Class 11 NCERT Math Notes
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Read 11th Maths Notes.One of the important functions amongst all is the bijective function which pairs every input with exactly one unique output. In the case of this function, each possible out comes exactly from one input. Bijective function is both injective and surjective. Considering the fact that it is an important topic, we have also shared an NCERT excercise of Relations and Functions since it is taught in both class 11th and 12th of the CBSE board. Let us now learn about this function in further detaill with examples.
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Revise 12th Math Topics.Unsure if your solutions are correct?
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Solve 12th NCERT Math Questions.A function will be considered as bijective if it is both injective (one-to-one) and surjective (onto) in nature.
Injective (one-to-one):
There are no distinct inputs in A that are sharing the same output in B.
Surjective (onto):
Each element of B is hit by the function.
f is bijective when for every B, y ∈ B there is only one x ∈ A with f(x) = y
Let us understand what are injective and surjective functions.
The Bijective Function is a combination of both since it is a one-to-one function that works onto the co-domain element.
To recognize the bijection between two terms, we have first to understand the sets, map f: A → B, and conclude that |A| = |B|.
To prove a function f is bijective, we obtain either the inverse or prove that it is both injective and surjective.
If two sets A and B are not of the same size, then the functions aren’t bijective because bijection is pairing up of the elements in the two sets perfectly. If |A| = |B| = n, then there exists n! bijections between A and B. Let us understand this with an example. Do remember that IIT JAM and NEET are some of the entrance exams that ask questions related to bijective function.
Example: Show that the function f(x) = 4x – 5 is a bijective function from R to R.
Given, f(x) = 4x – 5
The given function should be both injective and surjective.
(i) To Prove: The function is injective
For this f(a)=c and f(b)=c then a=b must be proved.
Let us take, f(a)=c and f(b)=c
Therefore, it can be written as:
c = 4a-5 and c = 4b-5
Thus, it can be written as: 4a-5 = 4b -5
Simplify the equation; we will get: a = b; hence the given function is injective
(ii) To Prove: The function is surjective
First, we should show that for “a” in R there exists a point “b” in the domain such that f(b) =a
Let, a = 4x -5
Therefore, b must be (a+5)/4
Since this is in R, the function is surjective.
As the function is both injective and surjective, it is bijective.
Class 12 students learn this in Relations and Functions,, which has a 40% weightage in the board exams. This topic comes in the functions category and 2-4 mark questions. Even in entrance exams like JEE Main, the chapter holds decent weightage. Now let us move on to understanding the properties of a bijective function:
Let us understand different methods through which we can prove if a function is bijective:
Let us prove that
For Example:
Let us assume that .
We will first subtract 2 and then divide it by 3
Finally, we will get: .
Let us prove that:
x first
For example: .
Given
solve
Since
, every
Entrance exams like CUET often ask questions about different types of functions. Therefore, we are considering some basic questions so that you can further move on to advanced level questions.
1. Is the mapping injective or surjective?
(i) { (x, y): x is a person, y is the father of x }.
Solution.
(i) Here, y can be a father to two terms in the x domain as it is not specific. Hence, this is a surjective function and not injective.
2. Prove that the function f: A → b is invertible only if f is both one-one and onto.
Solution.
A function f :X → Y is defined to be invertible, if there exists a function g = Y → X such that g(f) = Ix and f(g) = Iy, the function is called the inverse of f and is denoted by f-1. This, if bijective, means f can be invertible.
3. If set a contains five elements, and the set b contains six elements, then the number of bijective mappings from a to b is
Solution.
0. Since the number of elements in B is more than A.
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