Types of functions

This chapter rigorously defines and explores key classifications of functions: one-one, onto, bijective, constant, and identity. A comprehensive grasp of these fundamental function types is essential for success in CMI examinations, as they form the bedrock for advanced mathematical concepts and problem-solving within algebra and analysis.

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Chapter Contents

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| Topic |

|---|-------| | 1 | One-one functions | | 2 | Onto functions | | 3 | Bijective functions | | 4 | Constant functions | | 5 | Identity function |

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We begin with One-one functions.

Part 1: One-one functions

One-one Functions

Overview

A one-one function is a function that never takes the same value at two different inputs. In exam problems, this idea appears through direct definition, algebraic verification, graph tests, monotonicity, inverse functions, and domain restrictions. The real skill is to recognise when injectivity comes from structure and when it fails because symmetry or repetition is present. ---

Learning Objectives

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Core Definition

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Basic Examples

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Direct Test from Definition

Example Test $f(x)=5x-7$ on $\mathbb{R}$. Assume $\qquad 5x_1-7=5x_2-7$ Then $\qquad 5x_1=5x_2$ So $\qquad x_1=x_2$ Hence $f$ is one-one. ---

Graph Interpretation

Examples:

• $y=x^3$ passes the horizontal line test

• $y=x^2$ fails it

• $y=|x|$ fails it on $\mathbb{R}$

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Monotonicity and Injectivity

Examples:

• $x^3+2x+1$ is strictly increasing on $\mathbb{R}$, so it is one-one

• $-e^x$ is strictly decreasing, so it is one-one

• $x^2$ is not monotone on all of $\mathbb{R}$, so it is not one-one there

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Role of Domain Restrictions

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One-one and Inverse Functions

If $f:X\to Y$ is one-one, then each value in the range comes from exactly one input, so the inverse mapping is well-defined on $f(X)$. Examples:

• $f(x)=2x+1$ has inverse

$\qquad f^{-1}(x)=\dfrac{x-1}{2}$

• $f(x)=x^2$ on $\mathbb{R}$ has no inverse as a function

• $f(x)=x^2$ on $[0,\infty)$ has inverse

$\qquad f^{-1}(x)=\sqrt{x}$ ---

Common Non-Injective Patterns

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Minimal Worked Examples

Example 1 Test whether $f(x)=\dfrac{x-1}{x+2}$ is one-one on $\mathbb{R}\setminus\{-2\}$. Assume $\qquad \dfrac{x_1-1}{x_1+2}=\dfrac{x_2-1}{x_2+2}$ Cross-multiply: $\qquad (x_1-1)(x_2+2)=(x_2-1)(x_1+2)$ Expand both sides: $\qquad x_1x_2+2x_1-x_2-2=x_1x_2+2x_2-x_1-2$ So $\qquad 2x_1-x_2=2x_2-x_1$ $\qquad 3x_1=3x_2$ $\qquad x_1=x_2$ Hence the function is one-one. --- Example 2 Test whether $f(x)=x^2-4x$ is one-one on $\mathbb{R}$. Rewrite: $\qquad f(x)=x^2-4x=(x-2)^2-4$ This is a parabola opening upward, so horizontal lines above $y=-4$ cut it twice. Hence it is not one-one on $\mathbb{R}$. But on $[2,\infty)$ it is one-one. ---

Algebraic Shortcuts

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following functions is one-one on $\mathbb{R}$?" options=["$x^2$","$|x|$","$x^3$","$\sin x$"] answer="C" hint="Think about symmetry and monotonicity." solution="The function $x^2$ is not one-one because $1$ and $-1$ have the same image. The function $|x|$ is not one-one because $|1|=|-1|$. The function $\sin x$ is periodic, so it is not one-one on $\mathbb{R}$. The function $x^3$ is strictly increasing on $\mathbb{R}$, so it is one-one. Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find the value of $k$ for which the linear function $f(x)=kx+5$ fails to be one-one on $\mathbb{R}$." answer="0" hint="When does a linear function become constant?" solution="A linear function $f(x)=kx+5$ is one-one on $\mathbb{R}$ whenever $k\ne 0$. It fails to be one-one only when it becomes a constant function. That happens when $k=0$. Therefore the answer is $\boxed{0}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f$ is strictly increasing on its domain, then $f$ is one-one.","If $f$ is one-one, then every horizontal line intersects its graph in at most one point.","$x^2$ is one-one on $[0,\infty)$.","A function can be both one-one and onto."] answer="A,B,C,D" hint="Separate injectivity, graph interpretation, and domain restriction carefully." solution="1. True. Strictly increasing functions are injective.

True. This is exactly the horizontal line test for one-one functions.

True. On $[0,\infty)$ the function $x^2$ is strictly increasing.

True. Such functions are called bijections when both properties hold.

Hence the correct answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Show that the function $f(x)=\dfrac{2x+1}{x-3}$ is one-one on its domain $\mathbb{R}\setminus\{3\}$." answer="The function is one-one on $\mathbb{R}\setminus\\{3\\}$." hint="Assume $f(x_1)=f(x_2)$ and cross-multiply." solution="Let $\qquad \dfrac{2x_1+1}{x_1-3}=\dfrac{2x_2+1}{x_2-3}$ Since $x_1,x_2\ne 3$, cross-multiplication is valid: $\qquad (2x_1+1)(x_2-3)=(2x_2+1)(x_1-3)$ Expand: $\qquad 2x_1x_2-6x_1+x_2-3 = 2x_1x_2-6x_2+x_1-3$ So $\qquad -6x_1+x_2 = -6x_2+x_1$ $\qquad 7x_1 = 7x_2$ $\qquad x_1=x_2$ Hence $\qquad f(x_1)=f(x_2)\Rightarrow x_1=x_2$ Therefore $f$ is one-one on $\boxed{\mathbb{R}\setminus\{3\}}$." ::: ---

Summary

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Part 2: Onto functions

Onto functions, also known as surjective functions, are fundamental in mathematics, ensuring that every element in a function's codomain is reached by at least one element from its domain. Mastering their properties is crucial for CMI problems involving function classification and counting.

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Core Concepts

1. Definition of an Onto Function

Worked Example 1: Proving Onto for Finite Sets
Consider the function $f: \{1, 2, 3\} \to \{a, b\}$ defined by $f(1)=a$, $f(2)=b$, $f(3)=b$. We determine if $f$ is an onto function.

Step 1: Identify codomain elements

> The codomain is $Y = \{a, b\}$. We must check if every element in $Y$ is an image of some element in $X$.

Step 2: Check for $y=a$

> We observe $f(1) = a$. Thus, $a$ is in the range of $f$.

Step 3: Check for $y=b$

> We observe $f(2) = b$. Thus, $b$ is in the range of $f$.

Answer: Since every element in the codomain $Y$ has at least one pre-image in the domain $X$, $f$ is an onto function.

Worked Example 2: Proving Onto for Real Functions
We examine if the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x+3$ is onto.

Step 1: Take an arbitrary element $y$ from the codomain

> Let $y \in \mathbb{R}$ be an arbitrary element in the codomain.

Step 2: Solve $f(x)=y$ for $x$

>

>
>

Step 3: Check if $x$ is in the domain

> For any $y \in \mathbb{R}$, the value $x = \frac{y-3}{2}$ is always a real number. Therefore, for every $y$ in the codomain $\mathbb{R}$, there exists an $x$ in the domain $\mathbb{R}$ such that $f(x)=y$.

Answer: The function $f(x) = 2x+3$ is onto.

Worked Example 3: Disproving Onto for Real Functions
We determine if the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x^2$ is onto.

Step 1: Identify codomain elements that might not have pre-images

> The codomain is $\mathbb{R}$. We know that squares of real numbers are always non-negative.

Step 2: Choose a $y$ from the codomain with no pre-image

> Consider $y = -1$. We need to find an $x \in \mathbb{R}$ such that $g(x) = x^2 = -1$.

Step 3: Attempt to solve for $x$ and check domain

> The equation $x^2 = -1$ has no real solutions for $x$.

Answer: Since there exists an element $y=-1$ in the codomain $\mathbb{R}$ that has no pre-image in the domain $\mathbb{R}$, the function $g(x) = x^2$ is not onto.

:::question type="MCQ" question="Let $f: \mathbb{Z} \to \mathbb{Z}$ be defined by $f(x) = 2x$. Is $f$ an onto function?" options=["Yes, because every integer $y$ can be written as $2x$.","No, because odd integers in the codomain do not have integer pre-images.","Yes, because for every $x$, $f(x)$ is an integer.","No, because $f$ is one-to-one."] answer="No, because odd integers in the codomain do not have integer pre-images." hint="Consider if every element in the codomain $\mathbb{Z}$ has a pre-image in the domain $\mathbb{Z}$." solution="Step 1: Understand the definition of an onto function.
> A function $f: X \to Y$ is onto if for every $y \in Y$, there exists an $x \in X$ such that $f(x)=y$.

Step 2: Consider an element in the codomain $\mathbb{Z}$.
> Let $y=1$ (an odd integer) be an element in the codomain $\mathbb{Z}$.

Step 3: Try to find a pre-image $x$ in the domain $\mathbb{Z}$.
> We set $f(x) = 1$, so $2x = 1$.
> Solving for $x$, we get $x = \frac{1}{2}$.

Step 4: Check if $x$ is in the domain.
> The value $x = \frac{1}{2}$ is not an integer, so it is not in the domain $\mathbb{Z}$. Therefore, $y=1$ has no pre-image in the domain $\mathbb{Z}$.

Answer: The function $f(x)=2x$ is not onto because odd integers in the codomain do not have integer pre-images."
:::

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2. Cardinality and Onto Functions

We establish a necessary condition for a function to be onto in the context of finite sets.

Worked Example 1: Impossibility of Onto Function
Consider sets $A = \{1, 2, 3\}$ and $B = \{a, b, c, d\}$. We determine if a function $f: A \to B$ can be onto.

Step 1: Compare cardinalities of domain and codomain

> We have $|A|=3$ and $|B|=4$.

Step 2: Apply the cardinality condition

> Since $|A| < |B|$ (i.e., $3 < 4$), it is impossible for $f: A \to B$ to be an onto function. There are more elements in the codomain than in the domain, so at least one element in the codomain will not have a pre-image.

Answer: A function $f: A \to B$ cannot be onto.

:::question type="MCQ" question="Let $X$ and $Y$ be finite sets. Which of the following statements about an onto function $f: X \to Y$ is always true?" options=["$|X| < |Y|$","There are infinitely many such functions.","$|X| \ge |Y|$","The function must also be one-to-one."] answer="$|X| \ge |Y|$" hint="Recall the definition of onto and its implications for the sizes of the sets." solution="Step 1: Understand the definition of an onto function.
> Every element in the codomain $Y$ must be mapped to by at least one element in the domain $X$.

Step 2: Consider the implication for cardinalities.
> If there are more elements in the codomain $Y$ than in the domain $X$, then by the Pigeonhole Principle, it is impossible for every element in $Y$ to receive a mapping from $X$. At least one element in $Y$ would be left unmapped.

Step 3: Formulate the condition.
> Therefore, for a function $f: X \to Y$ to be onto, the number of elements in the domain $X$ must be at least as large as the number of elements in the codomain $Y$.

Answer: The statement $|X| \ge |Y|$ is always true for an onto function $f: X \to Y$ between finite sets."
:::

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3. Counting Onto Functions

We use the Principle of Inclusion-Exclusion to count the number of onto functions between two finite sets.

Worked Example:
We calculate the number of onto functions from a set $A$ with $|A|=3$ to a set $B$ with $|B|=2$.

Step 1: Identify $m$ and $n$

> Here, $m = |A|=3$ and $n = |B|=2$.

Step 2: Apply the formula for $N_{\text{onto}}(m,n)$

>

>

Step 3: Calculate each term

>

>
>

Answer: There are 6 onto functions from a set of size 3 to a set of size 2.

:::question type="NAT" question="How many onto functions are there from a set $P = \{p_1, p_2, p_3, p_4\}$ to a set $Q = \{q_1, q_2\}$?" answer="14" hint="Use the inclusion-exclusion principle for counting onto functions, where $m=4$ and $n=2$." solution="Step 1: Identify the sizes of the domain and codomain.
> The domain $P$ has $|P|=m=4$. The codomain $Q$ has $|Q|=n=2$.

Step 2: Apply the formula for the number of onto functions.
>

> Substitute $m=4$ and $n=2$:
>

Step 3: Expand and calculate the terms.
> For $k=0$: $(-1)^0 \binom{2}{0} (2-0)^4 = (1)(1)(2)^4 = 16$
> For $k=1$: $(-1)^1 \binom{2}{1} (2-1)^4 = (-1)(2)(1)^4 = -2$
> For $k=2$: $(-1)^2 \binom{2}{2} (2-2)^4 = (1)(1)(0)^4 = 0$

Step 4: Sum the terms.
>

Answer: 14"
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Advanced Applications

We consider functions involving multiple properties or more complex domains/codomains.

Worked Example: Onto and Injective Properties
We analyze the function $f: \mathbb{Z} \to \mathbb{N} \cup \{0\}$ defined by $f(x) = |x|$. We determine if $f$ is onto and if it is one-to-one.

Step 1: Check if $f$ is onto

> Let $y \in \mathbb{N} \cup \{0\}$ be an arbitrary element in the codomain.
> We need to find an $x \in \mathbb{Z}$ such that $f(x)=y$, i.e., $|x|=y$.
> Since $y \ge 0$, we can choose $x=y$. Then $x \in \mathbb{Z}$ (as $\mathbb{N} \cup \{0\} \subset \mathbb{Z}$) and $|x|=|y|=y$.
> Thus, for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x)=y$.

Step 2: Conclude about onto property

> The function $f(x)=|x|$ is onto.

Step 3: Check if $f$ is one-to-one

> A function is one-to-one if distinct elements in the domain map to distinct elements in the codomain.
> Consider $x_1 = 2$ and $x_2 = -2$. Both $x_1, x_2 \in \mathbb{Z}$ and $x_1 \ne x_2$.
> However, $f(x_1) = |2| = 2$ and $f(x_2) = |-2| = 2$.
> Since $f(x_1) = f(x_2)$ for $x_1 \ne x_2$, the function is not one-to-one.

Answer: The function $f(x)=|x|$ is onto but not one-to-one.

:::question type="MSQ" question="Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \sin(x)$. Which of the following statements are true?" options=["$f$ is onto.","$f$ is one-to-one.","The range of $f$ is $[-1, 1]$.","The codomain of $f$ is $[-1, 1]$."] answer="The range of $f$ is $[-1, 1]$" hint="Recall the definition of onto, one-to-one, range, and codomain for trigonometric functions." solution="Step 1: Analyze if $f$ is onto.
> A function $f: X \to Y$ is onto if its range equals its codomain. For $f(x) = \sin(x)$, the domain is $\mathbb{R}$ and the codomain is $\mathbb{R}$. The range of $\sin(x)$ is $[-1, 1]$. Since $[-1, 1] \ne \mathbb{R}$, $f$ is not onto. Thus, 'f is onto' is false.

Step 2: Analyze if $f$ is one-to-one.
> A function is one-to-one if distinct inputs map to distinct outputs. For $f(x) = \sin(x)$, we know that $\sin(0) = 0$ and $\sin(\pi) = 0$. Since $0 \ne \pi$ but $f(0)=f(\pi)$, $f$ is not one-to-one. Thus, 'f is one-to-one' is false.

Step 3: Analyze the range of $f$.
> The range of the sine function is indeed $[-1, 1]$. Thus, 'The range of $f$ is $[-1, 1]$' is true.

Step 4: Analyze the codomain of $f$.
> The function is defined as $f: \mathbb{R} \to \mathbb{R}$, which means its codomain is explicitly stated as $\mathbb{R}$. The option states 'The codomain of $f$ is $[-1, 1]$', which is incorrect.

Answer: The range of $f$ is $[-1, 1]$"
:::

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Problem-Solving Strategies

We outline effective approaches for determining if a function is onto.

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Common Mistakes

Avoiding common pitfalls is crucial for accuracy in CMI exams.

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Practice Questions

:::question type="MCQ" question="Let $f: \mathbb{N} \to \mathbb{N}$ be defined by $f(x) = x+1$. Is $f$ an onto function?" options=["Yes, because every natural number has a successor.","No, because $1$ in the codomain has no pre-image in $\mathbb{N}$.","Yes, because for every $x$, $f(x)$ is a natural number.","No, because $f$ is one-to-one."] answer="No, because $1$ in the codomain has no pre-image in $\mathbb{N}$." hint="Check if every natural number in the codomain can be expressed as $x+1$ for some natural number $x$." solution="Step 1: Understand the function and its domain/codomain.
> $f: \mathbb{N} \to \mathbb{N}$ means the domain $X = \{1, 2, 3, \ldots\}$ and the codomain $Y = \{1, 2, 3, \ldots\}$.
> The function is $f(x) = x+1$.

Step 2: Test for an arbitrary $y$ in the codomain.
> Let $y=1$ (the smallest element in the codomain). We need to find an $x \in \mathbb{N}$ such that $f(x)=1$.
> Setting $x+1=1$, we get $x=0$.

Step 3: Check if $x$ is in the domain.
> The value $x=0$ is not an element of the natural numbers $\mathbb{N}$.
> Therefore, $y=1$ in the codomain has no pre-image in the domain $\mathbb{N}$.

Answer: No, because $1$ in the codomain has no pre-image in $\mathbb{N}$."
:::

:::question type="NAT" question="Let $X = \{1, 2, 3, 4, 5\}$ and $Y = \{a, b, c\}$. If $f: X \to Y$ is an onto function, what is the minimum number of elements in $X$ that must map to a specific element in $Y$?" answer="1" hint="An onto function requires at least one pre-image for every element in the codomain." solution="Step 1: Understand the definition of an onto function.
> A function $f: X \to Y$ is onto if for every element $y \in Y$, there exists at least one element $x \in X$ such that $f(x)=y$.

Step 2: Apply to the question.
> The question asks for the minimum number of elements in $X$ that must map to a specific element in $Y$.
> The definition of onto directly states 'at least one'. There is no requirement for more than one pre-image for any element for the function to be onto.

Step 3: Consider an example.
> Let $Y = \{a,b,c\}$. If $f(1)=a, f(2)=a, f(3)=b, f(4)=c, f(5)=c$. Here, 'a' has 2 pre-images, 'b' has 1, 'c' has 2. The function is onto. The minimum number of pre-images for any element (like 'b') is 1.

Answer: 1"
:::

:::question type="MCQ" question="Let $f: \mathbb{R} \setminus \{1\} \to \mathbb{R} \setminus \{2\}$ be defined by $f(x) = \frac{2x+1}{x-1}$. Is $f$ an onto function?" options=["Yes, because for any $y$ in the codomain, $x = \frac{y+1}{y-2}$ is in the domain.","No, because $f(x)$ can never be $2$.","Yes, because the domain and codomain are properly restricted.","No, because the function is not one-to-one."] answer="Yes, because for any $y$ in the codomain, $x = \frac{y+1}{y-2}$ is in the domain." hint="To prove onto, take an arbitrary $y$ from the codomain and solve for $x$. Then check if this $x$ is always in the domain." solution="Step 1: Take an arbitrary $y$ from the codomain.
> Let $y \in \mathbb{R} \setminus \{2\}$ be an arbitrary element in the codomain. This means $y \ne 2$.

Step 2: Solve $f(x)=y$ for $x$.
>

>
>
>
>
>

Step 3: Check if $x$ is in the domain.
> The domain is $\mathbb{R} \setminus \{1\}$. We need to ensure that $x = \frac{y+1}{y-2}$ is never equal to $1$ for $y \in \mathbb{R} \setminus \{2\}$.
> If $x=1$, then $1 = \frac{y+1}{y-2}$. This implies $y-2 = y+1$, which simplifies to $-2 = 1$, a contradiction.
> Therefore, $x$ can never be $1$.
> Also, since $y \ne 2$ (by definition of the codomain), the denominator $y-2$ is never zero, so $x$ is always well-defined.
> Thus, for every $y \in \mathbb{R} \setminus \{2\}$, the corresponding $x = \frac{y+1}{y-2}$ is a real number and $x \ne 1$, so $x$ is in the domain $\mathbb{R} \setminus \{1\}$.

Answer: Yes, because for any $y$ in the codomain, $x = \frac{y+1}{y-2}$ is in the domain."
:::

:::question type="MCQ" question="Which of the following functions $f: \mathbb{R} \to \mathbb{R}$ is onto?" options=["$f(x) = |x|$","$f(x) = e^x$","$f(x) = x^3$","$f(x) = \cos(x)$"] answer="$f(x) = x^3$" hint="For a function $f: \mathbb{R} \to \mathbb{R}$ to be onto, its range must be $\mathbb{R}$. Check the range of each function." solution="Step 1: Analyze $f(x) = |x|$.
> The range of $f(x) = |x|$ is $[0, \infty)$. Since $[0, \infty) \ne \mathbb{R}$, this function is not onto.

Step 2: Analyze $f(x) = e^x$.
> The range of $f(x) = e^x$ is $(0, \infty)$. Since $(0, \infty) \ne \mathbb{R}$, this function is not onto.

Step 3: Analyze $f(x) = x^3$.
> For any real number $y$, we can find a real number $x = \sqrt[3]{y}$ such that $f(x) = (\sqrt[3]{y})^3 = y$. The cube root of any real number is a real number. Thus, the range of $f(x) = x^3$ is $\mathbb{R}$, which equals the codomain. This function is onto.

Step 4: Analyze $f(x) = \cos(x)$.
> The range of $f(x) = \cos(x)$ is $[-1, 1]$. Since $[-1, 1] \ne \mathbb{R}$, this function is not onto.

Answer: $f(x) = x^3$"
:::

:::question type="MSQ" question="Let $f: [0, \infty) \to [0, \infty)$ be defined by $f(x) = x^2$. Which of the following statements are true?" options=["$f$ is onto.","$f$ is one-to-one.","$f$ is bijective.","The equation $f(x)=y$ has exactly one solution for $x$ in the domain for every $y$ in the codomain."] answer="$f$ is onto.,$f$ is one-to-one.,$f$ is bijective.,The equation $f(x)=y$ has exactly one solution for $x$ in the domain for every $y$ in the codomain." hint="Carefully consider the restricted domain and codomain. Analyze injectivity and surjectivity under these conditions." solution="Step 1: Check if $f$ is onto.
> Let $y \in [0, \infty)$ be an arbitrary element in the codomain. We need to find $x \in [0, \infty)$ such that $f(x)=y$, i.e., $x^2=y$.
> Solving for $x$, we get $x = \pm \sqrt{y}$. Since the domain is $[0, \infty)$, we must choose $x = \sqrt{y}$.
> For any $y \in [0, \infty)$, $\sqrt{y}$ is a real number and $\sqrt{y} \ge 0$, so $x=\sqrt{y}$ is in the domain $[0, \infty)$.
> Thus, $f$ is onto.

Step 2: Check if $f$ is one-to-one.
> Let $x_1, x_2 \in [0, \infty)$ such that $f(x_1) = f(x_2)$.
> Then $x_1^2 = x_2^2$. This implies $x_1 = \pm x_2$.
> Since both $x_1$ and $x_2$ are in $[0, \infty)$, they must be non-negative. Therefore, $x_1 = x_2$.
> Thus, $f$ is one-to-one.

Step 3: Check if $f$ is bijective.
> Since $f$ is both onto and one-to-one, it is bijective.

Step 4: Check the number of solutions for $f(x)=y$.
> For any $y \in [0, \infty)$, we found that $x=\sqrt{y}$ is the unique solution in the domain $[0, \infty)$.
> Thus, the equation $f(x)=y$ has exactly one solution for $x$ in the domain for every $y$ in the codomain.

Answer: $f$ is onto.,$f$ is one-to-one.,$f$ is bijective.,The equation $f(x)=y$ has exactly one solution for $x$ in the domain for every $y$ in the codomain."
:::

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Summary

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What's Next?

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Part 3: Bijective functions

Bijective Functions

Overview

A bijective function is one that is both one-one and onto. In CMI-style problems, bijections are important not only as definitions, but as the correct language for invertibility, counting arguments, composition, and constructing functions between sets. The main idea is simple: every output is hit exactly once. ---

Learning Objectives

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Core Definitions

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Test for Injectivity

Example For $f:\mathbb{R}\to\mathbb{R}$ given by $\qquad f(x)=3x-5$ Suppose $\qquad f(x_1)=f(x_2)$ Then $\qquad 3x_1-5=3x_2-5$ So $\qquad 3x_1=3x_2$ Hence $\qquad x_1=x_2$ Thus $f$ is injective. ---

Test for Surjectivity

Example For $f:\mathbb{R}\to\mathbb{R}$ given by $\qquad f(x)=3x-5$ Take any $y\in\mathbb{R}$. Solve $\qquad 3x-5=y$ Then $\qquad x=\dfrac{y+5}{3}$ Since this $x$ is real, every real $y$ is hit. So $f$ is surjective. Hence $f$ is bijective. ---

Bijectivity and Inverse Functions

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Finite Set View

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Graphical View on Real Intervals

Example

• $f(x)=x^3$ from $\mathbb{R}$ to $\mathbb{R}$ is bijective

• $f(x)=x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is not injective and not surjective

• $f(x)=x^2$ from $[0,\infty)$ to $[0,\infty)$ is bijective

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Domain and Codomain Matter

This is one of the most tested conceptual traps. ---

Composition of Bijective Functions

Also, $\qquad (g\circ f)^{-1}=f^{-1}\circ g^{-1}$ ---

Standard Examples

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Minimal Worked Examples

Example 1 Check whether $\qquad f:\mathbb{R}\to\mathbb{R},\ f(x)=x^3+1$ is bijective. Since $x^3+1$ is strictly increasing on $\mathbb{R}$, it is injective. For any $y\in\mathbb{R}$, solve $\qquad x^3+1=y$ so $\qquad x=\sqrt[3]{y-1}$ This is real for every real $y$. Hence $f$ is surjective and therefore bijective. --- Example 2 Check whether $\qquad f:\mathbb{R}\to\mathbb{R},\ f(x)=x^2+1$ is bijective. It is not injective because $\qquad f(1)=f(-1)=2$ It is not surjective onto $\mathbb{R}$ because values less than $1$ are never attained. So it is not bijective. ---

CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Which of the following functions is bijective?" options=["$f:\mathbb{R}\to\mathbb{R},\\ f(x)=x^2$","$f:\mathbb{R}\to\mathbb{R},\\ f(x)=x^3$","$f:\mathbb{R}\to\mathbb{R},\\ f(x)=\sin x$","$f:\mathbb{R}\to\mathbb{R},\\ f(x)=x^2+1$"] answer="B" hint="Check both injectivity and surjectivity." solution="Among the given functions, $f(x)=x^3$ is strictly increasing on $\mathbb{R}$, so it is injective. Also every real number is attained, so it is surjective onto $\mathbb{R}$. Hence it is bijective. The others fail injectivity, surjectivity, or both. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Let $f:[0,\infty)\to[0,\infty)$ be given by $f(x)=x^2$. Find $f^{-1}(16)$." answer="4" hint="On $[0,\infty)$, $x^2$ is bijective." solution="Since the domain and codomain are both $[0,\infty)$, the function $f(x)=x^2$ is bijective. Its inverse is $\qquad f^{-1}(y)=\sqrt{y}$ Therefore $\qquad f^{-1}(16)=\sqrt{16}=4$ Hence the answer is $\boxed{4}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["Every bijective function is injective","Every bijective function is surjective","If $A$ and $B$ are finite sets with $|A|=|B|$, then every injective map $A\to B$ is bijective","If a function has an inverse function, then it must be bijective"] answer="A,B,C,D" hint="Use the definitions and the inverse-function theorem." solution="1. True, by definition of bijection.

True, by definition of bijection.

True, for finite sets of equal size, injective automatically implies surjective.

True, a function has an inverse if and only if it is bijective.

Hence all four statements are true. Therefore the answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Show that the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=2x-3$ is bijective, and find its inverse." answer="$f^{-1}(y)=\dfrac{y+3}{2}$" hint="Check injective and surjective separately." solution="To prove injectivity, suppose $\qquad f(x_1)=f(x_2)$ Then $\qquad 2x_1-3=2x_2-3$ So $\qquad 2x_1=2x_2$ and hence $\qquad x_1=x_2$ Therefore $f$ is injective. To prove surjectivity, let $y\in\mathbb{R}$. Solve $\qquad 2x-3=y$ Then $\qquad x=\dfrac{y+3}{2}$ This is a real number, so every real $y$ has a preimage. Thus $f$ is surjective. Hence $f$ is bijective. Now solve $\qquad y=2x-3$ for $x$: $\qquad x=\dfrac{y+3}{2}$ Therefore $\qquad f^{-1}(y)=\dfrac{y+3}{2}$ So the inverse is $\boxed{f^{-1}(y)=\dfrac{y+3}{2}}$." ::: ---

Summary

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Part 4: Constant functions

Constant Functions

Overview

A constant function is one of the simplest types of functions, but in exam questions it is often hidden inside logical language, range-based statements, injectivity, surjectivity, and composition. The main idea is that every input gets mapped to the same output. In CMI-style questions, the challenge is usually not the formula itself, but recognizing equivalent ways of saying that a function is constant. ---

Learning Objectives

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Core Definition

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Most Important Characterization

Why the word unique appears If all outputs are equal to one fixed value, then that value is the only element of $Y$ that works in the statement. So:

• existence means there is at least one such output value

• uniqueness means there is exactly one such output value

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Range of a Constant Function

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Constant Function vs One-One and Onto

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Graphical Understanding

Examples

• $f(x)=5$ gives the line $y=5$

• $f(x)=-2$ gives the line $y=-2$

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Algebraic Examples

Example 1 $\qquad f:\mathbb{R}\to\mathbb{R},\quad f(x)=7$ This is constant because every input maps to $7$. Its range is $\qquad \{7\}$ --- Example 2 $\qquad g:\mathbb{Z}\to\mathbb{Z},\quad g(n)=0$ This is also constant. Even though the domain is infinite, the output is still only one value. ---

Equivalent Ways to Recognize a Constant Function

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Constant Functions and Composition

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Constant Polynomial Functions

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Minimal Worked Examples

Example 1 Let $f:X \to Y$ satisfy $\qquad f(x)=a \text{ for every } x \in X$ Then $f$ is constant by definition, and its range is $\{a\}$. --- Example 2 Suppose $|X|>1$ and $f:X \to Y$ is constant. Choose distinct $u,v \in X$. Then $\qquad f(u)=f(v)$ So $f$ cannot be one-one. ---

Common Traps

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Let $f:X \to Y$ be a function with nonempty domain $X$. Which of the following is equivalent to $f$ being a constant function?" options=["For every $x_1,x_2 \in X$, $f(x_1)=f(x_2)$","$f$ is one-one","$f$ is onto","$\text{Codomain}(f)$ has exactly one element"] answer="A" hint="A constant function gives the same output for every input." solution="A function is constant exactly when all inputs give the same output. So the correct condition is $\qquad \forall x_1,x_2 \in X,\ f(x_1)=f(x_2)$ The function need not be one-one or onto, and the codomain need not have one element. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=5$. Find the number of elements in the range of $f$." answer="1" hint="A constant function on a nonempty domain has exactly one output value." solution="Since $\qquad f(x)=5 \text{ for all } x \in \mathbb{R}$, the set of actual outputs is $\qquad \{5\}$ So the range has exactly one element. Therefore, the answer is $\boxed{1}$." ::: :::question type="MSQ" question="Assume $|X|>1$, $|Y|>1$, and $f:X \to Y$ is a constant function. Which of the following statements are true?" options=["$f$ is not one-one","$f$ is onto","The range of $f$ has exactly one element","There exists a unique $y \in Y$ such that for every $x \in X$, $f(x)=y$"] answer="A,C,D" hint="Use the definition and compare range with codomain." solution="Since $f$ is constant, all elements of $X$ map to one fixed value.

True. Because $|X|>1$, distinct inputs get the same output, so $f$ is not one-one.

False. Since $|Y|>1$ but the range has only one element, $f$ is not onto.

True. A constant function on a nonempty domain has range of size $1$.

True. This is exactly a logical characterization of a constant function.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Prove that for a function $f:X \to Y$ with nonempty domain $X$, the following are equivalent: (i) $f$ is constant, (ii) there exists a unique $y \in Y$ such that for every $x \in X$, $f(x)=y$." answer="The two statements are equivalent." hint="Prove each direction separately." solution="We prove both directions. First, assume that $f$ is constant. Then by definition, there exists some element $c \in Y$ such that $\qquad f(x)=c \text{ for every } x \in X$ So there exists at least one $y \in Y$ with the required property, namely $y=c$. Now we show uniqueness. Suppose $y_1,y_2 \in Y$ both satisfy $\qquad f(x)=y_1 \text{ for every } x \in X$ and $\qquad f(x)=y_2 \text{ for every } x \in X$ Since $X$ is nonempty, choose any $a \in X$. Then $\qquad y_1=f(a)=y_2$ Hence the required element is unique. So (i) implies (ii). Now assume that there exists a unique $y \in Y$ such that for every $x \in X$, $\qquad f(x)=y$ Then all elements of the domain have the same image, namely $y$. Therefore $f$ is a constant function. So (ii) implies (i). Hence the two statements are equivalent, and the result is proved. Therefore, the answer is $\boxed{\text{(i) and (ii) are equivalent}}$." ::: ---

Summary

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Part 5: Identity function

Identity Function

Overview

The identity function is the simplest possible function on a set: it sends every element to itself. Even though it looks elementary, it plays a central role in composition, inverse functions, bijections, and function algebra. In abstract questions, the identity function behaves like the number $1$ does for multiplication: it is the neutral element for composition. ---

Learning Objectives

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Definition

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Basic Properties

This is exactly why the identity function is called the neutral element under composition. ---

Identity and Inverses

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Graph of the Identity Function

This line is important because:

• it is the graph of the identity function,

• the graph of an inverse function is the reflection of the graph of $f$ in the line

$\qquad y=x$ ::: ---

Identity Function Is Bijective

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Fixed Points and Identity

Every element of $A$ is a fixed point of $\operatorname{id}_A$. So the identity function is the function with all points fixed. ---

Distinguishing Identity from "Looks Like Identity"

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Composition Examples

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Minimal Worked Examples

Example 1 Let $f:\mathbb{R}\to\mathbb{R}$ be given by $\qquad f(x)=x+1$ Then $\qquad f\circ \operatorname{id}_{\mathbb{R}} = f$ and $\qquad \operatorname{id}_{\mathbb{R}}\circ f = f$ --- Example 2 Let $\qquad f(x)=2x+3$ Then $\qquad f^{-1}(x)=\dfrac{x-3}{2}$ Now $\qquad f^{-1}(f(x))=\dfrac{(2x+3)-3}{2}=x=\operatorname{id}_{\mathbb{R}}(x)$ and similarly $\qquad f(f^{-1}(x))=x$ So composition with the inverse gives the identity function. ---

Functional Equations Involving Identity

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CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="The identity function on $\mathbb{R}$ is" options=["$f(x)=0$","$f(x)=1$","$f(x)=x$","$f(x)=x^2$"] answer="C" hint="Identity sends each element to itself." solution="The identity function on $\mathbb{R}$ sends each real number to itself. So it is given by $\qquad f(x)=x$ Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="If $f(x)=2x-5$ on $\mathbb{R}$, find $(f^{-1}\circ f)(7)$." answer="7" hint="Composition of a bijection with its inverse gives identity." solution="Since $f(x)=2x-5$ is bijective on $\mathbb{R}$, its inverse exists. Also, $\qquad f^{-1}\circ f = \operatorname{id}_{\mathbb{R}}$ Hence $\qquad (f^{-1}\circ f)(7)=\operatorname{id}_{\mathbb{R}}(7)=7$ Therefore the answer is $\boxed{7}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["For every set $A$, the identity function $\operatorname{id}_A$ is bijective","$\operatorname{id}_A^{-1}=\operatorname{id}_A$","If $f:A\to B$, then $f\circ \operatorname{id}_B=f$","If $f:A\to B$, then $\operatorname{id}_B\circ f=f$"] answer="A,B,D" hint="Check domain compatibility in composition." solution="1. True. Identity is always injective and surjective.

True. Identity is its own inverse.

False in general. The composition $f\circ \operatorname{id}_B$ is not even defined unless the output of $\operatorname{id}_B$ lies in the domain of $f$, which would require $B=A$ in the usual setting.

True. Since $\operatorname{id}_B$ acts on the codomain of $f$, we have

$\qquad \operatorname{id}_B\circ f=f$ Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Show that if a function $f:A\to A$ satisfies $f(x)=x$ for every $x\in A$, then $f=\operatorname{id}_A$." answer="Such a function is exactly the identity function on $A$." hint="Compare the definition directly." solution="By definition, the identity function on $A$ is the function $\qquad \operatorname{id}_A:A\to A$ given by $\qquad \operatorname{id}_A(x)=x$ for every $x\in A$ Now the given function $f:A\to A$ satisfies $\qquad f(x)=x$ for every $x\in A$ So for each element of $A$, the functions $f$ and $\operatorname{id}_A$ give the same value. Therefore they are the same function. Hence $\qquad f=\operatorname{id}_A$ So the required result is proved." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="Which of the following functions $f: \mathbb{R} \to \mathbb{R}$ is injective but not surjective?" options=["$f(x) = x^2$", "$f(x) = e^x$", "$f(x) = x^3$", "$f(x) = \sin x$"] answer="$f(x) = e^x$" hint="Consider the range of each function relative to the codomain $\mathbb{R}$." solution="The function $f(x) = x^2$ is neither injective (e.g., $f(-1)=f(1)=1$) nor surjective (range is $[0, \infty)$). The function $f(x) = e^x$ is injective ($e^{x_1} = e^{x_2} \implies x_1 = x_2$) but not surjective because its range is $(0, \infty)$, which is a proper subset of $\mathbb{R}$. The function $f(x) = x^3$ is both injective and surjective (bijective). The function $f(x) = \sin x$ is neither injective nor surjective."
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:::question type="NAT" question="Let $A = \{1, 2, 3\}$ and $B = \{a, b, c, d\}$. How many distinct constant functions can be defined from $A$ to $B$?" answer="4" hint="A constant function maps all elements of its domain to a single element in its codomain." solution="For a function $f: A \to B$ to be constant, all elements in $A$ must map to the same element in $B$. Since there are 4 choices for this single element in $B$ (either $a$, $b$, $c$, or $d$), there are 4 distinct constant functions."
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:::question type="MCQ" question="If $f: A \to B$ is a bijective function, which of the following statements must be true?" options=["$|A| < |B|$", "$|A| > |B|$", "$|A| = |B|$", "The function is necessarily $f(x)=x$"] answer="$|A| = |B|$" hint="Recall the cardinality implications of injective and surjective functions for finite sets." solution="A bijective function is both injective and surjective. For finite sets, injectivity implies $|A| \le |B|$, and surjectivity implies $|A| \ge |B|$. Therefore, for a bijective function, it must be that $|A| = |B|$. The function is not necessarily $f(x)=x$; that is the identity function, a specific type of bijective function."
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:::question type="NAT" question="Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = (2k-1)x + (m+3)$. If $f$ is the identity function, find the value of $k+m$." answer="1" hint="The identity function is defined by $f(x) = x$ for all $x$ in its domain." solution="For $f(x) = (2k-1)x + (m+3)$ to be the identity function, it must satisfy $f(x) = x$ for all $x \in \mathbb{R}$. This means the coefficient of $x$ must be 1, and the constant term must be 0.
So, $2k-1 = 1 \implies 2k = 2 \implies k = 1$.
And $m+3 = 0 \implies m = -3$.
Therefore, $k+m = 1 + (-3) = -2$."
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What's Next?