One-One and Onto Functions – Types of Functions in Class 12 Maths Explained

Once you have understood what a function is, the next question naturally is: what kind of function is it? Class 12 Maths Chapter 1 introduces three critical types — one-one (injective), onto (surjective), and bijective — that describe how a function maps its domain to its co-domain.

These distinctions matter enormously in higher mathematics and appear consistently in CBSE board exams and competitive entrance tests like CUET and IPMAT.

What Is a One-One (Injective) Function?

A function f : X → Y is called one-one (or injective) if distinct inputs always produce distinct outputs.

Formally:

f(x₁) = f(x₂) ⟹ x₁ = x₂, for all x₁, x₂ ∈ X

Equivalently, if x₁ ≠ x₂, then f(x₁) ≠ f(x₂). No two different elements of the domain can share the same image in the co-domain.

A function that is not one-one is called many-one — two or more inputs map to the same output.

Checking Injectivity: The Algebraic Method

Step 1: Assume f(x₁) = f(x₂).
Step 2: Use algebra to deduce whether x₁ = x₂ must follow.
Step 3: If it must, the function is one-one; if you can find a counterexample, it is many-one.

Examples

One-one: f : N → N defined by f(x) = 2x.
Suppose f(x₁) = f(x₂). Then 2x₁ = 2x₂, so x₁ = x₂. ✓ Injective.

Many-one: f : R → R defined by f(x) = x².
f(−1) = f(1) = 1, but −1 ≠ 1. ✗ Not injective.

What Is an Onto (Surjective) Function?

A function f : X → Y is called onto (or surjective) if every element of the co-domain Y is the image of at least one element of X.

Formally:

For every y ∈ Y, there exists x ∈ X such that f(x) = y.

In other words, the range of the function equals the entire co-domain. If even one element of Y has no pre-image in X, the function is not onto (into).

Checking Surjectivity: The Algebraic Method

Step 1: Take an arbitrary y in the co-domain Y.
Step 2: Solve f(x) = y for x in terms of y.
Step 3: Check whether the resulting x actually belongs to the domain X.

Examples

Onto: f : R → R defined by f(x) = 2x.
For any y ∈ R, choose x = y/2 ∈ R. Then f(y/2) = y. ✓ Surjective.

Not onto: f : N → N defined by f(x) = 2x.
The element 1 ∈ N (co-domain) has no pre-image: 2x = 1 gives x = 1/2 ∉ N. ✗ Not surjective.

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One-One and Onto Functions – Types of Functions in Class 12 Maths Explained

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One-One vs Onto: A Key Observation

The function f(x) = 2x illustrates that injectivity and surjectivity depend on the domain and co-domain, not just the rule.

Domain & Co-domainOne-One?Onto?
f : N → N, f(x) = 2xYesNo
f : R → R, f(x) = 2xYesYes
f : R → R, f(x) = x²NoNo
f : N → N, f(1)=f(2)=1, f(x)=x−1 for x>2NoYes

This table makes an important NCERT point visible: for infinite sets, a function can be injective without being surjective, and vice versa.

An Important Property for Finite Sets

For finite sets, the situation is more symmetric. If f : X → X is a function from a finite set to itself, then:

  • f is one-one if and only if f is onto.

This means that on a finite domain, you only need to verify one of the two properties — the other follows automatically. This is not true for infinite sets, as the examples above show.

Solved Example from NCERT

Question: Let A be the set of 50 students in a class. Define f : A → N by f(x) = roll number of student x. Is f one-one? Is it onto?

One-one: No two students share a roll number. So f is one-one. ✓

Onto: Roll numbers run from 1 to 50. The natural number 51 ∈ N has no pre-image. So f is not onto. ✗

This simple example reinforces why understanding the co-domain is essential when classifying functions.

What is the difference between one-one and onto functions?

A one-one function ensures distinct inputs give distinct outputs. An onto function ensures every element of the co-domain has at least one pre-image in the domain.

Can a function be one-one but not onto?

Yes. f : N → N defined by f(x) = 2x is one-one (distinct inputs give distinct outputs) but not onto (odd numbers have no pre-image).

What is a many-one function?

A function is many-one if two or more distinct inputs produce the same output. For example, f(x) = x² is many-one since f(−2) = f(2) = 4.

How do you prove a function is not onto?

Find a single element in the co-domain that has no pre-image. For f : N → N, f(x) = 2x, the element 1 has no pre-image, so f is not onto.

Is f : R → R, f(x) = x³ one-one and onto?

Yes. It is one-one (f(x₁) = f(x₂) implies x₁³ = x₂³ implies x₁ = x₂) and onto (every real number has a real cube root). So it is bijective.

🎯 Can You Classify These Functions?

Put your knowledge to work with curated practice questions on one-one and onto functions — with step-by-step solutions — on ChampionsPrep.

Practice Function Classification on ChampionsPrep →

Test Your Knowledge

Interactive Invertibility & Bijection Simulator

Adjust slope 'a' of function f(x) = ax + b. A non-zero slope guarantees bijection and an inverse, while slope a = 0 makes f(x) constant (many-one, not onto) and destroys the inverse.

Function f(x) = ax + b

Slope a2
Intercept b1
Input x3
Output f(x)7
Inverse f⁻¹(x) 1
Inverse Exists? ⚠ Yes

Function g(x) = cx + d

Slope c1
Intercept d-2
Input x3
Output g(x)1

Compositions

(f ∘ g)(x) = f(g(x)) 3
(g ∘ f)(x) = g(f(x)) 5
Is Commutative? (f ∘ g = g ∘ f) No ✗
Syllabus Checkpoints & Exam Watch-Outs
  • When slope a = 0, f(x) becomes a horizontal constant function. It maps all x to the same value (many-one) and never covers the co-domain (not onto); therefore, its inverse does not exist.
WORKED EXAMPLE · Algebraic Proof: Proving Injective, Surjective & Bijective Mapping
Let f : \mathbb{R} \to \mathbb{R} be defined by f(x) = 2x. Prove that f is both one-one (injective) and onto (surjective), and therefore bijective.
Follow the standard NCERT algebraic verification procedure step by step:
Step 1: Test Injectivity (One-One)
f(x1)=f(x2)  ⟹  2x1=2x2  ⟹  x1=x2,∀x1,x2∈Rf(x_1) = f(x_2) \implies 2x_1 = 2x_2 \implies x_1 = x_2, \quad \forall x_1, x_2 \in \mathbb{R}
Assuming equal images forces equal inputs with no ambiguity. Distinct elements in the domain map to distinct elements in the co-domain. Hence, f is one-one.
Step 2: Test Surjectivity (Onto)
∀y∈R (co-domain),y=2x  ⟹  x=y2∈R (domain)\forall y \in \mathbb{R} \text{ (co-domain)}, \quad y = 2x \implies x = \frac{y}{2} \in \mathbb{R} \text{ (domain)}
For any arbitrary real number y, there exists a unique pre-image x = y/2 in the real domain such that f(y/2) = 2(y/2) = y. Hence, range equals co-domain, and f is onto.
Step 3: Bijective Conclusion & Invertibility
f is one-one and onto  ⟹  f is bijective  ⟹  f−1(x)=x2f \text{ is one-one and onto} \implies f \text{ is bijective} \implies f^{-1}(x) = \frac{x}{2}
Because f satisfies both conditions on ℝ → ℝ, it is a bijection. An inverse function exists and is given by f⁻¹(x) = x/2.
Solution Complete!
All 3 steps revealed and verified.
Q1.Review: Which of the following is the most important concept emphasized in this chapter?

Frequently Asked Questions

What is the difference between one-one and onto functions? +

A one-one function ensures distinct inputs give distinct outputs. An onto function ensures every element of the co-domain has at least one pre-image in the domain.

Can a function be one-one but not onto? +

Yes. f : N → N defined by f(x) = 2x is one-one but not onto, since odd numbers like 1 have no pre-image.

What is a many-one function? +

A function is many-one if two or more distinct inputs produce the same output. For example, f(x) = x² is many-one since f(−2) = f(2) = 4.

How do you prove a function is not onto? +

Find a single element in the co-domain that has no pre-image in the domain. For f : N → N, f(x) = 2x, the element 1 has no pre-image, proving it is not onto.

Is f : R → R, f(x) = x³ one-one and onto? +

Yes. f(x) = x³ is one-one (cube root is unique for real numbers) and onto (every real number has a real cube root), so it is bijective.

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