A one-to-one function, is also known as injective function. It is a function where the mapping of each domain element is done to a unique element in the codomain. Or we can say it the other way round that no two different elements in the domain map to the same element in the codomain. Having an insight into the concept of one-to-one functions is significant in various branches of mathematics, specifically in algebra and calculus.
In this article, we will explore the concept of one-to-one functions, a key topic within the broader category of relations and functions. This concept is crucial for board exams and competitive exams like the Joint Entrance Examination (JEE Main), as well as other entrance tests such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the past decade (2013-2023) in the JEE Main exam, a total of six questions have addressed this concept, with one question in 2017, two in 2019, one in 2021, one in 2022, and one in 2023.
A relation from a set $A$ to a set $B$ is said to be a function from $A$ to $B$ if every element of set $A$ has one and only one image in set $B$. One-to-one function is one of the type of functions in which every element in domain has a unique image in codomain. Now, let us look in detail about what is one-to-one function.
An injective function, mostly referred to as a one-to-one function, is one in which distinct elements of $A$ have distinct images with $B$. If a function has distinct images, it can only be one-to-one if the pre-images are different. On similar logic, if the elements in $B$ set differ, it can only be one-to-one if the elements in $A$ set had different pre-images.
A function $f: X \rightarrow Y$ is called a one-one (or injective) function, if different elements of $X$ have different images in $B$. i.e. no two elements of set $X$ can have the same image.
Consider,
$f: X \longrightarrow Y$, function given by $y=f(x)=x$, and
$X=\{-2,2,4,6\}$ and $Y=\{-2,2,4,6\}$
The relevant one-to-one function graphs are formed below:
Now, consider, $\mathrm{X}_1=\{1,2,3\}$ and $\mathrm{X}_2=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}$
$
\mathrm{f}: \mathrm{X}_1 \longrightarrow \mathrm{X}_2
$
A one-to-one function can be identified by verifying the condition of one-to-one function, $f\left(x_1\right)=f\left(x_2\right) \Longrightarrow x_1=x_2$.
For example, Let $f(x)=2 x+3$. We'll check if it's one-to-one using the definition:
Let us assume that $f\left(x_1\right)=f\left(x_2\right)$, meaning two outputs of the function are equal.
$
f\left(x_1\right)=f\left(x_2\right) \Longrightarrow 2 x_1+3=2 x_2+3
$
$
\Longrightarrow 2 x_1=2 x_2
$
$
\Longrightarrow x_1=x_2
$
Therefore, the function $f(x) = x+3$ is a one-to-one function.
Another way to identify one-to-one function is by horizontal line test.
An injective or one to one function can be identified by the horizontal line test which is also known famously as geometric test.The process is explained below:
Here are a few famous one-to-one function examples:
Identity Function: The identity function is a simple example of a one-to-one function. The input and output values remain the same.In other words, we get the same output as the input itself, with no change at all. For any real number $x$, the identity function is defined as:
$f(x)=x$
Note: Every distinct input x corresponds to a unique output $f(x)$, making it a one-to-one function.
Absolute Value Function: The absolute value function $f(x)=∣x∣$ is also a one-to-one function. For any natural number $x$, the absolute value function returns a non-negative output, and if we continuously change the values of $x$, we get different positive outputs each time, being unique every single time. But this doesn't work in the case of integers as $-x$ and $x$ will have the same output.
No, a parabola is not a one-to-one function. This conclusion can be proved by the horizontal line test.
Representation: $f(x) = x^2$
Now, if we draw the horizontal lines, then it will intersect the parabola at exactly two points in the graph upon observation. Hence, for each value of $x$, there will be two outputs for a single input.
1. Uniqueness: Each element of the domain maps to a unique element in the codomain.That means there is exactly one unique value in the output range(y) for every single value of input range(x).
2. Inverse Function: If f is a one-to-one function, then there always exists an inverse function $f^{-1}$ for it, which is also a function.
3. Horizontal Line Test: A function $f$ is one-to-one if and only if no horizontal line intersects the graph of $f$ more than once.
For every one-to-one function equation, there exist a inverse. If $f$ is a function defined as $y=f(x)$, then the inverse function of $f$ is $x$ $=f^{-1}(y)$ that is $f^{-1}$ defined from $y$ to $x$. In the inverse function, the codomain of $f$ is the domain of $f^{-1}$ and the domain of $f$ is the co-domain of $f^{-1}$.
Only one-to-one functions have its inverse as these functions have one to one mappings, that is each element from the range corresponds to one and only one domain element.
Let a function $\mathrm{f}: \mathrm{A}\rightarrow B$ is defined, then f is said to be invertible if there exists a function $\mathrm{g}: \mathrm{B}\rightarrow A$ in such a way that if we operate $\mathrm{f}\{\mathrm{g}(\mathrm{x})\}$ or $\mathrm{g}\{\mathrm{f}(\mathrm{x})\}$ we get the starting point or value.
Example:
Show that the function $f: X\rightarrow Y$, such that $f(x)=30 x+7$,
For all $x, y \in N$ is invertible.
Solution:
Let $y \in N \rightarrow y=f(x)=30 x+2$ for $x \in N$
$
x=(y-7) / 30
$
If we define $h: Y \rightarrow X$ by $h(y)=\frac{(y-2)}{30}$
Again $h \circ f(x)=h[f(x)]=h\{30 x+2\}=\frac{30(y-2)}{30+2=x}$
And $f \circ h(y)=f[h(y)]=f(\frac{(y-2)}{30})=\frac{30(y-2)}{30+2=y}$
Hence $f$ is an invertible function and $h$ is the inverse of $f$.
Here are the properties of the inverse of one to one function:
Feature | One-to-One Function (Injection) | Many-to-One Function |
---|---|---|
Definition | Each element in the domain maps to a unique element in the codomain. | At least two distinct elements in the domain map to the same element in the codomain. |
Relationship between inputs and outputs | Precise one-to-one correspondence. No two inputs produce the same output. | Multiple inputs can produce the same output. |
Horizontal Line Test | No horizontal line intersects the graph more than once. | A horizontal line can intersect the graph multiple times. |
Example | $f(x)=x, f(x)=x^3$, etc. | $(x)=x^2, f(x)=\sin (x) \text {, etc. }$ |
One-to-one function examples in real life includes
Example1: The function
$f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right]_{\text {defined as }} f(x)=\frac{x}{1+x^2}$, is
1) injective but not surjective
2) surjective but not injective
3) neither injective nor surjective
4) invertible
Solution:
$
\begin{aligned}
& f(x)=\frac{x}{1+x^2} \\
& f: R \rightarrow\left[-\frac{1}{2}, \frac{1}{2}\right] \\
& f^{\prime}(x)=\frac{\left(1+x^2\right) \times 1-x \times 2 x}{\left(1+x^2\right)^2}=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2}=\frac{1-x^2}{\left(1+x^2\right)^2} \\
& \therefore \text { So that } \frac{-\left(x^2-1\right)}{\left(x^2+1\right)^2}
\end{aligned}
$
So that it is not strictly increasing or decreasing function.
So that it is not one-one.
So, the given function is surjective but not injective.
Hence, the answer is the option 2.
Example 2: Which of the following functions are one - one functions?
1) $f(x)=x^2$
2) $f(x)=x^4$
3) $f(x)=\cos x$
4) $f(x)=\sqrt{x}$
Solution:
A line parallel to the $x$-axis cuts the curve of one-one function at at most one point.
$
f(x)=\sqrt{x}
$
Clearly, this function is one - one function.
Hence, the answer is the option (4).
Example 3: Which of the following functions are one-one functions?
1) $\sin (\cos x)$
2) $\cos (\sin x)$
3) $\sin (\tan x)$
4) None of these
Solution:
As we learned
In the case of composite functions,
If both $f(x)$ and $g(x)$ are one-one functions, then fog $(x)$ and $g \circ f(x)$ are both one-one functions.
Hence, the answer is the option 4 .
Example 4: Let $A=x \in \mathrm{R}: \mathrm{x}$ is not a positive integer. Define a function $f: A \rightarrow R_{\text {as }}$ $f(x)=\frac{2 x}{x-1}$ then f is:
1) injective but not surjective
2) neither injective nor surjective
3) not injective
4) surjective but not injective
Solution:
One - One or Injective functions -
A function in which every element of the range of function corresponds to exactly one element.
- wherein
A line parallel to the $x$-axis cuts the curve at most one point.
$
f(x)=\frac{2 x}{x-1}
$
This can be written as
$
\begin{aligned}
f(x) & =2\left(1+\frac{1}{x-1}\right) \\
f^{\prime}(x) & =-\frac{2}{(x-1)^2}
\end{aligned}
$
$\Rightarrow \mathrm{f}$ is one-one i.e injective but not surjective.
Hence, the answer is option 1 .
Example 5: Which of the functions $\mathrm{f}(\mathrm{x})$ will be one-one functions if $f^{\prime}(x)$ is given
1) $f^{\prime}(x)=\sin x, x \equiv R$
2) $f^{\prime}(x)=e^x, x \equiv R$
3) $f^{\prime}(x)=\ln x, x \equiv R^{+}$
4) $f^{\prime}(x)=\cos x, x \equiv R$
Solution:
When $f^{\prime}=e^x>0$
Then $f(x)$ is an increasing function, so it is a one-one function.
Hence, the answer is option 2.
A function $\mathrm{f}: \mathrm{X} \rightarrow \mathrm{Y}$ is called a one-one (or injective) function, if different elements of X have different images in B. i.e. no two elements of set $X$ can have the same image. The example of one-to-one function is $f(x) = x+5$.
One-one function has a single value for the single domain but many-one function has multiple values for a single input.
One-to-one function test is the horizontal line test which is used to identify whether the given function is one-to-one or not. In this one-to-one function test, a horizontal line is drawn in the graph of the function. If the horizontal line intersects the graph of the function at only one point, then it is a one-to-one function.
Logarithmic function, constant function, odd degree polynomial, etc are some one-one functions.
One-to-one function graph is the graphical representation of an one-to-one function.
Some not one-to-one function examples are $f(x) = x^2, f(x) = \sin x, f(x) = \cos x, f(x) = |x|,$ etc.