Classification of Functions | Mathematics (Maths) for JEE Main & Advanced PDF Download

One-One Function (Injective mapping)

Definition: A function f: A → B is called a one-one function or injective mapping if distinct elements of A have distinct images in B. Equivalently, for x₁, x₂ ∈ A,

f(x₁) = f(x₂) ⇔ x₁ = x₂.

In other words, x₁ ≠ x₂ implies f(x₁) ≠ f(x₂).

Diagrammatically an injective mapping can be shown as

OR

Remark:

• Any function which is strictly increasing or strictly decreasing on its domain is one-one.
• If any line parallel to the x-axis (a horizontal line) cuts the graph of the function at most once, then the function is one-one.

Q.1. Determine if the function given below is one to one. 

(i) To each state of India assign its Capital

Ans. This is not one to one function because each state of India has not different capital.

(ii) Function = {(2, 4), (3, 6), (-1, -7)}

Ans. The above function is one to one because each value of range has different value of domain.

(iii) f(x) = |x|

Ans. To check whether f(x) = |x| is one-one, evaluate values for some x in the domain and observe their images.

From the table we see that the range value 1 is obtained from both x = 1 and x = -1. Hence the mapping is not one-one.

Q.2. Without using graph prove that the function

F : R → R defiend by f (x) = 4 + 3x is one-to-one

Ans.

Assume f(x₁) = f(x₂) for arbitrary x₁, x₂ ∈ R.

4 + 3x₁ = 4 + 3x₂.

Subtract 4 from both sides: 3x₁ = 3x₂.

Divide both sides by 3: x₁ = x₂.

Thus f(x₁) = f(x₂) implies x₁ = x₂, so f is one-one.

Many-One Function

Definition: A function f: A → B is called a many-one function if two or more distinct elements of A have the same image in B. That is, there exist x₁, x₂ ∈ A with x₁ ≠ x₂ but f(x₁) = f(x₂).

Diagrammatically a many-one mapping can be shown as

OR

Remark:

• A continuous function that has at least one local maximum or minimum is many-one on an interval containing those points. Equivalently, if a horizontal line cuts the graph at two or more points, the function is many-one.
• If a function is one-one, it cannot be many-one and vice versa.
• If f and g are both one-one (injective) then f ∘ g and g ∘ f (when defined) are also one-one.

Q.3. Without using graph check the following function f: R → R defined by f (x) = x² is one-to-one or not ?

Ans.

If x = 1 then f(1) = 1.

If x = -1 then f(-1) = 1.

Different domain elements 1 and -1 give the same image 1. Hence f(x) = x² is not one-one; it is many-one.

Thus any function that is not one-one may be called many-one.

Worked Example

Example 16. Show that the function f(x) = (x² - 8x + 18)/(x² + 4x + 30) is not one-one.

Solution.

To test for one-one, check whether f(x₁) = f(x₂) implies x₁ = x₂.

From the algebraic manipulation (or by exhibiting distinct x values with the same image), we find that f(x₁) = f(x₂) does not necessarily imply x₁ = x₂. Hence f is not one-one.

Functional Equation Example

Example 17. Let f be an injective function such that f(x) f(y) + 2 = f(x) + f(y) + f(xy) ∀ x, y ∈ R. If f(4) = 65 and f(0) ≠ 2, then show that f(x) - 1 = x³ for x ∈ R

Solution.

Given f(x)f(y) + 2 = f(x) + f(y) + f(xy). (i)

Put x = y = 0 in (i): f(0)f(0) + 2 = f(0) + f(0) + f(0).

Hence (f(0))² + 2 = 3f(0).

Rearrange: (f(0) - 2)(f(0) - 1) = 0.

Since f(0) ≠ 2, we get f(0) = 1.

Put x = y = 1 in (i) and perform similar steps to obtain (f(1) - 2)(f(1) - 1) = 0.

Injectivity excludes f(1) = 1, so f(1) = 2.

Now put y = 1/x in (i) (x ≠ 0) and simplify to obtain a relation leading to

Let g(x) = f(x) - 1.

From the transformed relation we get g(x) g(1/x) = 1.

This forces g(x) to be of the form ± xⁿ for some integer n, so f(x) = ± xⁿ + 1.

Given f(4) = 65, we solve 65 = ± 4ⁿ + 1.

Therefore 4ⁿ = 64 = 4³, so n = 3.

With sign chosen to match values, we get f(x) = x³ + 1, i.e. f(x) - 1 = x³.

Onto Function (Surjective mapping)

Definition: A function f: A → B is called onto or surjective if every element of the codomain B is the image of at least one element of A. That is, for every b ∈ B, there exists a ∈ A such that f(a) = b.

Diagrammatically surjective mapping can be shown as

OR

Note: If the range of f equals its codomain, then f is onto.

Q.4. If f: R → R is defined as f (x)= 3x + 7, x ∈ R, then show that f is an onto function.

Ans. 

Let b ∈ R be arbitrary. Solve for x such that f(x) = b.

3x + 7 = b.

x = (b - 7)/3.

Since b ∈ R, x ∈ R. Thus for every b in R there exists x = (b - 7)/3 with f(x) = b.

Hence f is onto.

Into Function

Definition: A function f: A → B is called into if there exists at least one element of the codomain B that is not an image of any element of A.

Diagrammatically into function can be shown as

Remark:

• If a function is onto, it cannot be into (with the same codomain), and vice versa.
• If f and g are both onto, then g ∘ f and f ∘ g (when defined) are onto.
• Thus a function may be categorised into four types:

(a) One-one onto (injective & surjective)

(b) One-one into (injective but not surjective)

(c) Many-one onto (surjective but not injective)

(d) Many-one into (neither surjective nor injective)

Remark:

• If f is both injective and surjective, then f is called bijective. Bijective functions are also called invertible, non-singular, or one-to-one correspondences.
• If a set A contains n distinct elements then the total number of functions from A to A is nⁿ, and out of these, n! are one-one (bijective from A to A).
• The composite of two bijections is a bijection; if f and g are bijections and their composition is defined, then g ∘ f is bijective.

Q. Prove that
F : R → R defined by f (x) = 4x³ - 5 is a bijection

Ans. 

First show injectivity: assume f(x₁) = f(x₂).

4x₁³ - 5 = 4x₂³ - 5.

Therefore x₁³ = x₂³.

So x₁³ - x₂³ = 0, which factors as (x₁ - x₂)(x₁² + x₁x₂ + x₂²) = 0.

Since the quadratic factor is non-negative and zero only when x₁ = x₂, we get x₁ = x₂. Thus f is injective.

Now show surjectivity: given arbitrary y ∈ R, solve y = 4x³ - 5 for x.

4x³ = y + 5.

x = ((y + 5)/4)^1/3, which is a real number for every real y.

Hence every y has a preimage; f is onto.

Therefore f is both one-one and onto, i.e. bijective.

Additional Examples

Example 18. A function is defined as , f: D → R f(x) = cot^-1 (sgn x) + sin^-1(x - {x}) (where {x} denotes the fractional part function) Find the largest domain and range of the function. State with reasons whether the function is injective or not. Also draw the graph of the function.

Solution.

Analysis shows that such a function takes a finite set of values repeated on intervals because both signum and fractional part produce piecewise-constant or bounded behaviours. Therefore the function is many-one.

Example 19. Find the linear function(s) which map the interval [0, 2] onto [1, 4].

Solution.

Let f(x) = ax + b be linear and map [0, 2] onto [1, 4].

Either f(0) = 1 and f(2) = 4, or f(0) = 4 and f(2) = 1 (linear map reversing the interval).

Case 1: f(0) = b = 1 and f(2) = 2a + b = 4 ⇒ 2a + 1 = 4 ⇒ a = 3/2.

So f(x) = (3/2)x + 1.

Case 2: f(0) = b = 4 and f(2) = 2a + 4 = 1 ⇒ 2a = -3 ⇒ a = -3/2.

So f(x) = 4 - (3/2)x.

Thus the two linear maps are f(x) = 3x/2 + 1 and f(x) = 4 - 3x/2.

Example 20. 

(i) Find whether f(x) = x + cos x is one-one.

(ii) Identify whether the function f(x) = -x³ + 3x² - 2x + 4 ; R → R is ONTO or INTO

(iii) f(x) = x² - 2x + 3; [0, 3] → A. Find whether f(x) is injective or not. Also find the set A, if f(x) is surjective.

Solution.

(i)

Compute derivative: f'(x) = 1 - sin x.

Since -1 ≤ sin x ≤ 1, we have f'(x) ≥ 0 for all x and equality holds only at isolated points where sin x = 1.

Therefore f is strictly increasing except at isolated points and is one-one on R.

(ii)

The cubic polynomial -x³ + 3x² - 2x + 4 has codomain R and as a continuous odd-degree polynomial it attains all real values. Hence, as given with codomain = R, the function is ONTO.

(iii)

For f(x) = x² - 2x + 3 on [0, 3], derivative f'(x) = 2(x - 1).

f is not monotonic on [0, 3] (it decreases on [0,1] and increases on [1,3]) so it is not injective.

To find A for surjectivity, compute the range on [0,3]. Minimum occurs at x = 1 with f(1) = 2 and maximum at an endpoint x = 3 with f(3) = 9 - 6 + 3 = 6.

Thus the range is [2, 6], so A = [2, 6].

Example 21. If f and g be two linear functions from [-1, 1] onto [0, 2] and

be defined by

Solution.

Let h be a linear function from [-1,1] onto [0,2], write h(x) = ax + b.

If a > 0 (increasing) we require h(-1) = 0 and h(1) = 2 giving -a + b = 0 and a + b = 2.

Solving: a = 1, b = 1, so h(x) = x + 1.

If a < 0 (decreasing) we require h(-1) = 2 and h(1) = 0 giving -a + b = 2 and a + b = 0.

Solving: a = -1, b = 1, so h(x) = 1 - x.

Thus the two onto linear maps are f(x) = 1 + x and g(x) = 1 - x (or swapped).

In both cases the required algebraic combination (as given in the original problem statement) can be evaluated; the image placeholders above indicate the steps and graphs as in a worked solution.

Summary

This chapter categorises functions by how elements of the domain map to the codomain: injective (one-one), surjective (onto), many-one, and into. A function that is both injective and surjective is bijective. For each concept, test by algebraic conditions or derivatives (for continuous functions) and use graphs and example functions to verify. Examples above illustrate functional equations, algebraic tests for injectivity/surjectivity, and construction of linear maps between intervals.