Functions: one - Mathematics

📖Topic Explanations

🌐 Overview

Hello students! Welcome to Functions: One! Get ready to unlock a fundamental concept that forms the backbone of advanced mathematics!

Imagine you're at a concert, and every ticket has a unique seat number, and every seat number corresponds to only one ticket holder. No two people share the same seat, and no one ticket gets two seats. This idea of a unique, exclusive pairing is at the very core of what we call a one-to-one function!

In the broader world of functions, we've learned that each input from the domain maps to exactly one output in the codomain. But what if we add another, more stringent condition? What if we insist that each distinct input must always produce a distinct output? This is precisely what a one-to-one function, also known as an injective function, defines. It's a special kind of function where no two different elements in the domain ever map to the same element in the codomain. Think of it as a perfect match – no sharing of partners!

Understanding one-to-one functions is not just an academic exercise; it's a critical stepping stone in your mathematical journey. This concept is absolutely vital for comprehending inverse functions, understanding differentiability, grasping integrability, and many other advanced topics in Calculus and Algebra. For your IIT JEE and board exams, questions involving the identification and properties of one-to-one functions are very common and can often be the key to solving more complex problems efficiently.

In this section, we will delve into the precise definition of one-to-one functions, learn how to rigorously test if a given function is one-to-one both algebraically and graphically (using the powerful horizontal line test), and explore their significant properties. You'll gain the tools to classify functions with confidence and precision.

So, prepare to sharpen your analytical skills and build a robust foundation in functions. Let's explore the fascinating world of one-to-one functions and discover how this seemingly simple condition opens up a whole new dimension in mathematical analysis!

📚 Fundamentals

Namaste, future engineers and mathematicians! 👋 Welcome back to our exciting journey into the world of Functions. So far, we've understood what a function is – essentially, a special kind of relationship where every input has exactly one output. Think of it like a well-organized vending machine: put in a specific code (input), and you get *one* specific item (output). You can't put in code A and get both a chocolate bar AND a bag of chips!

Today, we're going to dive into a special characteristic that some functions possess, making them even more... well, *unique*! We're talking about One-to-One Functions.

### What Makes a Function "One-to-One"? The Intuition!

Let's start with an analogy to build our intuition. Imagine a classroom where each student is assigned a unique roll number.
* Can one student have two roll numbers? No, that's not how roll numbers work. (This is the definition of a function itself: one input, one output).
* Can two *different* students have the *same* roll number? Again, no! Roll numbers are meant to uniquely identify each student. If two students had the same roll number, it would cause utter chaos!

This second point is exactly what a one-to-one function is all about!

A function is called one-to-one (or injective) if every element in the domain maps to a *unique* element in the codomain. In simpler terms, no two different inputs can have the same output.

Think of it like a fingerprint matching system. Each person has a unique set of fingerprints. No two people in the world have identical fingerprints. So, if you match a fingerprint, you know exactly which person it belongs to. That's a one-to-one relationship!

Contrast this with something that is *not* one-to-one. Consider a "children-to-parents" mapping. Can one child have two parents? Yes (biological parents). Can two *different* children have the *same* parent? Absolutely! Siblings share parents. So, this mapping wouldn't be one-to-one because two different inputs (children) can lead to the same output (parent).

### The Formal Definition of a One-to-One Function

Mathematically, we define a one-to-one function as follows:

A function f: A → B is said to be one-to-one or injective if, for every x₁, x₂ ∈ A (elements in the domain):

If f(x₁) = f(x₂), then it must imply that x₁ = x₂.

This is the most common way we test for one-to-one algebraically.

Alternatively, we can state it this way:

If x₁ ≠ x₂, then it must imply that f(x₁) ≠ f(x₂).

This means if you pick two different inputs, their outputs must also be different. They cannot 'collide' and produce the same output.

Let's break this down:
* We start by *assuming* that two different inputs (let's call them x₁ and x₂) have produced the *same* output (f(x₁) = f(x₂)).
* If, after some algebraic manipulation, this assumption forces us to conclude that x₁ and x₂ *must* actually be the same value, then the function is one-to-one.
* If we can find a scenario where f(x₁) = f(x₂) but x₁ ≠ x₂, then the function is *not* one-to-one.

### Visualizing One-to-One Functions

#### 1. Using Arrow Diagrams (for Discrete Sets)

Let's use sets A = {1, 2, 3} and B = {a, b, c, d}.

Example 1: A One-to-One Function

Arrow diagram showing 1->a, 2->b, 3->c

Here, each element in A maps to a *distinct* element in B.
* 1 → a
* 2 → b
* 3 → c
No two inputs share the same output. So, this is a one-to-one function.

Example 2: A Function That is NOT One-to-One

Arrow diagram showing 1->a, 2->b, 3->b

Here:
* 1 → a
* 2 → b
* 3 → b
Notice that both 2 and 3 (two different inputs) map to the same output 'b'. Since f(2) = f(3) = b, but 2 ≠ 3, this function is NOT one-to-one. It's often called a many-to-one function.

#### 2. The Horizontal Line Test (for Graphs of Functions)

This is an incredibly powerful and intuitive graphical test!

For a function y = f(x), if any horizontal line intersects the graph of the function at most once, then the function is one-to-one.
* If a horizontal line intersects the graph at *more than one point*, it means there are at least two different x-values (inputs) that produce the same y-value (output). In this case, the function is not one-to-one.

Let's look at some common graphs:

Example 3: y = x (Identity Function)

Imagine the graph of y = x, a straight line passing through the origin at a 45-degree angle. If you draw any horizontal line, it will intersect the graph at exactly one point.
Conclusion: y = x is a one-to-one function.

Example 4: y = x² (Parabola)

Consider the graph of y = x², a parabola opening upwards.
If you draw a horizontal line, say y = 4, it intersects the parabola at two points: x = 2 and x = -2.
Here, f(2) = 2² = 4 and f(-2) = (-2)² = 4.
Since f(2) = f(-2) but 2 ≠ -2, the function y = x² is NOT one-to-one.
Remember: The Horizontal Line Test is a quick visual check for one-to-one functions. If a horizontal line cuts the graph more than once, it fails!

### Step-by-Step Examples: Proving One-to-One Algebraically

Let's solidify this with some algebraic examples. This is where the formal definition really shines, especially for JEE aspirants.

Example 5: Is f(x) = 2x + 3 a one-to-one function?
Assume f: R → R (domain and codomain are all real numbers).

Step 1: Assume f(x₁) = f(x₂)
Let x₁ and x₂ be any two elements in the domain such that their outputs are equal.
So, 2x₁ + 3 = 2x₂ + 3

Step 2: Simplify the equation to check if x₁ = x₂
Subtract 3 from both sides:
2x₁ = 2x₂
Divide by 2:
x₁ = x₂

Step 3: Conclude
Since f(x₁) = f(x₂) implies x₁ = x₂, the function f(x) = 2x + 3 is indeed one-to-one.
This makes sense graphically too: y = 2x + 3 is a straight line, and any horizontal line will cross it only once.

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Example 6: Is f(x) = x² a one-to-one function?
Assume f: R → R.

Step 1: Assume f(x₁) = f(x₂)
x₁² = x₂²

Step 2: Simplify the equation
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0

This implies either:
* x₁ - x₂ = 0 ⇒ x₁ = x₂
* OR x₁ + x₂ = 0 ⇒ x₁ = -x₂

Step 3: Conclude
Here's the crucial part: We found that f(x₁) = f(x₂) does not necessarily mean x₁ = x₂. It could also mean x₁ = -x₂.
For instance, if x₁ = 2, then x₂ could be 2 (where x₁ = x₂) or x₂ could be -2 (where x₁ = -x₂).
Since f(2) = 4 and f(-2) = 4, but 2 ≠ -2, the function f(x) = x² is NOT one-to-one over the domain of all real numbers.

CBSE Focus: For board exams, understanding the definition, arrow diagrams, and basic algebraic proofs like these two examples is key. You'll often be asked to prove or disprove if a given simple function is one-to-one.

JEE Focus: For JEE, you'll encounter functions with more complex domains and codomains (e.g., f: [0, ∞) → R for f(x) = x²). In such cases, the domain plays a vital role. If the domain of f(x) = x² were restricted to non-negative real numbers, i.e., f: [0, ∞) → R, then f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = ±x₂. But since x₁, x₂ ∈ [0, ∞), we must have x₁ ≥ 0 and x₂ ≥ 0. Thus, x₁ = -x₂ is only possible if x₁ = x₂ = 0. Otherwise, it forces x₁ = x₂. So, on this restricted domain, f(x) = x² *is* one-to-one! This highlights the importance of the domain.

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Example 7: Is f(x) = 1/x a one-to-one function?
Assume f: R - {0} → R. (Domain excludes 0 as 1/0 is undefined).

Step 1: Assume f(x₁) = f(x₂)
1/x₁ = 1/x₂

Step 2: Simplify the equation
Cross-multiply:
x₂ = x₁

Step 3: Conclude
Since f(x₁) = f(x₂) implies x₁ = x₂, the function f(x) = 1/x is one-to-one.
Graphically, the hyperbola y = 1/x also passes the horizontal line test.

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### Why is "One-to-One" Important?

The concept of a one-to-one function is fundamental because it's a prerequisite for a function to have an inverse. If a function is not one-to-one (meaning it's many-to-one), then when you try to reverse the mapping, one output would correspond to multiple inputs, which would violate the definition of a function itself. We'll explore inverse functions in detail later, but for now, remember that one-to-one functions are special because they are "invertible" in a clean, unambiguous way!

### Summary: Your Checklist for One-to-One Functions

1. Definition: Each input must have a unique output. Different inputs *must* lead to different outputs.
2. Algebraic Test: Assume f(x₁) = f(x₂). If this *always* leads to x₁ = x₂, then it's one-to-one. If you find even one case where f(x₁) = f(x₂) but x₁ ≠ x₂, then it's *not* one-to-one.
3. Graphical Test: Use the Horizontal Line Test. If any horizontal line intersects the graph of the function at more than one point, it's not one-to-one.
4. Domain Matters: Always pay close attention to the specified domain of the function. Changing the domain can change whether a function is one-to-one or not!

You've taken a crucial step today in understanding a key property of functions. Keep practicing these concepts, and you'll build a strong foundation for more advanced topics! Next up, we'll explore 'Into' and 'Onto' functions, which deal with the codomain and range. Stay tuned!

🔬 Deep Dive

Welcome, aspiring engineers! Today, we're diving deep into one of the fundamental classifications of functions: the One-to-One Function. This concept is absolutely crucial for understanding inverse functions, injectivity, and a host of other topics in higher mathematics, making it a cornerstone for your JEE preparation.

Before we jump into "one-to-one," let's quickly refresh our memory on what a function is.

Understanding Functions: A Quick Recap

Recall that a function is a special kind of relation between two sets, say set A (the domain) and set B (the codomain), such that every element in the domain A is mapped to exactly one element in the codomain B. We often denote a function as (f: A o B).

Every element in A must be used.

Each element in A must map to only *one* element in B.

The set of all actual output values in B that elements from A map to is called the range of the function.

The One-to-One Function: Injective Mapping

Imagine a concert hall where each person gets a unique seat number. No two people can occupy the same seat. This is the essence of a one-to-one relationship!

1. Intuitive Definition

A function (f: A o B) is said to be one-to-one, or injective, if every distinct element in the domain A maps to a distinct element in the codomain B. In simpler terms, different inputs always give different outputs.

Think of it this way: if you pick any two different people from the domain, they must go to two different seats in the codomain. No sharing allowed!

2. Formal Definition

Mathematically, a function (f: A o B) is defined as one-to-one if:

For any two distinct elements (x_1, x_2 in A), if (x_1
eq x_2), then their images under (f) must also be distinct, i.e., (f(x_1)
eq f(x_2)).

Equivalently, and more commonly used for proofs: If the images of two elements are the same, i.e., (f(x_1) = f(x_2)), then the elements themselves must be the same, i.e., (x_1 = x_2).

This second form is usually easier to use for proving injectivity algebraically.

JEE Focus: Understanding this definition is paramount. Many JEE problems test your ability to apply this definition to various types of functions, especially those involving algebraic manipulation or restricted domains.

Methods to Check if a Function is One-to-One

There are primarily three powerful methods to determine if a function is one-to-one: the algebraic method, the graphical method (Horizontal Line Test), and the calculus method (monotonicity).

Method 1: The Algebraic Method (Using the Definition)

This is the most fundamental and universally applicable method. It directly uses the formal definition of a one-to-one function.

Assume (f(x_1) = f(x_2)) for any two elements (x_1, x_2) in the domain of (f).

Perform algebraic manipulation to show that this assumption necessarily implies (x_1 = x_2).

If you successfully deduce (x_1 = x_2), the function is one-to-one. If you find even one instance where (x_1
eq x_2) but (f(x_1) = f(x_2)), then the function is not one-to-one.

Example 1: Proving a Linear Function is One-to-One

Let's consider the function (f: mathbb{R} o mathbb{R}) defined by (f(x) = 2x + 3).

Step-by-step Solution:

Assume (f(x_1) = f(x_2)) for some (x_1, x_2 in mathbb{R}).

Substitute the function definition:
(2x_1 + 3 = 2x_2 + 3)

Subtract 3 from both sides:
(2x_1 = 2x_2)

Divide by 2:
(x_1 = x_2)

Since (f(x_1) = f(x_2)) implies (x_1 = x_2), the function (f(x) = 2x + 3) is indeed one-to-one.

Example 2: Proving a Quadratic Function is NOT One-to-One (on (mathbb{R}))

Consider (f: mathbb{R} o mathbb{R}) defined by (f(x) = x^2).

Step-by-step Solution:

Assume (f(x_1) = f(x_2)) for some (x_1, x_2 in mathbb{R}).

Substitute the function definition:
(x_1^2 = x_2^2)

This implies:
(x_1^2 - x_2^2 = 0)
((x_1 - x_2)(x_1 + x_2) = 0)

From this, we get two possibilities:
(x_1 - x_2 = 0 Rightarrow x_1 = x_2)
OR
(x_1 + x_2 = 0 Rightarrow x_1 = -x_2)

Since (f(x_1) = f(x_2)) does not *necessarily* imply (x_1 = x_2) (it could also be (x_1 = -x_2), where (x_1
eq x_2) if (x_1
eq 0)), the function (f(x) = x^2) is not one-to-one on its domain (mathbb{R}).

For instance, (f(2) = 4) and (f(-2) = 4), but (2
eq -2).

Example 3: Making a Quadratic Function One-to-One by Restricting the Domain

Let's modify Example 2. Consider (g: [0, infty) o mathbb{R}) defined by (g(x) = x^2).

Step-by-step Solution:

Assume (g(x_1) = g(x_2)) for some (x_1, x_2 in [0, infty)).

Substitute: (x_1^2 = x_2^2)

This again gives ((x_1 - x_2)(x_1 + x_2) = 0), so either (x_1 = x_2) or (x_1 = -x_2).

However, since (x_1, x_2 in [0, infty)), both (x_1) and (x_2) must be non-negative.
If (x_1 = -x_2) and both are non-negative, this is only possible if (x_1 = 0) and (x_2 = 0), which means (x_1 = x_2).
Otherwise, if (x_1 > 0), then (-x_2) would be negative (unless (x_2=0)), which contradicts (x_1 in [0, infty)).

Therefore, given the restricted domain, (g(x_1) = g(x_2)) must imply (x_1 = x_2). Thus, (g(x) = x^2) is one-to-one on the domain ( [0, infty) ).

Important Note: The domain of a function plays a critical role in determining if it is one-to-one. Always pay close attention to the specified domain!

Method 2: The Graphical Method (Horizontal Line Test)

This is a quick visual test, especially useful for functions whose graphs you can easily sketch.

A function (f(x)) is one-to-one if and only if no horizontal line intersects its graph at more than one point.

If a horizontal line intersects the graph at *two or more* points, it means there are different input values ((x_1
eq x_2)) that produce the same output value ((f(x_1) = f(x_2))), so the function is NOT one-to-one.

If every horizontal line intersects the graph at *at most one* point, then the function IS one-to-one.

Example 4: Using the Horizontal Line Test

Function Graph	Horizontal Line Test	One-to-One?
(y = x^3) (A strictly increasing curve)	Any horizontal line intersects the graph at only one point.	Yes
(y = x^2) (A parabola opening upwards)	A horizontal line (e.g., (y=4)) intersects the graph at two points (((2,4)) and ((-2,4))).	No
(y = sin x) (A periodic wave)	Many horizontal lines intersect the graph at infinitely many points (e.g., (y=0) intersects at (..., -2pi, -pi, 0, pi, 2pi, ...)).	No

Method 3: The Calculus Method (Monotonicity Test)

This method is particularly useful for differentiable functions and is frequently employed in JEE problems.

If a function (f(x)) is strictly monotonic (either strictly increasing or strictly decreasing) over its entire domain, then it is a one-to-one function.

Strictly Increasing: If (f'(x) > 0) for all (x) in the domain (except possibly at isolated points where (f'(x) = 0)).

Strictly Decreasing: If (f'(x) < 0) for all (x) in the domain (except possibly at isolated points where (f'(x) = 0)).

Reasoning:

If a function is strictly increasing, as you move along the x-axis to the right, the y-values are always increasing. This means no two different x-values can ever have the same y-value. The same logic applies to strictly decreasing functions.

Example 5: Using Monotonicity Test

Consider (f: mathbb{R} o mathbb{R}) defined by (f(x) = x^3 + 2x - 1).

Step-by-step Solution:

Find the first derivative of (f(x)):
(f'(x) = frac{d}{dx}(x^3 + 2x - 1) = 3x^2 + 2)

Analyze the sign of (f'(x)):
For any real number (x), (x^2 ge 0).
Therefore, (3x^2 ge 0).
Adding 2, we get (3x^2 + 2 ge 2).
So, (f'(x) = 3x^2 + 2 > 0) for all (x in mathbb{R}).

Conclusion: Since (f'(x) > 0) for all (x) in its domain, (f(x)) is strictly increasing throughout (mathbb{R}). Hence, (f(x)) is a one-to-one function.

Example 6: A function that is one-to-one but (f'(x)) is zero at a point

Consider (f: mathbb{R} o mathbb{R}) defined by (f(x) = x^3).

Step-by-step Solution:

Find the first derivative:
(f'(x) = 3x^2)

Analyze the sign of (f'(x)):
For (x
eq 0), (f'(x) = 3x^2 > 0).
At (x = 0), (f'(0) = 0).

Even though (f'(0) = 0), (f(x) = x^3) is still strictly increasing (it just momentarily flattens out at (x=0)). It does not change its increasing nature. Therefore, (f(x) = x^3) is one-to-one.

Important Distinction for JEE: A function is one-to-one if it's strictly monotonic. However, a function can be one-to-one even if its derivative is zero at isolated points, as long as the derivative does not *change sign* around those points. If (f'(x)) changes sign (e.g., from positive to negative, indicating a local max/min), then the function is definitely *not* one-to-one.

Advanced Concepts and JEE Applications

1. Composition of One-to-One Functions:

If (f: A o B) and (g: B o C) are both one-to-one functions, then their composition ((g circ f): A o C) is also one-to-one.

Proof Sketch:
Assume ((g circ f)(x_1) = (g circ f)(x_2)).
This means (g(f(x_1)) = g(f(x_2))).
Since (g) is one-to-one, (f(x_1) = f(x_2)).
Since (f) is one-to-one, (x_1 = x_2).
Thus, ((g circ f)) is one-to-one.

2. Piecewise Functions:

Determining if a piecewise function is one-to-one requires checking injectivity within each piece and also across the boundaries. Ensure that the ranges of different pieces do not overlap for different domains. For example, if (f(x) = {x ext{ for } x<0; x^2 ext{ for } x ge 0}), this function is not one-to-one as (f(-2) = -2) and (f(1/2) = 1/4), but if (f(x) = {x ext{ for } x<0; x+1 ext{ for } x ge 0}), it might be one-to-one if the transition is handled correctly and both pieces are monotonic.

3. Relation to Inverse Functions:

A function can only have an inverse function if it is both one-to-one (injective) and onto (surjective). The one-to-one property ensures that each output corresponds to a unique input, making the inverse mapping well-defined.

Summary Table of One-to-One Function Checks

Method	Description	When to Use	Pros	Cons
Algebraic Method	Assume (f(x_1)=f(x_2)) and prove (x_1=x_2).	Always reliable, especially for complex functions.	Rigorous proof, works for all functions.	Can be algebraically intensive.
Horizontal Line Test	Draw horizontal lines; they should intersect the graph at most once.	When the graph is easy to draw or visualize.	Quick visual check, intuitive.	Not a formal proof, limited by graphing ability.
Calculus Method (Monotonicity)	Check if (f'(x)) is consistently (>0) or (<0) across the domain.	For differentiable functions.	Fast and efficient for smooth functions.	Requires differentiation, might need careful analysis for (f'(x)=0).

Mastering the concept of one-to-one functions is not just about memorizing definitions; it's about developing the analytical skills to apply these methods effectively. Practice a variety of problems using all three techniques, and you'll be well-prepared for any challenge JEE throws your way!

🎯 Shortcuts

Understanding and quickly identifying one-to-one functions is crucial for both board exams and JEE Main. These mnemonics and short-cuts will help you recall the definition and apply tests efficiently.

What is a One-to-One Function?

A function $f: A o B$ is said to be one-to-one (or injective) if every distinct element in the domain A has a distinct image in the codomain B. In simpler terms, no two different inputs map to the same output.

Mnemonic for Definition:

DIDO: Distinct Inputs, Distinct Outputs. This captures the essence perfectly.

"No Shared Addresses": Imagine elements of the domain sending letters. In a one-to-one function, no two letters (inputs) go to the same house (output).

Short-Cuts & Tests for One-to-One Functions

1. Algebraic Test (The "Default" Method)

To prove a function $f(x)$ is one-to-one, assume $f(x_1) = f(x_2)$ for any $x_1, x_2$ in the domain, and then algebraically show that this implies $x_1 = x_2$. If $x_1$ can be unequal to $x_2$ while $f(x_1) = f(x_2)$, then it's not one-to-one.

Mnemonic: AEPEI (Assume Equal Outputs, Prove Equal Inputs).

JEE Tip: This method is fundamental and always applicable. Practice it thoroughly.

2. Graphical Test: The Horizontal Line Test (HLT)

A function is one-to-one if and only if every horizontal line intersects its graph at most once.

Mnemonic: HLT - "Hits Line Only Once" (or "Horizontal Line Touches Once").

CBSE & JEE Relevance: This is a powerful visual short-cut. If a horizontal line cuts the graph at more than one point, it means multiple x-values have the same y-value, thus not one-to-one.

Example: The graph of $f(x) = x^2$ fails the HLT because a horizontal line like $y=4$ intersects it at $x=2$ and $x=-2$. Hence, $f(x)=x^2$ is not one-to-one on $mathbb{R}$.

3. Calculus Short-Cut: Monotonicity Test

If a function is strictly monotonic (either strictly increasing or strictly decreasing) throughout its entire domain, then it is guaranteed to be one-to-one.

Mnemonic: "Monotonic $implies$ Injective" (M $implies$ I).

How to apply:
1. Find the first derivative, $f'(x)$.
2. If $f'(x) > 0$ for all $x$ in the domain, the function is strictly increasing, hence one-to-one.
3. If $f'(x) < 0$ for all $x$ in the domain, the function is strictly decreasing, hence one-to-one.
4. If $f'(x)$ changes sign or is zero over an interval, the function is not strictly monotonic and thus likely not one-to-one (e.g., $f(x)=x^3$ is one-to-one, even though $f'(0)=0$, because it doesn't change sign and is strictly increasing). Be careful with points where $f'(x)=0$ if it doesn't form an interval.

JEE Tip: This is a very common and efficient short-cut in JEE problems, especially for functions involving trigonometry, exponentials, or polynomials where differentiation is straightforward.

By mastering these mnemonics and short-cuts, you can quickly analyze and determine if a function is one-to-one, saving valuable time in exams!

💡 Quick Tips

Quick Tips: Identifying One-to-One (Injective) Functions

Understanding and identifying one-to-one functions is fundamental for advanced topics in functions. These quick tips will help you master the concept for both board exams and JEE.

1. Core Definition (Injective Function)

A function (f: A o B) is one-to-one (or injective) if every distinct element in the domain (A) maps to a distinct element in the codomain (B).

In simpler terms, no two different inputs give the same output.

Key Idea: If (x_1
eq x_2), then (f(x_1)
eq f(x_2)). Equivalently, if (f(x_1) = f(x_2)), then (x_1 = x_2).

2. Methods to Check Injectivity

a. Algebraic Method (Universal)

This is the most rigorous method and works for all functions:

Assume (f(x_1) = f(x_2)) for any (x_1, x_2) in the domain of (f).

Solve this equation. If the only solution is (x_1 = x_2), then the function is one-to-one.

If you find any instance where (x_1
eq x_2) but (f(x_1) = f(x_2)), the function is not one-to-one.

Example Sketch: For (f(x) = x^3), assume (x_1^3 = x_2^3 implies x_1 = x_2). So, it's one-to-one.
For (f(x) = x^2), assume (x_1^2 = x_2^2 implies x_1 = pm x_2). Since (x_1 = x_2) or (x_1 = -x_2) (and (x_1
eq x_2) is possible), it's not one-to-one over (mathbb{R}).

b. Graphical Method (Horizontal Line Test)

This is quick for functions whose graphs you can visualize:

Draw the graph of the function (y = f(x)).

Draw any horizontal line (y = k) (where (k) is a constant) that intersects the graph.

If every horizontal line intersects the graph at at most one point, then the function is one-to-one.

If any horizontal line intersects the graph at more than one point, the function is not one-to-one.

JEE Tip: This is great for quickly eliminating options or confirming simple functions.

c. Monotonicity Test (JEE Specific)

For differentiable functions, this is often the fastest method:

If a function is strictly monotonic (i.e., strictly increasing or strictly decreasing) throughout its domain, then it is always one-to-one.

To check for strict monotonicity, find the derivative (f'(x)):
- If (f'(x) > 0) for all (x) in the domain, the function is strictly increasing (and thus one-to-one).
- If (f'(x) < 0) for all (x) in the domain, the function is strictly decreasing (and thus one-to-one).

If (f'(x)) changes sign in the domain (e.g., from positive to negative or vice versa), or is zero over an interval, the function is not one-to-one.

Caution: A function can be one-to-one without being strictly monotonic if its domain is restricted or if it's not differentiable everywhere (e.g., (f(x) = |x|) for (x in [0, infty)) is one-to-one but not strictly monotonic over a point if the derivative is not defined there).

3. Important Considerations

Domain Restriction: A function that is not one-to-one over its natural domain can become one-to-one if its domain is restricted. For example, (f(x) = x^2) is not one-to-one on (mathbb{R}), but (f(x) = x^2) for (x in [0, infty)) *is* one-to-one. Always check the given domain!

Piecewise Functions: For piecewise functions, check injectivity for each piece *and* across the transition points. Ensure no two inputs from different pieces map to the same output.

CBSE vs. JEE: CBSE typically expects clear algebraic steps. JEE often demands quicker recognition using graphical or derivative methods, especially with restricted domains or composite functions.

Practice applying these methods diligently to solidify your understanding of one-to-one functions!

🧠 Intuitive Understanding

Intuitive Understanding: One-One Functions (Injective Functions)

In mathematics, particularly when studying functions, understanding the nature of how elements from one set (domain) map to elements in another set (codomain) is crucial. A one-one function, also known as an injective function, represents a specific type of mapping where each distinct input from the domain is associated with a distinct output in the codomain. Simply put, no two different inputs will ever give you the same output.

The Core Idea: Unique Pairing

Imagine you have a group of students and a set of unique roll numbers. If you assign a roll number to each student such that:

Every student gets a roll number.

No two different students share the same roll number.

This assignment represents a one-one function. Each student (input) is uniquely identified by their roll number (output). If two students were to have the same roll number, this mapping would not be one-one, as distinct inputs (different students) would lead to the same output (roll number).

Mathematical Interpretation

Formally, a function f: A → B is said to be one-one (or injective) if for every x₁, x₂ ∈ A,

If f(x₁) = f(x₂), then it must imply that x₁ = x₂.

This is the standard definition used in both CBSE board exams and JEE. It essentially means that if two outputs are the same, then their corresponding inputs must also have been the same. There's no possibility of different inputs yielding the identical output.

Graphical Understanding: The Horizontal Line Test

For functions represented graphically, determining if a function is one-one is straightforward using the Horizontal Line Test.

Rule: If any horizontal line drawn across the graph of the function intersects the graph at most one point, then the function is one-one.

Intuition: A horizontal line represents a constant y-value (output). If it intersects the graph at more than one point, it means that the same y-value (output) is produced by different x-values (inputs), violating the one-one condition.

Consider the graph of y = x³. Any horizontal line intersects it at only one point, indicating it's a one-one function. In contrast, for y = x², a horizontal line like y = 4 intersects the graph at two points (x = 2 and x = -2), meaning y = x² is NOT a one-one function.

Why is it important?

Understanding one-one functions is fundamental in higher mathematics and particularly for JEE. It's a prerequisite for understanding inverse functions, as only one-one functions (along with being onto) can have inverses. Questions in JEE often test your ability to determine if a given function is one-one, either algebraically or by analyzing its graph. Mastering this concept intuitively will significantly help in solving such problems efficiently.

🌍 Real World Applications

While the study of functions often feels abstract, the concept of a one-to-one function (also known as an injective function) is surprisingly pervasive and fundamental in numerous real-world systems and technologies. Understanding these applications can provide a deeper appreciation for the mathematical concept beyond typical problem-solving.

A function $f: A o B$ is said to be one-to-one if every element in the domain $A$ maps to a unique element in the codomain $B$. In simpler terms, no two distinct inputs produce the same output. This property is crucial when uniqueness and non-ambiguity are required.

Real-World Applications of One-to-One Functions:

Identification Systems:

Perhaps the most intuitive example. Consider national identification numbers (like Aadhar in India, Social Security Number in the USA, or a student's roll number in an institution). Each individual is assigned a unique identification number. This forms a one-to-one function where the domain is the set of individuals and the codomain is the set of identification numbers. The injectivity ensures that no two people share the same ID, preventing ambiguity and ensuring correct identification.
- JEE Relevance: This concept underpins database design, which is indirectly related to the practical application of discrete mathematics principles.

Database Primary Keys:

In database management systems, a primary key uniquely identifies each record (row) in a table. For instance, in a 'Students' table, a 'StudentID' column would serve as a primary key, ensuring that each student record has a distinct ID. This is a one-to-one mapping from the set of records to the set of primary key values.

Biometric Authentication:

Systems like fingerprint scanners, facial recognition, or iris scans rely on one-to-one mappings. Each unique biometric pattern (input) is associated with a single individual (output). If two different individuals produced the same biometric output, the system would fail its authentication purpose.

Encryption Algorithms:

Effective encryption requires that each unique plaintext message maps to a unique ciphertext. If a many-to-one mapping occurred (i.e., different plaintexts producing the same ciphertext), decryption would be ambiguous, making the encryption scheme insecure. While complex, the underlying principle often involves one-to-one transformations.

Telephone Numbers:

Each active telephone number in a geographical region is uniquely assigned to a single subscriber or line. This prevents calls from being misdirected and ensures proper communication. The mapping from active phone lines to phone numbers is one-to-one.

Library Call Numbers/ISBNs:

In libraries, each book has a unique call number (e.g., Dewey Decimal Classification or Library of Congress Classification) and a unique ISBN (International Standard Book Number). This allows for unambiguous identification and location of specific books. The mapping from books to their unique identifiers is one-to-one.

Understanding these real-world applications helps reinforce the importance of properties like injectivity in function theory. While you won't be asked to solve specific real-world problems in your JEE exam, recognizing these connections can deepen your conceptual understanding.

🔄 Common Analogies

Understanding abstract mathematical concepts like one-one functions can be significantly simplified by relating them to everyday scenarios. These common analogies help build an intuitive grasp, making it easier to recall and apply the definition in problem-solving.

What is a One-One (Injective) Function?

A function f: A → B is called one-one (or injective) if distinct elements of set A (domain) are mapped to distinct elements of set B (codomain). In simpler terms, no two different elements in the domain have the same image in the codomain. Mathematically, for any x₁, x₂ ∈ A, if f(x₁) = f(x₂), then it must imply x₁ = x₂. Conversely, if x₁ ≠ x₂, then f(x₁) ≠ f(x₂).

Common Analogies for One-One Functions:

Unique Fingerprints for Each Person:
- Domain (Set A): All people in the world.
- Codomain (Set B): The set of all possible unique fingerprint patterns.
- Function (f): Assigning each person their fingerprint.
- One-One Aspect: Each person has a unique set of fingerprints. No two different people in the world have identical fingerprints. Therefore, if you identify a fingerprint, you uniquely identify the person it belongs to. This perfectly illustrates a one-one mapping – distinct people map to distinct fingerprints.
- Not One-One Scenario: If, hypothetically, two different people had the exact same fingerprint, this mapping would not be one-one.

Examination Roll Numbers for Students:
- Domain (Set A): The set of students appearing for an examination.
- Codomain (Set B): The set of available roll numbers.
- Function (f): Assigning a roll number to each student.
- One-One Aspect: For a fair examination, each student is assigned a distinct and unique roll number. No two different students are given the same roll number. If you know a roll number, you can uniquely identify the student it belongs to. This makes the assignment of roll numbers a one-one function.
- Not One-One Scenario: If two different students were accidentally assigned the same roll number, the mapping would no longer be one-one.

Dedicated Parking Spots:
- Domain (Set A): Cars in a parking lot.
- Codomain (Set B): Specific, numbered parking spots.
- Function (f): Each car being assigned to a parking spot.
- One-One Aspect: If each car gets its own distinct parking spot, and no two cars are allowed to occupy the same spot, then this is a one-one assignment. Distinct cars are mapped to distinct spots.
- Not One-One Scenario: If, theoretically, two different cars could somehow occupy the exact same parking spot (which isn't how parking works, but helps visualize), or if a valet service allowed different cars to be parked in the same "logical" spot at different times (confusing the instantaneous mapping), it would not be one-one.

By using these analogies, you can quickly verify whether a given mathematical mapping behaves like one of these scenarios to determine if it's a one-one function. Keep these pictures in mind, especially for JEE problems where identifying function types is crucial!

📋 Prerequisites

To effectively grasp the concept of a one-to-one function (or injective function), it is crucial to have a solid understanding of the foundational concepts of set theory and the general definition of a function. These prerequisites form the bedrock upon which the properties of one-to-one functions are built.

Before diving into one-to-one functions, ensure you are comfortable with the following:

Sets and Elements:
- Understand what a set is (a well-defined collection of distinct objects).
- Be familiar with set notation (e.g., A = {1, 2, 3}), elements, and basic set operations like union, intersection, and Cartesian product (A x B).
- Relevance: Functions map elements from one set (domain) to another (codomain). A clear understanding of sets is fundamental to define these mappings.

Relations:
- Recall that a relation from set A to set B is a subset of the Cartesian product A x B.
- Understand how relations are represented (e.g., set of ordered pairs, arrow diagrams).
- Relevance: A function is a specific type of relation that satisfies certain conditions. Knowing what a general relation is helps appreciate the specific constraints that define a function.

Definition of a Function:
- A relation f from set A to set B is called a function if every element of set A has one and only one image in set B. This is the most critical prerequisite.
- Clearly distinguish between the Domain (the set of all possible input values), Codomain (the set of all possible output values), and Range (the set of all actual output values which are images of domain elements).
- Relevance: Understanding the basic definition of a function is non-negotiable. One-to-one is a *property* that functions may or may not possess, so you must first understand what a function is.

Representation of Functions:
- Be able to represent functions using ordered pairs, arrow diagrams, equations (e.g., y = f(x)), and graphs.
- Relevance: Visual and algebraic understanding helps in identifying and proving function types. For JEE, interpreting functions from various representations is key.

Graphical Analysis (Vertical Line Test):
- Understand how to use the Vertical Line Test to determine if a given graph represents a function. (A vertical line must intersect the graph at most once).
- Relevance: This visual test reinforces the "one and only one image" condition for a function.

Basic Algebraic Manipulation:
- Proficiency in solving basic algebraic equations and inequalities is essential. When proving a function is one-to-one, you often start by assuming f(x₁) = f(x₂) and need to algebraically show that x₁ = x₂.
- Relevance: This skill is vital for the analytical proof of a function being one-to-one, which is frequently tested in both CBSE and JEE.

For JEE Main, a robust understanding of these foundational concepts is not just about definitions but about applying them quickly and accurately to solve problems. Ensure you can differentiate between relations and functions and clearly identify the domain, codomain, and range of a given function before proceeding to its injectivity.

Mastering these prerequisites will make your journey through one-to-one functions significantly smoother and more understandable.

🔗 Related Topics

Understanding one-one functions (also known as injective functions) is foundational in the study of relations and functions. Several other key concepts are intimately linked with it, either as prerequisites, extensions, or direct applications. Mastering these related topics will provide a holistic understanding and improve problem-solving capabilities in JEE Main and Board exams.

Basic Definition of a Function, Domain, Codomain, and Range:

Before classifying functions as one-one, a clear understanding of what constitutes a function and its fundamental components (domain, codomain, and range) is essential. Without these basics, the concept of mapping unique elements from the domain to unique elements in the codomain cannot be grasped. These are absolute prerequisites for any function-related topic.

Many-One Functions:

This is the direct opposite of a one-one function. A function is many-one if two or more distinct elements in the domain map to the same element in the codomain. Understanding many-one functions provides a clear contrast, helping to solidify the definition and properties of one-one functions. Identifying when a function is NOT one-one often means it is many-one.

Onto (Surjective) and Into Functions:

One-one functions classify functions based on how elements from the domain map to the codomain. Onto (surjective) and into functions classify them based on whether the entire codomain is covered by the range. These classifications are often discussed together to provide a complete picture of function types. A function can be one-one and onto, one-one and into, many-one and onto, or many-one and into.

Bijective Functions:

A function that is both one-one (injective) and onto (surjective) is called a bijective function. This is a critical concept, as bijective functions are essential for defining inverse functions. Questions in JEE often require checking for bijectivity, which inherently involves checking for one-oneness.

Inverse Functions:

A function has an inverse if and only if it is bijective. This means being one-one is a necessary condition for a function to have an inverse. Understanding one-oneness is therefore fundamental to the study of inverse functions, their graphs, and properties. (JEE Focus: Inverse functions are very important for calculus and other topics.)

Horizontal Line Test:

This is a graphical method to determine if a function is one-one. If any horizontal line intersects the graph of a function at more than one point, then the function is many-one (not one-one). If every horizontal line intersects the graph at most once, the function is one-one. This visual test is extremely practical for both JEE and Board exams.

Monotonicity of Functions:

Strictly increasing or strictly decreasing functions are always one-one. If a function is strictly monotonic over its domain, then distinct inputs will always lead to distinct outputs. This connection is crucial, especially in calculus, where derivatives are used to determine monotonicity. (JEE Focus: Linking one-oneness with monotonicity is a common problem type.)

Composition of Functions:

The one-one property extends to composite functions. For example, if $f$ and $g$ are both one-one functions, then their composition $(f circ g)$ is also one-one. Understanding how this property behaves under composition is vital for more complex function problems.

By connecting 'one-one functions' with these related concepts, you build a stronger foundation for solving a wider variety of problems and for understanding advanced topics in mathematics.

⚠️ Common Exam Traps

Understanding when a function is one-to-one (injective) is fundamental, yet students frequently fall into common traps during exams. These traps often stem from overlooking critical details or misapplying tests. Mastering these pitfalls can significantly boost your score.

Common Exam Traps for One-to-One Functions

Neglecting the Domain and Co-domain:
- JEE & CBSE Trap: The most frequent mistake is assuming a function's injectivity without explicitly considering its defined domain. A function can be non-injective over one domain but injective over a restricted domain.
- Example: The function `f(x) = x^2` is NOT one-to-one if its domain is `R` (since `f(-2) = 4` and `f(2) = 4`). However, if the domain is restricted to `[0, ∞)`, then `f(x) = x^2` IS one-to-one. Always read the domain and co-domain carefully.

Incorrect Application of the Horizontal Line Test (HLT):
- JEE Trap: For functions that are difficult to analyze algebraically, the HLT (if a horizontal line intersects the graph at most once, it's one-to-one) is crucial. The trap is to apply it incompletely or misinterpret the graph.
- Pitfall:
 - Piecewise Functions: Students might analyze each piece individually but fail to check if outputs from different pieces overlap. For example, if `f(x) = x` for `x < 0` and `f(x) = x^2` for `x ≥ 0`, `f(-1) = -1` and `f(1) = 1`. Each piece is injective, but for `y > 0`, a horizontal line could intersect both `y=x` (for `x>0`) and `y=x^2` (for `x>0`) parts if not careful with ranges. A more obvious case would be `f(x) = x` for `x < 0` and `f(x) = x-5` for `x ≥ 0`. Here `f(-1)=-1` and `f(4)=-1`, so it's not one-to-one.
 - Graph Truncation: Not considering the full extent of the graph based on the domain.

Incomplete Algebraic Proof:
- JEE & CBSE Trap: To prove injectivity algebraically, you set `f(x1) = f(x2)` and must show that this implies `x1 = x2`. The trap is to stop prematurely or ignore alternative solutions.
- Example: For `f(x) = x^4`, if you set `f(x1) = f(x2)`, then `x1^4 = x2^4`. This implies `x1^2 = x2^2`, which further implies `x1 = x2` OR `x1 = -x2`. If the domain allows `x1 ≠ -x2` (e.g., `R`), then the function is NOT one-to-one. You must show that `x1 = x2` is the *only* possibility.

Functions Involving Absolute Values or Even Powers:
- JEE Trap: Functions like `f(x) = |x|`, `f(x) = x^2`, `f(x) = cos(x)` (over R), or any function involving even powers, are inherently not one-to-one over domains symmetric about the origin (e.g., `R`, `[-a, a]`) because `f(a) = f(-a)` for `a ≠ 0`.
- Tip: Always be suspicious of even powers or absolute values; check if the domain restricts values to only positive or negative numbers to make it injective.

Misinterpreting "One-to-one" vs. "Onto":
- CBSE Trap: Sometimes students confuse the definitions of one-to-one (injective) and onto (surjective). A function can be one without being the other, or neither, or both. Focus solely on the injectivity criteria for one-to-one.

Key Takeaway: Always rigorously apply the definition of a one-to-one function, whether through the algebraic method or the horizontal line test. Pay meticulous attention to the given domain and co-domain, as they are crucial for determining injectivity.

⭐ Key Takeaways

Key Takeaways: One-to-One Functions (Injective Functions)

Understanding one-to-one functions (also known as injective functions) is fundamental in mathematics, especially for IIT JEE and Board exams. Here are the crucial points you must remember:

Definition Essence: A function f: A → B is one-to-one if every distinct element in the domain (set A) maps to a unique element in the codomain (set B). Simply put, no two different domain elements can have the same image.

Formal Condition: For a function f: A → B, it is one-to-one if for all x₁, x₂ ∈ A, the condition f(x₁) = f(x₂) &implies; x₁ = x₂ holds true.

Algebraic Test (Most Common for JEE):
1. Assume f(x₁) = f(x₂) for any arbitrary x₁, x₂ belonging to the function's domain.
2. Solve this equation. If you are consistently able to prove that this equality implies x₁ = x₂, then the function is one-to-one.
3. If, however, you find a scenario where f(x₁) = f(x₂) but x₁ ≠ x₂ (i.e., a counterexample), then the function is NOT one-to-one.
  
  Example thought process: For f(x) = x², if f(x₁) = f(x₂), then x₁² = x₂². This implies x₁ = ±x₂. If the domain allows both positive and negative values (like R), then x₁ could be -x₂, meaning x₁ ≠ x₂. So, f(x) = x² is not one-to-one on R.

Graphical Test (Horizontal Line Test):
- Graph the function.
- If any horizontal line intersects the graph at more than one point, the function is NOT one-to-one.
- If every horizontal line intersects the graph at at most one point, the function is one-to-one. This is a quick visual check.

Monotonicity Connection (JEE Focus):
- A function that is strictly monotonic (either strictly increasing or strictly decreasing) throughout its entire domain is always one-to-one. This is a powerful property often utilized with calculus (derivatives) in JEE problems.
- Caution: The converse is not always true for all functions. A one-to-one function is not necessarily monotonic (e.g., a discontinuous function that is injective). However, continuous injective functions are always monotonic.

Implication for Inverse Functions: A function must be one-to-one (and onto) to possess a well-defined inverse function. Injectivity ensures that each element in the range corresponds to a unique element in the domain for the inverse.

CBSE vs. JEE Perspective:
- CBSE: Generally focuses on direct application of the algebraic test for standard functions, often with clear domain restrictions.
- JEE: Often presents more complex scenarios, including piecewise functions, functions on specific or restricted domains, and problems where monotonicity (often determined using derivatives) is key to establishing injectivity.

Common Pitfall: Do not confuse 'one-to-one' with 'onto'. These are distinct properties of functions. A function can be one-to-one without being onto, and vice-versa.

Mastering these key takeaways will significantly strengthen your foundation for functions and related topics. Keep practicing! All the best!

🧩 Problem Solving Approach

A function is deemed one-to-one (injective) if distinct elements in the domain map to distinct elements in the codomain. In simpler terms, no two different inputs produce the same output. Mastering the problem-solving approaches to identify one-to-one functions is crucial for JEE Main.

I. Analytical Method (Algebraic Approach)

This is the most rigorous and fundamental method, especially important for JEE where graphical methods might be ambiguous or calculus might be too complex for certain functions.

Assume Equality of Outputs: Start by assuming that for any two elements, x₁ and x₂, in the domain of f, their function values are equal: f(x₁) = f(x₂).

Manipulate Algebraically: Use algebraic operations to simplify the equation f(x₁) = f(x₂).

Derive Equality of Inputs: If, after simplification, you can unequivocally show that x₁ = x₂, then the function is one-to-one.

Counterexample (if not one-to-one): If you find that x₁ = x₂ is not the only possibility (i.e., x₁ ≠ x₂ can also lead to f(x₁) = f(x₂)), then the function is not one-to-one. In such cases, provide a specific counterexample.

JEE Tip: Always pay close attention to the specified domain. The one-to-one nature of a function can change with its domain (e.g., f(x) = x² is not one-to-one on R, but is on [0, ∞)).

II. Graphical Method (Horizontal Line Test)

This method is intuitive and quick for functions whose graphs are easy to sketch. Useful for both CBSE and JEE, but often a preliminary check for complex functions.

Plot the Graph: Sketch the graph of the function y = f(x) over its given domain.

Draw Horizontal Lines: Imagine or draw any horizontal line (a line parallel to the x-axis).

Check Intersections:
- If no horizontal line intersects the graph at more than one point, the function is one-to-one.
- If any horizontal line intersects the graph at two or more points, the function is not one-to-one.

Caution: For functions with restricted domains or complex behavior, relying solely on a rough sketch might lead to errors. Always confirm with the analytical method if unsure.

III. Calculus Method (Monotonicity Test)

This method is highly effective for differentiable functions and is a powerful tool in JEE Main.

Find the Derivative: Calculate the first derivative, f'(x).

Analyze the Sign of the Derivative:
- If f'(x) > 0 for all x in the domain, the function is strictly increasing and thus one-to-one.
- If f'(x) < 0 for all x in the domain, the function is strictly decreasing and thus one-to-one.
- If f'(x) changes sign (i.e., is positive for some intervals and negative for others), the function is not monotonic over the entire domain and therefore not one-to-one.
- If f'(x) = 0 at isolated points but does not change sign (e.g., f(x)=x³ at x=0), the function can still be one-to-one. However, if f'(x) = 0 over an entire interval, it's not one-to-one.

Key Idea: A strictly monotonic function (strictly increasing or strictly decreasing) will always be one-to-one. If a function is not strictly monotonic over its entire domain, it cannot be one-to-one.

Example: Determine if `f(x) = x³ - 3x² + 3x - 1` is one-to-one on R.

Let's use the Calculus Method:

Find f'(x):
f'(x) = d/dx (x³ - 3x² + 3x - 1) = 3x² - 6x + 3

Simplify f'(x):
f'(x) = 3(x² - 2x + 1) = 3(x - 1)²

Analyze the Sign:
Since (x - 1)² ≥ 0 for all real x, it implies 3(x - 1)² ≥ 0 for all x ∈ R.
Moreover, f'(x) = 0 only at x = 1 (an isolated point).
Since f'(x) ≥ 0 for all x ∈ R and is zero only at an isolated point, the function is strictly increasing (non-decreasing in this case, but still injective).

Thus, f(x) = x³ - 3x² + 3x - 1 is a one-to-one function on R. (In fact, f(x) = (x-1)³, which is clearly one-to-one).

📝 CBSE Focus Areas

CBSE Focus Areas: One-One Functions

For CBSE board examinations, understanding and proving whether a function is one-one (or injective) is a fundamental concept. The emphasis is typically on algebraic proofs and clear conceptual understanding, rather than complex functions or intricate graphical analysis often seen in JEE. Mastery of this topic ensures you can tackle related concepts like inverse functions effectively.

Key Concepts & Definitions

Definition of a One-One Function: A function $f: A o B$ is said to be one-one if different elements of set $A$ have different images in set $B$. In other words, if $x_1, x_2 in A$ such that $x_1
eq x_2$, then $f(x_1)
eq f(x_2)$.

Equivalently, and more commonly used for proofs: If $f(x_1) = f(x_2)$ for some $x_1, x_2 in A$, then it must imply that $x_1 = x_2$.

Injective Function: "One-one function" is also known as an "injective function." This terminology is interchangeably used in CBSE.

Methods to Prove a Function is One-One (CBSE Perspective)

Algebraic Method (Most Crucial for CBSE):
- Assume $f(x_1) = f(x_2)$ for any $x_1, x_2$ in the domain of $f$.
- Algebraically manipulate the equation $f(x_1) = f(x_2)$ to show that this assumption necessarily leads to $x_1 = x_2$.
- If you arrive at $x_1 = x_2$, the function is one-one. If you find $x_1 = pm x_2$ (and the domain allows both positive and negative values), or any other possibility where $x_1
  eq x_2$ for $f(x_1)=f(x_2)$, then the function is not one-one.

Graphical Method (Horizontal Line Test):
- Draw the graph of the function $y = f(x)$.
- Draw any horizontal line $y=k$.
- If every horizontal line intersects the graph at most at one point, then the function is one-one. If even one horizontal line intersects the graph at two or more points, the function is not one-one (it is many-one).
- CBSE Note: While understanding the horizontal line test is important, board exam questions primarily expect the algebraic proof for formal answers.

Example for CBSE Board Exam

Question: Show that the function $f: mathbb{R} o mathbb{R}$ defined by $f(x) = 2x + 3$ is one-one.

Solution:

Let $x_1, x_2 in mathbb{R}$ such that $f(x_1) = f(x_2)$.

Then, $2x_1 + 3 = 2x_2 + 3$.

Subtracting 3 from both sides, we get $2x_1 = 2x_2$.

Dividing by 2 (since $2
eq 0$), we get $x_1 = x_2$.

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function $f$ is one-one.

Common Functions & Their One-One Nature (CBSE Context)

Function Type	Example	Is One-One?	Reason
Linear	$f(x) = ax+b$ ($a eq 0$)	Yes	Graph is a straight line, passes HLT. Algebraically $ax_1+b = ax_2+b Rightarrow x_1=x_2$.
Cubic (some)	$f(x) = x^3$	Yes	For any $x_1 eq x_2$, $x_1^3 eq x_2^3$. Graphically, it passes HLT.
Quadratic	$f(x) = x^2$ (on $mathbb{R}$)	No	$f(2) = 4$ and $f(-2) = 4$. Different inputs give same output. ($x_1^2 = x_2^2 Rightarrow x_1 = pm x_2$)
Trigonometric	$f(x) = sin x$ (on $mathbb{R}$)	No	$f(0) = 0$ and $f(pi) = 0$. Periodic functions are generally not one-one over their natural domain.

CBSE Tip: Always mention the domain and codomain when discussing functions, as a function's one-one nature can change depending on its domain (e.g., $f(x)=x^2$ is one-one on $mathbb{R}^+ o mathbb{R}$). Clearly state your assumptions about $x_1, x_2$ belonging to the domain.

🎓 JEE Focus Areas

Understanding and identifying one-to-one functions (also known as Injective Functions) is a fundamental skill frequently tested in JEE Main. This concept forms the backbone for many advanced topics, especially inverse functions.

What is a One-to-One Function?

A function (f: A o B) is said to be one-to-one or injective if different elements of domain (A) have different images in codomain (B). In simpler terms, no two distinct elements in the domain map to the same element in the codomain.

Algebraic Definition: For any (x_1, x_2 in A), if (f(x_1) = f(x_2)), then it must imply (x_1 = x_2). Conversely, if (x_1
eq x_2), then (f(x_1)
eq f(x_2)).

Key Methods to Check Injectivity for JEE Main

JEE problems often require you to determine if a given function is one-to-one. Here are the primary methods:

1. Algebraic Method:
- Assume (f(x_1) = f(x_2)) for (x_1, x_2) in the domain.
- Solve this equation. If the only solution is (x_1 = x_2), then the function is one-to-one. If other solutions exist (e.g., (x_1 = -x_2) for (x_1
  e x_2)), it is not one-to-one.
- JEE Tip: Be careful with domain restrictions. For example, (f(x) = x^2) is not one-to-one on (mathbb{R}) (since (f(2) = f(-2) = 4)), but it is one-to-one on ([0, infty)).

2. Graphical Method (Horizontal Line Test):
- Draw the graph of the function (y = f(x)).
- If any horizontal line intersects the graph at more than one point, then the function is NOT one-to-one.
- If every horizontal line intersects the graph at most at one point, then the function is one-to-one.

3. Calculus Method (Monotonicity):
- A function is one-to-one if it is strictly monotonic (i.e., strictly increasing or strictly decreasing) throughout its domain.
- Find the first derivative, (f'(x)).
- If (f'(x) > 0) for all (x) in the domain (strictly increasing) OR (f'(x) < 0) for all (x) in the domain (strictly decreasing), then (f(x)) is one-to-one.
- Caution: A function can be one-to-one even if (f'(x) = 0) at isolated points (e.g., (f(x) = x^3), where (f'(0)=0), but it's strictly increasing). However, if (f'(x)) changes sign in the domain, it's not one-to-one.

JEE Main Specific Focus Areas

Inverse Functions: A function must be bijective (both one-to-one and onto) to have an inverse. Checking injectivity is the first step.

Impact of Domain/Codomain: Always pay close attention to the specified domain. Changing the domain can change whether a function is one-to-one (e.g., (f(x) = x^2) on (mathbb{R}) vs. ([0, infty))).

Common Functions:
- Always One-to-One: Linear functions ((ax+b, a
  e 0)), Exponential functions ((a^x, a > 0, a
  e 1)), Logarithmic functions ((log_a x, a > 0, a
  e 1)).
- One-to-One on Restricted Domains: Quadratic functions, Trigonometric functions.
- Never One-to-One: Periodic functions (e.g., (sin x, cos x, an x) on their natural domains).

Piecewise Functions: For piecewise functions, check injectivity for each piece and then ensure that no two values from different pieces map to the same image.

Finding Parameters: Problems often ask to find values of a parameter for which a function becomes one-to-one (e.g., restricting the domain or modifying the function definition).

CBSE vs. JEE Main Perspective

Aspect	CBSE Board Exams	JEE Main
Complexity	Basic definitions, checking injectivity for simple functions (e.g., linear, quadratic on restricted domains).	More complex functions (piecewise, composite, functions with parameters), understanding the impact of domain.
Problem Type	Direct questions: "Show that (f(x)) is one-to-one."	Indirect questions: "Find values of (k) for which (f(x)) is injective," or involving inverse functions.

Mastering the concept of one-to-one functions is crucial. Practice a variety of problems, especially those involving domain restrictions and piecewise definitions, to ace this topic in JEE Main.

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🌐 Overview

One-One (Injective) Functions

- A function f: A→B is one-one if different inputs always give different outputs.
- Formal: f(x₁) = f(x₂) ⇒ x₁ = x₂ (equivalently, x₁≠x₂ ⇒ f(x₁)≠f(x₂)).
- Examples: f(x)=2x is injective on R; f(x)=x² is not injective on R (many-one).
- Tests: algebraic equality, Horizontal Line Test (HLT), and monotonicity (strictly ↑/↓).

📚 Fundamentals

Fundamentals

- Injective: ∀x₁≠x₂ in A, f(x₁)≠f(x₂).
- HLT: any horizontal line intersects y=f(x) at most once.
- Strict monotonicity on a domain ⇒ injectivity on that domain.
- Counterexample disproves injectivity (e.g., f(2)=f(-2) for x²).

🔬 Deep Dive

Deep dive

- Inverses exist on the range for injective functions.
- Bijections enable full inverses on codomain; link to cardinality arguments.
- Injectivity in linear maps ↔ trivial kernel.

🎯 Shortcuts

Mnemonics

- Inject → unique: injecting distinct inputs yields distinct outputs.
- H in one hints Horizontal line test for one-one.

💡 Quick Tips

Quick tips

- Suspect non-injective if even powers or |x| present on R.
- Restrict domain to fix injectivity when needed.
- Always state method (algebraic/HLT/derivative) in proofs.

🧠 Intuitive Understanding

Intuition

- Think unique ID assignment: no two different people share the same ID.
- In arrow diagrams, no two distinct domain points point to the same codomain point.
- HLT: no horizontal line crosses the graph more than once.

🌍 Real World Applications

Applications

- Unique identifiers (roll numbers, serial numbers).
- Cryptography: collision resistance (idealized).
- Invertibility of transformations (linear algebra, coding).

🔄 Common Analogies

Analogies

- Students and roll numbers (unique).
- Locks and unique keys (simplified).
- Fingerprints to persons mapping (one-to-one ideally).

📋 Prerequisites

Prerequisites

- Sets, functions, domain/codomain/range.
- Basic algebra and graph reading.
- Understanding even/odd symmetry pitfalls (e.g., x²).

🔗 Related Topics

Related & next

- Onto (surjective) and bijective functions.
- Function inverses (require bijection for full inverse).
- Monotonicity and derivatives (for continuous functions).

⚠️ Common Exam Traps

Common exam traps

- Ignoring domain restrictions and concluding wrongly.
- Claiming without proof; skipping the key injective step.
- Misusing HLT by checking only a few lines.

⭐ Key Takeaways

Key takeaways

- Injectivity ensures uniqueness of outputs and enables inverse on range.
- Linear f(x)=ax+b (a≠0) are injective on R.
- Domain choice is crucial (piecewise/limited intervals).

🧩 Problem Solving Approach

Problem-solving approach

1) Parse domain/codomain carefully.
2) Attempt algebraic test (often shortest rigorous route).
3) If complex: use derivative sign or HLT.
4) To disprove: provide explicit distinct inputs with same output.

📝 CBSE Focus Areas

CBSE focus

- Proving injectivity for given f(x).
- HLT and domain specificity.
- Simple counterexamples for non-injective cases.

🎓 JEE Focus Areas

JEE focus

- Monotonic intervals via derivatives.
- Piecewise definitions and continuity points.
- Invertibility and restricted domains.

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📝 Reviewer Notes:
Topic label interpreted as One-One (Injective) functions.

🌐 Overview

Functions are fundamental mathematical objects that map elements from one set (the domain) to another set (the codomain). One-to-one (injective), into, and onto (surjective) functions are key classifications that determine uniqueness and coverage properties. These concepts are critical for set theory, abstract algebra, and higher mathematics in both CBSE Class 12 and IIT-JEE. Understanding these distinctions forms the foundation for inverse functions, bijective functions, and permutations, which appear extensively in competitive examinations. The concepts bridge abstract set theory to concrete problem-solving, revealing structural properties of mathematical objects and their relationships.

📚 Fundamentals

Function Definition: A function ( f: A o B ) assigns to each element ( a in A ) (domain) exactly one element ( f(a) in B ) (codomain). The set of all outputs is the range: ( ext{Range}(f) = {f(a) : a in A} subseteq B ).

One-to-One (Injective) Function: A function ( f: A o B ) is injective if different inputs map to different outputs:
• Definition: ( f(a_1) = f(a_2) implies a_1 = a_2 ) for all ( a_1, a_2 in A )
• Contrapositive: ( a_1
eq a_2 implies f(a_1)
eq f(a_2) )
• Graphical test: every horizontal line intersects the graph at most once (horizontal line test)

Into Function: A function ( f: A o B ) is into if the range is a proper subset of the codomain:
• Definition: ( ext{Range}(f) subset B ) and ( ext{Range}(f)
eq B )
• Equivalently: at least one element in B is not the image of any element in A
• Into is the negation of onto (surjective)

Onto (Surjective) Function: A function ( f: A o B ) is onto if every element in B is mapped from at least one element in A:
• Definition: for every ( b in B ), there exists ( a in A ) such that ( f(a) = b )
• Equivalently: ( ext{Range}(f) = B )
• Graphical test: every horizontal line intersects the graph at least once

Key Relations:
• Bijective = Injective + Surjective (both one-to-one and onto)
• If ( f: A o B ) is bijective, then ( |A| = |B| ) (for finite sets)
• Bijective functions have inverses; into/onto functions may not

🔬 Deep Dive

The concepts of injective and surjective functions reveal deep structural properties in mathematics. Injective functions preserve distinctness: if two elements are different in the domain, their images are different in the codomain. This property is fundamental in algebra—injective homomorphisms (like embedding a subgroup into a larger group) preserve structure without collapsing distinctions.

Surjective functions achieve full coverage: every element in the codomain is "reached" from the domain. In linear algebra, a surjective linear transformation maps the vector space onto the target space, determining its rank: for ( T: V o W ), if T is surjective, rank(T) = dim(W).

Bijective functions establish isomorphisms—structure-preserving one-to-one correspondences. Two sets connected by a bijection have the same cardinality. In abstract algebra, bijective homomorphisms are isomorphisms, showing that two algebraic structures are essentially identical despite different representations.

Counting injective functions: from a finite set A (|A| = m) to set B (|B| = n), the number of injective functions is P(n, m) = n!/(n-m)! (only if m ≤ n). Each choice of an image for the first element gives n options; the second gives n-1 options (to maintain injectivity), and so on.

Counting surjective functions: from A (m elements) to B (n elements), using inclusion-exclusion:

[ ext{surjections} = sum_{k=0}^{n} (-1)^k inom{n}{k} (n-k)^m ]

For small sets:
• From {1, 2, 3} to {a, b}: surjections exist if 3 ≥ 2 (necessary condition |A| ≥ |B|). Count = 2³ - 2·1³ = 8 - 2 = 6.
• From {1, 2} to {a, b, c}: surjections don't exist because |A| < |B|; at most 2 elements in the codomain can be covered.

The relationship ( | ext{injections from A to B}| imes | ext{left inverses}| ) and ( | ext{surjections from A to B}| imes | ext{right inverses}| ) connect to the choice axiom in set theory. Bijections have both left and right inverses, and they coincide: ( f^{-1} circ f = ext{id}_A ) and ( f circ f^{-1} = ext{id}_B ).

🎯 Shortcuts

INJECTIVE-1to1: INJECtive is 1-to-1 (one input per output). ONTO-RANGE: Onto = Range = Codomain (full coverage). INTO-NOT: INTo = NOT onto (opposite). BIJ=BOTH: Bijective = Both injective AND onto. HORIZONTAL-TEST: Horizontal line test checks for injectivity.

💡 Quick Tips

To verify injectivity algebraically: assume ( f(a_1) = f(a_2) ), solve to show ( a_1 = a_2 ). To verify surjectivity: take an arbitrary element b in the codomain, set ( f(x) = b ), and solve for x; if a solution exists for every b, it's surjective. Use graphs: horizontal line test for injectivity (at most 1 crossing), then check if every horizontal line crosses (surjectivity). Don't confuse "into" with "injective"—into means not onto (coverage issue), not about uniqueness. For finite sets, use counting: if injective, ( |A| leq |B| ); if surjective, ( |A| geq |B| ).

🧠 Intuitive Understanding

Think of a function as a "matching system" between two groups. Imagine students (domain A) assigned to lockers (codomain B). (1) One-to-one: each student gets a unique locker, no two students share. This ensures you can identify who used which locker. (2) Into: some lockers remain empty; not all lockers are occupied. (3) Onto (or surjective): every locker has at least one student's belongings; no empty lockers. A bijection (both one-to-one and onto) means perfect pairing: each student has exactly one locker, each locker has exactly one student.

For real functions like ( f(x) = 2x ) on ℝ: it's one-to-one (different x values give different 2x values) and onto (every real number y can be written as 2x for some x, namely x = y/2). This is bijective.

For ( g(x) = x^2 ) on ℝ: it's not one-to-one (both 2 and -2 map to 4) and not onto (negative numbers like -1 are never outputs). It is into ℝ.

Visualize on graphs: the horizontal line test: if every horizontal line touches the curve at most once, the function is injective. If every horizontal line touches at least once, it's surjective.

🌍 Real World Applications

Database design: primary keys must be injective (one-to-one mapping of records to keys); foreign keys often use surjective mappings to link related tables. Hash functions should be injective to avoid collisions (though true injections on large domains are impossible; designs aim for injectivity over practical ranges).

Cryptography: encryption functions should be bijective, allowing decryption. If the encryption is not injective, different plaintexts might map to the same ciphertext, making decryption ambiguous. RSA encryption uses bijective functions from message space to ciphertext space.

Network addressing: IPv4 assigns unique IP addresses (injective) to devices; the space of all possible addresses (4.3 billion) exceeds the current number of connected devices, demonstrating an injective but not surjective mapping. IPv6 resolves this with a vastly larger address space.

Biology and genetics: DNA encoding uses bijective mappings: codons (triplets of nucleotides) map to amino acids. The code is nearly bijective (with minor redundancy, making some mappings 2-to-1; hence not quite injective).

Social networks: follower relations are often neither injective nor surjective; influencers may have millions of followers (many-to-one), and some users follow no one (surjective but not injective).

Economics: price functions might be neither injective (multiple goods at the same price) nor surjective (some prices never occur due to minimum markups). Inverse demand functions often aim to be injective to invert quantity-price relationships uniquely.

🔄 Common Analogies

Injective — Unique Seat Assignments: Imagine assigning seats to students. An injective assignment means no two students share a seat (each student gets a unique seat). You can always identify which student occupied which seat.

Surjective (Onto) — Full Auditorium: Every seat in the auditorium is occupied by at least one person. No empty seats remain. The audience (domain) "covers" all seats (codomain).

Into — Theater with Empty Seats: Some seats remain unoccupied; the audience doesn't fill the entire theater. The seating arrangement is into the set of seats.

Bijective — Perfect Dance Partner Pairing: At a formal dance, each male dancer has exactly one female partner, and each female has exactly one male partner. Perfect one-to-one correspondence. If one person arrives late or leaves early, the bijection breaks.

Horizontal Line Test — Horizontal Scan: Imagine scanning the graph horizontally (left to right) at various heights. If your scan line always hits the curve at most once, the function is injective (one output per input). If every scan line hits at least once, it's surjective (all outputs are covered).

📋 Prerequisites

Set theory fundamentals: understanding sets, subsets, and set notation. Concept of relations and how functions are special relations. Understanding domain, codomain, and range. Basic algebraic manipulations and solving equations. Familiarity with coordinate geometry and graphical representations of functions. Basic knowledge of different types of functions (linear, quadratic, etc.).

🔗 Related Topics

Composition of functions: determining injectivity/surjectivity of f ∘ g. Inverse functions: bijective functions have inverses. Relations and their properties: functions are special relations. Set cardinality and countability: bijections between sets imply equal cardinality. Permutations: bijections from a finite set to itself. Homomorphisms in algebra: injective/surjective homomorphisms (embeddings and quotients). Bijections in combinatorics: counting principle with one-to-one correspondence. Function spaces: sequences of functions with injective/surjective properties.

⚠️ Common Exam Traps

Trap 1: Confusing "one-to-one" (injective) with "into" (not onto). They address different properties: one-to-one is about uniqueness, into is about coverage. A function can be one-to-one and onto, one-to-one and into, etc.

Trap 2: Assuming bijection without verification. A function is bijective only if BOTH injective AND surjective. Checking one property is insufficient.

Trap 3: Misapplying the horizontal line test. Injective: each horizontal line intersects at most once (≤ 1). Surjective: each horizontal line intersects at least once (≥ 1). Getting these reversed is a common error.

Trap 4: Forgetting to specify the codomain. A function's onto property depends on the stated codomain. f: ℝ → ℝ defined by f(x) = x² is not onto, but f: ℝ → [0, ∞) is onto (same rule, different codomain).

Trap 5: Calculus error: differentiating to prove injectivity. Not all injective functions are strictly monotonic (they need not have a continuous derivative). Use the definition directly.

Trap 6: Cardinality misinterpretation. For finite sets, if |A| < |B|, no surjective function from A to B exists. If |A| > |B|, no injective function exists. These are necessary conditions.

Trap 7: Incorrectly counting functions. Injections from A (m elements) to B (n elements): P(n, m) = n(n-1)⋯(n-m+1), not n^m. The latter counts all functions.

⭐ Key Takeaways

1. Injective (one-to-one): no two domain elements map to the same codomain element. Test: ( f(a_1) = f(a_2) implies a_1 = a_2 ).
2. Surjective (onto): every codomain element is the image of at least one domain element. Test: for every ( b in B ), solve ( f(x) = b ) and verify a solution exists.
3. Into: at least one codomain element is not in the range. Equivalent to: not surjective.
4. Bijective: both injective and surjective; establishes one-to-one correspondence and allows inverse function definition.
5. Horizontal line test (graphs): injective if ≤ 1 intersection per line; surjective if ≥ 1 intersection per line.
6. For finite sets: injective requires |A| ≤ |B|; surjective requires |A| ≥ |B|; bijective requires |A| = |B|.
7. Only bijective functions have inverses.
8. Terminology clarification: "into" specifically means "not onto"; it refers to range properties, not one-to-one-ness.

🧩 Problem Solving Approach

Step 1: Identify the function f: A → B, noting domain A, codomain B, and the rule for f(x).

Step 2: To verify injectivity:
• Assume ( f(a_1) = f(a_2) ) for arbitrary ( a_1, a_2 in A ).
• Simplify the equation algebraically.
• Show that this implies ( a_1 = a_2 ).
• If successful for all cases, f is injective.
• Alternatively, show that ( a_1
eq a_2 ) implies ( f(a_1)
eq f(a_2) ) (contrapositive).

Step 3: To verify surjectivity (onto):
• Take an arbitrary element ( b in B ).
• Solve the equation ( f(x) = b ) for x.
• Demonstrate that a solution ( x in A ) exists for every b.
• If successful for all b in the codomain, f is surjective.
• Alternatively, show that the range of f equals B.

Step 4: To verify "into" property:
• Show that there exists at least one element in B that is not in the range of f.
• Equivalently, prove that f is not surjective.

Step 5: Graphical verification (for real functions):
• Plot the function or sketch its graph.
• Apply the horizontal line test: draw horizontal lines at various heights.
• Count intersections: if at most 1 intersection for each line, f is injective.
• If every horizontal line intersects at least once, f is surjective.
• If some horizontal line doesn't intersect, f is into.

Step 6: For finite set functions:
• Compute or count using combinatorial formulas if needed.
• Verify cardinality constraints: |A| ≤ |B| for injections to be possible; |A| ≥ |B| for surjections.

Step 7: State conclusions clearly:
• "f is injective" or "f is one-to-one"
• "f is surjective" or "f is onto"
• "f is bijective" (if both)
• "f is into" (if not onto)
• Justify based on the verification above.

📝 CBSE Focus Areas

CBSE Class 12 emphasizes: (1) Definitions of injective (one-to-one), surjective (onto), and bijective functions with clear examples. (2) Identification of these properties for given functions (simple algebraic, linear, quadratic). (3) Graphical identification using the horizontal line test. (4) Verification using algebraic definitions (assuming f(a₁) = f(a₂) and proving injectivity, or solving f(x) = b for surjectivity). (5) Understanding "into" as the negation of onto. (6) Connection to inverse functions and existence conditions (bijective ⟹ inverse exists). (7) Simple numerical examples and real-valued function analysis. CBSE avoids heavy counting and abstract set theory.

🎓 JEE Focus Areas

IIT-JEE emphasizes: (1) Rigorous proofs of injectivity/surjectivity from first principles. (2) Counting injective functions: P(n, m) = n!/(n-m)! for m ≤ n. (3) Counting surjective functions using inclusion-exclusion principle. (4) Complex function compositions: proving injectivity and surjectivity of f ∘ g from properties of f and g. (5) Bijections and permutations: counting bijections on a finite set (n!). (6) Applications in group theory: injective homomorphisms as embeddings, surjective homomorphisms as quotient maps. (7) Inverse function existence and uniqueness for bijections. (8) Problems combining function classification with calculus (derivatives to test injectivity).

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📝CBSE 12th Board Problems (12)

Problem 255

Easy 2 Marks

Show that the function f: R → R defined by f(x) = 3x + 2 is one-one.

Show Solution

1. Let x₁, x₂ ∈ R such that f(x₁) = f(x₂). 2. Substitute the function definition: 3x₁ + 2 = 3x₂ + 2. 3. Subtract 2 from both sides: 3x₁ = 3x₂. 4. Divide by 3: x₁ = x₂. 5. Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one.

Final Answer: The function f(x) = 3x + 2 is one-one.

Problem 255

Easy 3 Marks

Examine whether the function f: N → N given by f(x) = x² is one-one.

Show Solution

1. Let x₁, x₂ ∈ N such that f(x₁) = f(x₂). 2. Substitute the function definition: x₁² = x₂². 3. Since x₁, x₂ ∈ N (natural numbers), x₁ and x₂ must be positive. Therefore, taking the positive square root, x₁ = x₂. 4. Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one.

Final Answer: The function f(x) = x² is one-one when the domain is N (Natural Numbers).

Problem 255

Easy 2 Marks

Verify whether the function f: R → R defined by f(x) = |x| is one-one.

Show Solution

1. A function is one-one if distinct elements have distinct images. 2. Consider two distinct elements in the domain R, say x₁ = 1 and x₂ = -1. 3. Calculate their images under f: f(1) = |1| = 1 and f(-1) = |-1| = 1. 4. Here, x₁ ≠ x₂ (1 ≠ -1) but f(x₁) = f(x₂) (1 = 1). 5. Since distinct elements do not have distinct images, the function f is not one-one.

Final Answer: The function f(x) = |x| is not one-one.

Problem 255

Easy 3 Marks

Is the function f: R → R given by f(x) = x³ one-one? Justify your answer.

Show Solution

1. Let x₁, x₂ ∈ R such that f(x₁) = f(x₂). 2. Substitute the function definition: x₁³ = x₂³. 3. Taking the cube root of both sides: x₁ = x₂. 4. Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one.

Final Answer: Yes, the function f(x) = x³ is one-one.

Problem 255

Easy 2 Marks

Check whether the function f: R → R defined as f(x) = 1/x is one-one.

Show Solution

1. Let x₁, x₂ ∈ R-{0} such that f(x₁) = f(x₂). 2. Substitute the function definition: 1/x₁ = 1/x₂. 3. Cross-multiply: x₂ = x₁. 4. Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one.

Final Answer: The function f(x) = 1/x (for x ≠ 0) is one-one.

Problem 255

Easy 2 Marks

Determine whether the function f: R → R defined by f(x) = sin(x) is one-one.

Show Solution

1. A function is one-one if distinct elements have distinct images. 2. Consider two distinct elements in the domain R, say x₁ = 0 and x₂ = π. 3. Calculate their images under f: f(0) = sin(0) = 0 and f(π) = sin(π) = 0. 4. Here, x₁ ≠ x₂ (0 ≠ π) but f(x₁) = f(x₂) (0 = 0). 5. Since distinct elements do not have distinct images, the function f is not one-one.

Final Answer: The function f(x) = sin(x) is not one-one.

Problem 255

Hard 4 Marks

Determine if the function f: R → R, defined by f(x) = x + cos x, is one-one. Justify your answer.

Show Solution

1. To check if f(x) is one-one, we analyze its derivative f'(x). 2. Calculate f'(x): f'(x) = d/dx (x + cos x) = 1 - sin x. 3. Analyze the range of f'(x). Since -1 ≤ sin x ≤ 1 for all x ∈ R, it implies that 0 ≤ 1 - sin x ≤ 2. 4. Therefore, f'(x) ≥ 0 for all x ∈ R. 5. The derivative f'(x) = 0 only when sin x = 1, which occurs at isolated points, i.e., x = (2n + 1/2)π for integer n. 6. A function is one-one if it is strictly monotonic. Since f'(x) ≥ 0 for all x and f'(x) = 0 only at isolated points, the function f(x) is strictly increasing throughout its domain. 7. Hence, the function f(x) is one-one.

Final Answer: Yes, the function f(x) = x + cos x is one-one.

Problem 255

Hard 6 Marks

Show that the function f: R → R defined by f(x) = x|x| is one-one.

Show Solution

1. Define f(x) piecewise: f(x) = x^2 if x ≥ 0, and f(x) = -x^2 if x < 0. 2. Case 1: x_1, x_2 ≥ 0. Assume f(x_1) = f(x_2). 3. Then x_1^2 = x_2^2. Since x_1, x_2 ≥ 0, this implies x_1 = x_2. 4. Case 2: x_1, x_2 < 0. Assume f(x_1) = f(x_2). 5. Then -x_1^2 = -x_2^2, which implies x_1^2 = x_2^2. Since x_1, x_2 < 0, this implies x_1 = x_2. 6. Case 3: x_1 ≥ 0 and x_2 < 0. Assume f(x_1) = f(x_2). 7. Then x_1^2 = -x_2^2, which can be rewritten as x_1^2 + x_2^2 = 0. 8. Since x_1 ≥ 0 and x_2 < 0, x_1^2 ≥ 0 and x_2^2 > 0. The sum x_1^2 + x_2^2 can only be 0 if both x_1 = 0 and x_2 = 0. However, x_2 is strictly less than 0. Therefore, x_1^2 + x_2^2 cannot be 0 under the given conditions. 9. This means f(x_1) ≠ f(x_2) if x_1 ≥ 0 and x_2 < 0, unless x_1=0. If x_1=0, then f(0)=0. If x_2<0, f(x_2)=-x_2^2. For f(x_2)=0, x_2 must be 0, which contradicts x_2<0. Thus, f(x_1) = f(x_2) implies x_1 = x_2. 10. Combining all cases, f(x) is one-one.

Final Answer: The function f(x) = x|x| is one-one.

Problem 255

Hard 4 Marks

Consider the function f: R → R defined as f(x) = x3 - 3x2 + 3x + 5. Is f a one-one function? Justify your answer using calculus.

Show Solution

1. To check if f(x) is one-one, we analyze its derivative f'(x). 2. Calculate f'(x): f'(x) = d/dx (x3 - 3x2 + 3x + 5) = 3x2 - 6x + 3. 3. Factorize f'(x): f'(x) = 3(x2 - 2x + 1) = 3(x - 1)2. 4. Analyze the sign of f'(x). Since (x - 1)2 ≥ 0 for all x ∈ R, it implies that f'(x) = 3(x - 1)2 ≥ 0 for all x ∈ R. 5. The derivative f'(x) = 0 only when (x - 1)2 = 0, which means x = 1. This is an isolated point. 6. Since f'(x) ≥ 0 for all x and f'(x) = 0 only at an isolated point (x=1), the function f(x) is strictly increasing throughout its domain. 7. Therefore, the function f(x) is one-one.

Final Answer: Yes, the function f(x) = x3 - 3x2 + 3x + 5 is one-one.

Problem 255

Hard 6 Marks

Examine if the function f: R → R defined by f(x) = x / (1 + |x|) is one-one.

Show Solution

1. Express f(x) as a piecewise function: - If x ≥ 0, |x| = x, so f(x) = x / (1 + x). - If x < 0, |x| = -x, so f(x) = x / (1 - x). 2. Case 1: x_1, x_2 ≥ 0. Assume f(x_1) = f(x_2). x_1 / (1 + x_1) = x_2 / (1 + x_2) x_1(1 + x_2) = x_2(1 + x_1) x_1 + x_1x_2 = x_2 + x_1x_2 x_1 = x_2. (One-one for x ≥ 0). 3. Case 2: x_1, x_2 < 0. Assume f(x_1) = f(x_2). x_1 / (1 - x_1) = x_2 / (1 - x_2) x_1(1 - x_2) = x_2(1 - x_1) x_1 - x_1x_2 = x_2 - x_1x_2 x_1 = x_2. (One-one for x < 0). 4. Case 3: x_1 ≥ 0 and x_2 < 0. Assume f(x_1) = f(x_2). The range for f(x) when x ≥ 0 is [0, 1). To see this, as x increases from 0, f(x) = 1 - 1/(1+x) increases from 0 towards 1. So f(x_1) ≥ 0. The range for f(x) when x < 0 is (-1, 0). To see this, let x = -y where y > 0. Then f(-y) = -y / (1 - (-y)) = -y / (1 + y) = -(1 - 1/(1+y)). As y increases from 0, f(x) decreases from 0 towards -1. So f(x_2) < 0. Since f(x_1) ≥ 0 and f(x_2) < 0 (if x_1 ≠ 0 or x_2 ≠ 0), it is impossible for f(x_1) = f(x_2) unless both are 0. If f(x_1) = 0, then x_1 = 0. If f(x_2) = 0, then x_2 = 0. This contradicts x_2 < 0. Thus, f(x_1) ≠ f(x_2) if x_1 ≥ 0 and x_2 < 0 (unless x_1=x_2=0, but that is not this case). 5. Combining all cases, f(x) is one-one.

Final Answer: Yes, the function f(x) = x / (1 + |x|) is one-one.

Problem 255

Hard 6 Marks

Let f: [-1, 1] → R be defined by f(x) = x / (x+2). Is f one-one? Justify your answer. Also, find the range of f.

Show Solution

1. To check if f(x) is one-one: Assume f(x_1) = f(x_2) for x_1, x_2 ∈ [-1, 1]. 2. x_1 / (x_1+2) = x_2 / (x_2+2) 3. x_1(x_2+2) = x_2(x_1+2) 4. x_1x_2 + 2x_1 = x_1x_2 + 2x_2 5. 2x_1 = 2x_2 ⇒ x_1 = x_2. Thus, f is one-one. 6. To find the range, analyze the monotonicity of f(x) on the domain [-1, 1]. 7. Calculate the derivative f'(x): f'(x) = [(x+2)(1) - x(1)] / (x+2)2 = 2 / (x+2)2. 8. For x ∈ [-1, 1], x+2 is always positive, so (x+2)2 > 0. Thus, f'(x) = 2 / (x+2)2 > 0 for all x ∈ [-1, 1]. 9. Since f'(x) > 0, f(x) is strictly increasing on [-1, 1]. 10. The minimum value of f(x) occurs at x = -1: f(-1) = -1 / (-1+2) = -1 / 1 = -1. 11. The maximum value of f(x) occurs at x = 1: f(1) = 1 / (1+2) = 1 / 3. 12. The range of f is [f(-1), f(1)] = [-1, 1/3].

Final Answer: Yes, f is one-one. The range of f is [-1, 1/3].

Problem 255

Hard 4 Marks

Let f: [0, π/2] → R be a function defined by f(x) = x sin x. Is f an injective (one-one) function? Justify your answer.

Show Solution

1. To check if f(x) is one-one, we analyze its derivative f'(x). 2. Calculate f'(x) using the product rule: f'(x) = d/dx (x sin x) = (1)sin x + x(cos x) = sin x + x cos x. 3. Analyze the sign of f'(x) for x ∈ [0, π/2]. 4. For x ∈ [0, π/2]: - sin x ≥ 0 - cos x ≥ 0 - x ≥ 0 5. Therefore, sin x + x cos x ≥ 0 for all x ∈ [0, π/2]. 6. f'(x) = 0 only when sin x = 0 AND x cos x = 0 simultaneously. This happens only at x = 0. 7. Since f'(x) ≥ 0 for all x ∈ [0, π/2] and f'(x) = 0 only at an isolated point x=0, the function f(x) is strictly increasing on its domain. 8. Hence, the function f(x) is injective (one-one).

Final Answer: Yes, the function f(x) = x sin x is injective (one-one).

🎯IIT-JEE Main Problems (6)

Problem 255

Easy 4 Marks

Determine if the function f: R → R defined by f(x) = 3x - 5 is one-to-one.

Show Solution

1. Assume f(x₁) = f(x₂). 2. This means 3x₁ - 5 = 3x₂ - 5. 3. Add 5 to both sides: 3x₁ = 3x₂. 4. Divide by 3: x₁ = x₂. 5. Since f(x₁) = f(x₂) implies x₁ = x₂, the function is one-to-one.

Final Answer: Yes, the function is one-to-one.

Problem 255

Easy 4 Marks

Check if the function f: [2, ∞) → R defined by f(x) = x^2 - 4x + 7 is one-to-one.

Show Solution

1. Calculate the derivative of f(x): f'(x) = 2x - 4. 2. Analyze the sign of f'(x) over the given domain [2, ∞). 3. For x ∈ [2, ∞), 2x ∈ [4, ∞), so 2x - 4 ∈ [0, ∞). 4. Since f'(x) ≥ 0 for all x in the domain and f'(x) > 0 for x > 2, the function is strictly increasing. 5. A strictly increasing function is always one-to-one.

Final Answer: Yes, the function is one-to-one.

Problem 255

Easy 4 Marks

Is the function f: R → R defined by f(x) = x^3 + 2x one-to-one?

Show Solution

1. Calculate the derivative of f(x): f'(x) = 3x^2 + 2. 2. Analyze the sign of f'(x) over the domain R. 3. Since x^2 ≥ 0 for all real x, 3x^2 ≥ 0. 4. Therefore, 3x^2 + 2 ≥ 2, which means f'(x) > 0 for all x ∈ R. 5. Since f'(x) is always strictly positive, the function is strictly increasing. 6. A strictly increasing function is always one-to-one.

Final Answer: Yes, the function is one-to-one.

Problem 255

Easy 4 Marks

Determine whether the function f: R → (0, ∞) defined by f(x) = e^(2x) is one-to-one.

Show Solution

1. Assume f(x₁) = f(x₂). 2. This means e^(2x₁) = e^(2x₂). 3. Taking natural logarithm on both sides: ln(e^(2x₁)) = ln(e^(2x₂)). 4. This simplifies to 2x₁ = 2x₂. 5. Divide by 2: x₁ = x₂. 6. Since f(x₁) = f(x₂) implies x₁ = x₂, the function is one-to-one.

Final Answer: Yes, the function is one-to-one.

Problem 255

Easy 4 Marks

Is the function f: R → R defined by f(x) = { x+1, if x < 0; x-1, if x ≥ 0 } one-to-one?

Show Solution

1. To check if it's one-to-one, we look for two distinct inputs that give the same output. 2. Consider x₁ = -1 (which falls under x < 0). 3. f(x₁) = f(-1) = -1 + 1 = 0. 4. Consider x₂ = 1 (which falls under x ≥ 0). 5. f(x₂) = f(1) = 1 - 1 = 0. 6. Since f(-1) = f(1) = 0, but -1 ≠ 1, the function is not one-to-one.

Final Answer: No, the function is not one-to-one.

Problem 255

Easy 4 Marks

Determine if the function f: [1, ∞) → R defined by f(x) = |x-1| is one-to-one.

Show Solution

1. Analyze the function definition over the given domain. 2. For the domain x ∈ [1, ∞), we have x - 1 ≥ 0. 3. Therefore, for x ∈ [1, ∞), |x-1| simplifies to x-1. 4. So, f(x) = x-1 for x ∈ [1, ∞). 5. This is a linear function with a positive slope (slope = 1). 6. A linear function with a non-zero slope is strictly monotonic (increasing in this case). 7. A strictly monotonic function is always one-to-one.

Final Answer: Yes, the function is one-to-one.

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📐Important Formulas (3)

Definition of One-to-One (Injective) Function

( f(x_1) = f(x_2) implies x_1 = x_2 )

Text: A function f: A → B is one-to-one (injective) if, for any two elements x₁ and x₂ in the domain A, whenever their images under f are equal, the elements themselves must be equal.

This is the fundamental definition of an injective function. It means that distinct elements in the domain must map to distinct elements in the codomain. No two different inputs can produce the same output.

Variables: To rigorously prove or understand the concept of a one-to-one function. This definition forms the basis for all practical tests.

Algebraic Test for Injectivity

( ext{To prove } f ext{ is injective, assume } f(x_1) = f(x_2) ext{ and derive } x_1 = x_2 )

Text: To verify if a function f(x) is one-to-one algebraically: 1. Assume f(x₁) = f(x₂) for any x₁, x₂ in the domain. 2. Algebraically manipulate this equation. 3. If this consistently leads to x₁ = x₂, then the function is one-to-one. If it leads to other possibilities (e.g., x₁ = -x₂), then it is not.

This is the most common method for proving injectivity in both board exams and competitive exams. JEE Tip: Always state your assumption clearly and show each step of the derivation. If you find even one pair of distinct x₁ and x₂ such that f(x₁) = f(x₂), the function is not injective.

Variables: Applicable to all types of functions, especially those defined by algebraic expressions (polynomials, rational, trigonometric, exponential, logarithmic functions).

Calculus Test for Injectivity (Monotonicity)

( ext{If } f'(x) > 0 ext{ or } f'(x) < 0 ext{ throughout its domain, then } f(x) ext{ is injective.} )

Text: For a differentiable function f(x), if its first derivative f'(x) is strictly positive (f is strictly increasing) or strictly negative (f is strictly decreasing) for all x in its domain, then f(x) is one-to-one.

This method relies on the property that a strictly monotonic function (either always increasing or always decreasing) will never repeat its output values. Caution: This test is for differentiable functions. If f'(x) changes sign or is zero over an interval, the function is not necessarily one-to-one. For example, f(x)=x³ has f'(x)=3x² ≥ 0, which is zero at x=0, but it is still injective. The condition should strictly be that f'(x) is non-zero everywhere or zero at isolated points for it to be strictly monotonic.

Variables: Highly effective for differentiable functions, especially in JEE Main/Advanced problems to quickly determine injectivity.

📚References & Further Reading (10)

Book

Integral Calculus for JEE Main & Advanced

By: G. Tewani

N/A (available in print/e-book)

A comprehensive book specifically designed for JEE Main & Advanced preparation. It covers functions in detail, including one-one functions, with numerous solved examples and practice problems tailored for competitive exams. The treatment is rigorous and problem-oriented.

Note: Highly relevant for JEE-level understanding and problem-solving techniques related to one-one functions.

Book

By:

Website

Injective Function (One-to-One Function)

By: Brilliant.org Community

https://brilliant.org/wiki/injective-function/

Brilliant.org offers interactive lessons and problem-solving challenges. Their wiki page on injective functions provides a concise definition, properties, examples, and a strong emphasis on applying the concept in problem-solving scenarios, suitable for JEE aspirants.

Note: Good for concise definitions, properties, and practical application examples for competitive exams.

Website

By:

PDF

Lecture Notes on Functions

By: Prof. B.V. Limaye (IIT Bombay, typically)

N/A (often available on academic course pages, e.g., math.iitb.ac.in/courses/MA105/notes.pdf for similar content)

University-level lecture notes on functions from a reputable institution. These notes provide a formal and rigorous treatment of functions, including the definition and properties of injective (one-one) functions, often with proofs and a higher level of mathematical precision suitable for advanced learners.

Note: Provides a rigorous mathematical foundation, useful for JEE Advanced aspirants seeking deeper understanding.

PDF

By:

Article

Understanding Injective, Surjective, and Bijective Functions

By: Mathema.tips

https://mathema.tips/posts/injective-surjective-bijective-functions/

This online article directly addresses the definitions and characteristics of injective (one-to-one), surjective (onto), and bijective functions with clear examples and graphical illustrations, making it a very focused and useful resource for the topic.

Note: Directly relevant to the topic, offering clear explanations and examples of one-one functions.

Article

By:

Research_Paper

Injective functions of a variable over a finite field (or related topic in discrete math)

By: Various authors in Number Theory/Discrete Mathematics

N/A (often found on academic databases like arXiv, MathSciNet, or direct journal publisher sites)

Research papers in abstract algebra, number theory, or discrete mathematics often utilize the concept of injective functions as a fundamental tool or property. While the specific context may be advanced (finite fields, group theory), it showcases the application and importance of injectivity in higher mathematics.

Note: Very advanced, illustrating the application of injective functions in higher mathematics. Not directly for JEE syllabus but shows the foundational importance of the concept. Useful for students with a keen interest beyond the syllabus.

Research_Paper

By:

⚠️Common Mistakes to Avoid (60)

Minor Other

❌ Confusing Monotonicity with One-to-One Property

Students often incorrectly assume that a function must be strictly monotonic (either strictly increasing or decreasing) to be one-to-one (injective). This is a common overgeneralization.

💭 Why This Happens:

Most standard examples of one-to-one functions encountered in textbooks (like polynomials, exponentials) are continuous and monotonic. This leads to an implicit belief that monotonicity is a necessary condition for injectivity.

✅ Correct Approach:

A function f is one-to-one if for any x₁, x₂ in its domain, f(x₁) = f(x₂) implies x₁ = x₂. Graphically, this means every horizontal line intersects the function's graph at most once (Horizontal Line Test). While strict monotonicity on an interval guarantees injectivity (for continuous functions), injectivity does not universally imply strict monotonicity for all functions, especially piecewise or discontinuous ones.

📝 Examples:

❌ Wrong:

Incorrectly concluding that the function
f(x) = x (for x < 0)
f(x) = x + 1 (for x ≥ 0)
is NOT one-to-one simply because it's discontinuous at x=0 and thus not strictly monotonic across its entire domain.

✅ Correct:

The function from the wrong example is indeed one-to-one. If f(x₁) = f(x₂), it implies x₁ = x₂. For instance, if x₁ < 0 and x₂ ≥ 0, then f(x₁) < 0 and f(x₂) ≥ 1, so f(x₁) can never equal f(x₂). This function clearly passes the Horizontal Line Test, demonstrating injectivity without strict monotonicity.

💡 Prevention Tips:

Always apply the fundamental definition: f(x₁) = f(x₂) &implies; x₁ = x₂.
Use the Horizontal Line Test for graphical verification.
JEE Advanced Specific: Do not equate injectivity with strict monotonicity for piecewise-defined functions or those with discontinuities.

JEE_Advanced

Minor Conceptual

❌ Confusing One-to-One with Strictly Monotonic or Misapplying Tests

Students often incorrectly assume a function is one-to-one (injective) simply because it's strictly monotonic over a partial domain, or they misapply the horizontal line test/derivative test without considering the entire domain or the precise definition. This leads to errors in determining injectivity.

💭 Why This Happens:

This mistake primarily stems from a conceptual misunderstanding of the definition of a one-to-one function: distinct elements in the domain must map to distinct elements in the codomain. Students might:

Haphazardly apply the derivative test (f'(x) > 0 or f'(x) < 0) without ensuring it holds for the entire domain.
Not consider the full domain when using the horizontal line test, especially for functions defined piecewise or on restricted domains.
Confuse injectivity with surjectivity (onto function).

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one (injective):

Definition-Based: Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A. If this implies x₁ = x₂, then the function is one-to-one.
Graphical (Horizontal Line Test): Draw the graph of the function. If any horizontal line intersects the graph at at most one point, the function is one-to-one.
Calculus (Monotonicity for JEE): If a function is strictly increasing [f'(x) > 0] or strictly decreasing [f'(x) < 0] throughout its entire domain (where differentiable), then it is one-to-one. Be cautious with non-differentiable points.

📝 Examples:

❌ Wrong:

Consider f(x) = x², where f: R → R. A student might incorrectly conclude it is one-to-one because f'(x) = 2x, which is positive for x > 0. This overlooks the negative domain.

✅ Correct:

For f(x) = x² with f: R → R:

Applying Definition: Let f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = ±x₂. Since x₁ can be equal to -x₂ (e.g., f(-2) = 4 and f(2) = 4, but -2 ≠ 2), it is not one-to-one.
Horizontal Line Test: A horizontal line like y = 4 intersects the graph at x = -2 and x = 2. Thus, it fails the horizontal line test.
Calculus: f'(x) = 2x. This derivative changes sign (negative for x < 0, positive for x > 0), indicating the function is not strictly monotonic over R. Hence, not one-to-one.

💡 Prevention Tips:

Master the Definition: Always start by understanding the definition of one-to-one thoroughly.
Check the Entire Domain: Whether using graphical, analytical, or calculus methods, ensure you consider the function's behavior across its complete specified domain.
Use Counterexamples: If you suspect a function is NOT one-to-one, try to find two distinct inputs that map to the same output (e.g., for f(x) = x², f(-1) = f(1)).
Practice Diverse Problems: Work on problems involving polynomials, trigonometric functions, piecewise functions, and functions with restricted domains to build a robust understanding.

JEE_Main

Minor Calculation

❌ Algebraic Errors in Proving One-to-One Nature

Students often make algebraic mistakes or fail to be rigorous when proving a function is one-to-one using the definition $f(x_1) = f(x_2) implies x_1 = x_2$. Common errors include incorrectly handling square roots, squaring both sides without considering sign changes, or overlooking division by expressions that could be zero, leading to an incorrect conclusion about injectivity.

💭 Why This Happens:

This mistake stems from a lack of complete understanding of algebraic rules and the precise definition of a one-to-one function. Haste, overlooking domain restrictions, and an assumption that simple manipulation is sufficient contribute to these errors. Students often forget that the implication $f(x_1) = f(x_2) implies x_1 = x_2$ must hold true for ALL $x_1, x_2$ in the domain.

✅ Correct Approach:

To correctly prove a function is one-to-one, begin by assuming $f(x_1) = f(x_2)$ for any $x_1, x_2$ in the function's domain. Then, through careful and step-by-step algebraic manipulation, demonstrate that this assumption necessarily leads to $x_1 = x_2$. Pay close attention to operations like squaring (which can introduce extraneous solutions) or dividing (ensure the divisor is non-zero). Always keep the given domain in mind, as it critically affects the validity of certain steps.

📝 Examples:

❌ Wrong:

Consider $f(x) = x^2$ for $x in mathbb{R}$.
A student might incorrectly reason:
$f(x_1) = f(x_2) implies x_1^2 = x_2^2 implies x_1 = x_2$.
This is wrong because $x_1 = -x_2$ is also a possibility (e.g., $f(2) = 4$ and $f(-2) = 4$, but $2
eq -2$). This demonstrates a failure to consider all possible solutions when taking a square root.

✅ Correct:

Consider $f(x) = x^2$ for $x in [0, infty)$.
Assume $f(x_1) = f(x_2)$ for $x_1, x_2 in [0, infty)$.
$x_1^2 = x_2^2$
$x_1^2 - x_2^2 = 0$
$(x_1 - x_2)(x_1 + x_2) = 0$
This implies either $x_1 - x_2 = 0$ (i.e., $x_1 = x_2$) OR $x_1 + x_2 = 0$.
Since $x_1, x_2 in [0, infty)$, both $x_1 ge 0$ and $x_2 ge 0$.
For $x_1 + x_2 = 0$ to be true, it must be that $x_1 = 0$ and $x_2 = 0$. In this case, $x_1 = x_2$ still holds.
For any other positive $x_1, x_2$, $x_1+x_2 > 0$, so the only possibility is $x_1 - x_2 = 0 implies x_1 = x_2$.
Thus, $f(x) = x^2$ is one-to-one on $[0, infty)$.

💡 Prevention Tips:

Always start with $f(x_1) = f(x_2)$ and rigorously work towards $x_1 = x_2$.
Be extremely careful with algebraic manipulations. Understand when operations like squaring or taking roots introduce or eliminate solutions.
Pay meticulous attention to the domain and codomain of the function. These are crucial for determining the validity of algebraic steps.
Practice with a variety of function types to build confidence in algebraic rigor.

JEE_Main

Minor Formula

❌ Ignoring non-trivial solutions when applying the <code>f(x1) = f(x2)</code> test for injectivity.

Students frequently use the algebraic test for a function to be one-to-one (injective), which states: If f(x₁) = f(x₂) implies x₁ = x₂ for all x₁, x₂ in the domain, then the function is one-to-one. The common mistake is failing to consider all possible solutions to the equation f(x₁) = f(x₂), especially when functions involve even powers or absolute values. This leads to an incorrect conclusion that the function is injective.

💭 Why This Happens:

This mistake stems from a lack of thoroughness in solving algebraic equations or forgetting fundamental properties like a² = b² ⇒ a = ±b or |a| = |b| ⇒ a = ±b. Students often rush to conclude x₁ = x₂ without exploring other possibilities, particularly when the function's domain allows for such distinct values.

✅ Correct Approach:

To correctly apply the injectivity test, always solve the equation f(x₁) = f(x₂) completely. If after full simplification and considering all roots/cases, the only conclusion is x₁ = x₂ for all relevant x₁, x₂ in the domain, then the function is one-to-one. If you find even one instance where f(x₁) = f(x₂) but x₁ ≠ x₂, then the function is not one-to-one. Always be mindful of the function's domain and codomain.

📝 Examples:

❌ Wrong:

Consider f(x) = x² for f: R → R.
Student's thought process:
Let f(x₁) = f(x₂).
x₁² = x₂².
Therefore, x₁ = x₂ (incorrectly assuming only positive roots or ignoring x₁ = -x₂).
Conclusion: f(x) = x² is one-to-one.

✅ Correct:

Consider f(x) = x² for f: R → R.
Correct approach:
Let f(x₁) = f(x₂).
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This equation yields two possibilities:
1. x₁ - x₂ = 0 ⇒ x₁ = x₂
2. x₁ + x₂ = 0 ⇒ x₁ = -x₂
Since x₁ = -x₂ provides distinct values (e.g., x₁ = 2, x₂ = -2 where x₁ ≠ x₂) for which f(x₁) = f(x₂) = 4, the function is not one-to-one on R. (JEE Main Tip: Graphical method via horizontal line test is also quick for such functions).

💡 Prevention Tips:

Complete Algebraic Solution: Always ensure all possible solutions to f(x₁) = f(x₂) are considered. Do not take shortcuts with square roots, absolute values, or even powers.
Counter-Example Search: If you suspect a function is not one-to-one, try to find two distinct inputs (x₁ ≠ x₂) that produce the same output (f(x₁) = f(x₂)). This immediately proves it's not injective.
Domain and Codomain Check: Always refer to the specified domain. A function might be one-to-one on a restricted domain (e.g., f(x) = x² is injective on [0, ∞)) but not on its natural domain.

JEE_Main

Minor Unit Conversion

❌ Misapplication of Physical Unit Conversion Logic to Mathematical Function Properties

Students sometimes incorrectly attempt to apply the intuitive understanding of physical unit conversion (e.g., converting meters to centimeters or area units) to the properties of mathematical functions, particularly when determining injectivity (one-to-one property). This leads to confusion regarding how input values relate to output values, even though mathematical functions in Pure Mathematics typically operate on abstract numbers without physical units requiring such conversion.

💭 Why This Happens:

This confusion often stems from an overgeneralization of concepts learned in physics or chemistry to a purely mathematical context. Students might erroneously assume that if a function maps values from one set to another (domain to codomain), there's an inherent 'conversion' occurring that needs to be understood in a unit-like sense, rather than a direct mapping of elements or a transformation of numerical values.

✅ Correct Approach:

In mathematics, especially for functions like f: A → B, elements from set A are mapped to elements in set B. There are no 'units' to convert in the physical sense. To check if a function is one-to-one, one must ensure that f(x₁) = f(x₂) strictly implies x₁ = x₂ for all x₁, x₂ in the domain. The focus is purely on the algebraic or graphical relationship between numerical inputs and outputs, irrespective of any physical dimensions.

📝 Examples:

❌ Wrong:

When checking the injectivity of f(x) = x² for x ∈ R, a student might incorrectly reason: 'If x represents a length (e.g., in meters), then x² represents an area (e.g., in square meters). Since the 'units' are different, the function maps to different 'types' of quantities, implying it's not one-to-one.' This line of reasoning is flawed because it introduces physical units where they are not relevant to the mathematical definition of injectivity.

✅ Correct:

To correctly check f(x) = x² (for x ∈ R) for injectivity:
1. Assume f(x₁) = f(x₂) for some x₁, x₂ ∈ R.
2. This means x₁² = x₂².
3. Rearranging, we get x₁² - x₂² = 0, which factors to (x₁ - x₂)(x₁ + x₂) = 0.
4. This implies either x₁ = x₂ OR x₁ = -x₂.
Since x₁ is not necessarily equal to x₂ (e.g., f(2) = 4 and f(-2) = 4, where 2 ≠ -2), the function f(x) = x² is not one-to-one on the set of real numbers. No unit conversion logic is involved or relevant.

💡 Prevention Tips:

Always remember that mathematical functions in JEE context primarily deal with abstract numbers and their specified relationships, not physical quantities with units, unless explicitly stated in an applied problem.
Clearly distinguish between the mathematical definition of domain and codomain (sets of numbers) and physical dimensions or measurement units.
Stick rigorously to the algebraic (f(x₁) = f(x₂) &implies; x₁ = x₂) or graphical (horizontal line test) methods for checking the injectivity (one-to-one property) of a function, without introducing extraneous concepts like unit conversion.
JEE Specific Tip: Questions on functions in JEE Main Mathematics typically test core mathematical properties. Physical units are generally irrelevant to these fundamental concepts unless the problem is specifically from a physics application or explicitly defines units within the function context.

JEE_Main

Minor Sign Error

❌ Sign Error in Solving Equations to Check Injectivity

Students often make sign errors when solving the equation f(x₁) = f(x₂) to determine if a function is one-one, particularly with functions involving even powers (like x², x⁴) or absolute values (like |x|). They might overlook both positive and negative roots, incorrectly simplifying A² = B² to just A = B instead of A = ±B, leading to a false conclusion that the function is one-one.

💭 Why This Happens:

This error primarily stems from a lack of complete algebraic rigor. Students tend to rush or apply incomplete algebraic rules, especially when solving for variables in squared terms or absolute value equations. They might focus only on the principal (positive) root or simply forget that negative counterparts also satisfy the equation, thereby missing distinct domain elements mapping to the same codomain element.

✅ Correct Approach:

When checking for injectivity by setting f(x₁) = f(x₂), always solve the resulting algebraic equation meticulously for x₁ in terms of x₂ (or vice versa). Always consider all possible solutions, including both positive and negative roots. If, after solving, you find any instance where f(x₁) = f(x₂) for distinct values x₁ ≠ x₂, then the function is not one-one. Remember that A² = B² implies A = ±B and |A| = |B| implies A = ±B.

📝 Examples:

❌ Wrong:

Consider f(x) = x² for x ∈ R.
To check if f is one-one, assume f(x₁) = f(x₂).
x₁² = x₂²
Incorrect step: Students might mistakenly conclude x₁ = x₂ only, and therefore deduce that f(x) is one-one. This overlooks the case where x₁ = -x₂.

✅ Correct:

Consider f(x) = x² for x ∈ R.
To check if f is one-one, assume f(x₁) = f(x₂).
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies either x₁ - x₂ = 0 (so x₁ = x₂) OR x₁ + x₂ = 0 (so x₁ = -x₂).
Since x₁ = -x₂ allows for distinct values (e.g., if x₁ = 2, then x₂ = -2, where x₁ ≠ x₂ but f(2) = 4 and f(-2) = 4), the function f(x) = x² is not one-one on R.

CBSE vs JEE: While CBSE emphasizes the algebraic definition, JEE often tests the ability to quickly identify such functions using graphical insights (horizontal line test) or by quickly finding a counterexample.

💡 Prevention Tips:

Thorough Algebraic Solving: Always practice solving equations completely, especially those involving squares, roots, or absolute values. Ensure you account for all possible sign combinations.
Test with Counterexamples: If a function involves x², |x|, or other similar terms, mentally (or quickly on paper) test a positive and its corresponding negative value (e.g., x=1 and x=-1). If they yield the same output, the function is not one-one.
Graphical Check (JEE): For JEE, quickly visualize the graph of the function. If any horizontal line intersects the graph at more than one point, the function is not one-one. This is a very efficient way to avoid such sign errors.

JEE_Main

Minor Approximation

❌ Assuming Injectivity/Non-Injectivity Based on General Graph Shape or Function Type

Students often make a quick judgment about whether a function is one-to-one (injective) by simply visualizing its general graph shape or identifying it as an 'odd degree polynomial' or 'even degree polynomial' without a rigorous algebraic check or careful consideration of the specified domain and codomain. This leads to an 'approximate' understanding rather than a precise one, especially when the domain is restricted.

💭 Why This Happens:

This mistake usually happens due to a rush to solve problems quickly, a superficial understanding of the horizontal line test, or failing to appreciate that the domain of a function is crucial for determining injectivity. Students might correctly recall that y = x^2 is not one-to-one on R, but fail to recognize that it *is* one-to-one on a restricted domain like [0, ∞).

✅ Correct Approach:

The most reliable approach is to algebraically verify the definition of a one-to-one function: if f(x₁) = f(x₂), then it must imply x₁ = x₂ for all x₁, x₂ in the domain. For continuous functions, checking monotonicity over the entire domain is also a valid method (strictly increasing or strictly decreasing implies one-to-one). Always pay close attention to the given domain and codomain.

📝 Examples:

❌ Wrong:

Consider the function f: [-3, 3] → R defined by f(x) = x² - 2x.
Wrong thought: 'It's a quadratic (even degree polynomial), so it must not be one-to-one.'
This ignores the specific domain and the potential for the function to be injective over certain restricted intervals, leading to an incorrect approximation.

✅ Correct:

Consider the function f: [-3, 3] → R defined by f(x) = x² - 2x.
Correct approach (Algebraic Verification):
Let f(x₁) = f(x₂) for x₁, x₂ ∈ [-3, 3].
x₁² - 2x₁ = x₂² - 2x₂
x₁² - x₂² - 2x₁ + 2x₂ = 0
(x₁ - x₂)(x₁ + x₂) - 2(x₁ - x₂) = 0
(x₁ - x₂)(x₁ + x₂ - 2) = 0
This implies either x₁ - x₂ = 0 ⇒ x₁ = x₂ OR x₁ + x₂ - 2 = 0 ⇒ x₁ + x₂ = 2.
If x₁ + x₂ = 2, and x₁ ≠ x₂, then the function is not one-to-one. For example, if x₁ = 3 (in domain), then x₂ = 2 - 3 = -1 (also in domain).
f(3) = 3² - 2(3) = 9 - 6 = 3
f(-1) = (-1)² - 2(-1) = 1 + 2 = 3
Since f(3) = f(-1) but 3 ≠ -1, the function f(x) = x² - 2x is not one-to-one on the domain [-3, 3].

Correct approach (Calculus/Monotonicity):
f'(x) = 2x - 2. Setting f'(x) = 0 gives x = 1.
Since x = 1 lies within the domain [-3, 3], and the derivative changes sign, the function is not strictly monotonic over the entire domain. Hence, it is not one-to-one.
For example, f(x) decreases on [-3, 1] and increases on [1, 3], failing injectivity. For CBSE, this calculus method is often preferred for continuous functions.

💡 Prevention Tips:

Always check the domain: The domain of the function is critical for determining injectivity. A function not one-to-one on R might be one-to-one on a restricted domain.
Algebraic verification is paramount: Always rely on the definition: f(x₁) = f(x₂) ⇒ x₁ = x₂. This is the most foolproof method for JEE Main.
Don't overgeneralize: While certain function types have general properties, these might change for specific cases or restricted domains.
Sketch, but verify: Use graphical methods (like the horizontal line test) as a quick intuitive check, but always back it up with a rigorous algebraic proof, especially for competitive exams.

JEE_Main

Minor Other

❌ Confusing One-to-One with Strictly Monotonic Functions

Students often incorrectly assume that for a function to be one-to-one (injective), it must necessarily be strictly increasing or strictly decreasing throughout its domain. While strictly monotonic functions are indeed one-to-one, the converse is not always true, especially for discontinuous or piecewise-defined functions.

💭 Why This Happens:

This misunderstanding often arises because most introductory examples of one-to-one functions are continuous and strictly monotonic. The visual intuition from the Horizontal Line Test often aligns with smooth, monotonic graphs, leading students to generalize incorrectly to all types of functions.

✅ Correct Approach:

Always refer back to the fundamental definition of a one-to-one function: A function f: A → B is one-to-one if for any x₁, x₂ ∈ A, if f(x₁) = f(x₂), then x₁ = x₂. Alternatively, if x₁ ≠ x₂, then f(x₁) ≠ f(x₂). The Horizontal Line Test is a reliable graphical tool, applicable even to non-monotonic or discontinuous functions, to check injectivity.

📝 Examples:

❌ Wrong:

Assuming that a function like f(x) = { x, if x < 0; x+1, if x ≥ 0 } is not one-to-one because it exhibits a jump discontinuity at x=0, thus not being strictly increasing everywhere. This assumption is incorrect.

✅ Correct:

Consider the function f: R → R defined by f(x) = { x, if x < 0; x + 2, if x ≥ 0 }. This function is one-to-one (injective) but not strictly monotonic over its entire domain.

If x₁, x₂ < 0 and f(x₁) = f(x₂), then x₁ = x₂.
If x₁, x₂ ≥ 0 and f(x₁) = f(x₂), then x₁ + 2 = x₂ + 2 ⇒ x₁ = x₂.
If x₁ < 0 and x₂ ≥ 0, then f(x₁) = x₁ (which is < 0), and f(x₂) = x₂ + 2 (which is ≥ 2). Clearly, f(x₁) ≠ f(x₂).

Despite the jump, no horizontal line intersects its graph more than once, satisfying the one-to-one condition.

💡 Prevention Tips:

Focus on Definition: Always use the algebraic definition of one-to-one property first.
Practice Diverse Functions: Work with piecewise, discontinuous, and step functions to broaden your understanding.
Apply HLT Critically: Use the Horizontal Line Test for all types of graphs, not just continuous ones.

JEE_Main

Minor Other

❌ Overlooking Domain/Codomain and Incomplete Algebraic Proof for Injectivity

Students frequently make errors in determining if a function is one-to-one (injective) by either neglecting the specified domain and codomain, or by failing to perform a complete and rigorous algebraic proof. They might assume injectivity based on the function's appearance or make incomplete deductions from the `f(x1) = f(x2)` step, particularly when `x1` and `x2` could be related in multiple ways (e.g., `x1 = -x2`).

💭 Why This Happens:

This mistake often stems from a lack of careful attention to the function's definition, especially its domain. Students might rush algebraic steps, forgetting rules like `a^2 = b^2` implies `a = ±b`. Sometimes, they rely on memorized characteristics of common function types rather than understanding the fundamental definition of injectivity, which requires checking for all possible inputs in the domain.

✅ Correct Approach:

To correctly determine if a function `f: A → B` is one-to-one, always:

First, note the domain (A) and codomain (B). These are crucial.
Assume `f(x1) = f(x2)` for any `x1, x2 ∈ A`.
Algebraically manipulate this equation to rigorously prove that this assumption must lead to `x1 = x2`.
If `f(x1) = f(x2)` allows for possibilities other than `x1 = x2` (e.g., `x1 = -x2`) within the given domain, then the function is not one-to-one.
Alternatively, demonstrate a specific counterexample: find two distinct elements `x1 ≠ x2` in the domain such that `f(x1) = f(x2)`.

📝 Examples:

❌ Wrong:

Consider `f: R → R` defined by `f(x) = x^2`. A student might write: "Let `f(x1) = f(x2)`. Then `x1^2 = x2^2`. Taking square root on both sides gives `x1 = x2`. Thus, `f` is one-to-one." (This is incorrect as it ignores the `x1 = -x2` possibility for real numbers.)

✅ Correct:

Consider `f: R → R` defined by `f(x) = x^2`.
To check injectivity:
Assume `f(x1) = f(x2)` for `x1, x2 ∈ R`.
`x1^2 = x2^2`
`x1^2 - x2^2 = 0`
`(x1 - x2)(x1 + x2) = 0`
This implies `x1 - x2 = 0` or `x1 + x2 = 0`.
So, `x1 = x2` or `x1 = -x2`.
Since for distinct values like `x1 = 2` and `x2 = -2`, we have `f(2) = 4` and `f(-2) = 4`, the function is not one-to-one over the domain `R`.
CBSE Tip: Clearly showing the `x1 = ± x2` step and explaining why it means 'not one-to-one' is key.

💡 Prevention Tips:

Read the Question Carefully: Always identify the domain and codomain before starting the solution.
Master Algebraic Manipulations: Be precise, especially with squares, roots, and absolute values, remembering all possible solutions.
Think of Counterexamples: If you suspect a function is not one-to-one, try to find two different inputs that give the same output.
Practice Rigorous Proofs: Consistently apply the `f(x1) = f(x2) ⇒ x1 = x2` method for all types of functions.

CBSE_12th

Minor Approximation

❌ Assuming one-to-one nature from partial observation or visual approximation

Students frequently conclude that a function is one-to-one based on observing a few input-output pairs or by an informal visual inspection of its graph. They might apply the horizontal line test *approximately* to a small segment of the graph or verify injectivity for only a limited set of domain values, without performing a rigorous proof across the entire defined domain.

💭 Why This Happens:

This mistake stems from a lack of rigorous application of the definition of a one-to-one function. Students may be pressured for time, lack confidence in algebraic manipulation to prove injectivity universally, or confuse 'it looks one-to-one' or 'it works for these values' with a formal proof valid for the entire domain. It's an 'approximation' of the analytical process.

✅ Correct Approach:

To prove a function f: A → B is one-to-one (injective), always start by assuming f(x₁) = f(x₂) for any arbitrary x₁, x₂ ∈ A. Then, through algebraic manipulation, rigorously demonstrate that this assumption *must* lead to x₁ = x₂. For graphical methods, ensure the horizontal line test is applied exhaustively over the function's entire domain, confirming that no horizontal line intersects the graph at more than one point.

📝 Examples:

❌ Wrong:

Consider the function f: R → R defined by f(x) = x². A student might observe that f(1)=1, f(2)=4, f(3)=9, and conclude it's one-to-one, or might only sketch the graph for x > 0 and assume injectivity for the entire real domain.

✅ Correct:

For the function f: R → R, defined by f(x) = x², let's test for one-to-one:
Assume f(x₁) = f(x₂)
Then x₁² = x₂²
This implies x₁ = ±x₂.
Since x₁ can be equal to -x₂ (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the condition x₁ = x₂ is not universally met. Hence, f(x) = x² is NOT one-to-one on R.
However, if the domain is restricted to f: [0, ∞) → R, then assuming x₁² = x₂² and knowing x₁, x₂ ≥ 0, we must have x₁ = x₂. Thus, f(x) = x² IS one-to-one on [0, ∞).

💡 Prevention Tips:

Always rely on the formal definition and perform algebraic proofs consistently.
For CBSE exams, analytical proof is almost always required.
When using graphical analysis, consider the entire domain, especially looking for any symmetries that might lead to f(x₁) = f(x₂) for x₁ ≠ x₂.
JEE Tip: While quick sketches and derivative analysis (f'(x) > 0 or f'(x) < 0 implies one-to-one) are useful for speed, ensure your understanding of their rigorous implications.

CBSE_12th

Minor Sign Error

❌ Ignoring Negative Solutions in One-to-One Function Proofs

Students frequently make a sign error by overlooking the negative root or solution when solving equations like f(x₁) = f(x₂), particularly for functions involving even powers (e.g., x²) or absolute values (e.g., |x|). This leads to an incorrect conclusion that the function is one-to-one.

💭 Why This Happens:

This error stems from incomplete algebraic manipulation or a rush to conclude x₁ = x₂. For instance, when presented with x₁² = x₂², students often only consider x₁ = x₂ and forget the equally valid solution x₁ = -x₂. This oversight is a fundamental misunderstanding of how even powers or absolute values behave with respect to signs.

✅ Correct Approach:

To correctly prove or disprove a function's one-to-one nature, always perform a complete algebraic solution for f(x₁) = f(x₂). This involves factoring expressions (like difference of squares) and considering all possible real solutions. If you can find distinct x₁ and x₂ such that f(x₁) = f(x₂), then the function is not one-to-one.

📝 Examples:

❌ Wrong:

Consider f: R → R given by f(x) = x².
Student's Approach:
Assume f(x₁) = f(x₂).
x₁² = x₂²
Therefore, x₁ = x₂.
Conclusion: f is one-to-one. (Incorrect because it ignored x₁ = -x₂)

✅ Correct:

Consider f: R → R given by f(x) = x².
Correct Approach:
Assume f(x₁) = f(x₂).
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies either x₁ - x₂ = 0 (so x₁ = x₂) OR x₁ + x₂ = 0 (so x₁ = -x₂).
Since we can find distinct values (e.g., x₁ = 2, x₂ = -2, where f(2)=4 and f(-2)=4) such that f(x₁) = f(x₂) but x₁ ≠ x₂, the function is NOT one-to-one.
Conclusion: f is NOT one-to-one.

💡 Prevention Tips:

Factorize Completely: Always factor expressions like (a² - b²) into (a - b)(a + b) to reveal all roots.
Remember Absolute Value Rules: If |a| = |b|, then a = b or a = -b.
Verify with Counterexamples: If unsure, pick simple numerical values (e.g., 2 and -2) from the domain to test if f(x₁) = f(x₂) for x₁ ≠ x₂.
Contextualize with Domain: Pay close attention to the function's domain. The same function can be one-to-one on a restricted domain (e.g., f(x)=x² on [0, ∞)) but not on R.

CBSE_12th

Minor Unit Conversion

❌ Ignoring Domain/Codomain in Injectivity Proofs

Students often perform algebraic steps correctly for f(x1) = f(x2) but overlook the function's specified domain (e.g., N, Z, R). This can lead to incorrect conclusions about whether a function is one-to-one (injective).

💭 Why This Happens:

Algebraic Focus: Prioritizing algebraic manipulation over considering that x1, x2 must belong to the function's defined domain.

Implicit Assumptions: Implicitly assuming the domain is the set of all real numbers (R) even when it's explicitly specified as Natural numbers (N) or Integers (Z).

Rushed Proofs: Skipping the crucial step of explicitly defining x1, x2 from the given domain in the proof.

✅ Correct Approach:

Always start your injectivity proof by stating "Let x1, x2 be elements of the function's domain." When assuming f(x1) = f(x2), ensure the conclusion x1 = x2 holds true only for values within that domain. If distinct x1, x2 in the specified domain yield f(x1) = f(x2), the function is not injective.

📝 Examples:

❌ Wrong:

Consider f: Z → Z (function from integers to integers) defined by f(x) = x².
Student's Attempt (Wrong):
Assume f(x1) = f(x2).
⇒ x1² = x2²
⇒ x1 = x2 (Incorrect, as x1 = -x2 is also a valid possibility for integers and leads to different inputs having the same output.)

✅ Correct:

Consider f: Z → Z (function from integers to integers) defined by f(x) = x².
Correct Approach:
Let x1, x2 ∈ Z.
Assume f(x1) = f(x2)
⇒ x1² = x2²
⇒ (x1 - x2)(x1 + x2) = 0
This implies either x1 = x2 OR x1 = -x2.
Since x1, x2 ∈ Z, we can choose x1 = 1 and x2 = -1. Here, x1 ≠ x2, but f(1) = 1² = 1 and f(-1) = (-1)² = 1.
Thus, as f(1) = f(-1) for 1 ≠ -1 (both in Z), the function f(x) = x² is not one-to-one for the domain Z.

JEE Tip: For JEE, understanding how domain restriction impacts injectivity is crucial, especially for piecewise functions or functions defined on specific intervals.

💡 Prevention Tips:

State Domain First: Always explicitly state "Let x1, x2 belong to the domain of f" at the beginning of your proof.

Thorough Algebra: When solving f(x1) = f(x2), consider all valid algebraic solutions within the specified domain and do not jump to conclusions.

Counterexamples: If your analysis suggests the function is not injective, always find concrete counterexamples (distinct inputs mapping to the same output) from the domain.

CBSE Rigor: For CBSE exams, showing all steps clearly, including initial assumptions and explicit reference to the domain, is essential for full marks.

CBSE_12th

Minor Formula

❌ Incomplete Algebraic Analysis for One-to-One Condition

Students often make a minor error in algebraically proving a function is one-to-one (injective) by not thoroughly investigating all possibilities when setting f(x₁) = f(x₂). They might incorrectly assume that only x₁ = x₂ is the sole outcome, overlooking other valid relationships like x₁ = -x₂, which indicate the function is not one-to-one for distinct x₁ and x₂.

💭 Why This Happens:

Over-simplification: Rushing through algebraic steps, especially when dealing with squares, absolute values, or even powers, and prematurely concluding x₁ = x₂.
Lack of Rigor: Not explicitly considering that for the function to be one-to-one, f(x₁) = f(x₂) must only imply x₁ = x₂.
Domain Neglect: Forgetting to consider the function's domain when solving the equation; sometimes, extra solutions might fall outside the domain, but other valid distinct solutions might exist within it.

✅ Correct Approach:

To prove a function f: A → B is one-to-one, one must take any two elements x₁, x₂ ∈ A such that f(x₁) = f(x₂) and then algebraically demonstrate that this equality strictly implies x₁ = x₂. If any other possibility arises (e.g., x₁ = -x₂, where x₁ ≠ x₂ for some values in the domain), then the function is not one-to-one.

JEE Tip: For competitive exams, quickly check for symmetry (e.g., if f(x) = f(-x) for x ≠ 0) or mentally apply the Horizontal Line Test to quickly spot non-injective functions.

📝 Examples:

❌ Wrong:

Problem: Check if f(x) = x² defined on R → R is one-to-one.
Student's Wrong Approach:

Let f(x₁) = f(x₂)
x₁² = x₂²
Taking square root on both sides:
x₁ = x₂
Therefore, f(x) is one-to-one.

Mistake: Ignoring the negative root (x₁ = -x₂), which indicates two distinct inputs can map to the same output.

✅ Correct:

Problem: Check if f(x) = x² defined on R → R is one-to-one.
Correct Approach:

Let f(x₁) = f(x₂) for some x₁, x₂ ∈ R.
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies either (x₁ - x₂) = 0  OR  (x₁ + x₂) = 0
So,  x₁ = x₂  OR  x₁ = -x₂

Since we found a case where x₁ = -x₂ (for example, if x₁ = 2, then x₂ = -2, and f(2) = 4, f(-2) = 4, but 2 ≠ -2), the function f(x) = x² is not one-to-one on R.

CBSE Note: Clearly showing all algebraic possibilities and explaining why they lead to or against injectivity is crucial for full marks in descriptive answers.

💡 Prevention Tips:

Be Thorough with Algebra: When solving f(x₁) = f(x₂), treat it like any equation. Don't forget ± signs when taking square roots or properties of absolute values.
Factorization: Always try to bring all terms to one side and factorize (e.g., a² - b² = (a-b)(a+b)) to reveal all possible solutions for x₁ and x₂.
Test with Values: If unsure, pick simple distinct values (e.g., x=1, x=-1) from the domain and check if f(x₁) = f(x₂) even when x₁ ≠ x₂.
Practice: Work through problems involving functions like x², |x|, cos(x), or those with even powers, as these are common traps for injectivity.

CBSE_12th

Minor Calculation

❌ Incorrect Algebraic Deduction for Even Powers

A common error in proving one-one functions involves algebraic manipulation, particularly when dealing with even powers. Students often incorrectly conclude that if f(x₁) = f(x₂) leads to x₁ⁿ = x₂ⁿ (where n is an even integer), then necessarily x₁ = x₂. This oversight ignores the possibility of x₁ = -x₂, which is crucial for determining injectivity.

💭 Why This Happens:

Hasty Conclusion: Students often jump to conclusions without considering all algebraic solutions to an equation.
Incomplete Algebraic Knowledge: A lack of thorough understanding of solving non-linear equations, especially those with even exponents.
Ignoring Domain Context: Failing to check if x₁ = -x₂ is a valid scenario within the function's given domain.

✅ Correct Approach:

When proving injectivity using f(x₁) = f(x₂):

If you reach an equation like x₁ⁿ = x₂ⁿ (for even n, e.g., n=2, 4), always deduce x₁ = x₂ OR x₁ = -x₂.
Then, verify if x₁ = -x₂ (with x₁ ≠ x₂) is possible for values within the given domain.
If it is, and f(x₁) = f(x₂) for distinct x₁ and x₂, the function is not one-one. If the domain restriction prevents x₁ = -x₂ (e.g., domain is [0, ∞)), then the function might be one-one.

📝 Examples:

❌ Wrong:



Function: f: R → R, f(x) = x²

To check one-one: Assume f(x₁) = f(x₂)

x₁² = x₂²

Therefore, x₁ = x₂.

Conclusion: f(x) is one-one. (INCORRECT)

✅ Correct:



Function: f: R → R, f(x) = x²

To check one-one: Assume f(x₁) = f(x₂)

x₁² = x₂²

This implies x₁ = x₂  OR  x₁ = -x₂.

Since the domain is R, we can choose x₁ = 2 and x₂ = -2.

Here, x₁ ≠ x₂, but f(2) = 2² = 4 and f(-2) = (-2)² = 4.

Since f(2) = f(-2) for 2 ≠ -2, the function f(x) = x² is NOT one-one.

CBSE/JEE Tip: Always verify the domain restrictions when dealing with squares or absolute values.

💡 Prevention Tips:

Rigorous Algebra: Always solve equations comprehensively, considering all possible roots and scenarios (e.g., x² = a² implies x = ±a).
Domain Awareness: Explicitly check if the 'extra' solutions (like x₁ = -x₂) are permissible within the function's defined domain.
Counter-Examples: If you suspect a function is not one-one, try to find two distinct values in the domain that map to the same image. This is a quick way to disprove injectivity.

CBSE_12th

Minor Conceptual

❌ Misinterpreting Injectivity by Incomplete Domain Analysis

Students often incorrectly conclude a function is one-to-one by evaluating it for only a few values or over a limited interval, failing to rigorously apply the definition across the entire specified domain. This leads to errors, especially with functions that 'fold back' like parabolas or absolute value functions.

💭 Why This Happens:

This error arises from a superficial understanding of the 'one-to-one' criterion, assuming distinct outputs for tested inputs implies injectivity for all inputs. Inadequate graphical analysis over the full domain or a reluctance to apply the formal algebraic proof also contributes.

✅ Correct Approach:

To prove a function f: A → B is one-to-one (injective), assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A and demonstrate that this algebraically implies x₁ = x₂. To disprove injectivity, find at least one pair of distinct inputs x₁ &neq; x₂ such that f(x₁) = f(x₂). The Horizontal Line Test (HLT) is a useful visual tool: if any horizontal line intersects the graph of the function at more than one point, the function is not one-to-one.

📝 Examples:

❌ Wrong:

Statement: 'For the function f: R → R defined by f(x) = x², it is one-to-one because f(1) = 1 and f(2) = 4 (distinct outputs for distinct inputs).'
Mistake: Overlooking that f(-1) = 1 and f(-2) = 4, meaning f(1) = f(-1) and f(2) = f(-2) despite inputs being different. Only considering positive inputs leads to this error.

✅ Correct:

Analysis of f(x) = x² on R → R for injectivity:
1. Choose two distinct inputs, say x₁ = -2 and x₂ = 2. Clearly, x₁ &neq; x₂.
2. Evaluate the function at these inputs:
f(x₁) = f(-2) = (-2)² = 4
f(x₂) = f(2) = (2)² = 4
3. Since f(-2) = f(2) but -2 &neq; 2, the function f(x) = x² is not one-to-one on the domain of all real numbers (R).
Alternatively, if f(x₁) = f(x₂) &implies; x₁² = x₂² &implies; x₁ = ±x₂. Since x₁ is not necessarily equal to x₂ (it could be -x₂), it's not one-to-one.

💡 Prevention Tips:

Always use the algebraic definition: To prove injectivity, derive x₁ = x₂ from f(x₁) = f(x₂).
For functions with even powers (e.g., x², x⁴) or absolute values (e.g., |x|), explicitly check both positive and negative inputs to find counterexamples.
When using the Horizontal Line Test (HLT), ensure your visualization or sketch of the graph accurately covers the function's entire specified domain.
CBSE Specific: Clearly state the domain and codomain at the beginning of your solution. Marks are often awarded for demonstrating the condition over the specified domain.

CBSE_12th

Minor Approximation

❌ Assuming Injectivity from Partial or Approximate Monotonicity

Students often make the error of concluding a function is one-to-one by observing its behavior over a subinterval or by roughly estimating its monotonicity (e.g., 'mostly increasing') from a quick sketch. This superficial or 'approximate' understanding leads to an incorrect inference of injectivity without formally checking the strict definition or verifying strict monotonicity over the *entire* domain.

💭 Why This Happens:

This mistake stems from over-reliance on visual intuition without rigorous analytical checks. Students might focus on a part of the graph where the function appears monotonic and generalize this to the entire domain, failing to consider the comprehensive behavior or edge cases where monotonicity might change or points where f(x1) = f(x2) for x1 ≠ x2 exist.

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one, always apply one of the following rigorous methods:

Formal Definition: Assume f(x1) = f(x2) for any x1, x2 ∈ A. If this necessarily implies x1 = x2, then the function is one-to-one.
Strict Monotonicity: Prove that the function is either strictly increasing (x1 < x2 ⇒ f(x1) < f(x2)) or strictly decreasing (x1 < x2 ⇒ f(x1) > f(x2)) over its entire domain.
Derivative Test (for differentiable functions): Show that f'(x) > 0 for all x in the domain (or f'(x) < 0 for all x), allowing f'(x)=0 at isolated points only. If f'(x) changes sign, it is not one-to-one.

📝 Examples:

❌ Wrong:

Consider f(x) = x^2 with domain [-3, 3]. A student might observe: 'The function is increasing for x ∈ [0, 3], and decreasing for x ∈ [-3, 0]. It looks like it just turns around, so it must be one-to-one after considering its turning point.' This 'approximate' reasoning is incorrect.

✅ Correct:

For f(x) = x^2 with domain [-3, 3]:
Let x1 = 2 and x2 = -2. Both x1, x2 ∈ [-3, 3].
Then f(x1) = f(2) = 2^2 = 4.
And f(x2) = f(-2) = (-2)^2 = 4.
Here, f(x1) = f(x2) but x1 ≠ x2. Therefore, f(x) = x^2 on [-3, 3] is not one-to-one. The 'approximation' based on parts of the domain is insufficient; the horizontal line test fails.

💡 Prevention Tips:

Always specify the domain and codomain: Injectivity depends critically on the domain.
Rigorous Proof, Not Intuition: While sketching helps visualize, JEE Advanced demands analytical proofs.
Apply Definition Systematically: Work through f(x1) = f(x2) ⇒ x1 = x2 for all relevant x1, x2 in the domain.
Horizontal Line Test: Mentally (or graphically) ensure no horizontal line intersects the graph more than once *across the entire domain*.
CBSE vs. JEE Advanced: While CBSE might accept simpler arguments, JEE Advanced requires meticulous justification.

JEE_Advanced

Minor Sign Error

❌ Sign Errors in Absolute Value or Square Root Simplification during Injectivity Check

Students frequently make sign errors when testing if a function is one-one (injective) by solving `f(x1) = f(x2)`. This typically occurs when the function involves absolute values (e.g., |x|) or square roots of squared terms (e.g., sqrt(x^2)). The error lies in incorrectly simplifying expressions like |A| = |B| to just A = B, or sqrt(A^2) = A instead of |A|, neglecting the potential negative counterpart.

💭 Why This Happens:

This mistake stems from a lack of careful algebraic manipulation and sometimes a hurried approach. Students might forget the fundamental properties that |A| = |B| implies A = B or A = -B, and that sqrt(A^2) is identically equal to |A|, not just A, unless A is explicitly known to be non-negative. Forgetting to consider both positive and negative cases leads to an erroneous conclusion about the function's injectivity.

✅ Correct Approach:

Always apply the definitions and properties of absolute values and square roots rigorously. When solving |A| = |B|, write down both possibilities: A = B and A = -B. Similarly, when encountering sqrt(expression^2), replace it with |expression|. Then, evaluate if `x1 = x2` is the only possibility or if other distinct values of x1 and x2 exist that satisfy f(x1) = f(x2) within the function's domain.

📝 Examples:

❌ Wrong:

Consider the function f(x) = |x - 3| for x ∈ R.
Incorrect approach to check injectivity:
Assume f(x1) = f(x2).
|x1 - 3| = |x2 - 3|
Student incorrectly assumes x1 - 3 = x2 - 3
x1 = x2
Conclusion: The function is one-one. (This is wrong!)

✅ Correct:

Consider the function f(x) = |x - 3| for x ∈ R.
Correct approach to check injectivity:
Assume f(x1) = f(x2).
|x1 - 3| = |x2 - 3|
This implies two possibilities:
1. x1 - 3 = x2 - 3 => x1 = x2 (This is one condition for injectivity)
2. x1 - 3 = -(x2 - 3) => x1 - 3 = -x2 + 3 => x1 = 6 - x2 (This is the critical sign consideration)
Since x1 = 6 - x2 allows for x1 ≠ x2 (e.g., if x1 = 1, then x2 = 5; f(1) = |1-3| = 2 and f(5) = |5-3| = 2, but 1 ≠ 5), the function f(x) = |x - 3| is not one-one. (JEE Advanced often tests these nuanced cases.)

💡 Prevention Tips:

Be Meticulous: Always write down all possible cases when dealing with absolute values or equations involving even powers.
Remember Identities: Specifically, sqrt(A^2) = |A| and |A| = |B| <=> A = B or A = -B.
Graphical Check: For simple functions, a quick mental graph (horizontal line test) can often reveal if multiple inputs map to the same output, thereby catching such algebraic slips.
Domain Awareness: Always consider the function's domain when evaluating possible solutions for x1 and x2.

JEE_Advanced

Minor Conceptual

❌ Incorrectly Applying Injectivity Test for Restricted Domains

Students often fail to properly account for the specified domain of a function when determining if it is one-to-one (injective). They might apply the injectivity test (algebraic or horizontal line test) over the entire real line (R) instead of strictly adhering to the given, often restricted, domain.

💭 Why This Happens:

This mistake stems from a lack of attention to detail regarding the function's domain. Students may:

Over-rely on generalized properties of functions learned for R.
Perform incomplete algebraic analysis when solving f(x₁)=f(x₂), not checking if all potential solutions for x₁ and x₂ are valid within the given domain.
Not visualize the graph correctly for the restricted domain when using the horizontal line test.

✅ Correct Approach:

To correctly determine if a function f: D → C is one-to-one:

Algebraic Method: Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ D. If this algebraically implies x₁ = x₂, and crucially, no other valid distinct pair (x₁, x₂) exists within the domain D that satisfies f(x₁)=f(x₂), then the function is one-to-one.
Graphical Method (Horizontal Line Test): Draw the graph of the function only for its specified domain D. If any horizontal line intersects the graph at most once, the function is one-to-one.

📝 Examples:

❌ Wrong:

Concluding that f(x) = x² is not one-to-one, even when its domain is explicitly given as [0, 5]. This common error ignores that x = -x is not possible for non-zero values in the non-negative domain.

✅ Correct:

Consider the function f(x) = x² with domain D = [0, 5].
Let f(x₁) = f(x₂) for x₁, x₂ ∈ [0, 5].
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ = x₂ OR x₁ = -x₂.
Since x₁, x₂ ∈ [0, 5], both are non-negative. The condition x₁ = -x₂ can only be satisfied if x₁ = x₂ = 0 (which is a case of x₁ = x₂). For any other x₁ > 0, -x₂ would be negative or zero, meaning x₂ would have to be negative or zero. However, x₂ must be in [0, 5].
Therefore, the only valid conclusion within the domain [0, 5] is x₁ = x₂. Hence, f(x) = x² is one-to-one on [0, 5].
JEE Advanced Note: Be extra careful with questions involving trigonometric functions on specific intervals (e.g., sin x on [0, π/2] vs [0, π]).

💡 Prevention Tips:

Always Check the Domain First: Before starting any analysis, clearly write down and understand the function's domain.
Thorough Algebraic Scrutiny: When solving f(x₁) = f(x₂), systematically check if all derived possibilities for x₁ and x₂ are valid and distinct within the given domain.
Visualize Carefully: If using the horizontal line test, sketch the graph of the function only over the specified domain.
Practice with Restricted Domains: Solve problems where injectivity depends entirely on the domain restriction.

JEE_Advanced

Minor Calculation

❌ Ignoring Multiple Solutions in Algebraic Simplification of f(x₁) = f(x₂)

Students often make calculation errors when simplifying the equation f(x₁) = f(x₂) to check for injectivity. A common mistake is to assume a unique solution for x₁ in terms of x₂, especially when dealing with even powers, square roots, or absolute values. For instance, concluding x₁ = x₂ directly from x₁² = x₂² or |x₁| = |x₂|, thereby overlooking the possibility of x₁ = -x₂ (where x₁ ≠ x₂). This oversight leads to incorrectly classifying a non-injective function as injective.

💭 Why This Happens:

This mistake stems from a lack of careful algebraic manipulation and a hurried approach to solving equations. Students might forget to consider all possible roots or branches of solutions (e.g., both positive and negative square roots, or conditions for absolute values). It's a 'minor' error because the conceptual understanding of injectivity might be present, but the execution of the algebraic test is flawed.

✅ Correct Approach:

To correctly determine if a function is one-to-one, always solve the equation f(x₁) = f(x₂) meticulously. Ensure that all possible solutions for x₁ in terms of x₂ are considered. Only if f(x₁) = f(x₂) uniquely implies x₁ = x₂ for all x₁, x₂ in the domain, is the function one-to-one. If other possibilities like x₁ = -x₂ (where x₁ ≠ x₂) arise, the function is not one-to-one.

📝 Examples:

❌ Wrong:

Let f(x) = x² for x ∈ ℝ.
Wrong Calculation: Assume f(x₁) = f(x₂).
x₁² = x₂²
Conclude x₁ = x₂ (by taking square root directly without considering ±).
Based on this, incorrectly deduce that f(x) is one-to-one.

✅ Correct:

Let f(x) = x² for x ∈ ℝ.
Correct Calculation: Assume f(x₁) = f(x₂).
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂)=0
This implies x₁ - x₂ = 0 (i.e., x₁ = x₂) OR x₁ + x₂ = 0 (i.e., x₁ = -x₂).
Since x₁ = -x₂ for x₁ ≠ 0 means x₁ ≠ x₂ (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the function f(x) = x² is not one-to-one. (This is a crucial distinction for JEE Advanced, as domain restrictions often define injectivity).

💡 Prevention Tips:

Be Meticulous: Treat algebraic simplification of f(x₁) = f(x₂) with utmost care.
Consider All Roots: For equations involving even powers (like x²) or absolute values, remember to account for both positive and negative possibilities.
Factorization: Use factorization (e.g., a² - b² = (a-b)(a+b)) to correctly identify all potential relationships between x₁ and x₂.
Test Counter-examples: If you find a possibility where x₁ ≠ x₂ but f(x₁) = f(x₂), mentally or explicitly test a numerical example to confirm non-injectivity.

JEE_Advanced

Minor Formula

❌ Ignoring Domain Restrictions when Applying <code>f(x1) = f(x2)</code> for Injectivity

Students often correctly recall the algebraic condition for a function to be one-to-one (injective): f(x₁) = f(x₂) ⇒ x₁ = x₂. However, a common minor error is to apply this rigidly without fully solving for all possible x₁, x₂ values or, critically, without verifying if these potential solutions lie within the specified domain. This leads to incorrect conclusions about the function's injectivity.

💭 Why This Happens:

Over-reliance on the initial step of setting f(x₁) = f(x₂) without completing the algebraic simplification to ensure x₁ = x₂ is the *only* possibility.
Neglecting the given domain for the function. A function that is not one-to-one over its natural domain (e.g., R) might become one-to-one over a restricted domain.
Rushing algebraic steps, especially with functions involving squares, absolute values, or trigonometric expressions, where multiple solutions might exist before considering the domain.

✅ Correct Approach:

Start by assuming f(x₁) = f(x₂) for any x₁, x₂ belonging to the function's given domain.
Carefully solve this equation algebraically. This often involves factoring or taking roots.
If, after solving, you find that x₁ = x₂ is the *only* possible conclusion for all x₁, x₂ within the domain, then the function is one-to-one.
If you find an additional possibility where x₁ ≠ x₂ is also valid (e.g., x₁ = -x₂), you must check if both x₁ and x₂ (with x₁ ≠ x₂) can simultaneously exist within the function's domain. If they can, the function is not one-to-one.

📝 Examples:

❌ Wrong:

Consider f(x) = x² with domain [0, ∞).

Student's Mistake: A student might write: x₁² = x₂² ⇒ x₁ = ±x₂. Concluding immediately that since x₁ could be -x₂ (and x₁ ≠ x₂ in this case), the function is not one-to-one. This overlooks the domain constraint that x₁, x₂ ≥ 0.

✅ Correct:

Consider f(x) = x² with domain [0, ∞).

Let f(x₁) = f(x₂) for x₁, x₂ ∈ [0, ∞).
Then x₁² = x₂².

This implies x₁ = ±x₂.

Since the domain is [0, ∞), both x₁ and x₂ must be non-negative.
If x₁ = -x₂, and both are non-negative, this is only possible if x₁ = x₂ = 0.
If x₁, x₂ > 0, then x₁ = -x₂ is not possible because it would mean one is positive and the other is negative (contradicting the domain). Thus, x₁ = x₂ is the only valid solution.

Therefore, for x₁, x₂ ∈ [0, ∞), f(x₁) = f(x₂) ⇒ x₁ = x₂. The function is one-to-one.

💡 Prevention Tips:

Always write down the domain explicitly at the start of your solution before applying the injectivity test.
Be rigorous in algebraic manipulations. Do not jump to conclusions about x₁ = ±x₂ without thoroughly evaluating if both possibilities are allowed by the domain.
For JEE Advanced, practice extensively with functions having restricted domains or those involving absolute values/piecewise definitions, as these are common traps.
CBSE vs JEE: While CBSE might focus more on standard functions over R, JEE Advanced will frequently test your understanding with specific, non-standard domains, making this mistake more crucial to avoid.

JEE_Advanced

Minor Unit Conversion

❌ Attempting Unit Conversion for Abstract Function Properties

A common conceptual error is to mistakenly associate 'unit conversion' with the abstract mathematical property of a function being 'one-to-one' (injective). The one-to-one property is about the mapping of elements between sets, not about physical units or their transformations. Unit conversion is irrelevant to determining if a function is one-to-one.

💭 Why This Happens:

This mistake stems from a confusion between pure mathematical concepts and their applications in physics or engineering, where physical quantities often carry units. Students might overgeneralize the importance of units, applying it even where it's not applicable, especially in problems that might superficially involve variables that *could* represent physical quantities but are being examined for their abstract functional properties.

✅ Correct Approach:

Understand that the definition of a one-to-one function is purely based on its mapping rule: 'If f(x₁) = f(x₂), then x₁ = x₂' for all x₁, x₂ in the domain. This definition is unit-agnostic. Whether the domain or codomain elements have units (e.g., meters, seconds) or are dimensionless numbers, the injectivity of the function is determined solely by this mapping criterion. Focus on the algebraic or graphical properties of the function, not on units.

📝 Examples:

❌ Wrong:

A student considering f(x) = x³ and thinking 'if x is in meters, then x³ is in cubic meters, and these units are different, so how does unit conversion affect whether it's one-to-one?' This line of reasoning is fundamentally flawed. The units of x or f(x) have no bearing on the injectivity of f(x) = x³.

✅ Correct:

To check if f(x) = x³ is one-to-one: Assume f(x₁) = f(x₂). This means x₁³ = x₂³. Taking the cube root, we get x₁ = x₂. Since assuming f(x₁) = f(x₂) leads to x₁ = x₂, the function f(x) = x³ is one-to-one for real numbers. This determination does not involve any unit conversion.

💡 Prevention Tips:

Recall Definitions: Always refer back to the fundamental definition of a one-to-one function.
Separate Concepts: Clearly distinguish between abstract mathematical properties of functions and the physical interpretation or units of variables in applied problems.
JEE Advanced Focus: For JEE Advanced, function properties are often tested abstractly or within specified domains, where unit considerations are typically absent unless explicitly part of a physics-related problem.

JEE_Advanced

Important Sign Error

❌ Incorrect Sign Analysis of Derivative for Injectivity

Students frequently commit sign errors when analyzing the first derivative, f'(x), to ascertain if a function is strictly monotonic over its domain. An incorrect determination of intervals where f'(x) > 0 or f'(x) < 0 directly leads to wrong conclusions about whether the function is one-to-one (injective).

💭 Why This Happens:

Algebraic Manipulation Errors: Carelessness during differentiation or simplification of f'(x).
Misinterpretation of Inequalities: Incorrectly solving quadratic or other complex inequalities (e.g., assuming x² - 4 > 0 implies x > 2 instead of x > 2 or x < -2).
Ignoring Critical Points: Failing to identify all points where f'(x) changes sign or is undefined.
Domain Overlook: Analyzing f'(x) without strictly adhering to the specified domain of the function.

✅ Correct Approach:

To correctly determine injectivity using derivatives:

Step 1: Accurately calculate the first derivative, f'(x).
Step 2: Find all critical points by setting f'(x) = 0 or identifying where f'(x) is undefined.
Step 3: Use a sign chart (number line method) to rigorously determine the sign of f'(x) in the intervals created by the critical points.
Step 4: Restrict this analysis strictly to the given domain of the function.
Step 5: If f'(x) maintains a constant sign (either always positive or always negative) throughout the entire domain, the function is one-to-one. If f'(x) changes sign within the domain, it is not one-to-one.

📝 Examples:

❌ Wrong:

Consider f(x) = x³ - 3x. A common mistake is to differentiate to f'(x) = 3x² - 3 and then incorrectly assume that since x² is generally positive, 3x²-3 will also mostly be positive, leading to the erroneous conclusion that f(x) is always increasing and thus one-to-one over R. This overlooks the negative interval for f'(x).

✅ Correct:

For f(x) = x³ - 3x defined on the domain R (all real numbers):
1. Calculate f'(x) = 3x² - 3 = 3(x² - 1) = 3(x - 1)(x + 1).
2. Critical points are where f'(x) = 0, which are x = -1 and x = 1.
3. Create a sign chart for f'(x):

Interval	Sign of (x-1)	Sign of (x+1)	Sign of f'(x)	Monotonicity
x < -1	-	-	+	Increasing
-1 < x < 1	-	+	-	Decreasing
x > 1	+	+	+	Increasing

4. Since f'(x) changes sign within the domain R (from positive to negative at x = -1 and negative to positive at x = 1), the function f(x) is NOT one-to-one over R. For JEE, accurately using this derivative test is crucial for injectivity.
Note: If the domain was restricted, for instance, to [1, ∞), then f'(x) would be consistently positive, making the function one-to-one in that specific domain.

💡 Prevention Tips:

Master Inequality Solving: Practice solving quadratic and rational inequalities to correctly identify intervals of positive/negative values.
Always Use Sign Charts: For determining the sign of a polynomial or rational derivative, a number line/sign chart is invaluable for avoiding errors.
Verify Differentiation: Double-check your derivative calculations before proceeding to sign analysis.
Domain-Specific Analysis: Always perform the sign analysis strictly within the given domain of the function, as injectivity often depends on it (JEE focus).

JEE_Main

Important Approximation

❌ Misidentifying One-to-One Functions (Injectivity)

Students often incorrectly apply the algebraic test or misinterpret the horizontal line test, leading to misidentification of one-to-one functions. Crucially, they often overlook the impact of the specified domain and codomain on injectivity.

💭 Why This Happens:

Poor grasp of the injectivity definition: 'distinct elements in the domain must map to distinct elements in the codomain.'
Ignoring the specified domain and codomain of the function.
Algebraic errors when solving f(x₁) = f(x₂) for x₁ = x₂.
Misinterpreting the Horizontal Line Test (e.g., confusing 'at most one intersection' with 'exactly one intersection').

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one:

Algebraic Method (Crucial for JEE): Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A. If this rigorously implies x₁ = x₂, the function is one-to-one. If x₁ ≠ x₂ is possible, it's NOT one-to-one.
Graphical Method (Horizontal Line Test): Every horizontal line must intersect the graph at at most one point. If any horizontal line intersects the graph at more than one point, the function is not one-to-one.
Key Reminder: Always check the domain carefully.

📝 Examples:

❌ Wrong:

Consider f: R → R given by f(x) = x².
Wrong Reasoning: 'If I take x=2, f(x)=4. If x=3, f(x)=9. Different inputs yield different outputs for these values, so it must be one-to-one.' This reasoning is flawed as it doesn't consider all possible inputs or the definition of injectivity universally.

✅ Correct:

Consider f: R → R given by f(x) = x².
Correct Approach:

Algebraic Test: Let f(x₁) = f(x₂). Then x₁² = x₂², which implies x₁ = ±x₂. Since x₁ can be equal to -x₂ (e.g., 2 ≠ -2, but f(2)=4 and f(-2)=4), the function is NOT one-to-one on R.
Horizontal Line Test: A horizontal line like y=4 intersects the graph y=x² at two points (x=2 and x=-2). Thus, it is NOT one-to-one on R.

Note: If the domain was restricted to f: [0, ∞) → R, then f(x) = x² would be one-to-one.

💡 Prevention Tips:

Understand Core Definition: Internalize that 'one-to-one' means no two distinct inputs produce the same output.
Prioritize Domain: Always check the domain; it fundamentally impacts injectivity. Many functions are one-to-one only on restricted domains.
Practice Algebraic Proofs: For JEE, algebraic methods are key. Master solving f(x₁) = f(x₂) ⇒ x₁ = x₂.
Avoid Partial Checks: The definition must hold for *all* elements in the domain, not just a few examples.

JEE_Main

Important Other

❌ Ignoring Domain and Codomain when Checking Injectivity (One-One Property)

Students frequently overlook or misinterpret the specified domain and codomain when determining if a function is one-one (injective). They tend to apply standard tests (like f(x₁) = f(x₂) ⇒ x₁ = x₂) or the horizontal line test without carefully considering the actual set of input values (domain) and potential output values (codomain) for the given function. This often leads to incorrect conclusions, especially in JEE problems where domains are frequently restricted.

💭 Why This Happens:

Over-reliance on general definitions: Students might memorize the definition for R → R functions and apply it universally without adapting to specific domains.
Lack of attention to notation: Not carefully reading the function declaration f: A → B, where A is the domain and B is the codomain.
Assuming default domain: Assuming the natural domain or R as the domain, even if a restricted domain is explicitly given.
Conceptual gap: Not fully understanding that injectivity is a property highly dependent on the function's domain.

✅ Correct Approach:

Always begin by identifying the precise domain (A) and codomain (B) of the function f: A → B.

For Algebraic Functions: When testing f(x₁) = f(x₂) ⇒ x₁ = x₂, ensure that both x₁ and x₂ are elements of the specified domain A. If the equality x₁ = x₂ holds only for values within A, then the function is one-one.
For Graphical Functions (Horizontal Line Test): A horizontal line should intersect the graph of y = f(x) at most once only for the portion of the graph defined by the given domain A. Do not consider parts of the graph outside the domain.
Monotonicity: A function that is strictly increasing or strictly decreasing over its entire domain A is always one-one. Verify strict monotonicity over the exact given domain.

JEE Tip: JEE problems often test this exact understanding by providing familiar functions with unusual or restricted domains.

📝 Examples:

❌ Wrong:

Consider f(x) = x² from R → R.

Mistake: Declaring it one-one because 'most simple functions are'.
Reason: Fails to acknowledge that f(2) = 4 and f(-2) = 4, where 2 ≠ -2, and both 2, -2 ∈ R.

✅ Correct:

Consider f(x) = x² from [0, ∞) → R.

Correct Approach: If f(x₁) = f(x₂), then x₁² = x₂². This implies x₁ = ±x₂. However, since the domain is [0, ∞), both x₁ and x₂ must be non-negative. Therefore, x₁ = x₂ is the only possibility within the domain. Thus, the function is one-one over the domain [0, ∞).

💡 Prevention Tips:

Always read the function definition carefully: Pay close attention to f: A → B.
Visualize the graph over the specified domain: Sketching or mentally visualizing the function's behavior only for the given domain can prevent errors.
Practice with varying domains: Solve problems where standard functions are given with non-standard domains.
Don't jump to conclusions: Verify the one-one condition (f(x₁) = f(x₂) ⇒ x₁ = x₂) thoroughly for all relevant values in the domain.

JEE_Main

Important Unit Conversion

❌ Misinterpreting the 'Distinct Mapping' Rule (Conceptual Conversion Error)

Students often make an error in understanding the fundamental 'conversion' or mapping rule for one-to-one functions. Instead of ensuring each distinct input (from the domain) 'converts' to a distinct output (in the codomain), they might incorrectly allow two different inputs to 'convert' to the same output. This is a conceptual misunderstanding of injectivity, confusing it with a general function or a many-to-one mapping, rather than a literal unit conversion.

💭 Why This Happens:

This mistake stems from a lack of clarity on the formal definition of a one-to-one (injective) function. Students might confuse it with the definition of an 'onto' function or simply misunderstand the implication of 'unique mapping'. Over-reliance on visual cues without algebraic verification or a weak grasp of the horizontal line test for graphs also contributes to this conceptual 'conversion' error.

✅ Correct Approach:

A function f: A → B is one-to-one if and only if for any x₁, x₂ ∈ A, if f(x₁) = f(x₂), then it must imply x₁ = x₂. This means no two distinct domain elements 'convert' to the same codomain element. Graphically, this implies that any horizontal line intersects the graph of the function at most once. For JEE, it's crucial to rigorously apply this algebraic test.

📝 Examples:

❌ Wrong:

Consider the function f(x) = x² for f: ℜ → ℜ. A common mistake is to incorrectly claim it is one-to-one. For instance, f(-2) = 4 and f(2) = 4. Here, two distinct inputs (-2 and 2) 'convert' to the same output (4). This violates the one-to-one condition.

✅ Correct:

Consider the function f(x) = x³ for f: ℜ → ℜ. To check if it's one-to-one, assume f(x₁) = f(x₂). This means x₁³ = x₂³, which algebraically implies x₁ = x₂. Since f(x₁) = f(x₂) only if x₁ = x₂, each distinct input 'converts' to a distinct output, confirming it is a one-to-one function.

💡 Prevention Tips:

Master the Definition: Thoroughly understand the algebraic definition: f(x₁) = f(x₂) &implies; x₁ = x₂.
Horizontal Line Test (Graphical): For functions whose graphs can be easily drawn, use the horizontal line test. For JEE, be careful with domain/codomain restrictions.
Algebraic Proofs: Always practice proving injectivity algebraically for complex functions. This is critical for JEE Main.
Distinguish Function Types: Clearly differentiate between one-to-one, many-to-one, onto, and into functions.

JEE_Main

Important Conceptual

❌ Incorrect Application of Algebraic/Graphical Tests for One-One Functions

Students frequently make errors by either blindly applying derivative tests (f'(x) > 0 or f'(x) < 0) without verifying the function's domain or points of non-differentiability, or by incorrectly using the algebraic definition (f(x₁) = f(x₂) ⇒ x₁ = x₂) without considering all possible cases or the specified domain. Another common pitfall is the misinterpretation of the Horizontal Line Test, especially for piecewise functions or functions with a restricted domain/codomain.

💭 Why This Happens:

This mistake stems from a lack of thorough conceptual understanding of a one-one function's definition and the conditions under which different tests (algebraic, graphical, derivative) are applicable. Students often focus solely on the 'rule' of the function (e.g., x², sin x) without paying crucial attention to the given domain and codomain, which fundamentally define the function's behavior regarding injectivity. Over-reliance on rote formulas without understanding their underlying principles leads to these errors.

✅ Correct Approach:

Always begin by clearly stating the definition: a function f: A → B is one-one if for every x₁, x₂ ∈ A, if f(x₁) = f(x₂), then x₁ = x₂.

For differentiable functions, check if f'(x) is strictly positive or strictly negative throughout the *entire* domain. Be cautious at points of non-differentiability or discontinuities.
For non-differentiable or piecewise functions, the algebraic definition or the Horizontal Line Test (visualizing the graph within the given domain) is more reliable.
Crucially, always consider the specified domain and codomain. A function that is not one-one over R might be one-one over a restricted domain.

📝 Examples:

❌ Wrong:

Consider the function f: R → R given by f(x) = x². A student might incorrectly conclude it's one-one by thinking f'(x) = 2x, which changes sign, but fail to explicitly show f(x₁) = f(x₂) leads to x₁ = x₂ only after considering signs. They might overlook that f(2) = 4 and f(-2) = 4, but 2 ≠ -2.

✅ Correct:

For f: R → R defined by f(x) = x²: If f(x₁) = f(x₂), then x₁² = x₂², which implies x₁ = ±x₂. Since x₁ can be equal to -x₂ (e.g., f(2)=4, f(-2)=4 where 2 ≠ -2), the function is NOT one-one on R.

However, if the function is defined as f: [0, ∞) → R given by f(x) = x², then if f(x₁) = f(x₂), x₁² = x₂². Since x₁, x₂ ∈ [0, ∞), both must be non-negative. Thus, x₁ = x₂ is the only possibility, making the function one-one on this restricted domain.

💡 Prevention Tips:

Always write down the function's domain and codomain before attempting to determine injectivity.
Master the definition and practice applying it algebraically for various function types.
For the derivative test, confirm that f'(x) maintains a consistent sign (strictly positive or strictly negative) *throughout the entire open interval of the domain*.
Sketch the graph of the function (or its relevant part) and apply the Horizontal Line Test to visually confirm.
Pay special attention to piecewise functions; analyze each piece and their conjunction points separately.

JEE_Advanced

Important Other

❌ Ignoring Domain/Codomain and Partial Monotonicity in Injectivity Checks

Students frequently make the mistake of assuming a function is one-to-one (injective) based on its behavior over a limited part of its domain, or by failing to fully consider the specified domain and codomain. They might observe that a function is strictly increasing or decreasing in certain intervals and incorrectly conclude injectivity over the entire domain. Similarly, an incomplete application of the horizontal line test (e.g., looking only at a restricted graph) leads to errors.

💭 Why This Happens:

Incomplete Understanding of Definition: Not fully grasping the formal definition that if f(x₁) = f(x₂) for two domain elements, then x₁ must necessarily equal x₂.
Over-reliance on Partial Graphs: Many students tend to visualize or graph functions only for positive values or familiar intervals, missing critical behavior over the full domain (e.g., negative values, periodic behavior).
Confusion with Monotonicity: Assuming that if a function is monotonic in parts, it implies overall injectivity. A function must be strictly monotonic (either strictly increasing or strictly decreasing) over its entire domain to be one-to-one.
Neglecting Domain Specification: The domain is crucial. A function's one-to-one property often depends heavily on how its domain is defined.

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one:

Algebraic Method: Take any two elements x₁, x₂ ∈ A. Assume f(x₁) = f(x₂). If this assumption necessarily implies x₁ = x₂, then the function is one-to-one. If you can find x₁ ≠ x₂ for which f(x₁) = f(x₂), it's not one-to-one. (Crucial for both CBSE and JEE)
Calculus Method (for differentiable functions): If the derivative f'(x) is either strictly positive (f'(x) > 0) for all x ∈ A or strictly negative (f'(x) < 0) for all x ∈ A, then the function is strictly monotonic and thus one-to-one. If f'(x) changes sign within the domain, the function is not one-to-one. (Highly relevant for JEE Advanced)
Graphical Method (Horizontal Line Test): Draw horizontal lines across the entire graph of the function. If any horizontal line intersects the graph at more than one point, the function is not one-to-one. (Apply over the full domain)

Always explicitly consider the given domain.

📝 Examples:

❌ Wrong:

Function: f(x) = x², defined from R → R.

Student's Incorrect Thought Process: "I'll look at the graph of y = x². For x > 0, it's increasing, so it must be one-to-one." Or, "If I just look at the right half (x ≥ 0), it passes the horizontal line test."

Incorrect Conclusion: "f(x) = x² is a one-to-one function from R to R."

✅ Correct:

Function: f(x) = x², defined from R → R.

Correct Analysis:

Algebraic: Let f(x₁) = f(x₂). Then x₁² = x₂². This implies x₁ = ±x₂. Since x₁ can be equal to x₂ or -x₂ (and these are different if x₁ ≠ 0), the condition x₁ = x₂ is not necessarily met. For instance, f(2) = 4 and f(-2) = 4, but 2 ≠ -2. Therefore, it is NOT one-to-one.
Graphical (Horizontal Line Test): A horizontal line like y = 4 intersects the parabola y = x² at two distinct points, x = 2 and x = -2. This clearly violates the horizontal line test.

Contrast: If the function were f(x) = x² defined from [0, ∞) → R, then for f(x₁) = f(x₂), we'd have x₁² = x₂². Since both x₁ and x₂ must be ≥ 0, it would necessarily imply x₁ = x₂. In this restricted domain, the function IS one-to-one.

💡 Prevention Tips:

Always Start with the Definition: For JEE Advanced, always attempt the algebraic proof (f(x₁) = f(x₂) ⇒ x₁ = x₂) first. It is the most rigorous method.
Pay Close Attention to Domain: The domain of the function is paramount. A function's injectivity can change dramatically with a change in its domain.
Use Calculus Wisely: If f'(x) changes sign (i.e., crosses the x-axis) within the domain, the function is generally not one-to-one. Look for points where f'(x) = 0, as these often indicate a change in monotonicity.
Avoid Partial Graphing: If using the horizontal line test, ensure you visualize or sketch the graph over its entire specified domain, not just a convenient portion.
Distinguish Functions: Remember that one-to-one (injectivity) is different from onto (surjectivity) and bijection. Don't mix up their definitions.

JEE_Advanced

Important Sign Error

❌ Sign Error in Monotonicity Analysis for One-One Functions

Students frequently make critical sign errors when analyzing the derivative, f'(x), or the difference f(x₁) - f(x₂) to determine if a function is strictly increasing or strictly decreasing. A function is one-one if it is strictly monotonic (either strictly increasing or strictly decreasing) over its domain, or intervals within its domain. Incorrectly determining the sign of f'(x) directly leads to a wrong conclusion about injectivity.

💭 Why This Happens:

This error primarily stems from

Carelessness in Algebraic Manipulation: Especially when factoring, expanding, or solving inequalities.
Incorrect Inequality Handling: Forgetting to reverse the inequality sign when multiplying or dividing by a negative number.
Improper Sign Analysis: Misinterpreting the sign of terms in a product or quotient, particularly around critical points or roots.
Ignoring Domain Constraints: Not considering the given domain while determining the sign of an expression.

✅ Correct Approach:

For differentiable functions, first calculate f'(x). Then, perform a meticulous sign analysis of f'(x) over the function's domain using methods like the number line (Wavy Curve Method). If f'(x) is consistently positive or consistently negative (except for isolated points where f'(x)=0) over the entire domain, the function is one-one. For non-differentiable functions, or as an alternative, use the definition: assume f(x₁) = f(x₂) and rigorously prove x₁ = x₂. This proof often involves algebraic steps where sign errors can occur.

📝 Examples:

❌ Wrong:

Consider f(x) = x² - 4x + 3. A student finds f'(x) = 2x - 4. Then, they might mistakenly conclude that since 'x' can be positive, 2x - 4 is always positive, or they might hastily assume a constant sign without proper analysis. This leads to the incorrect conclusion that f(x) is strictly increasing and thus one-one for all real x.

✅ Correct:

For f(x) = x² - 4x + 3, f'(x) = 2x - 4.
To analyze the sign:

If 2x - 4 < 0, then 2x < 4, which means x < 2. So, f'(x) < 0 for x ∈ (-∞, 2).
If 2x - 4 > 0, then 2x > 4, which means x > 2. So, f'(x) > 0 for x ∈ (2, ∞).

Since f'(x) changes sign (from negative to positive) at x = 2, f(x) is decreasing for x < 2 and increasing for x > 2. Therefore, f(x) is not one-one over its entire domain R. This careful sign analysis avoids the error.

💡 Prevention Tips:

Systematic Sign Analysis: Always use the number line method for f'(x) to identify intervals where it's positive or negative.
Double-Check Inequalities: Be extremely careful when manipulating inequalities, especially reversing signs for multiplication/division by negative numbers.
Factor Carefully: Ensure all factors are correctly identified and their individual signs are considered.
Practice, Practice, Practice: Solve numerous problems involving sign analysis to build proficiency and avoid common pitfalls.
Review Basics: Refresh your understanding of basic algebraic operations and inequality rules.

JEE_Advanced

Important Unit Conversion

❌ Misapplying Unit Conversion to Abstract Mathematical Functions

Students sometimes incorrectly attempt to apply concepts of unit conversion or unit consistency when dealing with abstract mathematical functions, such as determining if a function is one-one. For purely mathematical functions (e.g., f: R → R), where the domain and codomain are sets of numbers without associated physical units, unit conversion is irrelevant.

💭 Why This Happens:

This mistake stems from an overgeneralization of problem-solving strategies common in Physics or Chemistry, where units are crucial. Students may fail to differentiate between abstract mathematical concepts and their practical applications in physical sciences, leading them to search for units even when none exist or are relevant.

✅ Correct Approach:

For abstract functions, focus exclusively on the mathematical definition of a one-one function: If f(x₁) = f(x₂), then x₁ = x₂. No consideration of units or unit conversion is necessary. The analysis should be purely algebraic or graphical, based on the properties of numbers and the function itself.

📝 Examples:

❌ Wrong:

A student checking if f(x) = x² is one-one for f: R → R might mistakenly ponder whether 'x' represents a length and 'x²' represents an area, attempting to reconcile these 'units' rather than directly applying the injectivity definition.

✅ Correct:

To check if f(x) = eˣ is one-one for f: R → R:
1. Assume f(x₁) = f(x₂).
2. This means eˣ₁ = eˣ₂.
3. Taking the natural logarithm on both sides, ln(eˣ₁) = ln(eˣ₂), which simplifies to x₁ = x₂.
4. Since x₁ = x₂ is derived from f(x₁) = f(x₂), the function is one-one.
No unit considerations are involved.

💡 Prevention Tips:

Differentiate Contexts: Clearly distinguish between problems in pure mathematics (where unit conversion is typically absent) and applied problems in physics or chemistry (where units are fundamental).
Focus on Definitions: For function properties like injectivity, adhere strictly to the mathematical definitions. Understand that 'one-one' is about the mapping of elements, not their physical dimensions.
JEE Advanced Tip: In JEE Advanced (Mathematics), questions about functions are generally abstract unless explicitly stated otherwise. Do not introduce extraneous concepts like units unless the problem context clearly demands it.

JEE_Advanced

Important Formula

❌ Ignoring Domain Restrictions and Algebraic Rigor in One-to-One Function Test

Students often apply the algebraic test for one-to-one functions, i.e., assuming if f(x₁) = f(x₂) then x₁ = x₂, but fail to rigorously solve for all possible solutions of x₁ and x₂ or overlook the specified domain of the function. This leads to incorrect conclusions about injectivity, especially with functions involving even powers, absolute values, or periodic functions.

💭 Why This Happens:

This mistake stems from a superficial understanding of the one-to-one definition. Students tend to hastily conclude x₁ = x₂ without considering other potential values that satisfy f(x₁) = f(x₂), or they forget that the injectivity property is heavily dependent on the function's domain. A common trap is assuming algebraic simplification (e.g., cancelling terms) inherently means x₁ = x₂ without checking if other roots exist or if the cancellation is valid under the given conditions.

✅ Correct Approach:

To correctly test if a function f: A → B is one-to-one (injective):

Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A (the domain).
Algebraically manipulate this equation to rigorously prove that the only possible conclusion is x₁ = x₂. If you find even one instance where f(x₁) = f(x₂) but x₁ ≠ x₂, the function is not one-to-one.
Always consider the domain (A) of the function while solving. Solutions outside the domain must be discarded.
For JEE Advanced, be particularly careful with functions like f(x) = x², f(x) = |x|, f(x) = sin(x), or composite functions, where multiple inputs can yield the same output if the domain is not restricted appropriately.

📝 Examples:

❌ Wrong:

Function: f(x) = x² on domain R (all real numbers)

Student's Incorrect Approach:
Assume f(x₁) = f(x₂)
x₁² = x₂²
Taking square root on both sides:
x₁ = x₂
Conclusion: f(x) = x² is one-to-one on R.

Error: Incorrectly assuming that x₁² = x₂² only implies x₁ = x₂. It also implies x₁ = -x₂. The domain R allows for both positive and negative values.

✅ Correct:

Function: f(x) = x² on domain R

Correct Approach:
1. Assume f(x₁) = f(x₂) for x₁, x₂ ∈ R.
2. x₁² = x₂²
3. x₁² - x₂² = 0
4. (x₁ - x₂)(x₁ + x₂) = 0
5. This implies either x₁ - x₂ = 0 (i.e., x₁ = x₂) OR x₁ + x₂ = 0 (i.e., x₁ = -x₂).

Since x₁, x₂ are from R, we can find distinct values such that x₁ = -x₂ (e.g., x₁ = 2, x₂ = -2). In this case, f(2) = 2² = 4 and f(-2) = (-2)² = 4. Here, f(x₁) = f(x₂) but x₁ ≠ x₂.

Conclusion: Since there exist x₁ ≠ x₂ such that f(x₁) = f(x₂), the function f(x) = x² is NOT one-to-one on R.

💡 Prevention Tips:

Master the Definition: Understand that one-to-one means 'distinct inputs yield distinct outputs'.
Rigorous Algebra: When solving f(x₁) = f(x₂), factorize and find all possible relationships between x₁ and x₂, not just the obvious one.
Domain Awareness: Always keep the function's domain in mind. Solutions for x₁ and x₂ must belong to the domain.
Test with Counterexamples: If you suspect a function is NOT one-to-one, try to find two different inputs that give the same output. This is a quick way to disprove injectivity.
Graphical Analysis (JEE Specific): Use the Horizontal Line Test. If any horizontal line intersects the graph at more than one point, the function is not one-to-one. This is a visual aid but algebraic proof is often required for rigor.

JEE_Advanced

Important Calculation

❌ Algebraic Errors in Deductions for One-One Functions

Students frequently make algebraic calculation mistakes when attempting to prove a function is one-one. This typically involves incorrectly solving the equation f(x₁) = f(x₂) to derive x₁ = x₂, or misinterpreting the implications of algebraic steps. Such errors can lead to an erroneous conclusion about the injectivity (one-one nature) of the function.

💭 Why This Happens:

Premature Division: Dividing by a term (e.g., x₁ - x₂) without first factoring it out and considering the case where it might be zero.
Ignoring Multiple Solutions: Forgetting that operations like squaring or taking even roots can introduce extraneous solutions or lose valid ones.
Incomplete Factorization: Not factoring expressions fully, which hides alternative possibilities for x₁ and x₂.
Domain Neglect: Failing to filter algebraic solutions based on the function's defined domain, which is crucial for determining injectivity.

✅ Correct Approach:

When proving a function f: A → B is one-one, assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A. Then, meticulously perform algebraic manipulations to show that this assumption necessarily implies x₁ = x₂. Always factorize expressions completely and consider all possible solutions arising from the algebra, evaluating each within the context of the given domain. If f(x₁) = f(x₂) can lead to x₁ ≠ x₂ for some x₁, x₂ ∈ A, then the function is not one-one.

📝 Examples:

❌ Wrong:

Consider the function f(x) = x + 1/x on the domain (0, ∞).
Student's flawed reasoning:
Assume f(x₁) = f(x₂)
x₁ + 1/x₁ = x₂ + 1/x₂
x₁ - x₂ = 1/x₂ - 1/x₁
x₁ - x₂ = (x₁ - x₂) / (x₁x₂)
Then, the student might incorrectly divide both sides by (x₁ - x₂), concluding 1 = 1/(x₁x₂) which implies x₁x₂ = 1. From this, they might incorrectly conclude that since x₁x₂ = 1 doesn't always mean x₁ = x₂, the function is not one-one.
The mistake is in prematurely dividing by (x₁ - x₂) without accounting for the case where x₁ - x₂ = 0.

✅ Correct:

For f(x) = x + 1/x on (0, ∞):
Assume f(x₁) = f(x₂) for x₁, x₂ ∈ (0, ∞).
x₁ + 1/x₁ = x₂ + 1/x₂
x₁ - x₂ = 1/x₂ - 1/x₁
x₁ - x₂ = (x₁ - x₂) / (x₁x₂)
Now, move all terms to one side and factorize:
(x₁ - x₂) - (x₁ - x₂) / (x₁x₂) = 0
(x₁ - x₂) [1 - 1 / (x₁x₂)] = 0
This equation holds if either x₁ - x₂ = 0 OR 1 - 1 / (x₁x₂) = 0.
1. If x₁ - x₂ = 0, then x₁ = x₂.
2. If 1 - 1 / (x₁x₂) = 0, then 1 = 1 / (x₁x₂), which implies x₁x₂ = 1.
Since f(x₁) = f(x₂) implies x₁ = x₂ OR x₁x₂ = 1, it is not guaranteed that x₁ = x₂. For instance, if x₁ = 2 and x₂ = 1/2, then x₁x₂ = 1. Here, f(2) = 2 + 1/2 = 2.5 and f(1/2) = 1/2 + 2 = 2.5. So, f(2) = f(1/2) but 2 ≠ 1/2.
Therefore, the function f(x) = x + 1/x is not one-one on (0, ∞).

💡 Prevention Tips:

Factorize Completely: Always aim to factor out common terms to reveal all possible conditions for the equation to hold.
Avoid Premature Division: Do not divide by an expression involving variables unless you are certain it cannot be zero, or you explicitly consider the case where it is zero.
Scrutinize Every Step: After each algebraic step, pause and check for potential loss of information or introduction of extraneous solutions.
Verify with Domain: Ensure that any derived conditions for x₁ and x₂ are consistent with the function's defined domain.
Test for Counterexamples: If the algebraic solution gives conditions other than x₁ = x₂, try to find concrete values within the domain that satisfy those conditions to prove the function is not one-one.

JEE_Advanced

Important Formula

❌ Incorrect Application of the One-One Test (f(x1) = f(x2) ⇒ x1 = x2)

Students frequently fail to rigorously prove that f(x₁) = f(x₂) necessarily implies x₁ = x₂. This is particularly common when functions involve non-linear terms (like squares, absolute values), trigonometric expressions, or when the domain is not explicitly considered during algebraic manipulation. They might prematurely conclude injectivity without exhausting all possibilities or properly accounting for the domain restrictions.

💭 Why This Happens:

Incomplete Algebraic Manipulation: Not thoroughly solving f(x₁) = f(x₂) to find all possible relationships between x₁ and x₂.
Ignoring Domain/Codomain: Overlooking how the specified domain restricts possible values for x₁ and x₂.
Graphical Misinterpretation: Relying solely on a rough sketch of the graph or misapplying the horizontal line test without formal algebraic backing.
Assuming Uniqueness: Believing that if a function 'looks' increasing/decreasing, it must be one-one, without formal proof.

✅ Correct Approach:

To prove a function f: A → B is one-one (injective):

Assume any two elements x₁, x₂ ∈ A such that f(x₁) = f(x₂).
Perform rigorous algebraic steps to show that this assumption must lead to x₁ = x₂.
If the algebraic process yields x₁ = x₂ OR x₁ = 'some other value' (e.g., x₁ = -x₂), and if 'some other value' is distinct from x₁ and within the domain, then the function is NOT one-one.
For JEE, always provide a clear, step-by-step algebraic proof or a precise counterexample if the function is not one-one.

📝 Examples:

❌ Wrong:

Consider: f(x) = x² + 1 defined on R.
Wrong thought process: A student might only consider positive x values, or think 'if x₁²+1 = x₂²+1, then x₁² = x₂² implies x₁=x₂'. This ignores the negative root possibility.

✅ Correct:

Function: f(x) = x² + 1, Domain = R, Codomain = R

Correct Analysis: Assume f(x₁) = f(x₂) for x₁, x₂ ∈ R.

x₁² + 1 = x₂² + 1

x₁² = x₂²

x₁² - x₂² = 0

(x₁ - x₂)(x₁ + x₂) = 0

This implies either (x₁ - x₂) = 0 ⇒ x₁ = x₂ OR (x₁ + x₂) = 0 ⇒ x₁ = -x₂.

Since x₁ = x₂ is not the ONLY possibility (e.g., if x₁ = 2, then x₂ can be -2, and f(2) = 5, f(-2) = 5, but 2 ≠ -2), the function is NOT one-one on R.

💡 Prevention Tips:

Master the Definition: Continuously remind yourself: f is one-one iff f(x₁) = f(x₂) uniquely implies x₁ = x₂.
Algebraic Vigilance: When solving f(x₁) = f(x₂), handle square roots, absolute values, and trigonometric inverses with extreme care, considering all possible signs and branches.
Counterexample Search: If the function involves even powers, absolute values, or is periodic, immediately try to find two distinct inputs that map to the same output. This is a quick way to disprove injectivity.
Domain-Specific Analysis: Always verify if the 'other' solutions (like x₁ = -x₂) are valid within the given domain. A function not one-one on R might be one-one on a restricted domain (e.g., x² is one-one on [0, ∞)).

JEE_Main

Important Calculation

❌ Ignoring All Algebraic Possibilities When Proving Injectivity

When testing for injectivity (one-to-one property) by setting f(x₁) = f(x₂) and aiming to prove x₁ = x₂, students frequently make calculation errors. A critical mistake is to overlook or incorrectly simplify equations involving non-linear terms like squares, absolute values, or terms that can be factored. For instance, incorrectly assuming that A² = B² strictly implies A = B, or cancelling a variable factor without considering the possibility of it being zero. This oversight leads to a false conclusion about the function's injectivity.

💭 Why This Happens:

Incomplete Algebraic Knowledge: Lack of thorough understanding of solving equations with non-linear terms (e.g., properties of square roots, absolute values, and factorization).
Over-simplification: Rushing through algebraic steps and making assumptions, such as taking only the principal root or cancelling terms without checking for zero.
Domain Neglect: Not considering how the function's domain restricts or allows for solutions where x₁ ≠ x₂.

✅ Correct Approach:

To rigorously determine injectivity using the algebraic method, follow these steps:

Set up the equation f(x₁) = f(x₂) based on the function definition.
Carefully manipulate the equation, ensuring all algebraic properties are correctly applied. For equations like A² = B², always consider both A = B and A = -B.
If you factor out a term (e.g., (x₁ - x₂)), analyze both scenarios: where the factor is zero (leading to x₁ = x₂) and where it is non-zero (leading to other potential relationships).
If any solution other than x₁ = x₂ exists within the function's domain, then the function is not one-to-one. If x₁ = x₂ is the only possible solution, then it is one-to-one.
JEE Tip: Always be meticulous with signs and consider all cases arising from squares, roots, and absolute values.

📝 Examples:

❌ Wrong:

Consider f(x) = x² with domain R (real numbers).
Wrong Calculation:

f(x₁) = f(x₂)

x₁² = x₂²

(Incorrectly assumes only positive root)

x₁ = x₂

Wrong Conclusion: Function f(x) = x² is one-to-one.

✅ Correct:

Consider f(x) = x² with domain R (real numbers).
Correct Calculation:

f(x₁) = f(x₂)

x₁² = x₂²

x₁² - x₂² = 0

(x₁ - x₂)(x₁ + x₂) = 0

This implies two possibilities:

x₁ - x₂ = 0 → x₁ = x₂
x₁ + x₂ = 0 → x₁ = -x₂

Since x₁ = -x₂ allows for x₁ ≠ x₂ (e.g., f(2) = 4 and f(-2) = 4, so f(2) = f(-2) but 2 ≠ -2), the function f(x) = x² is NOT one-to-one on R.

💡 Prevention Tips:

Thorough Algebraic Review: Regularly practice solving various types of equations, paying close attention to operations involving squares, square roots, absolute values, and fractions.
Always Factor Completely: When possible, factor expressions to clearly see all potential solutions or relationships between x₁ and x₂.
Test Edge Cases: After deriving potential solutions, mentally or explicitly test simple values from the domain to ensure your conclusions hold.
Understand Definitions: A solid grasp of the definition of one-to-one functions (f(x₁) = f(x₂) ⇒ x₁ = x₂) will guide your algebraic manipulation.

JEE_Main

Important Conceptual

❌ Confusing One-to-One with Strict Monotonicity

Students frequently assume that a function must be strictly increasing or strictly decreasing throughout its domain to be one-to-one (injective). They fail to recognize functions that are one-to-one but not strictly monotonic, or functions where injectivity is valid only over a specific interval, not necessarily implying strict monotonicity everywhere.

💭 Why This Happens:

This conceptual error often stems from an over-reliance on graphical intuition for simple functions, where monotonicity strongly correlates with injectivity. Students might not deeply understand the algebraic definition of a one-to-one function (f(x₁) = f(x₂) ⇒ x₁ = x₂) and its implications, especially for functions whose derivatives are zero at isolated points or for piecewise definitions. For JEE Main, a rigorous understanding is crucial.

✅ Correct Approach:

Always revert to the fundamental definition for injectivity: A function f: A → B is one-to-one if for every x₁, x₂ ∈ A, f(x₁) = f(x₂) implies x₁ = x₂.
For differentiable functions, calculate f'(x). If f'(x) > 0 for all x (strictly increasing) or f'(x) < 0 for all x (strictly decreasing) over the domain, the function is one-to-one. However, if f'(x) ≥ 0 or f'(x) ≤ 0 and f'(x) = 0 only at isolated points, the function can still be one-to-one.

📝 Examples:

❌ Wrong:

Consider f(x) = x^3 - 3x^2 + 3x + 1. A student might compute f'(x) = 3x^2 - 6x + 3 = 3(x-1)^2. Seeing that f'(x) = 0 at x = 1, they might incorrectly conclude that the function is not strictly monotonic, and therefore not one-to-one. This reasoning is flawed as f'(x) ≥ 0 for all x, and it is zero only at a single point, not over an interval.

✅ Correct:

For f(x) = x^3 - 3x^2 + 3x + 1:
1. Algebraic method: Rewrite f(x) = (x^3 - 3x^2 + 3x - 1) + 2 = (x-1)^3 + 2. If f(x₁) = f(x₂), then (x₁-1)^3 + 2 = (x₂-1)^3 + 2, which implies (x₁-1)^3 = (x₂-1)^3. Taking the cube root, x₁-1 = x₂-1, so x₁ = x₂. Hence, it is one-to-one.
2. Derivative method: As calculated, f'(x) = 3(x-1)^2. Since 3(x-1)^2 ≥ 0 for all real x and f'(x) = 0 only at the isolated point x = 1, the function is non-decreasing and strictly increasing overall, making it one-to-one.

💡 Prevention Tips:

Master the Definition: Always start with the definition f(x₁) = f(x₂) ⇒ x₁ = x₂.
Apply the Horizontal Line Test: Graphically, if any horizontal line intersects the graph at most once, the function is one-to-one.
Differentiate Carefully: Understand that f'(x) ≥ 0 (or ≤ 0) with f'(x) = 0 at isolated points, still implies injectivity.
Practice Diverse Examples: Work with piecewise functions and functions that are not simple polynomials to broaden your understanding beyond strictly monotonic cases.

JEE_Main

Important Conceptual

❌ Confusing One-to-One Condition (Injectivity) and Ignoring Domain/Codomain

Students often incorrectly assume a function is one-to-one (injective) by just looking at a few distinct input-output pairs or by confusing it with the definition of a function itself. A common error is failing to rigorously apply the condition f(x₁) = f(x₂) ⇒ x₁ = x₂ for all elements in the domain, especially when the function involves squares, absolute values, or trigonometric functions.

💭 Why This Happens:

This mistake stems from a lack of thorough conceptual understanding of injectivity and not fully grasping the implications of the domain and codomain. Students might:

Misinterpret the definition: Confuse 'every element in the domain has a unique image' (definition of a function) with 'every distinct element in the domain has a distinct image' (definition of one-to-one).
Insufficient algebraic proof: Fail to start with f(x₁) = f(x₂) and logically deduce x₁ = x₂.
Ignoring domain restrictions: Assume a function's injectivity based on its general form without considering its specified domain (e.g., x² is not one-to-one on R but is on [0, ∞)).
Over-reliance on intuition: Sometimes students guess based on simple graphs without algebraic verification, which is crucial for CBSE exams.

✅ Correct Approach:

To prove a function f: A → B is one-to-one for CBSE (and JEE), always follow the algebraic method:

Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A.
Perform algebraic manipulations to show that this assumption necessarily leads to x₁ = x₂.
If you find any instance where f(x₁) = f(x₂) but x₁ ≠ x₂, then the function is not one-to-one. Remember to always check for counterexamples.

📝 Examples:

❌ Wrong:

Consider the function f: R → R given by f(x) = x².
A student might argue: "If x = 2, f(x) = 4. If x = 3, f(x) = 9. Since different inputs give different outputs here, it's one-to-one."
This reasoning is incorrect because it uses only specific examples and fails to consider all possible inputs from the domain R. It does not prove the general condition.

✅ Correct:

Consider the function f: R → R given by f(x) = x².
To check if it's one-to-one, let x₁, x₂ ∈ R such that f(x₁) = f(x₂).
⇒ x₁² = x₂²
⇒ x₁² - x₂² = 0
⇒ (x₁ - x₂)(x₁ + x₂) = 0
⇒ x₁ - x₂ = 0 OR x₁ + x₂ = 0
⇒ x₁ = x₂ OR x₁ = -x₂
Since x₁ can be equal to -x₂ (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the condition x₁ = x₂ is not uniquely met. Therefore, f(x) = x² is not one-to-one on R.

If the domain were f: [0, ∞) → R, then x₁, x₂ ∈ [0, ∞). In this case, x₁ = -x₂ is only possible if x₁ = x₂ = 0. For any non-zero x₁, x₂ in [0, ∞), if x₁ = -x₂, then one must be negative, which contradicts the domain. So, we are left with only x₁ = x₂. Thus, f(x) = x² is one-to-one on [0, ∞). This highlights the importance of the domain.

💡 Prevention Tips:

Always use the definition: For injectivity, always start with f(x₁) = f(x₂) and aim to prove x₁ = x₂.
Consider the domain carefully: The domain significantly impacts whether a function is one-to-one. Always read the function definition, including its domain and codomain.
Look for counterexamples: If you suspect a function is not one-to-one, try to find two distinct inputs (x₁ ≠ x₂) that produce the same output (f(x₁) = f(x₂)).
Practice with varied functions: Work through examples involving quadratic, absolute value, and trigonometric functions on different domains to build robust understanding.
JEE Tip: While the horizontal line test can quickly tell if a function is one-to-one from its graph, for Board exams, a rigorous algebraic proof is mandatory.

CBSE_12th

Important Calculation

❌ Ignoring ± (plus-minus) sign during algebraic manipulation for injectivity

Students frequently overlook the possibility of both positive and negative values when performing algebraic steps like `x1^2 = x2^2` or `|x1| = |x2|` while attempting to prove if a function is one-to-one (injective). This often leads to an incorrect conclusion that `x1 = x2` directly, thereby falsely asserting injectivity.

💭 Why This Happens:

This error primarily stems from a lack of thorough algebraic precision and an incomplete understanding of implications. The common simplification error of assuming `a^2 = b^2` implies only `a = b` (instead of `a = ±b`) is prevalent. Exam pressure can also lead to hasty conclusions without considering all mathematical possibilities.

✅ Correct Approach:

When proving a function `f` is one-to-one, begin by assuming `f(x1) = f(x2)`. Carefully execute all algebraic steps. If you arrive at an equation like `x1^2 = x2^2` or `|x1| = |x2|`, you must deduce that `x1 = x2` OR `x1 = -x2`. If `x1 = -x2` leads to a distinct `x1` and `x2` within the domain (i.e., `x1 ≠ x2`), then the function is NOT one-to-one. For CBSE boards, showing this complete algebraic breakdown is crucial for securing full marks. For JEE, this critical insight helps in quickly identifying non-injective functions.

📝 Examples:

❌ Wrong:

Let `f(x) = x^2` for `f: R → R` (real numbers to real numbers).
Assume `f(x1) = f(x2)`. Then `x1^2 = x2^2`. Incorrectly deduce: Therefore, `x1 = x2`. Hence, `f` is one-to-one.
This ignores the case where `x1 = -x2`.

✅ Correct:

Let `f(x) = x^2` for `f: R → R`.
Assume `f(x1) = f(x2)` for `x1, x2 ∈ R`.
Array
This implies either `x1 - x2 = 0` (i.e., `x1 = x2`) OR `x1 + x2 = 0` (i.e., `x1 = -x2`).
Since we found a possibility where `x1 = -x2` (e.g., if `x1 = 2`, then `x2 = -2`), and for these distinct values (`2 ≠ -2`), `f(2) = 4` and `f(-2) = 4`, the function `f(x) = x^2` is NOT one-to-one on `R`.
If the domain was restricted to `[0, ∞)`, then `x1 = -x2` would only be possible if `x1 = x2 = 0`, making the function one-to-one in that specific domain.

💡 Prevention Tips:

Always be meticulous with algebraic manipulations. When an equation involves even powers or absolute values, actively consider all possible solutions.
Remember the fundamental algebraic rule: `a^2 = b^2` implies `a = b` OR `a = -b`. Similarly, `|a| = |b|` implies `a = b` OR `a = -b`.
Before concluding, quickly test with simple non-equal values (e.g., `x` and `-x`) from the function's domain to see if they yield the same function output.
Practice problems involving functions with even powers (e.g., `x^2, x^4`), absolute values (e.g., `|x|, |x-a|`), or piecewise definitions, as these are common traps for injectivity.

CBSE_12th

Important Formula

❌ Misapplication of the Algebraic Test for One-to-One Functions

Students often understand the definition of a one-to-one (injective) function: 'If f(x₁) = f(x₂), then x₁ = x₂.' However, a common mistake is making algebraic errors or stopping prematurely during the proof, failing to rigorously show that the equality of function values *necessarily* implies the equality of input values. This leads to incorrect conclusions about injectivity. They might overlook alternative solutions (e.g., negative roots, absolute value implications) where x₁ ≠ x₂ but f(x₁) = f(x₂).

💭 Why This Happens:

Algebraic Inaccuracies: Errors in factoring, taking square roots, or handling absolute values.
Lack of Rigor: Not fully exploring all mathematical consequences of f(x₁) = f(x₂), assuming the conclusion prematurely.
Ignoring Domain Restrictions: Forgetting that the test must hold for all elements within the specified domain, or that domain changes can affect injectivity.
Insufficient Practice: Limited experience with diverse function types and their algebraic properties.

✅ Correct Approach:

To correctly prove a function f: A → B is one-to-one:

Assume Equality: Begin by assuming f(x₁) = f(x₂) for arbitrary x₁, x₂ ∈ A (the domain).
Rigorous Manipulation: Perform careful algebraic steps to demonstrate that this assumption *must logically lead* to x₁ = x₂.
Identify Counter-Scenarios: If, at any point, your manipulations show that it's possible for x₁ ≠ x₂ while still having f(x₁) = f(x₂), then the function is not one-to-one.
Domain Consideration (JEE & CBSE): Always ensure your conclusion is consistent with the given domain. Injectivity is highly dependent on the domain.

📝 Examples:

❌ Wrong:

Wrong Attempt:

Let f(x) = x² with domain R (all real numbers).

Student's proof:

Assume f(x₁) = f(x₂)
x₁² = x₂²
x₁ = x₂ (Incorrect step: Assuming only the positive root)
Conclusion: f(x) = x² is one-to-one.

This is wrong because if x₁ = 2 and x₂ = -2, then f(2) = 4 and f(-2) = 4, so f(x₁) = f(x₂) but x₁ ≠ x₂.

✅ Correct:

Correct Approach:

Let f(x) = x² with domain R.

Assume f(x₁) = f(x₂) for x₁, x₂ ∈ R.
x₁² = x₂²
Rearrange: x₁² - x₂² = 0
Factor: (x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ - x₂ = 0 OR x₁ + x₂ = 0.
So, x₁ = x₂ OR x₁ = -x₂.

Since it's possible for x₁ = -x₂ (e.g., 2 and -2 are different inputs but f(2)=4 and f(-2)=4), the assumption f(x₁) = f(x₂) does not *necessarily* lead to x₁ = x₂. Therefore, f(x) = x² on domain R is not one-to-one.

(CBSE Tip: Clearly showing the 'x₁ = -x₂' case is crucial for full marks.)

💡 Prevention Tips:

Thorough Algebraic Practice: Master solving equations involving squares, absolute values, and other functions, always considering all possible solutions.
Seek Counterexamples: If you suspect a function is not one-to-one, actively look for two different inputs that produce the same output. One such example is enough to disprove injectivity.
Rigor Over Speed: Avoid rushing through algebraic steps. Each deduction must be logically sound and fully explored.
Verify Domain & Codomain: Always keep the function's domain in mind, as it dictates the validity of x₁ and x₂.
Understand Graphical Test: Use the Horizontal Line Test as a quick check (especially for JEE), but for CBSE proofs, the algebraic method is generally required.

CBSE_12th

Important Unit Conversion

❌ Misinterpreting Domain/Codomain 'Units' for Injectivity

Students often make a critical mistake by not fully understanding how the specified domain and codomain sets (e.g., Natural Numbers (N), Integers (Z), Real Numbers (R)) impact the one-to-one (injective) property of a function. They might incorrectly 'convert' or apply algebraic rules that are valid for one type of set to another, leading to an erroneous conclusion about injectivity. This is a conceptual 'mis-conversion' of the allowed input/output properties based on the set type.

💭 Why This Happens:

This error stems from a lack of careful consideration of the specific properties of the domain and codomain sets. Students often over-rely on generic algebraic simplifications (e.g., thinking x² = y² always implies x = y) without checking their validity within the specified domain. They might rush to conclude injectivity based on a limited set of positive examples or an incomplete algebraic argument, ignoring the implications of negative numbers or other types of elements allowed by the domain.

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one, always begin by clearly identifying the sets A (domain) and B (codomain). Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A, and then rigorously deduce if this necessarily implies x₁ = x₂, considering all constraints and properties of the elements within set A. If you can find a counterexample where x₁ ≠ x₂ but f(x₁) = f(x₂), then the function is not one-to-one.

📝 Examples:

❌ Wrong:

Consider the function f: Z → Z defined by f(x) = x².

Wrong thought process: A student might think, 'If x₁² = x₂², then taking the square root gives x₁ = x₂. So, it must be one-to-one.'

✅ Correct:

Consider the function f: Z → Z defined by f(x) = x².

Correct approach:

Assume f(x₁) = f(x₂) for some x₁, x₂ ∈ Z.
This means x₁² = x₂².
Rearranging gives x₁² - x₂² = 0, which factors as (x₁ - x₂)(x₁ + x₂) = 0.
This implies either x₁ - x₂ = 0 (so x₁ = x₂) OR x₁ + x₂ = 0 (so x₁ = -x₂).
Since x₁, x₂ are integers, we can choose x₁ = 1 and x₂ = -1.
Here, x₁ ≠ x₂ (because 1 ≠ -1), but f(1) = 1² = 1 and f(-1) = (-1)² = 1.
Since f(1) = f(-1) but 1 ≠ -1, the function is NOT one-to-one.

💡 Prevention Tips:

Always read the domain and codomain carefully: These are fundamental 'units' that define the function's behavior and determine which values are valid inputs and outputs.
Test with diverse values: If the domain allows, test negative numbers, zero, and fractions (for real numbers) in addition to positive integers.
Perform complete algebraic analysis: When assuming f(x₁) = f(x₂), ensure all possible solutions for x₁ and x₂ are considered within the specified domain, not just the most obvious one.
CBSE vs. JEE: This careful analysis is crucial for both CBSE board exams (for descriptive proofs) and JEE (for correctly identifying function properties in MCQs).

CBSE_12th

Important Sign Error

❌ Sign Errors in Proving One-One Functions (CBSE 12th)

Students frequently make sign errors when attempting to prove a function is one-one using the condition f(x₁)=f(x₂) ⇒ x₁=x₂. This often occurs during algebraic manipulation, especially when dealing with even powers, square roots, or absolute values. They might prematurely conclude x₁=x₂, overlooking other valid solutions like x₁=-x₂, which would make the function not one-one if these values are within the domain.

💭 Why This Happens:

This mistake stems from a combination of factors:

Algebraic Oversight: Forgetting that equations like x₁² = x₂² lead to (x₁-x₂)(x₁+x₂) = 0, meaning x₁=x₂ OR x₁=-x₂.
Lack of Domain Consideration: Not checking if the 'extra' solutions (e.g., x₁=-x₂) are permissible within the given domain of the function. If they are, and x₁ ≠ x₂, then the function is not one-one.
Rushing Steps: Hurrying through algebraic derivations without carefully considering all potential roots or implications of signs.

✅ Correct Approach:

To correctly prove a function is one-one:

Assume f(x₁)=f(x₂) for any x₁, x₂ in the domain.
Carefully perform algebraic manipulation to simplify the equation.
Always consider all possible solutions for x₁ in terms of x₂, especially when squaring/taking square roots or dealing with absolute values.
If you arrive at solutions like x₁=x₂ AND x₁=-x₂ (where x₁ ≠ x₂), then for the function to be one-one, you must show that the case x₁=-x₂ is impossible or irrelevant based on the function's domain. If x₁=-x₂ is possible for distinct x₁, x₂, the function is not one-one.
The ultimate goal is to rigorously show that only x₁=x₂ is possible within the function's domain.

📝 Examples:

❌ Wrong:

Consider f: R → R given by f(x) = x².
Let f(x₁) = f(x₂)
x₁² = x₂²
Incorrect Deduction: x₁ = x₂
Based on this, a student might wrongly conclude f(x) = x² is one-one.

✅ Correct:

Consider f: R → R given by f(x) = x².
Let f(x₁) = f(x₂) for x₁, x₂ ∈ R.
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ = x₂ OR x₁ = -x₂.
Since for real numbers, it is possible to have x₁ = -x₂ where x₁ ≠ x₂ (e.g., f(2)=4 and f(-2)=4, but 2 ≠ -2), the function is NOT one-one. The sign error here is ignoring the x₁ = -x₂ possibility.

💡 Prevention Tips:

Factorize Completely: Always factorize expressions like x₁² - x₂² = 0 to reveal all roots.
Absolute Value Awareness: Remember that √(a²) = |a|. When solving |x₁| = |x₂|, it implies x₁ = x₂ OR x₁ = -x₂.
Domain Check: After finding all possible relations between x₁ and x₂, verify if non-equal possibilities are allowed by the function's domain.
Practice Variety: Solve problems involving various function types (polynomials with even/odd powers, rational functions, absolute value functions) to build confidence in handling signs.

CBSE_12th

Important Approximation

❌ Confusing One-One with Onto or Incorrect Application of the One-One Definition

Students frequently misunderstand the precise definition of a one-one function (injectivity), leading to errors in proving or disproving it. A common error is assuming 'distinct inputs give distinct outputs' is automatically satisfied, or confusing it with the definition of an onto function (surjectivity). This often happens when the domain or codomain is not carefully considered.

💭 Why This Happens:

Lack of formal definition understanding: Students often don't rigorously apply the definition: 'If f(x₁) = f(x₂), then x₁ = x₂'.
Ignoring domain/codomain: The nature of the domain (e.g., N, Z, R) significantly impacts whether a function is one-one, but this is frequently overlooked.
Graphical misinterpretation: Relying solely on the horizontal line test without understanding its algebraic basis or limits for specific domains.
Careless algebraic manipulation: Making errors while solving f(x₁) = f(x₂) for x₁ and x₂ (e.g., forgetting ± roots).

✅ Correct Approach:

To prove a function f: A → B is one-one, always follow these steps:

Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A.
Perform algebraic manipulation to show that this assumption necessarily implies x₁ = x₂.

To disprove a function f: A → B is one-one, find at least one pair of distinct elements x₁, x₂ ∈ A (i.e., x₁ ≠ x₂) such that f(x₁) = f(x₂). This is called finding a counterexample.

📝 Examples:

❌ Wrong:

Function: f: R → R given by f(x) = x²

Incorrect Proof (CBSE Common Mistake):

Assume f(x₁) = f(x₂)

x₁² = x₂²

Taking square root on both sides, x₁ = x₂. Thus, f is one-one.

Reason for error: Forgetting that x² = a² implies x = ±a. The step 'x₁ = x₂' is not the only possibility from x₁² = x₂².

✅ Correct:

Function: f: R → R given by f(x) = x²

Correct Approach (Disproving One-One):

Let x₁ = 2 and x₂ = -2. Clearly, x₁ ≠ x₂.

Now, f(x₁) = f(2) = 2² = 4.

And, f(x₂) = f(-2) = (-2)² = 4.

Since f(x₁) = f(x₂) but x₁ ≠ x₂, the function f(x) = x² is not one-one.

💡 Prevention Tips:

Master the Definition: Commit the definition of one-one (injectivity) to memory and practice applying it rigorously.
Check Domain/Codomain: Always explicitly write down and consider the given domain and codomain. They are crucial for determining injectivity.
Beware of Square Roots/Absolute Values: When solving f(x₁) = f(x₂), remember all possible roots (e.g., x² = y² implies x = ±y) or cases for absolute values.
Practice Counterexamples: For disproving, actively try to find two different inputs that yield the same output.

CBSE_12th

Important Other

❌ Incomplete One-to-One Proofs & Domain Neglect

Students often stop their algebraic proof for f(x1) = f(x2) => x1 = x2 prematurely, ignoring alternative solutions (e.g., x1 = -x2 for x^2). They also fail to check if these alternatives are valid within the given function domain, which is crucial for determining injectivity.

💭 Why This Happens:

Incomplete understanding of injectivity's definition.
Insufficient algebraic practice with equations yielding multiple solutions (e.g., quadratics).
Overlooking critical domain constraints specified in the question.

✅ Correct Approach:

To prove a function f: A -> B is one-to-one (injective):

Assume f(x1) = f(x2) for any arbitrary x1, x2 ∈ A.
Rigorously solve the equation for x1 in terms of x2. This process may yield x1 = x2 OR an alternative solution, say x1 = k.
Crucially, confirm if this alternative k can exist in the domain A and be different from x2. If such a k ≠ x2 exists (where f(k) = f(x2)), then the function is not one-to-one.

📝 Examples:

❌ Wrong:

Function: f: R -> R, f(x) = x^2
Assume f(x1) = f(x2). This leads to x1^2 = x2^2. Students might incorrectly conclude only x1 = x2.
Error: Ignoring the possibility x1 = -x2. For example, f(2)=4 and f(-2)=4, but 2 ≠ -2. So, it's not one-to-one.

✅ Correct:

Function: f: [0, ∞) -> R, f(x) = x^2
Assume f(x1) = f(x2) for x1, x2 ∈ [0, ∞).
x1^2 = x2^2 implies x1 = x2 or x1 = -x2.
Since both x1, x2 must be non-negative (from the domain [0, ∞)), the condition x1 = -x2 only holds if x1 = 0 and x2 = 0. In all other cases, for `x1, x2 ≥ 0`, `x1 = -x2` is not possible unless both are zero, which is `x1=x2`.
Thus, x1 = x2 is the only valid conclusion in this domain. Hence, f(x) = x^2 is one-to-one on [0, ∞).

💡 Prevention Tips:

Master Definitions: Clearly distinguish injectivity (one-to-one) from surjectivity (onto).
Full Algebraic Analysis: Always explore all algebraic solutions when solving f(x1) = f(x2).
Domain Check: Verify if any alternative solutions (other than x1 = x2) are valid within the given function domain.
CBSE Requirement: For board exams, always present complete, step-by-step algebraic proofs.

CBSE_12th

Critical Calculation

❌ Assuming x² = y² implies x = y during One-One verification

A critical error in determining if a function is one-one often arises during the algebraic manipulation of f(x₁) = f(x₂). Students frequently conclude that if x₁² = x₂², then necessarily x₁ = x₂. This overlooks the possibility that x₁ = -x₂, which is equally valid algebraically and can lead to a function being incorrectly identified as one-one, especially when the domain includes both positive and negative numbers or zero.

💭 Why This Happens:

This mistake stems from a lack of careful algebraic analysis and an incomplete understanding of the implications of even powers or absolute values. Students often forget to consider all possible solutions when solving equations involving squares or modulus functions. It's also linked to neglecting the specified domain of the function, which is crucial for one-one mapping.

✅ Correct Approach:

To correctly verify if f(x) is one-one, start by assuming f(x₁) = f(x₂). Proceed with algebraic manipulation to solve for x₁ in terms of x₂. If the only possibility is x₁ = x₂, then the function is one-one. However, if you find other possibilities like x₁ = -x₂ (and these distinct values are within the domain), then the function is NOT one-one. Always ensure your algebraic steps consider all roots and solutions, and crucially, check if these solutions are valid within the given domain.

📝 Examples:

❌ Wrong:

Question: Check if f: R → R given by f(x) = x² is one-one.

Student's Wrong Approach:

Assume f(x₁) = f(x₂)
x₁² = x₂²
Taking square root on both sides: x₁ = x₂
Conclusion: Since x₁ = x₂, the function is one-one. (Incorrect!)

✅ Correct:

Question: Check if f: R → R given by f(x) = x² is one-one.

Correct Approach:

Assume f(x₁) = f(x₂) for x₁, x₂ ∈ R.
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies either x₁ - x₂ = 0 (i.e., x₁ = x₂) OR x₁ + x₂ = 0 (i.e., x₁ = -x₂).
Since the domain is R (real numbers), we can choose x₁ = 2 and x₂ = -2. Both are in R and x₁ ≠ x₂.
However, f(2) = 2² = 4 and f(-2) = (-2)² = 4.
So, f(2) = f(-2) even though 2 ≠ -2.
Conclusion: The function f(x) = x² is NOT one-one.

JEE & CBSE Tip: Always consider the domain carefully. If the domain was R⁺ (positive real numbers), then x₁ = -x₂ would not be possible for distinct x₁, x₂, and the function would be one-one in that restricted domain.

💡 Prevention Tips:

Be Vigilant with Squares and Modulus: Whenever you encounter x², |x|, or other even powers in f(x₁) = f(x₂), remember that x₁ = ±x₂ or x₁ = ±(expression). Don't blindly cancel terms or take only the positive root.
Always Check the Domain: After performing algebraic steps, critically examine if the derived relationship (e.g., x₁ = -x₂) is possible within the given domain of the function. If it is, and x₁ ≠ x₂, then the function is not one-one.
Look for Counter-examples: If you suspect a function is NOT one-one, actively try to find two distinct elements in the domain that map to the same image. This is often quicker and more convincing than a lengthy algebraic proof of non-injectivity.
Practice Diverse Problems: Work through problems involving various types of functions (polynomials, rational, trigonometric, exponential, logarithmic) to recognize common pitfalls and develop a systematic approach.

CBSE_12th

Critical Conceptual

❌ Misunderstanding the Role of Domain/Codomain and Flawed Algebraic Proof for One-One Functions

Students frequently overlook the significance of the specified domain and codomain when determining if a function is one-one (injective). They might also commit critical algebraic errors or present incomplete arguments while trying to prove injectivity using the definition: f(x₁) = f(x₂) ⇒ x₁ = x₂. This often leads to incorrect conclusions, especially for functions that exhibit different behaviors across various intervals (e.g., quadratic or trigonometric functions).

💭 Why This Happens:

Incomplete Conceptual Grasp: Students often memorize the definition without fully understanding its implications concerning the function's domain.
Algebraic Carelessness: Errors such as incorrectly simplifying expressions (e.g., √(x²) = x instead of |x|) or missing potential solutions (like ± in square roots).
Over-generalization: Applying rules valid for linear functions to all types of functions without considering their specific properties.
Lack of Practice: Insufficient exposure to functions with restricted domains or those that are not inherently one-one over the entire real number set.

✅ Correct Approach:

To rigorously prove a function f: A → B is one-one (injective):

Take any two arbitrary elements x₁, x₂ from the domain A.
Assume that f(x₁) = f(x₂).
Through correct and logical algebraic steps, demonstrate that this assumption necessarily implies x₁ = x₂.
If the algebraic steps yield results like x₁ = ±x₂ (or other multiple possibilities), then you must analyze if all these possibilities are valid within the given domain A. If other possibilities (like x₁ = -x₂) are allowed by the domain and x₁ ≠ x₂, then the function is not one-one.

📝 Examples:

❌ Wrong:

Function: f: R → R given by f(x) = x²

Student's Flawed Reasoning:
"Let f(x₁) = f(x₂). Then x₁² = x₂². Taking square root on both sides, we get x₁ = x₂. Therefore, f(x) = x² is a one-one function."

Mistake: Ignoring that x₁² = x₂² implies x₁ = ±x₂. For the domain R, both 2 and -2 are valid inputs, and f(2) = 4, f(-2) = 4, but 2 ≠ -2. Thus, the function is NOT one-one over R.

✅ Correct:

Function: f: [0, ∞) → R given by f(x) = x²

Correct Reasoning:
"Let f(x₁) = f(x₂) for any x₁, x₂ ∈ [0, ∞).
Then x₁² = x₂².
This implies x₁ = ±x₂.
Since the domain is [0, ∞), both x₁ and x₂ must be non-negative.
Therefore, the possibility x₁ = -x₂ is only true if x₁ = x₂ = 0. For any other positive x₁, x₂, x₁ = -x₂ is not allowed as it would imply one of them is negative, which is outside the domain.
Thus, within the domain [0, ∞), the only valid possibility is x₁ = x₂.
Hence, f(x) = x² is one-one for the domain [0, ∞)."

💡 Prevention Tips:

Prioritize Domain and Codomain: Always write down the domain and codomain first. They are fundamental to defining the function's properties.
Algebraic Precision is Key: Be meticulous with algebraic steps. Remember rules like √(a²) = |a|. (JEE Focus: This is a common trap.)
Test with Counter-Examples (JEE/Advanced CBSE): Before a formal proof, try to find two different inputs that give the same output. If you succeed, the function is not one-one, and this counter-example is a valid disproof.
Graphical Interpretation (CBSE): Understand the Horizontal Line Test: A function is one-one if no horizontal line intersects its graph more than once. This helps build intuition.

CBSE_12th

Critical Formula

❌ Incorrect Application of the Algebraic Test for One-One Functions

Students often correctly state the definition of a one-one (injective) function as: 'If f(x₁) = f(x₂) implies x₁ = x₂ for all x₁, x₂ in the domain.' However, a critical mistake arises in the algebraic application of this test. They frequently fail to exhaust all possibilities when solving f(x₁) = f(x₂), or incorrectly conclude x₁ = x₂, especially for functions involving squares, absolute values, or certain trigonometric expressions. The core issue is not rigorously proving that *only* x₁ = x₂ is possible.

💭 Why This Happens:

Incomplete Algebraic Steps: Rushing through calculations, particularly with squaring both sides, taking square roots, or dealing with absolute values, which can hide alternative solutions or lead to premature conclusions.
Ignoring Domain and Codomain: Overlooking the specified domain and codomain, which are crucial for defining the function's behavior and valid inputs.
Conceptual Gaps: Not fully internalizing that if even *one* instance of distinct inputs (x₁ ≠ x₂) maps to the same output (f(x₁) = f(x₂)) exists, the function is immediately not one-one.
Limited Practice: Practicing mostly with linear or simple polynomial functions where the condition x₁ = x₂ is easily met, leading to oversimplification for more complex cases.

✅ Correct Approach:

Assume f(x₁) = f(x₂) for any x₁, x₂ in the given domain of the function.
Rigorously manipulate this equation algebraically. Your goal is to see if this assumption uniquely forces x₁ = x₂.
If the algebraic manipulation leads to x₁ = x₂ as the only possibility within the domain, then the function is one-one.
If it leads to x₁ = x₂ AND other possibilities (e.g., x₁ = -x₂, x₁ = x₂ + 2πn, etc., where x₁ ≠ x₂ is possible), then you must find an example where x₁ ≠ x₂ but f(x₁) = f(x₂) to prove it is not one-one.
Always refer back to the function's domain to validate any solutions or counter-examples.

📝 Examples:

❌ Wrong:

Wrong Approach:

Consider f: R → R given by f(x) = x².

Student's steps:

1. Assume f(x₁) = f(x₂)

2. x₁² = x₂²

3. x₁ = x₂  (Student stops here, incorrectly concluding f is one-one. This step is incomplete as x₁ = -x₂ is also a possibility.)

✅ Correct:

Correct Approach:

Consider f: R → R given by f(x) = x².

1. Let f(x₁) = f(x₂) for x₁, x₂ ∈ R.

2. x₁² = x₂²

3. x₁² - x₂² = 0

4. (x₁ - x₂)(x₁ + x₂) = 0

5. This implies either x₁ = x₂ OR x₁ = -x₂.



Since the domain is R, we can find distinct values for x₁ and x₂ where x₁ = -x₂ (e.g., x₁ = 2, x₂ = -2). 

Here, x₁ ≠ x₂ (2 ≠ -2), but f(x₁) = f(2) = 2² = 4 and f(x₂) = f(-2) = (-2)² = 4.



Since distinct elements (2 and -2) map to the same image (4), the function f is NOT one-one.



CBSE Tip: For functions like x², |x|, cos(x) on their natural domains, always anticipate multiple inputs giving the same output.

💡 Prevention Tips:

Complete Algebraic Manipulation: Never leave the algebraic solution of f(x₁) = f(x₂) unfinished. Ensure all potential relationships between x₁ and x₂ are explored.
Test Counter-Examples: If your algebraic steps lead to x₁ = x₂ OR some other possibility, immediately try to construct a counter-example (x₁ ≠ x₂ but f(x₁) = f(x₂)) using that other possibility.
Domain-Specific Analysis: Always check if the domain restrictions prevent any of the 'other possibilities' from occurring. For instance, f(x) = x² is one-one if its domain is restricted to [0, ∞) because then x₁ = -x₂ is only possible if x₁ = x₂ = 0.
JEE Focus: For JEE, functions can be piecewise or involve greatest integer/fractional part functions. The algebraic test remains key, but often requires careful case analysis based on the function's definition over different intervals. Always visualize or sketch if unsure.

CBSE_12th

Critical Sign Error

❌ Ignoring Sign Differences in Algebraic Manipulation for One-One Functions

Students frequently make a critical sign error when proving or disproving if a function is one-one. When faced with equations like x₁² = x₂² or |x₁| = |x₂| during the one-one test (i.e., assuming f(x₁) = f(x₂) and trying to show x₁ = x₂), they often incorrectly conclude x₁ = x₂, completely overlooking the possibility that x₁ = -x₂. This leads to a false assertion that a function is one-one when it is not.

💭 Why This Happens:

This error stems from a combination of factors:

Lack of Fundamental Algebraic Recall: Forgetting that square roots have both positive and negative values, or that equations involving squares and absolute values have multiple solutions.
Haste and Carelessness: Rushing through derivations, particularly under exam pressure.
Misunderstanding of One-One Definition: Not fully grasping that f(x₁) = f(x₂) must uniquely imply x₁ = x₂ for all elements in the domain.

✅ Correct Approach:

When testing for one-one, if f(x₁) = f(x₂) leads to an equation like x₁² = x₂² or |x₁| = |x₂|, always consider both positive and negative possibilities.

For x₁² = x₂², rewrite it as x₁² - x₂² = 0, then factorize as (x₁ - x₂)(x₁ + x₂) = 0. This clearly shows x₁ = x₂ or x₁ = -x₂.
For |x₁| = |x₂|, it implies x₁ = x₂ or x₁ = -x₂.

If x₁ = -x₂ (where x₁ ≠ x₂) is possible for values within the domain, the function is not one-one. This rigorous approach is crucial for both CBSE board exams and JEE Advanced.

📝 Examples:

❌ Wrong:

Consider the function f: R → R given by f(x) = x².

Incorrect Step: Assume f(x₁) = f(x₂).
x₁² = x₂²
Therefore, x₁ = x₂. (Incorrect conclusion)
So, f(x) is one-one. (False)

✅ Correct:

Consider the function f: R → R given by f(x) = x².

Correct Approach: Assume f(x₁) = f(x₂).
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ - x₂ = 0 or x₁ + x₂ = 0.
So, x₁ = x₂ or x₁ = -x₂.

Since x₁ can be equal to -x₂ (e.g., f(2) = 4 and f(-2) = 4, where 2 ≠ -2), the function is not one-one. This highlights the importance of considering all sign possibilities.

💡 Prevention Tips:

Always Factorize: For quadratic expressions like a² = b², always rewrite as a² - b² = 0 and factorize into (a-b)(a+b)=0.
Understand Absolute Value: Recall that |a| = |b| implies a = b or a = -b.
Test with Simple Values: If unsure, mentally test with a positive and its corresponding negative value from the domain (e.g., x=2, x=-2) to see if they yield the same function output.
Review Domain: Always consider the function's domain. If the domain is restricted (e.g., f: R⁺ → R for f(x) = x²), then x₁ = -x₂ might not be possible, and the function could be one-one.

CBSE_12th

Critical Approximation

❌ <h3 style='color: #FF0000;'>Confusing Injectivity with Visual Monotonicity or Incomplete Algebraic Proof</h3>

Students often *approximate* whether a function is one-one by visually checking its graph for strict monotonicity, or by performing an incomplete algebraic proof, leading to incorrect conclusions. This is particularly problematic when the domain is not explicitly considered or when dealing with functions that are not globally monotonic (e.g., quadratic functions, trigonometric functions).

💭 Why This Happens:

Reliance on visual inspection of graphs without formal algebraic proof, assuming a visual pattern holds universally.
Lack of understanding that one-one means *every* distinct input maps to a distinct output, not just 'looks increasing/decreasing' on a segment.
Incomplete algebraic manipulation when setting `f(x1) = f(x2)` and solving for `x1` and `x2`.
Forgetting to consider the full domain or assuming standard domains (like real numbers) for functions where a restricted domain might make it one-one.

✅ Correct Approach:

The definition of a one-one (injective) function states: A function `f: A → B` is one-one if for every `x1, x2 ∈ A`, `f(x1) = f(x2) ⇒ x1 = x2`.

Step 1: Assume `f(x1) = f(x2)` for arbitrary `x1, x2` in the given domain.
Step 2: Perform rigorous algebraic manipulations to demonstrate that this assumption *must* imply `x1 = x2`.
Step 3: If `x1 = x2` is the *only* possible conclusion after considering the domain, the function is one-one. If other possibilities exist (e.g., `x1 = -x2` for `x^2`), then the function is not one-one, unless the domain specifically restricts these possibilities.

📝 Examples:

❌ Wrong:

Incorrect Logic: A student might consider `f(x) = x^2` on `R`. Seeing its graph increases for `x > 0`, they might wrongly approximate it as one-one. Algebraically, they might write `x1^2 = x2^2 ⇒ x1 = x2`, ignoring the possibility `x1 = -x2` which is valid for distinct `x1, x2` (e.g., 2 and -2) in the domain `R`.

✅ Correct:

Consider f: R → R defined by f(x) = x².

Assume f(x1) = f(x2) for x1, x2 ∈ R.
Then x1² = x2².
This implies x1² - x2² = 0.
Factoring gives (x1 - x2)(x1 + x2) = 0.
So, either x1 - x2 = 0 (which means x1 = x2) OR x1 + x2 = 0 (which means x1 = -x2).
Since x1 = -x2 is a possibility for distinct x1, x2 (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the function is not one-one on R.

Contrast with a restricted domain: If f: [0, ∞) → R defined by f(x) = x².

Assume f(x1) = f(x2) for x1, x2 ∈ [0, ∞).
Then x1² = x2², which implies x1 = x2 or x1 = -x2.
Since both x1, x2 ≥ 0, the only way x1 = -x2 can hold is if x1 = x2 = 0. For any other positive x1, x2 cannot be -x1 (as -x1 would be negative). Thus, the only valid conclusion is x1 = x2.
Therefore, f(x) = x² is one-one on [0, ∞).

💡 Prevention Tips:

Always use the definition: Start by assuming `f(x1) = f(x2)` and rigorously derive `x1 = x2`. This is the fundamental and most reliable method.
Be algebraic, not just visual: While graphs aid understanding, the formal proof for injectivity is strictly algebraic. Avoid making 'approximations' based on how a graph looks.
Pay attention to the domain: The domain of the function is absolutely crucial. Many functions are one-one on a restricted domain but not on their natural domain.
Consider all possibilities: When solving equations like `x1^2 = x2^2`, do not jump to `x1 = x2` immediately. Always consider all algebraic solutions (e.g., `x1 = ±x2`) and then filter them based on the given domain.
CBSE Specific: Clear, step-by-step application of the definition with proper reasoning for each algebraic step is essential for full marks.

CBSE_12th

Critical Other

❌ Ignoring the Specified Domain/Codomain when Checking Injectivity

Students frequently fail to properly account for the given domain and codomain of a function when determining if it is one-to-one (injective). They often apply general knowledge of a function's behavior over R (real numbers) even when the question specifies a restricted domain, leading to incorrect conclusions about injectivity.

💭 Why This Happens:

Lack of Attention: Overlooking the explicit domain and codomain mentioned in the problem statement.
Assumption of R: Defaulting to the assumption that the domain is always the set of real numbers (R) unless otherwise stated, which is a common but incorrect generalization.
Conceptual Gap: Not fully grasping that a function's injectivity is intrinsically tied to its specific domain; a function can be one-to-one on one domain but not on another.
Insufficient Practice: Limited exposure to problems where domains are explicitly restricted (e.g., natural numbers, positive real numbers, or specific intervals).

✅ Correct Approach:

Identify Domain and Codomain: Always begin by clearly noting the given function as f: A → B, where A is the domain and B is the codomain.
Apply Definition Correctly: To prove f is one-to-one, assume f(x₁) = f(x₂) for x₁, x₂ ∈ A (i.e., x₁ and x₂ must belong to the specified domain). Then, algebraically manipulate to show that this implies x₁ = x₂.
Disproving Injectivity: To show f is NOT one-to-one, find two distinct elements x₁ ≠ x₂, both belonging to the domain A, such that f(x₁) = f(x₂).

📝 Examples:

❌ Wrong:

Question: Is f: R → R given by f(x) = x² one-to-one?

Student's Common Mistake:
"Assume f(x₁) = f(x₂) &implies; x₁² = x₂² &implies; x₁ = x₂. So, it's one-to-one."
This reasoning is flawed because x₁² = x₂² also implies x₁ = -x₂. The student ignored the negative root.

✅ Correct:

Question: Is f: [0, ∞) → R given by f(x) = x² one-to-one?

Correct Approach:
Let x₁, x₂ ∈ [0, ∞) such that f(x₁) = f(x₂).
x₁² = x₂²
&implies; x₁ = ±x₂
Since x₁ and x₂ both belong to the domain [0, ∞), they must both be non-negative. Therefore, the case x₁ = -x₂ is only possible if x₁ = x₂ = 0. For any other positive values, x₁ = -x₂ would mean one is positive and one is negative, which is not allowed in the domain. Thus, we must have x₁ = x₂.

Hence, f(x) = x² is one-to-one on the domain [0, ∞). (This is different from the previous example where the domain was R).

💡 Prevention Tips:

CBSE/JEE Critical Tip: Always explicitly write down the domain and codomain at the beginning of your solution. This habit forces you to consider them throughout the problem.
Practice Diversity: Solve problems with various specified domains like N → N, Z → Z, R⁺ → R, or functions on specific intervals [a, b].
Algebraic Rigor: When solving f(x₁) = f(x₂), always consider all possible algebraic solutions for x₁ in terms of x₂ (e.g., both ± roots for even powers, all roots for cubic equations). Then, filter these solutions based on whether x₁ and x₂ can exist in the given domain such that x₁ ≠ x₂.
Horizontal Line Test: For graphical understanding, remember the horizontal line test must be applied only over the specified domain of the function.

CBSE_12th

Critical Conceptual

❌ Confusing Global Monotonicity with Local Behavior or Misapplying Tests

Students often make a critical conceptual error by assuming a function is one-to-one (injective) if it appears monotonic in certain regions or if they incorrectly apply the derivative test (f'(x) > 0 or f'(x) < 0) without considering the function's entire domain or piecewise nature. This leads to misidentifying functions as one-to-one when they are not, or vice-versa.

💭 Why This Happens:

This mistake stems from an incomplete understanding of the definition of a one-to-one function and the conditions under which tests (like the derivative test or Horizontal Line Test) can be reliably applied. Students often forget that injectivity requires each distinct input to map to a distinct output across the entire domain, not just a part of it. They might also overlook domain restrictions or discontinuities when evaluating injectivity.

✅ Correct Approach:

To determine if a function f: A → B is one-to-one, apply one of the following methods meticulously across its defined domain A:

Algebraic Method (Definition): Assume f(x₁ ) = f(x₂ ) for any x₁, x₂ ∈ A. If this assumption uniquely implies x₁ = x₂ , then the function is one-to-one.
Graphical Method (Horizontal Line Test - HLT): Draw the graph of the function. If any horizontal line intersects the graph at most one point, then the function is one-to-one.
Calculus Method (Monotonicity): If a function is strictly monotonic (i.e., strictly increasing or strictly decreasing) over its entire domain, then it is one-to-one. This is usually checked by examining the sign of f'(x). However, this method requires the function to be differentiable and continuous. Be cautious with functions that are not differentiable everywhere (e.g., |x|) or piecewise functions.

📝 Examples:

❌ Wrong:

Consider f(x) = x² defined on all real numbers (R). A student might incorrectly argue: 'f'(x) = 2x. For x > 0, f'(x) > 0, so it's increasing. For x < 0, f'(x) < 0, so it's decreasing. Since it's monotonic in parts, it must be one-to-one.' This is wrong because it's not strictly monotonic over the entire domain R. They ignore the fact that f(x) is not strictly monotonic over all R.

✅ Correct:

For f(x) = x² on R:

Algebraic: Let f(x₁ ) = f(x₂ ) ⇒ x₁² = x₂² ⇒ x₁ = ±x₂ . Since x₁ = -x₂ is possible (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the function is NOT one-to-one.
Graphical: The graph of y = x² is a parabola opening upwards. A horizontal line (e.g., y = 4) intersects the graph at two points (x = 2 and x = -2), violating the Horizontal Line Test. Hence, it's NOT one-to-one.

💡 Prevention Tips:

Always check the domain: The domain of the function is crucial. A function can be one-to-one on a restricted domain but not on a larger one.
Use the definition: When in doubt, revert to the algebraic definition f(x₁ ) = f(x₂ ) ⇒ x₁ = x₂. It is the most fundamental and universally applicable test.
Visualize: Sketching the graph (even roughly) can quickly reveal if a function passes the Horizontal Line Test, especially for common functions like polynomials, trigonometric functions, or piecewise functions.
Be careful with f'(x): The derivative test only works if the function is strictly monotonic over its entire domain. A zero derivative at isolated points (e.g., f(x) = x³ at x=0) does not necessarily imply non-monotonicity or failure of injectivity.

JEE_Main

Critical Other

❌ Ignoring Domain and Codomain When Checking for One-One Functions

A common and critical error in JEE Advanced is determining the one-one (injective) nature of a function without adequately considering its specified domain and codomain. Students often apply the algebraic test (f(x1) = f(x2) ⇒ x1 = x2) or the horizontal line test (HLT) assuming a default domain (like R) or a standard codomain, which can lead to incorrect conclusions.

💭 Why This Happens:

This mistake typically arises from a superficial understanding of function definitions. The one-one property is intrinsically linked to the set of inputs (domain) and the nature of outputs (which relate to the codomain). Rushing through problems, overlooking the function's complete definition, or failing to restrict the analysis to the given domain are primary reasons.

✅ Correct Approach:

Always begin by explicitly noting the domain and codomain of the function.

Algebraic Test: When solving f(x1) = f(x2), ensure that both x1 and x2 belong to the specified domain. If multiple solutions for x1 exist for a given x2 (e.g., x1 = ±x2), and both are in the domain, then the function is not one-one.
Graphical Test (HLT): Draw the graph of the function only for its given domain. A function is one-one if no horizontal line intersects its graph at more than one point within that specific domain.

📝 Examples:

❌ Wrong:

Consider f: R → R defined by f(x) = x².
A student might incorrectly reason: 'If f(x1) = f(x2), then x1² = x2², which means x1 = x2, so it's one-one.' This is wrong because x1 = ±x2. For example, f(2) = 4 and f(-2) = 4, but 2 ≠ -2. Since both 2 and -2 are in the domain R, the function is NOT one-one.

✅ Correct:

Consider g: [0, ∞) → R defined by g(x) = x².
Here, if g(x1) = g(x2), then x1² = x2². This means x1 = ±x2. However, since the domain is [0, ∞), both x1 and x2 must be non-negative. Therefore, x1 = -x2 is only possible if x1 = x2 = 0. For any other x1 > 0, -x1 is not in the domain. Thus, the only possibility is x1 = x2. Hence, g(x) = x² is one-one on the domain [0, ∞).

💡 Prevention Tips:

Read Carefully: Always highlight the domain and codomain when reading the question.
Visualize: Sketch the graph (or a mental image) of the function specifically over its given domain.
Test All Cases: For algebraic methods, ensure you consider all possible solutions to f(x1) = f(x2) and verify if they fall within the specified domain.
JEE Advanced Focus: Pay close attention to restricted domains in problems involving trigonometric functions, inverse functions, and piecewise definitions, as these are common areas for this mistake.

JEE_Advanced

Critical Approximation

❌ Confusing `f'(x) ≥ 0` with `f'(x) > 0` for One-One Functions (JEE Advanced)

Students frequently make the critical error of assuming a function is one-one if its derivative `f'(x)` is always non-negative (or non-positive), i.e., `f'(x) ≥ 0` (or `f'(x) ≤ 0`). This 'approximation' of monotonicity overlooks cases where `f'(x)` can be zero over a non-trivial interval. If `f'(x) = 0` across an interval, the function is constant over that segment, making it a many-one function, directly violating the injectivity condition. This is a severe conceptual misunderstanding.

💭 Why This Happens:

Oversimplification: Students often recall that 'monotonic implies one-one' but fail to internalize the critical distinction between 'monotonic' (which allows `f'(x)=0` over an interval) and 'strictly monotonic' (which requires `f'(x) > 0` or `f'(x) < 0`).
Graphical Misinterpretation: A quick, non-rigorous sketch or visual approximation might not clearly highlight a 'flat' segment, leading to an incorrect conclusion about injectivity.
Insufficient Rigor: Relying solely on derivative signs without performing the algebraic test `f(x1) = f(x2) ⇒ x1 = x2`, especially for piecewise functions or functions with regions of zero slope.

✅ Correct Approach:

For a function `f(x)` to be one-one, it must be strictly monotonic (either strictly increasing or strictly decreasing) over its entire domain.
If `f(x)` is differentiable on an interval, then:
- If `f'(x) > 0` for all `x` in the interval, `f(x)` is strictly increasing (and one-one).
- If `f'(x) < 0` for all `x` in the interval, `f(x)` is strictly decreasing (and one-one).
- If `f'(x) ≥ 0` but `f'(x) = 0` over an interval, then `f(x)` is NOT one-one. (CBSE vs JEE: JEE Advanced rigorously tests this nuance).
- If `f'(x) ≥ 0` but `f'(x) = 0` only at isolated points, then `f(x)` can still be one-one (e.g., `f(x) = x^3` where `f'(0) = 0`).
The most rigorous method is always to assume `f(x1) = f(x2)` and prove that `x1 = x2` for the given domain.

📝 Examples:

❌ Wrong:

Consider the function defined as:

f(x) = x     for x < 0

f(x) = 0     for 0 ≤ x ≤ 1

f(x) = x - 1 for x > 1

Student's Approximation: "The function is non-decreasing everywhere (`f'(x) = 1` for `x < 0` and `x > 1`, and it's flat/zero in `[0,1]`). So, since `f'(x) ≥ 0` everywhere it's defined, it must be a one-one function."

✅ Correct:

Analyzing the function `f(x)` from the wrong example:

We can observe that `f(0) = 0`.
Also, `f(0.5) = 0`.
And `f(1) = 0`.

Since `f(0) = f(0.5)` but `0 ≠ 0.5`, the function clearly violates the definition of a one-one function. The student's approximation failed because they overlooked the interval `[0, 1]` where the function is constant and `f'(x) = 0` (or undefined at endpoints but constant in between). This flat segment immediately makes the function many-one.

💡 Prevention Tips:

Always Refer to Definition: When in doubt, go back to the fundamental definition: `f(x1) = f(x2) ⇒ x1 = x2`.
Scrutinize `f'(x) = 0` Cases: If `f'(x)` is zero at any point, critically examine whether it's an isolated point or an entire interval. This is a common JEE trap.
Handle Piecewise Functions Carefully: Pay close attention to the definition of the function across different intervals, especially at boundary points, as injectivity can fail there.
Rigorous Graphing: If using graphs, plot them precisely, especially around critical points or intervals where the derivative might be zero or undefined, to avoid visual misjudgments.

JEE_Advanced

Critical Sign Error

❌ Critical Sign Error in Determining One-to-One Functions

Students frequently make sign errors or overlook alternative solutions when simplifying the equation f(x1) = f(x2) to prove x1 = x2 for one-to-one functions. This oversight often leads to misclassifying many-to-one functions (especially those with even powers or absolute values) as one-to-one.

💭 Why This Happens:

Algebraic Oversights: Forgetting that an equation like a^2 = b^2 implies a = ±b, not just a = b.
Insufficient Rigor: Rushing through simplification steps, missing non-obvious relationships between x1 and x2 where x1 ≠ x2 but f(x1) = f(x2).
Domain Misinterpretation: Not fully integrating domain restrictions into the algebraic solution of f(x1) = f(x2).

✅ Correct Approach:

When proving a function is one-to-one, meticulously solve f(x1) = f(x2). The function is one-to-one if and only if the only conclusion for all x1, x2 in the domain is x1 = x2. If you find any instance where f(x1) = f(x2) but x1 ≠ x2, the function is many-to-one. Always check for all possible roots and scenarios.

📝 Examples:

❌ Wrong:

Function: f(x) = x^2 (Domain: Real numbers)
Assumption: f(x1) = f(x2)
x1^2 = x2^2
Student's mistake: Concluding directly that x1 = x2.
Wrong Conclusion: Based on this, they might incorrectly state that f(x) is one-to-one.

✅ Correct:

Function: f(x) = x^2 (Domain: Real numbers)
Assumption: f(x1) = f(x2)
x1^2 = x2^2
x1^2 - x2^2 = 0
(x1 - x2)(x1 + x2) = 0
This equation implies two possibilities:
1. x1 - x2 = 0 ⇒ x1 = x2
2. x1 + x2 = 0 ⇒ x1 = -x2
Since x1 can be equal to -x2 (e.g., f(2) = 4 and f(-2) = 4, where 2 ≠ -2), the condition x1 = x2 is not the only outcome. Therefore, the function f(x) = x^2 is not one-to-one (it is many-to-one).

💡 Prevention Tips:

Factorize Completely: Always factorize f(x1) - f(x2) = 0 as much as possible to reveal all roots and relationships between x1 and x2.
Handle Even Powers/Absolute Values Carefully: Remember the ± sign when solving equations involving x^n (where n is even), |x|, or square roots.
Test Non-Trivial Points: For functions like x^2 or cos x, quickly verify if f(a) = f(-a) for some a ≠ 0. If so, it's many-to-one.
Graphical Intuition: Use the horizontal line test to quickly verify your algebraic conclusion. If a horizontal line intersects the graph at more than one point, the function is not one-to-one.

JEE_Advanced

Critical Formula

❌ Ignoring non-trivial solutions in the injectivity test (f(x1) = f(x2) ⇒ x1 = x2)

A critical mistake in JEE Advanced is the superficial application of the injectivity test, f(x1) = f(x2) ⇒ x1 = x2. Students often fail to explore all algebraic possibilities when equating f(x1) and f(x2), especially for functions involving even powers, absolute values, or trigonometric terms. They incorrectly conclude a function is one-to-one by only arriving at x1 = x2, without realizing that other distinct values of x1 and x2 might also satisfy f(x1) = f(x2).

💭 Why This Happens:

This error stems from:

Incomplete algebraic manipulation: Not solving equations thoroughly (e.g., overlooking ± roots).
Lack of conceptual clarity: Not fully grasping that only x1 = x2 must be the outcome for injectivity.
Ignoring function properties: Failing to recognize inherent symmetries (like even functions) or periodicity that violate injectivity.

✅ Correct Approach:

To correctly test for injectivity:

Assume f(x1) = f(x2) for x1, x2 in the domain.
Solve this equation completely, exploring all possible algebraic solutions for x1 in terms of x2 (or vice versa).
If the only solution obtained is x1 = x2, then the function is one-to-one.
If any solution other than x1 = x2 exists (e.g., x1 = -x2, x1 = 2π - x2), for distinct x1 and x2, then the function is not one-to-one (injective).

📝 Examples:

❌ Wrong:

Consider f(x) = x² on R. If f(x1) = f(x2), then x1² = x2². A student might hastily conclude x1 = x2 and declare it one-to-one, ignoring the possibility of x1 = -x2.

✅ Correct:

For f(x) = x² on R, if f(x1) = f(x2):
x1² = x2²
x1² - x2² = 0
(x1 - x2)(x1 + x2) = 0
This yields x1 = x2 OR x1 = -x2.
Since x1 = -x2 implies that for distinct values (e.g., x1 = 2, x2 = -2), f(2) = 4 and f(-2) = 4, but 2 ≠ -2. Therefore, f(x) = x² is NOT one-to-one on R.

💡 Prevention Tips:

Thorough Algebraic Solution: Always solve f(x1) = f(x2) completely, considering all roots, positive/negative possibilities, and general solutions for trigonometric equations.
Check for Symmetry/Periodicity: Functions that are even (f(x) = f(-x)) or periodic cannot be one-to-one on their entire standard domain.
Horizontal Line Test (JEE Advanced - Graphical aid): If a horizontal line intersects the graph of f(x) at more than one point, the function is not one-to-one.

JEE_Advanced

Critical Calculation

❌ Algebraic Errors and Misinterpretation in One-One Function Tests

Students frequently make critical algebraic mistakes when testing for one-one functions, particularly when solving f(x1) = f(x2). Common errors include:

Incorrectly handling even powers or square roots, leading to loss of potential distinct inputs (e.g., assuming x1^2 = x2^2 implies only x1 = x2).
Neglecting domain restrictions during simplification.
Misinterpreting the derivative test for injectivity; specifically, failing to understand that a function can still be one-one if its derivative is zero at only isolated points (e.g., f(x) = x^3 at x=0).

These calculation errors lead to incorrect conclusions about a function's injectivity.

💭 Why This Happens:

Incomplete Algebraic Knowledge: Lack of rigorous practice in solving equations involving powers, roots, and absolute values, especially when multiple solutions are possible.
Over-Generalization: Applying rules without considering edge cases or specific properties of the function and its domain.
Misunderstanding of Strict Monotonicity: Not fully grasping that a function is strictly monotonic (and thus one-one) even if its derivative is zero at isolated points, as long as it doesn't change sign around those points.
Carelessness: Rushing calculations, particularly in high-pressure exam environments, leading to oversight of critical algebraic steps.

✅ Correct Approach:

Method Correct Approach (JEE Advanced Focus)

Method	Correct Approach (JEE Advanced Focus)
1. Algebraic Test (f(x1) = f(x2))	Assume `f(x1) = f(x2)` for `x1, x2` in the domain. Solve the equation meticulously, considering all possible algebraic implications (e.g., `x^2=y^2` means `x=±y`). If the only valid conclusion is `x1 = x2` (for all `x1, x2` in the domain), the function is one-one. If you find distinct `x1, x2` such that `f(x1) = f(x2)`, it is not one-one.
2. Derivative Test (Monotonicity)	Calculate `f'(x)`. If `f'(x) > 0` for all `x` in the domain, the function is strictly increasing (one-one). If `f'(x) < 0` for all `x` in the domain, the function is strictly decreasing (one-one). JEE Advanced Nuance: If `f'(x) ≥ 0` (or `f'(x) ≤ 0`) for all `x` in the domain, and `f'(x) = 0` only at isolated points (i.e., not over an entire interval), the function is still strictly monotonic and thus one-one.

1. Algebraic Test (f(x1) = f(x2))

Assume f(x1) = f(x2) for x1, x2 in the domain.
Solve the equation meticulously, considering all possible algebraic implications (e.g., x^2=y^2 means x=±y).
If the only valid conclusion is x1 = x2 (for all x1, x2 in the domain), the function is one-one. If you find distinct x1, x2 such that f(x1) = f(x2), it is not one-one.

2. Derivative Test (Monotonicity)

Calculate f'(x).
If f'(x) > 0 for all x in the domain, the function is strictly increasing (one-one).
If f'(x) < 0 for all x in the domain, the function is strictly decreasing (one-one).
JEE Advanced Nuance: If f'(x) ≥ 0 (or f'(x) ≤ 0) for all x in the domain, and f'(x) = 0 only at isolated points (i.e., not over an entire interval), the function is still strictly monotonic and thus one-one.

📝 Examples:

❌ Wrong:

Function: f: R -> R defined by f(x) = x^2
Student's Incorrect Calculation:
Assume f(x1) = f(x2)
x1^2 = x2^2
x1 = x2 (Incorrect deduction, critical error in simplification)
Conclusion: Function is one-one. (Wrong)

Here, the mistake is in concluding x1 = x2; it should be x1 = ±x2. For example, f(2) = 4 and f(-2) = 4, but 2 ≠ -2. So f(x) = x^2 is not one-one.

✅ Correct:

Function: f: R -> R defined by f(x) = x^3
1. Using Algebraic Test:
Assume f(x1) = f(x2)
x1^3 = x2^3
x1^3 - x2^3 = 0
(x1 - x2)(x1^2 + x1x2 + x2^2) = 0
For real x1, x2, the term (x1^2 + x1x2 + x2^2) = (x1 + x2/2)^2 + 3x2^2/4 is always non-negative and is zero only if x1=0 and x2=0. If x1 ≠ x2, then (x1^2 + x1x2 + x2^2) > 0. Therefore, for the product to be zero, we must have x1 - x2 = 0, which means x1 = x2. Hence, f(x) = x^3 is one-one.

2. Using Derivative Test:
f'(x) = 3x^2.
For all x ∈ R, f'(x) ≥ 0.
f'(x) = 0 only at x = 0, which is an isolated point.
Since f'(x) ≥ 0 everywhere and equals zero only at an isolated point, f(x) is strictly increasing and thus one-one.

💡 Prevention Tips:

Master Algebraic Manipulations: Practice solving various types of equations, paying attention to all possible solutions, especially when dealing with even powers, roots, logarithms, and trigonometric functions.
Always Check Domain: The domain of the function is crucial. Ensure all algebraic steps and conclusions are valid within the given domain.
Rigor in Derivative Test: For JEE Advanced, understand the distinction that f'(x) = 0 at isolated points does not break strict monotonicity. Avoid hastily concluding 'not one-one' if f'(x) is merely zero at a single point.
Verify with Examples: If unsure, try testing with a few numerical values from the domain to confirm your algebraic conclusions.
(CBSE vs. JEE Advanced): While CBSE typically focuses on clear cases where f'(x) > 0 or f'(x) < 0, JEE Advanced questions often test the nuanced understanding of f'(x) = 0 at isolated points.

JEE_Advanced

Critical Conceptual

❌ Neglecting Domain Restrictions for Injectivity

Students often apply tests for one-to-one (injective) functions, whether algebraic or graphical, without rigorously considering the function's explicitly specified domain. This fundamental oversight frequently leads to incorrect conclusions about injectivity, especially in complex JEE Advanced problems.

💭 Why This Happens:

Over-reliance on general function knowledge: Many forget that a function's injectivity is intrinsically dependent on its domain (e.g., f(x) = x² is not one-to-one on R, but it is on [0, ∞)).
Incomplete algebraic analysis: Failing to validate all potential solutions of f(x₁) = f(x₂) within the given domain boundaries.
Conceptual gap: A lack of deep understanding that 'one-to-one' means distinct elements within the domain must map to distinct elements.

✅ Correct Approach:

To correctly determine if a function f: A → B is one-to-one:

Algebraic Method: Assume f(x₁) = f(x₂) for any x₁, x₂ ∈ A. Solve the equation. If the only valid solution within the domain A is x₁ = x₂, then the function is one-to-one.
Graphical Method (Horizontal Line Test): Draw the graph of y = f(x) only over the specified domain A. If any horizontal line intersects this restricted graph at most once, the function is one-to-one.

📝 Examples:

❌ Wrong:

Function: f: [0, ∞) → R defined by f(x) = x².
Wrong thought process: "f(x) = x² is a parabola. I know that f(2) = 4 and f(-2) = 4. Since 2 ≠ -2, but their function values are equal, it's not one-to-one."
Issue: The student is considering x = -2, which is not part of the given domain [0, ∞). They are applying the general properties of f(x)=x² over all real numbers, instead of the specified domain.

✅ Correct:

Function: f: [0, ∞) → R defined by f(x) = x².
Correct approach (Algebraic):

Assume x₁, x₂ ∈ [0, ∞) such that f(x₁) = f(x₂) &implies; x₁² = x₂².
This simplifies to (x₁ - x₂)(x₁ + x₂) = 0.
Thus, x₁ = x₂ or x₁ = -x₂.
Since both x₁ and x₂ must be non-negative (from the domain [0, ∞)), the condition x₁ = -x₂ is only possible if x₁ = x₂ = 0.
Therefore, the only valid conclusion for elements within the domain is x₁ = x₂.
Hence, f(x) = x² on [0, ∞) is indeed a one-to-one function.

💡 Prevention Tips:

Prioritize the Domain: Always begin by clearly noting the function's domain. It's the most critical piece of information for injectivity.
Restrict the Graph: When using the horizontal line test, only sketch or consider the portion of the graph that falls within the given domain.
Domain-Specific Validation: If solving f(x₁) = f(x₂) algebraically, ensure that any derived relationships between x₁ and x₂ are strictly permissible within the specified domain.
Practice with Restricted Domains: Actively practice problems where standard functions (like trigonometric or polynomial) have their domains restricted.

JEE_Advanced

Critical Calculation

❌ Algebraic Errors in One-One Function Verification

Students frequently make algebraic errors when testing if f(x₁) = f(x₂) implies x₁ = x₂. Incorrect factorization, sign changes, or mismanaging quadratic, absolute value, or square root expressions often lead to wrong injectivity conclusions.

💭 Why This Happens:

Rushed Calculations: Insufficient care during algebraic simplification.

Ignoring Domain: Failing to apply domain restrictions during algebraic steps.

Incomplete Solutions: Overlooking all possibilities from equations (e.g., x² = y² yields x = ±y).

✅ Correct Approach:

Begin with f(x₁) = f(x₂).

Systematically simplify, ensuring each step is algebraically sound and respects the function's domain.

Consider all outcomes from manipulations.

Conclude one-one only if x₁ = x₂ is the *sole* possibility.

📝 Examples:

❌ Wrong:

For f(x) = x² with domain R:
Student writes x₁² = x₂² ⇒ x₁ = x₂, concluding one-one.

Mistake: Ignores x₁ = -x₂ (e.g., f(2) = f(-2) = 4). Thus, not one-one on R.

✅ Correct:

For f(x) = x² with domain [0, ∞):
Let x₁, x₂ ∈ [0, ∞) such that x₁² = x₂².
This gives (x₁ - x₂)(x₁ + x₂) = 0.
Since x₁, x₂ ≥ 0, x₁ + x₂ ≥ 0. This forces x₁ - x₂ = 0 (so x₁ = x₂) or x₁ + x₂ = 0 (which means x₁ = x₂ = 0).
Thus, x₁ = x₂ is the only conclusion. Hence, one-one.

💡 Prevention Tips:

Verify Steps: Always recheck all algebraic transformations.

Account for All Solutions: Be aware of multiple outcomes from operations.

Respect Domain: Rigorously apply domain restrictions throughout your proof.

Practice: Consistent practice improves algebraic precision.

JEE_Main

Critical Formula

❌ Ignoring Domain Restrictions in Algebraic Proof of Injectivity

Students often correctly initiate the proof for one-to-one (injective) functions by setting f(x1) = f(x2) and attempting to show x1 = x2. However, a critical mistake arises when they perform algebraic manipulations without rigorously considering the function's specified domain. This leads to incorrect conclusions, especially for functions involving squares, absolute values, or trigonometric terms, where algebraic steps might suggest x1 = x2 while other distinct values also satisfy f(x1) = f(x2) within the domain.

💭 Why This Happens:

Incomplete Algebraic Reasoning: Overlooking that equations like x² = a² yield x = ±a, not just x = a.
Domain Neglect: Forgetting that x1 and x2 must belong to the specific domain given for the function.
Over-simplification: Rushing through algebraic steps without checking for all valid solutions or implications within the domain.
Lack of Rigor (JEE specific): JEE problems often test this exact understanding, differentiating between candidates who only know the formula and those who can apply it contextually.

✅ Correct Approach:

To correctly prove a function f: A → B is one-to-one:

1. Assume x1, x2 ∈ A (the domain) such that f(x1) = f(x2).
2. Perform algebraic manipulations carefully, ensuring each step is valid and accounts for all possibilities.
3. Crucially, at every step, verify that any derived conditions or potential solutions for x1 and x2 are consistent with the function's specified domain A.
4. If, and only if, the algebraic process rigorously leads to x1 = x2 as the only possibility under the given domain, then the function is one-to-one.
CBSE vs. JEE: While CBSE might be more lenient, JEE rigorously checks for domain awareness in such proofs.

📝 Examples:

❌ Wrong:

Consider f: R → R defined by f(x) = x².
Student's incorrect approach:
1. Assume f(x1) = f(x2)
2. x1² = x2²
3. Taking square root on both sides: x1 = x2.
4. Conclusion: Therefore, f(x) = x² is one-to-one. (INCORRECT!) The student ignored the possibility of x1 = -x2.

✅ Correct:

Consider f: R → R defined by f(x) = x².
To prove injectivity:
1. Let x1, x2 ∈ R (the domain) such that f(x1) = f(x2).
2. Then, x1² = x2².
3. This implies x1² - x2² = 0.
4. Factoring, we get (x1 - x2)(x1 + x2) = 0.
5. This gives two possibilities:
a) x1 - x2 = 0 ⇒ x1 = x2 OR
b) x1 + x2 = 0 ⇒ x1 = -x2.
6. Since x1 and x2 are from R, and x1 = -x2 is a valid possibility where x1 ≠ x2 (e.g., f(2) = 4 and f(-2) = 4, but 2 ≠ -2), the function is not one-to-one on R.

JEE Tip: If the domain was restricted to f: [0, ∞) → R, then for x1, x2 ∈ [0, ∞), x1 = -x2 would only be possible if x1 = x2 = 0. Otherwise, if x1, x2 > 0, then x1 + x2 > 0, forcing x1 - x2 = 0, i.e., x1 = x2. Thus, f(x) = x² is one-to-one on [0, ∞).

💡 Prevention Tips:

Explicitly Write Down Domain: Always start your proof by stating 'Let x1, x2 ∈ Domain'. This reinforces the importance of the domain.
Rigorous Factorization: Instead of directly taking square roots, factorize expressions like x1² - x2² to (x1 - x2)(x1 + x2) to reveal all possible solutions.
Test Cases/Counter-examples: If you're unsure, try to find two distinct values in the domain that map to the same image. If you find one, the function is not one-to-one.
Monotonicity Check (JEE specific): For differentiable functions, if f'(x) is strictly positive or strictly negative throughout the domain, the function is one-to-one. This is a powerful shortcut for JEE, but remember to consider points where f'(x) might be zero or undefined.

JEE_Main

Critical Unit Conversion

❌ Misapplying Unit Conversion to Abstract Mathematical Functions (e.g., One-One Functions)

A critical misconception arises when students attempt to apply principles of unit conversion to abstract mathematical concepts like one-one functions (injective functions). The definition and properties of a one-one function, which describe the mapping behavior between elements of a domain and codomain, are purely mathematical and do not involve physical quantities or their units.

💭 Why This Happens:

This mistake typically stems from a fundamental confusion between:

Abstract Mathematical Functions: Such as those studied in calculus and algebra (e.g., polynomial functions, trigonometric functions), where variables often represent dimensionless numbers.
Functions Representing Physical Phenomena: Common in Physics or Chemistry, where variables represent measurable quantities (e.g., distance, time, mass) and thus have units, making unit conversion relevant.

Students might mistakenly believe all functions must involve units, especially after extensive exposure to physics problems.

✅ Correct Approach:

When dealing with one-one functions in pure mathematics, focus solely on the mathematical definition of injectivity: A function f: A → B is one-one if for any x₁, x₂ ∈ A, f(x₁) = f(x₂) implies x₁ = x₂. There is no role for unit conversion here. The variables x and f(x) are treated as numbers or elements of a set, not physical quantities with units.

📝 Examples:

❌ Wrong:

Consider the function f(x) = x² from R to R. A student might incorrectly ponder:

"If x is measured in meters, then f(x) would be in meters². Should I convert f(x) to cm² before checking if it's one-one? How does unit conversion affect whether it's one-one?"

This line of thinking is irrelevant and distracting. Units have no bearing on the injectivity of f(x) = x².

✅ Correct:

To check if f(x) = x² (from R to R) is one-one:

Assume f(x₁) = f(x₂)
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ = x₂ OR x₁ = -x₂.

Since x₁ = x₂ is not the *only* possibility (e.g., f(2) = 4 and f(-2) = 4, but 2 &neq; -2), the function is not one-one. Notice, no units were involved, nor were they relevant.

💡 Prevention Tips:

Understand the Scope: Differentiate clearly between pure mathematical problems and applied science problems. JEE Main often tests both, but within their respective contexts.
Focus on Definitions: Always refer back to the fundamental definitions of mathematical concepts. For one-one functions, it's about unique mapping, not physical measurement.
JEE Main Context: For mathematical functions, assume variables are dimensionless numbers unless explicitly stated otherwise within an applied problem (which is rare for basic function properties).

JEE_Main

Critical Sign Error

❌ Critical Sign Error: Forgetting Both Positive and Negative Roots When Proving One-to-One Functions

Students often make a critical sign error when attempting to prove a function is one-to-one (injective) by setting f(x₁) = f(x₂). The mistake typically arises during algebraic manipulation, specifically when dealing with even powers or square roots, where they neglect to consider both the positive and negative possibilities, assuming x₁ = x₂ is the only outcome. This oversight leads to an incorrect conclusion that a function is one-to-one when it is not.

💭 Why This Happens:

This error stems from a fundamental misunderstanding or carelessness in basic algebra, particularly with operations like squaring or taking even roots.

Incomplete Algebraic Manipulation: Students simplify x₁² = x₂² directly to x₁ = x₂, ignoring the possibility of x₁ = -x₂.
Lack of Domain Consideration: The domain of the function is not fully considered. For instance, a function like f(x) = x² is not one-to-one on R but is on [0, ∞).
Rushing Steps: Under exam pressure, students might rush through algebraic steps, missing critical sign implications.

✅ Correct Approach:

To correctly determine if a function is one-to-one, always ensure all algebraic possibilities, including those involving signs, are thoroughly explored.

Systematic Algebraic Transformation: When f(x₁) = f(x₂), manipulate the equation carefully. Factorize wherever possible to reveal all potential solutions for x₁ in terms of x₂.
Consider All Roots: For equations like a² = b², remember that a = ±b. Convert it to a² - b² = 0 and factorize as (a-b)(a+b) = 0, which clearly yields a=b or a=-b.
Domain Check: After finding all potential relationships between x₁ and x₂, verify if x₁ = -x₂ (or other non-identical relations) is possible within the given domain of the function. If it is, the function is not one-to-one.

JEE Main Tip: Many JEE questions embed such sign traps. Always be vigilant about absolute values, square roots, and even powers.

📝 Examples:

❌ Wrong:

Problem: Is f(x) = x² + 1 one-to-one on R?
Wrong Approach:
Let f(x₁) = f(x₂)
x₁² + 1 = x₂² + 1
x₁² = x₂²
x₁ = x₂ (Mistake: only considering the positive root)
Conclusion: f(x) is one-to-one. (Incorrect)

✅ Correct:

Problem: Is f(x) = x² + 1 one-to-one on R?
Correct Approach:
Let f(x₁) = f(x₂)
x₁² + 1 = x₂² + 1
x₁² = x₂²
x₁² - x₂² = 0
(x₁ - x₂)(x₁ + x₂) = 0
This implies x₁ - x₂ = 0 OR x₁ + x₂ = 0.
So, x₁ = x₂ OR x₁ = -x₂.
Since the domain is R, we can pick x₁ = 2 and x₂ = -2. Then f(2) = 2² + 1 = 5 and f(-2) = (-2)² + 1 = 5.
Here, f(2) = f(-2) but 2 ≠ -2.
Conclusion: f(x) = x² + 1 is NOT one-to-one on R. (Correct)

💡 Prevention Tips:

✅ Always factorize a² - b² into (a-b)(a+b) instead of directly taking square roots.
✅ Pay close attention to the function's domain. If x₁ = -x₂ is possible within the domain for x₁ ≠ x₂, the function is not one-to-one.
✅ Practice with functions involving absolute values (e.g., f(x) = |x|) and even powers to train your algebraic rigor.
✅ After algebraic steps, verbally confirm: "Does this step cover all possible positive and negative outcomes?"

JEE_Main

Critical Approximation

❌ Approximating One-oneness without Rigorous Proof (Critical)

Students often make critical errors by 'approximating' whether a function is one-one, rather than applying rigorous algebraic or calculus-based methods. This usually manifests as:

Visual Approximation: Assuming a function is one-one by a quick glance at a graph or its general form, without performing the Horizontal Line Test or algebraic verification.
Incomplete Monotonicity Analysis: Forgetting that a function must be strictly monotonic (either strictly increasing or strictly decreasing) throughout its entire domain to be one-one. They might check monotonicity over only a part of the domain or incorrectly assume it based on the dominant term.
Ignoring Critical Points/Intervals: Failing to analyze points where the derivative f'(x) is zero or changes sign, or intervals where f'(x) = 0 (making the function constant over an interval).

💭 Why This Happens:

This critical mistake stems from:

Over-reliance on Intuition: Trusting visual intuition or general knowledge of basic function types without formal verification.
Lack of Conceptual Clarity: Not fully internalizing the definition of a one-one function (f(x1) = f(x2) ⇒ x1 = x2).
Haste and Time Pressure (JEE Main): Rushing through problems, especially in a time-constrained environment, leading to shortcuts instead of complete analysis.
Inadequate Domain Consideration: Forgetting to analyze the function's behavior across its entire given domain, especially for piecewise functions or functions with restricted domains.

✅ Correct Approach:

To avoid critical errors, always use one of these rigorous methods:

Algebraic Method (Definition-based): Assume f(x₁) = f(x₂) for any x₁, x₂ in the domain. Then, rigorously prove that this implies x₁ = x₂. If you find even one instance where x₁ ≠ x₂, the function is not one-one.
JEE Tip: This is the most fundamental and universally applicable method.
Calculus Method (for Differentiable Functions): Calculate the first derivative, f'(x).
- If f'(x) > 0 for all x in the domain (strictly increasing), or f'(x) < 0 for all x in the domain (strictly decreasing), then the function is one-one.
- Be extremely careful if f'(x) = 0 at some points. If f'(x) = 0 only at isolated points and doesn't change sign, it can still be one-one (e.g., f(x) = x³). However, if f'(x) = 0 over an entire interval, the function is constant over that interval and thus not one-one.
Horizontal Line Test (Graphical): Draw the graph of the function. If any horizontal line intersects the graph at more than one point, the function is not one-one. CBSE vs JEE: More common in CBSE for understanding; for JEE, analytical methods are usually required, but a quick sketch can confirm.

📝 Examples:

❌ Wrong:

Consider f(x) = x³ - 3x. A student might 'approximate' that since x³ is generally increasing, this function must be one-one. Or they might quickly calculate f'(x) = 3x² - 3 and notice it's positive for large |x|, incorrectly concluding it's always one-one.
This is incorrect because f'(x) = 3(x² - 1) = 3(x-1)(x+1). f'(x) changes sign at x = -1 and x = 1, meaning the function increases, then decreases, then increases again. For instance, f(0) = 0, f(√3) = 0, f(-√3) = 0, showing it is not one-one.

✅ Correct:

Let's check if f(x) = x³ - 3x is one-one on the domain [1, ∞).
1. Calculus Method: f'(x) = 3x² - 3 = 3(x² - 1).
For x ∈ [1, ∞), x² ≥ 1, so x² - 1 ≥ 0. Thus, f'(x) ≥ 0 for x ∈ [1, ∞).
More specifically, f'(x) = 0 only at x=1 in this domain. For x > 1, f'(x) > 0. Since f'(x) ≥ 0 and is strictly positive for x > 1, the function is strictly increasing on [1, ∞).
Therefore, f(x) = x³ - 3x is one-one on the domain [1, ∞).
2. Algebraic Method (briefly): Assume f(x₁) = f(x₂) for x₁, x₂ ∈ [1, ∞).
x₁³ - 3x₁ = x₂³ - 3x₂
x₁³ - x₂³ - 3(x₁ - x₂) = 0
(x₁ - x₂)(x₁² + x₁x₂ + x₂²) - 3(x₁ - x₂) = 0
(x₁ - x₂)(x₁² + x₁x₂ + x₂² - 3) = 0
Since x₁, x₂ ∈ [1, ∞), x₁² ≥ 1, x₂² ≥ 1, x₁x₂ ≥ 1.
So, x₁² + x₁x₂ + x₂² ≥ 1+1+1 = 3.
Thus, x₁² + x₁x₂ + x₂² - 3 ≥ 0.
For x₁² + x₁x₂ + x₂² - 3 to be zero, it must be x₁=x₂=1.
If x₁, x₂ > 1, then x₁² + x₁x₂ + x₂² - 3 > 0.
Therefore, for the product to be zero, (x₁ - x₂) must be zero, implying x₁ = x₂. Hence, it's one-one.

💡 Prevention Tips:

Master the Definition: Always start with the definition: 'f is one-one if f(x₁) = f(x₂) implies x₁ = x₂'. Practice applying this algebraically.
Analyze the Full Domain: Before concluding monotonicity, analyze f'(x) across the entire specified domain. Look for intervals where f'(x) changes sign or becomes zero over an interval.
Don't Skip Steps: Especially in JEE, rigorous justification is key. Don't rely on 'it looks like' or 'it generally behaves like'.
Practice with Piecewise Functions: Pay extra attention to piecewise functions. Check one-oneness within each piece and then ensure values from different pieces don't overlap (i.e., f(x) for x in domain A should not equal f(x) for x in domain B unless x is the boundary point where they are equal).
Utilize Graphs Strategically: Use mental graphs or quick sketches to visualize behavior, but always cross-verify with algebraic or calculus methods.

JEE_Main

Critical Other

❌ Misinterpreting Monotonicity for One-to-One Functions

Students often incorrectly conclude a function is not one-to-one (injective) if its derivative f'(x) becomes zero at some points, even if it doesn't stay zero over an interval. They mistakenly believe that any instance of f'(x) = 0 implies a loss of strict monotonicity and thus injectivity, ignoring the crucial distinction between f'(x) = 0 at isolated points versus over an interval.

💭 Why This Happens:

This mistake stems from a superficial understanding of the relationship between monotonicity, derivatives, and injectivity. Students often memorize that 'strictly monotonic implies one-to-one' but fail to grasp the nuance when f'(x) = 0. They confuse functions like f(x) = x³ (where f'(0) = 0 but is one-to-one) with functions like f(x) = constant or f(x) = x² on [-1, 1] (where the function is not one-to-one).

✅ Correct Approach:

A function f(x) is one-to-one if for any x₁, x₂ in its domain, f(x₁) = f(x₂) ⇒ x₁ = x₂.

Using the derivative test: If f'(x) ≥ 0 (or f'(x) ≤ 0) throughout the domain, and f'(x) = 0 only at isolated points (not over any interval), then the function is still strictly monotonic and therefore one-to-one.

JEE Tip: Always verify if f'(x) = 0 holds true for an entire interval, or just isolated points. This distinction is critical.

📝 Examples:

❌ Wrong:

Function: f: ℝ → ℝ defined by f(x) = x + sin(x).
Student's Mistake:
1. Calculate f'(x) = 1 + cos(x).
2. Observe that f'(x) = 0 when cos(x) = -1 (i.e., at x = π, 3π, ...).
3. Conclude: "Since f'(x) is zero at some points, the function is not strictly monotonic, and therefore not one-to-one."

✅ Correct:

Function: f: ℝ → ℝ defined by f(x) = x + sin(x).
Correct Approach:
1. Calculate f'(x) = 1 + cos(x).
2. Observe that -1 ≤ cos(x) ≤ 1 for all x ∈ ℝ. Therefore, 0 ≤ 1 + cos(x) ≤ 2, meaning f'(x) ≥ 0 for all x ∈ ℝ.
3. Note that f'(x) = 0 only occurs when cos(x) = -1, which happens at isolated points (x = (2n+1)π for integer n). It is never zero over an entire interval.
4. Conclusion: Since f'(x) ≥ 0 throughout the domain and f'(x) = 0 only at isolated points, the function f(x) is strictly increasing and thus one-to-one (injective) on ℝ.

💡 Prevention Tips:

Understand the Formal Definition: Always revert to f(x₁) = f(x₂) ⇒ x₁ = x₂ if unsure about derivative tests.
Derivative Test Nuance: Remember that a function is one-to-one if f'(x) ≥ 0 (or f'(x) ≤ 0) and f'(x) = 0 only at isolated points.
Graphical Test: Use the horizontal line test. If any horizontal line intersects the graph at most once, the function is one-to-one.
Practice Complex Cases: Focus on functions where the derivative becomes zero at discrete points or where the domain is restricted, as these are common JEE traps.
CBSE vs JEE: While CBSE might stick to simpler cases like f(x) = x³, JEE often tests your understanding with functions like f(x) = x + sin(x), where careful analysis of f'(x) = 0 is required.

JEE_Main

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📖Topic Explanations

Understanding Functions: A Quick Recap

The One-to-One Function: Injective Mapping

1. Intuitive Definition

2. Formal Definition

Methods to Check if a Function is One-to-One

Method 1: The Algebraic Method (Using the Definition)

Example 1: Proving a Linear Function is One-to-One

Example 2: Proving a Quadratic Function is NOT One-to-One (on (mathbb{R}))

Example 3: Making a Quadratic Function One-to-One by Restricting the Domain

Method 2: The Graphical Method (Horizontal Line Test)

Example 4: Using the Horizontal Line Test

Method 3: The Calculus Method (Monotonicity Test)

Reasoning:

Example 5: Using Monotonicity Test

Example 6: A function that is one-to-one but (f'(x)) is zero at a point

Advanced Concepts and JEE Applications

Summary Table of One-to-One Function Checks

What is a One-to-One Function?

Short-Cuts & Tests for One-to-One Functions

1. Algebraic Test (The "Default" Method)

2. Graphical Test: The Horizontal Line Test (HLT)

3. Calculus Short-Cut: Monotonicity Test

Quick Tips: Identifying One-to-One (Injective) Functions

1. Core Definition (Injective Function)

2. Methods to Check Injectivity

a. Algebraic Method (Universal)

b. Graphical Method (Horizontal Line Test)

c. Monotonicity Test (JEE Specific)

3. Important Considerations

Intuitive Understanding: One-One Functions (Injective Functions)

The Core Idea: Unique Pairing

Mathematical Interpretation

Graphical Understanding: The Horizontal Line Test

Why is it important?

Real-World Applications of One-to-One Functions:

What is a One-One (Injective) Function?

Common Analogies for One-One Functions:

Common Exam Traps for One-to-One Functions

Key Takeaways: One-to-One Functions (Injective Functions)

I. Analytical Method (Algebraic Approach)

II. Graphical Method (Horizontal Line Test)

III. Calculus Method (Monotonicity Test)

Example: Determine if f(x) = x³ - 3x² + 3x - 1 is one-to-one on R.

CBSE Focus Areas: One-One Functions

Key Concepts & Definitions

Methods to Prove a Function is One-One (CBSE Perspective)

Example for CBSE Board Exam

Common Functions & Their One-One Nature (CBSE Context)

What is a One-to-One Function?

Key Methods to Check Injectivity for JEE Main

JEE Main Specific Focus Areas

CBSE vs. JEE Main Perspective

📊 AI Generation Metadata

📊 AI Generation Metadata

📊 AI Generation Metadata

📝CBSE 12th Board Problems (12)

🎯IIT-JEE Main Problems (6)

📐Important Formulas (3)

📚References & Further Reading (10)

⚠️Common Mistakes to Avoid (60)

❌ Confusing Monotonicity with One-to-One Property

❌ Confusing One-to-One with Strictly Monotonic or Misapplying Tests

❌ Algebraic Errors in Proving One-to-One Nature

❌ Ignoring non-trivial solutions when applying the <code>f(x<sub>1</sub>) = f(x<sub>2</sub>)</code> test for injectivity.

❌ <strong>Misapplication of Physical Unit Conversion Logic to Mathematical Function Properties</strong>

❌ <strong>Sign Error in Solving Equations to Check Injectivity</strong>

❌ Assuming Injectivity/Non-Injectivity Based on General Graph Shape or Function Type

❌ Confusing One-to-One with Strictly Monotonic Functions

❌ Overlooking Domain/Codomain and Incomplete Algebraic Proof for Injectivity

❌ Assuming one-to-one nature from partial observation or visual approximation

❌ Ignoring Negative Solutions in One-to-One Function Proofs

❌ <span style='color: #FF0000;'>Ignoring Domain/Codomain in Injectivity Proofs</span>

❌ Incomplete Algebraic Analysis for One-to-One Condition

❌ <b><span style='color: #FF0000;'>Incorrect Algebraic Deduction for Even Powers</span></b>

❌ Misinterpreting Injectivity by Incomplete Domain Analysis

❌ <span style='color: #FF0000;'>Assuming Injectivity from Partial or Approximate Monotonicity</span>

❌ Sign Errors in Absolute Value or Square Root Simplification during Injectivity Check

❌ Incorrectly Applying Injectivity Test for Restricted Domains

❌ Ignoring Multiple Solutions in Algebraic Simplification of f(x₁) = f(x₂)

Example: Determine if `f(x) = x³ - 3x² + 3x - 1` is one-to-one on R.

❌ <span style='color: red;'>Incorrect Application of the One-One Test (f(x<sub>1</sub>) = f(x<sub>2</sub>) ⇒ x<sub>1</sub> = x<sub>2</sub>)</span>

❌ Ignoring non-trivial solutions in the injectivity test (f(x1) = f(x2) ⇒ x1 = x2)