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Bijective(One-one and Onto): Functions and Its Properties

Bijective(One-one and Onto): Functions and Its Properties

Can you recognize what is so special about this arrow diagram(mapping) ?    

Image           

  • Distinct elements in X are distinctly related to some element of Y
  • Every element of Y is related to some or the other element of X

These kinds of functions are given a special name i.e. Bijective functions.Let us learn how to check that the given function is bijective.

Table of Contents

Bijective Function

A function f : X  Y is said to be bijective, if F is both one-one and onto. 

So, distinct elements of X have distinct images & codomain = range.

For example, the mapping given below is a bijective function.

image2

The domain of this mapping is a,b,c,d

The codomain is 1,2,3,4

The range is 1,2,3,4

So, the codomain = range and every element has a unique image and pre-image.

Therefore, the function is both one-one and onto, hence bijective.

Note: There are various methods to prove one-one and onto.One such method to prove whether a function is one-one or not is using the concept of  Derivatives.

A continuous(and differentiable) function whose derivative is always positive or always negative (strictly increasing or decreasing) is a one-one function.

Properties of Bijective Function

  • The domain and codomain of a bijective function have an equal number of elements.
  • The codomain and range of a bijective function are the same.
  • Every bijective function has an inverse function.
  • The bijective function follows reflexive, symmetric, and transitive property.

Practice Problems of Bijective

Example:Determine whether the functionf: -1,0, given by f(x)=(4x+4)  is a bijective function.

Solution:

Given Function: f(x)=(4x+4) 

For a function to be bijective,the function should be both injective and surjective.

For one one:

Using the chain rule of differentiation we have,

image3

For the interval (-1,)  since, f'x>0 for all X on the interval (-1,) , we can clearly say that this function is one-one on this interval.

The domain of the function is the interval (-1,) , however f-1=0 which does not coincide with fx for any X in the interval (-1,) , so the function is one-one on its domain.

For onto:

Since, f'x>0f is an increasing function and the domain is  -1,

Now, f-1=0. Hence, Range of f=0,

So, Range of f= Co-domain of F

Therefore the function is onto.Thus, the given function satisfies the conditions of one-one function and onto function, thus the given function is bijective.

Example: Determine whether the function f: RR defined byimage4

is a bijection or not.

Solution:

Given, f: RR defined by  image5

Now we have to check both one-one and onto conditions.

For one-one:Let X and Y be any two elements in the domain R, such that fx= fy.

image6

image7

image8

So, F is not one-one.

For Onto: Let Y be any element in the codomain(R), such that fx= y for some element xR  (domain).

image9

image10

Clearly y ∉R

So, F is not onto.Hence F is not a bijection.

Note that if fx is not one-one,then we can conclude it is not bijective, irrespective of onto or not.

Example:Show that the function f: RR given by f(x)=[x]2+[x+1]-3 ,where [.] represents the Greatest Integer Function, is not bijective.

Solution:

Given: f(x)=[x]2+[x+1]-3 and [.] is a greatest integer function

Clearly we can observe that in [1,2) for any x, the value of (X)  will be zero

So, fx will give same value for different values of X.

Therefore, it is not a one-one function, it is a many-one function.

[x]2+[x+1] will always be an integer so range of fx will always be a subset of integers.

RangeZ codomain(R), therefore the function is not onto.

Hence the function is not bijective.

Example:Show that the function F: R → R, defined as f x= x2, is neither one-one nor onto.

Solution:

image11

For f(x)=x2, f(-1)=(-1)2=1 and f(1)=(1)2=1

Since two elements in the domain have the same image.Therefore, F is not a one-one function.

Also, the element -2 in the codomain R is not an image of any element X in the domain R as the square of any real number can’t be negative. Therefore F is not onto.

Example: The function g : 0, 3 1, 29 defined by gx= 2x3 + 36x  15x2  + 1 is

(a) Bijective

(b) Surjective but not injective

(c) Injective but not surjective

(d) neither injective nor surjective

Solution:

The function g : 0, 3 1, 29 defined by gx= 2x3 + 36x  - 15x2+ 1

g'x= 6x2 +36 -30x

              = 6x2 -30x + 36

              = 6x2  5x + 6

= 6x  2x  3

gx is decreasing in 2, 3 and increasing in [0,2] .

We know, if a function is strictly increasing or decreasing in its domain, then it is one-one. But the given function is increasing as well as decreasing.So, gx is a many-one function.

gx is increasing in [0,2]  and the values at extreme poins are g(0) = 1 &  g(2) = 29, so the minimum value of gx is 1 and the maximum value is 29 in interval 0,2

gx is decreasing in 2,3. and the values at extreme poins are g(2) = 29 & g(3) = 28, so the minimum value of gx is 28 and the maximum value is 29 in interval 2,3.

So, for the domain of the function, the maximum and minimum values are 29 and 1, respectively, Therefore, range is [1,29]..

Since, range = codomain, the function is onto.

Therefore, option (B) is the correct answer.

Example: fx=x+x2 is a function such that F  R, then fx is

  1. Injection
  2. Surjection
  3. Bijection
  4. None of these

Solution:

We have fx=x+x2=x+x, clearly F is not one-one as 

f-1=f-2=0 but -1-2..

Also, fx0 for all x   R as 

 

So, range of f=0, R

Range Codomain, therefore given function is not onto.

The function is neither one-one nor onto, so option (d) is correct.

Example:For xR, fx=x3+5x+1 then

  1. F is one-one but not onto on R
  2.  F is onto but not one-one on R
  3. F is one-one and onto on R
  4. F is neither one-one nor onto on R

Solution:

fx=x3+5x+1

f'x=3x2+5>0 for all xR

fx is strictly increasing

fx is one-one.

Clearly, fx is a continuous function and strictly increasing on R

image13

So, F takes all the values from - to .

Thus, fx is onto.

So, fx is both one-one and onto and hence a bijection.

Therefore option (c) is the correct answer.

FAQs of Bijective

Question1. Does one-one function and one-to-one correspondence mean the same?

Answer: No, one-one function means injective function and one-to-one correspondence means bijective function.

Question2. To prove a function to be bijective, what should be proved first, injective or surjective?

Answer: Function should be proved both as injective and surjective, order of proving it doesn't matter.

Question3. Is there any other way to prove a function bijective?

Answer: A function can be proved to be bijective using an arrow diagram also(if possible to draw)

Related Topics to Bijective Function In Maths

 

NCERT Class 11 Maths Chapters

Sets Relations and Functions Triginometric Functions
Mathematical Induction Numbers and Quadriatic Equations Linear Inequalities
Premutations and Combinations Binomial Theorem Sequence and Series
Straight Lines Conic Sections 3 D Geometry
Limits and Derivatives Mathematical Reasoning Statistics
Probability  

NCERT Class 12 Maths Chapters

 

Relations and Functions Continuity and Differentiability Differential Equations
Inverse Trigonometric Functions Applications of Derivatives Vector Algebra
Matrices Integrals Three Dimensional Geometry
Determinants Applications of Integrals Linear Programming
Probability    

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