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

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

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
Properties of Bijective Function
Practice Problems
FAQs

Bijective Function

A function $f : X \to Y$ 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.

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 function $f : [- 1, \infty) \to [0, \infty)$ given by $f (x) = \sqrt (4 x + 4)$ is a bijective function.

Solution:

Given Function: $f (x) = \sqrt (4 x + 4)$

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,

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

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

For onto:

Since, $f^{'} (x) > 0 \Rightarrow f$ is an increasing function and the domain is $[- 1, \infty)$

Now, $f (- 1) = 0$ . Hence, Range of $f = [0, \infty)$

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 $f : R \to R$ defined by

is a bijection or not.

Solution:

Given, $f : R \to R$ 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 $f (x) = f (y)$ .

So, F is not one-one.

For Onto: Let Y be any element in the codomain(R), such that $f (x) = y$ for some element $x \in R$ (domain).

Clearly y ∉R

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

Note that if $f (x)$ is not one-one,then we can conclude it is not bijective, irrespective of onto or not.

Example:Show that the function $f : R \to R$ 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, $f (x)$ 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 $f (x)$ will always be a subset of integers.

Range $(Z) \neq$ codomain(R), therefore the function is not onto.

Hence the function is not bijective.

Example:Show that the function F: R → R, defined as $f (x) = x^{2}$ , is neither one-one nor onto.

Solution:

For $f (x) = x^{2}, f (- 1) = (- 1)^{2} = 1 a n d f (1) = (1)^{2} = 1$

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 $g : [0, 3] \to [1, 29]$ defined by $g (x) = 2 x^{3} + 36 x - 15 x^{2} + 1$ is

(a) Bijective

(b) Surjective but not injective

(d) neither injective nor surjective

Solution:

The function $g : [0, 3] \to [1, 29]$ defined by $g (x) = 2 x^{3} + 36 x - 15 x^{2} + 1$

$\Rightarrow g^{'} (x) = 6 x^{2} + 36 - 30 x$

$= 6 x^{2} - 30 x + 36$

$= 6 (x^{2} - 5 x + 6)$

$= 6 (x - 2) (x - 3)$

$g (x)$ 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, $g (x)$ is a many-one function.

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

$g (x)$ is decreasing in $[2,3] .$ and the values at extreme poins are $g (2) = 29 & g (3) = 28$ , so the minimum value of $g (x)$ 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: $f (x) = x + \sqrt{x^{2}}$ is a function such that F R $\to$ R, then $f (x)$ is

Injection
Surjection
Bijection
None of these

Solution:

We have $f (x) = x + \sqrt{x^{2}} = x + |x|$ , clearly F is not one-one as

$f (- 1) = f (- 2) = 0$ but $- 1 \neq - 2 .$ .

Also, $f (x) \geq 0$ for all $x \in$ R as

So, range of $f = [0, \infty) \neq$ R

$\Rightarrow$ Range $\neq$ Codomain, therefore given function is not onto.

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

Example:For $x \in R, f (x) = x^{3} + 5 x + 1$ then

F is one-one but not onto on R
F is onto but not one-one on R
F is one-one and onto on R
F is neither one-one nor onto on R

Solution:

$f (x) = x^{3} + 5 x + 1$

$\Rightarrow f^{'} (x) = 3 x^{2} + 5 > 0$ for all $x \in R$

$f (x)$ is strictly increasing

$f (x)$ is one-one.

Clearly, $f (x)$ is a continuous function and strictly increasing on R

So, F takes all the values from $- \infty t o \infty$ .

Thus, $f (x)$ is onto.

So, $f (x)$ 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)

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