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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
A function 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.
The domain of this mapping is
The codomain is
The range is
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.
Example:Determine whether the function given by is a bijective function.
Solution:
Given Function:
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,
For the interval since, for all X on the interval , we can clearly say that this function is one-one on this interval.
The domain of the function is the interval , however which does not coincide with for any X in the interval , so the function is one-one on its domain.
For onto:
Since, is an increasing function and the domain is
Now, . Hence, Range of
So, Range of 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 defined by
is a bijection or not.
Solution:
Given, defined by
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 .
So, F is not one-one.
For Onto: Let Y be any element in the codomain(R), such that for some element (domain).
Clearly y ∉R
So, F is not onto.Hence F is not a bijection.
Note that if is not one-one,then we can conclude it is not bijective, irrespective of onto or not.
Example:Show that the function given by ,where [.] represents the Greatest Integer Function, is not bijective.
Solution:
Given: 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, will give same value for different values of X.
Therefore, it is not a one-one function, it is a many-one function.
will always be an integer so range of will always be a subset of integers.
Range codomain(R), therefore the function is not onto.
Hence the function is not bijective.
Example:Show that the function F: R → R, defined as , is neither one-one nor onto.
Solution:
For
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 defined by is
(a) Bijective
(b) Surjective but not injective
(c) Injective but not surjective
(d) neither injective nor surjective
Solution:
The function defined by
is decreasing in and increasing in .
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, is a many-one function.
is increasing in and the values at extreme poins are , so the minimum value of is 1 and the maximum value is 29 in interval
is decreasing in and the values at extreme poins are , so the minimum value of is 28 and the maximum value is 29 in interval
So, for the domain of the function, the maximum and minimum values are 29 and 1, respectively, Therefore, range is .
Since, range = codomain, the function is onto.
Therefore, option (B) is the correct answer.
Example: is a function such that F R R, then is
Solution:
We have , clearly F is not one-one as
but .
Also, for all R as
So, range of R
Range Codomain, therefore given function is not onto.
The function is neither one-one nor onto, so option (d) is correct.
Example:For then
Solution:
for all
is strictly increasing
is one-one.
Clearly, is a continuous function and strictly increasing on R
So, F takes all the values from .
Thus, is onto.
So, is both one-one and onto and hence a bijection.
Therefore option (c) is the correct answer.
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)