NDA Maths · Functions

What a Function Is, and How to Classify It

A function assigns each input exactly one output; classifying it as one-one, onto, or bijective is about how inputs and outputs are paired.

All Functions notes NDA Maths strategy

Why this matters

Eight PYQs, all EASY–MODERATE — the vocabulary the rest of the chapter is built on. The bank tests three things: whether a given rule even is a function (the vertical-line / well-defined test), whether it is one-one and/or onto, and how to count functions of a given type. Get the definitions exact and these are free marks; blur 'onto' and 'into' and you lose them.

Concept 1 of 4

Domain, codomain, range — the vocabulary

Intuition

Every function comes with three sets. The domain is what you are allowed to put in, the codomain is the declared set of possible outputs, and the range is the part of the codomain you actually hit. Range is always a subset of the codomain — the gap between them is exactly what 'onto' is about.

Definition

For $f:A\to B$ :

Domain $=A$ : the set of all valid inputs.
Codomain $=B$ : the declared target set.
Range $=f(A)=\{f(x):x\in A\}$ : the set of values actually taken. Always $f(A)\subseteq B$ .
Image of $x$ is $f(x)$ ; pre-image of $y$ is any $x$ with $f(x)=y$ .

Worked example

For

f:\mathbb{R}\to\mathbb{R}

f(x)=x^2

, state the domain, codomain and range.

Domain is the declared input set $\mathbb{R}$ — every real has a square.
Codomain is the declared target $\mathbb{R}$ .
Range is what is actually produced: squares are $\ge 0$ , so range $=[0,\infty)$ .
Range $[0,\infty)\subsetneq\mathbb{R}$ — the negatives are never hit, so this is not onto.

Answer:Domain

\mathbb{R}

, codomain

\mathbb{R}

, range

[0,\infty)

Practice this concept4 quick reps

Concept 2 of 4

Is it a function? Well-defined and the vertical-line test

Intuition

A rule is a function only if every input gives exactly one output — no input left out, none sent to two places. On a graph this is the vertical-line test: any vertical line meets the graph at most once.

Definition

$f:A\to B$ is a function iff for every $x\in A$ there is one and only one $y\in B$ with $f(x)=y$ . Failures: a value with no output (gap in domain) or a value with two outputs (relation, not a function). A piecewise rule must agree at the join to stay well-defined.

Worked example

Does

y^2=x

define

y

as a function of

x

for

x>0

Solve for $y$ : $y=\pm\sqrt{x}$ .
For $x=4$ this gives $y=2$ and $y=-2$ — two outputs for one input.
The vertical line $x=4$ meets the curve twice.

Answer:No — one input has two outputs, so it is not a function.

From the bank · past-year question

Example 2FunctionsMODERATE

Let

A = \{x \in \mathbb{R} : -1 \leq x \leq 1\}

B = \{y \in \mathbb{R} : -1 \leq y \leq 1\}

and S be the subset of

A \times B

, defined by

S = \{(x,y) \in A \times B : x^2 + y^2 = 1\}

. Which one of the following is correct?

[Q25 · Sep · 2018]

Piecewise rules must agree at the boundary

A two-piece rule like

f(x)=x^2

[0,4]

and

3x

[4,8]

is only a function if the pieces give the same value at the shared point

x=4

(here

16\neq12

, so it is not well-defined). Always check the join before declaring it a function.

Drill 1 more on is it a function? well-defined and the vertical-line test

Concept 3 of 4

One-one, onto, and bijective

Intuition

One-one (injective): different inputs give different outputs — no two arrows land together. Onto (surjective): every element of the codomain is hit — range equals codomain. Bijective: both at once, so inputs and outputs pair up perfectly (and the function is invertible).

Definition

Injective: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ (equivalently, every horizontal line meets the graph at most once).
Surjective: range $=$ codomain, i.e. every $y\in B$ has a pre-image.
Bijective: injective and surjective. Only bijections have an inverse.
Monotonic ⇒ injective: a strictly increasing or strictly decreasing function is automatically one-one — the fastest injectivity test. If it is also onto its codomain, it is a bijection. (You can often see monotonicity directly, without calculus: $x|x|$ is $-x^2$ then $x^2$ , both increasing and meeting at 0, so it is strictly increasing and one-one.)

Onto depends on the codomain you declare — shrinking the codomain to the range makes any function onto.

Worked example

Classify

f:\mathbb{R}\to\mathbb{R}

f(x)=x^2

, as one-one / onto.

One-one? $f(2)=f(-2)=4$ — two inputs, one output, so not one-one.
Onto? Range is $[0,\infty)$ , which is not all of $\mathbb{R}$ — negatives are missed, so not onto.
If instead $f:[0,\infty)\to[0,\infty)$ , it becomes both one-one and onto — a bijection. Domain/codomain matter.

Answer:On

\mathbb{R}\to\mathbb{R}

: neither one-one nor onto.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3FunctionsMODERATE

Consider the following statements: 1. A function

f:\mathbb{Z}\to\mathbb{Z}

, defined by

f(x)=x+1

, is one-one as well as onto. 2. A function

f:\mathbb{N}\to\mathbb{N}

, defined by

f(x)=x+1

, is one-one but not onto. Which of the above statements is/are correct?

[Q38 · Apr · 2021]

'Onto' is not absolute — it depends on the codomain

f:\mathbb{N}\to\mathbb{N}

f(x)=x+1

is one-one but not onto (nothing maps to 1). The same rule on

\mathbb{Z}\to\mathbb{Z}

is onto. Read the declared domain and codomain before deciding.

A non-monotone function usually fails one-one (and often onto)

On a symmetric interval, an even-ish function doubles back:

\cos(\pi x)

(-1,1)

hits its maximum at

x=0

, so it is neither one-one nor onto

(-1,1)

. Contrast

x|x|

, which is strictly increasing and is a bijection of

(-1,1)

. Check monotonicity before declaring a bijection.

Drill 4 more on one-one, onto, and bijective

Concept 4 of 4

Counting functions of a given type

Intuition

Counting how many functions exist from a finite set to another is just the multiplication principle: each input independently picks an output. Restricting to one-one or onto narrows the count.

Definition

Let $|A|=m$ , $|B|=n$ .

All functions $A\to B$ : $n^m$ (each of $m$ inputs has $n$ choices).
One-one (needs $n\ge m$ ): $n(n-1)\cdots(n-m+1)={}^{n}P_{m}$ .
Onto (general): inclusion–exclusion; for $n=2$ it is $2^m-2$ .

Number of functions A → B

\text{functions }A\to B=n^{m}\qquad \text{injections}={}^{n}P_{m}=\dfrac{n!}{(n-m)!}\qquad \text{bijections }(m=n)=n!\qquad \text{relations}=2^{mn}

Worked example

With

|A|=3

|B|=4

: how many functions, and how many are one-one?

All functions: $n^m=4^3=64$ .
One-one: ${}^{4}P_{3}=4\cdot3\cdot2=24$ .

Answer:64 functions in all, of which 24 are one-one.

Practice this concept4 quick reps

From the bank · past-year question

Example 4FunctionsEASY

Let

A=\{1,2,3,4,5\}

and

B=\{6,7\}

. What is the number of onto functions from

A

B

[Q35 · Apr · 2024]

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Counting functions of a given type
Number of functions A → B
$\text{functions }A\to B=n^{m}\qquad \text{injections}={}^{n}P_{m}=\dfrac{n!}{(n-m)!}\qquad \text{bijections }(m=n)=n!\qquad \text{relations}=2^{mn}$

Watch out for (3)

Piecewise rules must agree at the boundary
→ Is it a function? Well-defined and the vertical-line test
'Onto' is not absolute — it depends on the codomain
→ One-one, onto, and bijective
A non-monotone function usually fails one-one (and often onto)
→ One-one, onto, and bijective

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