

[4.56567] = 4  , [4.99] = 4

[50] = 50

[-3.010] = -4

Greatest Integer Function Properties

•  [x] = x, where x is an integer
• [x + n] = [x] + n, where n ∈ Z
• [-x] = –[x], if x ∈ Z
• [-x] =-[x] – 1, if x ∉ Z
• If [f(x)] ≥ Y, then f(x) ≥ Y

Smallest Integer Function:

Ceiling function f(x) = ⌈x⌉ and  floor function f(x)=⌊x ⌋

f(x) = ⌊x⌋ = Largest Nearest Integer of specified value

f(x) = ⌈x⌉ = Least Nearest successive Integer of specified value

A  Function f(x) : R à R is said to be Smallest integer function if y=f(x)= ⌊x ⌋

 

Domain of f = R

Range of f = Z

Codomain=R

Properties:

• ⌈x⌉ + ⌈y⌉ – 1 ≤ ⌈x + y⌉ ≤ ⌈x⌉ + ⌈y⌉
• ⌈x + a⌉ = ⌈x⌉ + a
• ⌈x⌉ = a; iff x ≤ a < x + 1
• ⌈x⌉ = a; iff x – 1 < a ≤ x
• a < ⌈x⌉ iff a < x
• a ≤ ⌈x⌉ iff x < a

Step Function:

A step function (also called as staircase function) is defined as a piecewise constant function, that has only a finite number of pieces. In other words, a function on the real numbers can be described as a finite linear combination of indicator functions of given intervals.

Domain of f = referred to as the set of input values

Range of f = referred to as the set of output values generated for the domain (input values)

Properties

The important properties of step functions are given below:

• The sum or product of two-step functions is also a step function.
• If a step function is multiplied by a number, then the result produced is again a step function. That indicates the step functions create an algebra over the real numbers
• A step function can take only a finite number of values
• Piecewise linear function is the definite integral of a step function

Indicator function / Characteristic function: 

An indicator function or a characteristic function of a subset of a set is a function that maps elements of the subset to one, and all other elements of the set to zero. The indicator function of a subset A of a set X maps X to the two-element set 

{ 0,1}; c_A(x)=1 if an element {displaystyle x}in X belongs to A

and c_A(x)=0 if x does not belong to A.

It may be denoted as 1_A , by I_A or by c_A

The indicator function of a subset, that is the function

c _A : X à {0,1}

which for a given subset A of X, has value 1 at points of A and 0 at points of X − A.

Fractional Part Function:

The fractional part function is the difference between a number and its integer part.

A  Function f(x) : R à R is said to be fractional part function if  y = f(x) = {x} = x- [x]

Domain of a fractional function is all the real numbers except the roots of denominator of the fraction.

Codomain =R

Range = [0,1)

{1}=1−[1]=1-1=0.

Properties;

Exponential Function:

For a > 1, the Exponential of b to base a is x if y =f(x)=  a^x = b. Thus, The  function is known as Exponential function

Exponential function having base 10 is known as a common exponential function.

i.e.,y=f(x)= 10^x

Exponential function having base e is known as a Natural exponential function.

y=f(x)= e^x

Domain =R

Codomain =R⁺

Range = R⁺

Properties:

• The graph passes through the point (0,1).
• The domain is all real numbers
• The range is y>0
• The graph is increasing
• The graph is asymptotic to the x-axis as x approaches negative infinity
• The graph increases without bound as x approaches positive infinity
• The graph is continuous
• The graph is smooth

If a>0, and  b>0, the following hold true for all the real numbers x and y:

• • a^x a^y = a^x+y
  • a^x/a^y = a^x-y
  • (a^x)^y = a^xy
  • a^xb^x=(ab)^x
  • (a/b)^x= a^x/b^x
  • a⁰=1
  • a^-x= 1/ a^x

Logarithmic Function:

For a > 1, the logarithm of b to base a is x if a^x = b. Thus, log_a b = x if a^x = b. This function is known as logarithmic function.

logarithm function having base 10 is known as a common logarithm function.

i.e.,y=f(x)= log₁₀ x

logarithm function having base e is known as a Natural logarithm function.

y=f(x)= log_e x= lnx

Domain = R⁺

Codomain =R

Range = R

Properties:

• The graph passes through the point (1,0).
• The domain is all +ve real numbers
• Log_ap = α, log_bp = β and log_ba = µ, then a^α = p, b^β = p and b^µ = a
• Log_bpq = Log_bp + Log_bq
• Log_bp^y = ylog_bp
• Log_b (p/q) = log_bp – log_bq
• When we plot the graph of log functions and move from left to right, the functions show increasing behaviour.
• The graph of log function never cuts the x-axis or y-axis, though it seems to tend toward them.

Sinusoidal Function:

A sinusoidal function is a function using the sine function. The basic form of a sinusoidal function is  where A is the amplitude or height of our function, B is the change in period defined by   the horizontal shift, and D the vertical shift.​​​​​​​

The sine and cosine functions have several distinct characteristics:

​​​​​​​

Even and Odd Function:

The definition of even and odd functions:

 Even function: A function, f(x) is even if f(x) = f(-x) 

Example:f(x)=cosx

 f(-x)=cos(-x)= cosx=f(x)

so f is even function.

examples of even functions are x⁴, cot x, y = x², etc.

Odd Function: A function, f(x) is odd if f(x) = -f(x) For example,

Check if function is even or odd: f (x) = tan x

f(-x)=tan(-x)= -tanx=-f(x)

so f is odd function.

The polynomial function f(x)=x2+x4+x6 is even. The polynomial function f(x)=x+x3+x5 is odd

Periodic Function:

A function y= f(x) is said to be a periodic function if there exists a positive real number P such that f(x + P) = f(x), for all x belongs to real numbers. The least value of the positive real number P is called the fundamental period of a function.

This fundamental period of a function is also called the period of the function, at which the function repeats itself. f(x + P) = f(x)

The period of the sine ,cosine function is 2π (units). All the trigonometric functions are periodic functions.

Composite Function:

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by g ∘ f, is defined as the function g ∘ f : A → C given by g ∘ f (x) = g(f (x)), ∀ x ∈ A.

The composite function gof(x) is read as “g of f of x”. If f(x)and g(x) are two functions then fog(x), gof(x), gog(x) and fof(x) are composite functions.

• fog(x) = f(g(x))
• gof(x) = g(f(x))
• gog(x) = g(g(x))
• fof(x) = f(f(x))
• fogoh(x) = f(g(h(x)))
• fofof(x) = f(f(f(x)))

The order of the function is important in a composite function since (fog)(x) is not equal to (gof)(x).