Functions are defined as the relations which give a particular output for a particular input value. A function has a domain and codomain (range). f(x) usually denotes a function where x is the input of the function. In general, a function is written as y = f(x).
Vertical line test is used to determine whether a curve is a function or not. If any curve cuts a vertical line at more than one point then the curve is not a function.
Function in algebra is an equation that gives a relation between a particular value of input say x with the particular value of output say y. A function is written as,
y = f(x)
where
Example: Some functions in algebra are:
The elements of set X are called the domain of f and the elements of set Y are called the codomain of f. The images of the elements of set X are called the range of function, which is always a subset of Y. The image given below demonstrates the domain, codomain, and range of the function.
The image demonstrates the domain, codomain, and range of the function. Remember the element which is mapped only will be counted in the range as shown in the image. The domain, codomain, and range of the above function are
There are several types of functions in maths. Some of the important types are:
A function f: X > Y is said to be a onetoone function if the images of distinct elements of X under f are distinct. Thus, f is one to one if f(x_{1}) = f(x_{2})
If the function is not one to one function then it should be many to one function means every element of the domain has more than one image at codomain after mapping.
A function f: X > Y is said to be an onto function if every element of Y is an image of some element of set X under f, i.e for every y ∈ Y there exists an element x in X such that f(x) = y.
Properties:
Condition to be Onto function: The range of function should be equal to the codomain.
As we see in the above two images, the range is equal to the codomain means that every element of the codomain is mapped with the element of the domain, as we know that elements that are mapped in the codomain are known as the range. So these are examples of the Onto function.
Properties:
Example 1: Check whether the function f(x) = 2x + 3, is onetoone or not if Domain = {1, 2, 1/2} and Codomain = {5, 7, 4}
Solution:
Putting 1, 2, 1/2 in place of x in f(x) = 2x + 3, we get
f(1) = 5,
f(2) = 7,
f(1/2) = 4
As, for every value of x we get a unique f(x) thus, we can conclude that our function f(x) is One to One.
Example 2: Check whether the function is onetoone or not: f(x) = 3x – 2
Solution:
To check whether a function is one to one or not, we have to check that elements of the domain have only a single preimage in codomain or not. For checking, we can write the function as,
f(x_{1}) = f(x_{2})
3x_{1} – 2 = 3x_{2} – 2
3x_{1} = 3x_{2}
x_{1} = x_{2}
Since both x_{1} = x_{2} which means that elements of the domain having a single preimage in its codomain. Hence the function f(x) = 3x – 2 is one to one function.
Example 3: Check whether the function is onetoone or not: f(x) = x^{2} + 3.
Solution:
To check whether the function is One to One or not, we will follow the same procedure. Now let’s check, we can write the function as,
f(x_{1}) = f(x_{2})
(x_{1})^{2} + 3 = (x_{2})^{2} + 3
(x_{1})^{2} = (x_{2})^{2}
Since (x_{1})^{2} = (x_{2})^{2} is not always true.
Hence the function f(x) = x^{2} + 3 is not one to one function.
Example 4: If N: > N, f(x) = 2x + 1 then check whether the function is injective or not.
Solution:
In question N > N, where N belongs to Natural Number, which means that the domain and codomain of the function is a natural number. For checking whether the function is injective or not, we can write the functions as,
Let, f(x_{1}) = f(x_{2})
2x_{1} + 1= 2x_{2} + 1
2x_{1} = 2x_{2 }
x_{1} = x_{2}
Since x_{1} = x_{2}, means all elements of the domain are mapped with a single element of the codomain. Hence function f(x) = 2x + 1 is Injective (One to One).
Example 5: f(x) = x^{2}, check whether the function is Many to One or not.
Solution:
Domain = {1, 1, 2, 2}, let’s put the elements of the domain in the function
f(1) = 1^{2 }= 1
f(1) = (1)^{2 }= 1
f(2) = (2)^{2 }= 4
f(2) = (2)^{2 }= 4
Thus, we can see that more than one element of the domain have similar image after mapping. So this is Many to One function.
Example 6: If f(x) = 2x + 1 is defined on R:> R. Then check whether the following function is Onto or not
Solution:
For checking the function is Onto or not, Let’s first put the function f(x) equal to y
f(x) = y
y = 2x + 1
y – 1 = 2x
x = (y – 1) / 2
Now put the value of x in the function f(x), we get,
f((y – 1) / 2) = 2 × [(y – 1) / 2] +1
Taking LCM 2, we get
= [2(y – 1) + 2] / 2
= (2y – 2 + 2) / 2
= y
Since we get back y after putting the value of x in the function. Hence the given function f(x) = 2x + 1 is Onto function.
Example 7: If f:N > N is defined by f(x) = 3x + 1. Then prove that function f(x) is Surjective.
Solution:
To prove that the function is Surjective or not, firstly we put the function equal to y. Then find out the value of x and then put that value in the function. So let’s start solving it.
Let f(x) = y
3x + 1 = y
3x = y – 1
x = (y – 1) / 3
Now put the value of x in the function f(x), we get
f((y – 1) / 3) = {3 (y – 1) / 3} + 1
= y – 1 + 1
= y
Since we get back y after putting the value of x in the function. Hence the given function f(x) = (3x + 1) is Onto function.
Example 8: If A = R – {3} and B = R – {1}. Consider the function f: A > B defined by f(x) = (x – 2)/(x – 3), for all x ∈ A. Then show that the function f is bijective.
Solution:
To show the function is bijective we have to prove the given function both One to One and Onto.
Let’s first check for One to One:
Let x_{1}, x_{2} ∈ A such that f(x_{1}) = f(x_{2})
Then, (x_{1} – 2) / (x_{1} – 3) = (x_{2} – 2) / (x_{2} – 3)
(x_{1} – 2) ( x_{2} – 3) = (x_{2} – 2) (x_{1} – 3)
x_{1} . x_{2} – 3x_{1} – 2x_{1} + 6 = x_{1} . x_{2} – 3x_{2} 2x_{1} + 6
3x_{1} – 2x_{2} = 3x_{2} – 2x_{1}
3( x_{1} – x_{2}) + 2( x_{1} – x_{2}) = 0
( x_{1} – x_{2}) = 0
x1 – x_{2} = 0
⇒ x_{1} = x_{2}
Thus, f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}, ∀ x1, x2 ∈ A
So, the function is a One to One
Now let us check for Onto:
Let y ∈ B = R – {1} be any arbitrary element.
Then, f(x) = y
⇒ (x – 2) / (x – 3) = y
⇒ x – 2 = xy – 3y
⇒ x – xy = 2 – 3y
⇒ x(1 – y) = 2 – 3y
⇒ x = (2 – 3y) / (1 – y) or x = (3y – 2) / (y – 1)
Now put the value of x in the function f(x)
f((3y – 2) / (y – 1)) = { (3y – 2) / (y – 1) } – 2 / { (3y – 2) / (y – 1) – 3 }
= (3y – 2 – 2y + 2) / (3y – 2 – 3y + 3)
= y
Hence f(x) is Onto function. Since we proved both One to One and Onto this implies that the function is Bijective.
Example 9: A = {1, 2, 3, 4}, B = {a, b, c, d} then the function is defined as f = {(1, a), (2, b), (3, c), (4, d)}. Check whether the function is One to One Onto or not.
Solution:
To check whether the function is One to One Onto or not. We have to check for both one by one.
Let’s check for One to One:
Now let’s check for Onto:
Example 10: A = {1, 2, 3, 4}, B = {a, b, c, d}. The function is defined as f = {(1, a), (2, b), (3, c), (4, c)}. Check whether the function is Many to One Into or not.
Solution:
To check the function is Many to One Into or not. We have to check for both one by one.
Let’s first check for Many to One function:
As we know the condition for Many to One function is that more than one element of domain should have more same image in codomain. From the above mapping we can see that the elements of A {3, 4 } are having same image in B { c }, so the function is Many to One.
Now let’s check for Into function:
As we know the condition for Into function is that the Range of function should be the subset of codomain and also not equal to codomain. Let’s check both the conditions are satisfied or not.
Range of function ≠ Codomain of function
As we check that the range of function is not equal to codomain of the function. Hence we can say that the function is Into function. As we prove that the function is Many to One and Into.
Hence the function is Many to One Into.
204 videos288 docs139 tests


Explore Courses for JEE exam
