Understanding Functions in Mathematics
Functions in mathematics establish a relationship between sets where each element in the first set has a unique image in the second set. This concept is crucial in various real-life scenarios, such as travel expenses, state capitals, and health data analysis. Not all relations are functions, and it is essential to differentiate one-one mappings from one-many relations to understand the function's characteristics fully.
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Topic: Topic: Functions Functions
Real life Examples:- Example:-1 Suppose I have to travel from Cuttack to Bhubaneswar . It has been seen if I travel by a bus then I have to spend Rs 30 , If I travel by scooter I have to spend Rs 100 , If I travel by car I have to spend Rs 500. It leads to a relationship between two sets as the mode of transport and expenditure for it . BUS BUS Rs 30 Scooter Scooter Rs 100 Car Car Rs 500
Let we have to make the relationship between set of capitals of states and set of states in India. i.e the capital capital- -state state relationship. Bhubaneswar Telangana Maharastra Hyderabad Andhrapradesh Kolkota Punjab Chandigard Westbengle Mumbai Haryana Odisha
Let a relationship between a set of some districts of Odisha and a set of number of corona patients in different districts till today. Khorda 52 37 3 Bhadrakh 10 Balasore 41 Keonjhar 2
The relation explained in example 1 is a function as every element of 1st set has a unique image in the second set. This is called a one-one mapping. but the relation given in example 2 is not a function as the capital Hyderabad is related to two states as Telangana and Andhra Pradesh . i.e it has no unique image in the second set. So it is a one-many relation but not a function. Also the relation explained in example 3 is not a function, as the district Keonjhar has no corona patient till today, i.e it has no image, i. e it is not related to any member of the second set. Note: may not be a function. Only one relations are functions. Note:- - Every function is a relation but every relation may not be a function. Only one- -one and many relations are functions. Every function is a relation but every relation one and many- -one one
A relation R from A to B is a function if it is not one-many as well as many-many . Also domain of the relation R is A itself . Ex: Let A={1,2,3} B={a,b,c,d} and R A X B R={(1,a),(2,c)} is not a function as domain of R is not the set A . (fig (fig- -1 1) R={(1,a),(2,d),(3,b),(1,c)}is not a function as it is an one- many relation. (fig (fig- -2) 2) R={(1,b),(2,c),(3,b),(2,d),(1,a)} is not a function as it is a many-many relation. (fig (fig- -3) 3) R={(1,c),(2,a),(3,b)} is a function as it is not one-many and domain of the relation R is A itself. (fig (fig- -4) 4)
1 1 a b c d a b c d 2 2 3 3 Fig- -1 1 Fig Fig-2 1 1 a b c d a b c d 2 2 3 3 Fig-3 Fig-4
A rule f which associates each element of a set A to a unique element of a set B is called a function. We write it as f : : A B . From definition it is clear that All the elements of A must be associated with elements of B . No element of A is associated with two or more elements of B . Here A=domain of the function B=Co-domain of the function Rf ={f(x): x A}=range of the function
Types of functions (Identity Function,Equal Function ,Injection and surjection). Methods to check injectivity and surjectivity of a function. Composition of a function. Theorems related to composition of a function. Invertible function and related theorems. Methods to find the inverse of a function.
If f : AB , g : C D are said to be equal if (i) A=C (ii) Range of f = range of g (iii) f(x)=g(x) for all x in the domain . We write f=g IX is said to be an identity function defined on the set X if IX (x) =x for all x X .
One-One Function ( Injective Function) Let f : A B If distinct elements of A are associated with distinct elements of B then the function is one-one or Injective. Methods to check: For x,y A , f(x)=f(y) x=y . For x,y A , x y f(x) f(y).
A Polynomial Function. 1. 2. A Piecewise defined function . No. of Injective Functions No. of Injective Functions If n(A)=m , n(B)=n , then no. of one-one functions from A to B are = 0 if m>n =m! Or n! if m=n =P(n,m) if m<n
a b 1 c 2 d A B Here two elements c and d of A can t be associated with any element of B . It can t be a function. So no.of one-one function is zero.
a a a 1 1 1 b b b 2 2 2 c c c a a a 1 1 1 b b b 2 2 2 c c c
a b c 1 a b c 1 a b c 1 2 2 2 3 3 3 a b c 1 a b c 1 a b c 1 2 2 2 3 3 3
Let f : A B If range of the function is same as the co-domain of the function , then function is onto . In other words , every element of B is the image of some element of A under f . Methods to check: (i)Range of f = Co-domain of f (ii)For every y B , there exists x A such that f(x)=y .
Example -01 Example-02 No. of onto functions If n(A)=n and n(B)=2 ,then no. of onto functions from A to B is 2 -2 . N.B. onto is a bijection N.B.:A function which is both one-one and bijection .
n(A)=4 , n(B)=2 . Total no. Of functions from A to B is 2 =16 . 1 1 a 2 a 2 3 3 b b 4 4
Let f :ABand g: C D be two functions . Then the composition of f and g denoted by gof is possible only when B C and is defined as gof : A C given by gof(x)=g(f(x)) for all x A . f g B A C D gof
Let f : {1,2,3} {3,4,5} and g : {3,4,5,6} {a,b,c,d} defined by f(1)=4 , f(2)=3 ,f(3)=5 , g(3)=a ,g(4)=c , g(5)=b, g(6)=d . Find gof . Ans Ans : : Now gof: {1,2,3} {a,b,c,d} gof(1)=g(f(1))=g(4)=c gof(2)=g(f(2))=g(3)=a gof(3)=g(f(3))=g(5)=b
Composition of functions is not Commutative. Composition of functions is Associative . If f :A B and g: B C are one-one functions , then gof : A C is also one-one . If f :A B and g: B C are onto functions , then gof : A C is also onto .
Let f : A B be a function . f is said to be invertible if there exists a function g: B A such that gof= IA and fog =IB . The function g is called the inverse function of f . We write it as N.B. N.B.:If f is invertible then f must be one-one and onto i.e. f must be a bijection .
f:RR , given by f(x)= sinx is not invertible as it is neither one-one nor onto . f: R [-1,1], given by f(x)= sinx is also not invertible as it is not one-one. f : [0,?] [-1,1] , given by f(x)= sinx is not invertible as it is neither one-one nor onto . f: [- ?/2 , ?/2] [-1,1] given by f(x)= sinx is invertible as it is both one-one and onto .
Inverse of a function is unique . Inverse of a bijection is also a bijection Inverse of a invertible function is the function itself. If f and g be two invertible functions , then gof is also invertible and Proofs:
Check the given function is a bijection Consider y=f(x) and express x as a function of y i.e. x=g(y) Then Examples:
A function f :AB is one-one if for x,y A , f(x)=f(y) x=y . A function f :A B is onto if for every y B there exists x A such that f(x)=y . A function f :A B is invertible if there exists g:B A such that gof=IAandfog=IB A function f : A B is invertible if and only if it is a bijection . Mind map Assignments (BASIC) Assignments (BASIC) Assignments (STANDARD) Assignments (STANDARD) Assignments (HOTS) Assignments (HOTS)
Students will be able to understand a function. know about one-one and onto function . learn about composition of functions find composition of two functions . know about the inverse of a function. find the inverse of a function.
