NDA Maths · Functions

Domain, Range, and the Standard Properties

Find where a function is allowed to live (domain), what values it produces (range), and whether it is even, odd, or periodic.

All Functions notes NDA Maths strategy

Why this matters

The biggest slice of the chapter — 48 PYQs, and most are EASY or MODERATE. The recurring asks are narrow and learnable: domain from square-roots / denominators / logs, range of a bounded rational or quadratic-on-an-interval, even-vs-odd, and period. A few standard functions (modulus, reciprocal, sign, exponential) show up again and again. Drill these and you bank the chapter's easy marks.

Concept 1 of 7

Reading domain and range from a graph

Intuition

On a graph, the domain is the shadow the curve casts on the x-axis (how far left–right it extends) and the range is its shadow on the y-axis (how far up–down). Sweep across, then sweep up.

Definition

Domain $=$ set of $x$ -values the graph covers; range $=$ set of $y$ -values it reaches. A filled dot / solid edge includes the endpoint (closed); an open dot / asymptote excludes it (open). Reading them off a sketch is often faster than algebra for standard curves.

Worked example

The graph of

f

is the upper semicircle of radius 2 centred at the origin. Domain and range?

Horizontally the semicircle runs from $-2$ to $2$ : domain $[-2,2]$ .
Vertically it runs from $0$ (the ends) up to $2$ (the top): range $[0,2]$ .

Answer:Domain

[-2,2]

, range

[0,2]

Practice this concept4 quick reps

Concept 2 of 7

Finding the domain (roots, denominators, logs)

Intuition

The domain is everything you are allowed to feed in. Three rules cover almost every NDA question: don't divide by zero, don't take an even root of a negative, and only take logs of positives. Apply each restriction and intersect.

Definition

Even root $\sqrt[\text{even}]{g(x)}$ : need $g(x)\ge 0$ .
Denominator: need denominator $\neq 0$ .
Logarithm $\log g(x)$ : need $g(x)>0$ ; base must be $>0,\neq1$ .

When several appear, the domain is the intersection of all the individual conditions.

Worked example

Find the domain of

f(x)=\dfrac{1}{\sqrt{5-x}}

Square root needs $5-x\ge 0\Rightarrow x\le 5$ .
But it sits in a denominator, so it must be $\neq 0$ : $5-x>0\Rightarrow x<5$ (strict).
Combine: $x<5$ .

Answer:Domain

=(-\infty,5)

Practice this conceptself-check

From the bank · past-year question

Example 2FunctionsMODERATE

The domain of the function

f(x)=\sqrt{(2-x)(x-3)}

[Q80 · Apr · 2019]

≥ 0 under a plain root, but > 0 when the root is a denominator

A root by itself allows equality (

\sqrt{g}

needs

g\ge0

). The moment that root is in a denominator — e.g.

\dfrac{1}{\sqrt{|x|-x}}

— the value 0 is banned too, so you need the strict inequality

g>0

. Missing this flips a closed bracket to an open one and loses the mark.

Drill 10 more on finding the domain (roots, denominators, logs)

Concept 3 of 7

Finding the range

Intuition

The range is the set of outputs. For bounded rationals, solve for

x

in terms of

y

and ask which

y

keep

x

real; for a quadratic on an interval, use the vertex and endpoints; for sine/cosine combinations, bound the amplitude.

Definition

Common techniques:

**Solve for $x$ **: rearrange $y=f(x)$ to $x=\dots$ ; the range is the $y$ for which $x$ is real/in-domain.
Quadratic on an interval: check the vertex and the endpoints; mind whether endpoints are included.
** $a\sin x+b\cos x+c$ ** lies in $[c-\sqrt{a^2+b^2},\,c+\sqrt{a^2+b^2}]$ .

Worked example

Find the range of

f(x)=\dfrac{1}{1+x^2}

x\in\mathbb{R}

$1+x^2\ge 1$ , and it grows without bound as $|x|\to\infty$ .
So the denominator runs over $[1,\infty)$ , hence $f=\dfrac{1}{1+x^2}$ runs over $(0,1]$ .
Max $1$ at $x=0$ (attained); $0$ is approached but never reached.

Answer:Range

=(0,1]

Practice this conceptself-check

From the bank · past-year question

Example 3FunctionsMODERATE

For

f(x)=\frac{x^2}{1+x^2}

, what is the range?

[Q72 · Apr · 2018]

Range is not the codomain

If a question declares

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

but the outputs only fill

[0,1)

, the range is

[0,1)

— not

\mathbb{R}

. 'Onto' questions are really 'shrink the codomain to the range' questions.

Drill 16 more on finding the range

Concept 4 of 7

Even and odd functions

Intuition

An even function is mirror-symmetric about the y-axis (

f(-x)=f(x)

); an odd function has half-turn symmetry about the origin (

f(-x)=-f(x)

). Most functions are neither — only special ones qualify.

Definition

Even: $f(-x)=f(x)$ for all $x$ (e.g. $x^2,\ \cos x,\ |x|$ ).
Odd: $f(-x)=-f(x)$ for all $x$ (e.g. $x^3,\ \sin x,\ x$ ). An odd function defined at 0 has $f(0)=0$ .
Test by computing $f(-x)$ and comparing. If it matches neither, the function is neither.

Even and odd tests

f\text{ even}\iff f(-x)=f(x)\qquad f\text{ odd}\iff f(-x)=-f(x)

Worked example

Classify

f(x)=x^3-x

as even, odd, or neither.

Compute $f(-x)=(-x)^3-(-x)=-x^3+x$ .
Factor: $-x^3+x=-(x^3-x)=-f(x)$ .
Matches the odd condition $f(-x)=-f(x)$ .

Answer:Odd.

Practice this conceptself-check

From the bank · past-year question

Example 4FunctionsMODERATE

f(x) = \ln(x+\sqrt{1+x^2})

, then which one of the following is correct?

[Q83 · Sep · 2022]

$f(0)=0$ is necessary for odd, not sufficient

Many odd functions pass through the origin, but passing through the origin does not make a function odd — you must verify

f(-x)=-f(x)

for all

x

. And a sum like even + odd is usually neither.

Drill 5 more on even and odd functions

Concept 5 of 7

Periodic functions and their period

Intuition

A periodic function repeats: its graph is one tile copied endlessly. The period is the width of the smallest tile. For combinations you scale the base period, then take the LCM.

Definition

$f$ is periodic with period $T>0$ if $f(x+T)=f(x)$ for all $x$ ; the smallest such $T$ is the period.

$\sin x,\cos x$ : period $2\pi$ ; $\tan x$ : period $\pi$ .
$\sin(kx)$ has period $\tfrac{2\pi}{|k|}$ ; $\sin^2 x$ has period $\pi$ .
Period of a sum is the LCM of the individual periods.

Period after scaling the argument

\text{period of }\sin(kx)=\dfrac{2\pi}{|k|}\qquad \text{period of }\tan(kx)=\dfrac{\pi}{|k|}\qquad \text{period of }f(ax+b)=\dfrac{T}{|a|}

Worked example

Find the period of

f(x)=\cos\!\left(\dfrac{x}{2}\right)

$\cos(kx)$ has period $\dfrac{2\pi}{|k|}$ ; here $k=\tfrac12$ .
Period $=\dfrac{2\pi}{1/2}=4\pi$ .

Answer:

4\pi

Practice this conceptself-check

From the bank · past-year question

Example 5FunctionsEASY

What is the period of the function

f(x)=\ln(2+\sin^{2}x)

[Q23 · Sep · 2021]

Drill 6 more on periodic functions and their period

Concept 6 of 7

The modulus function and distance

Intuition

|x|

measures distance from 0, so it is never negative and its V-shaped graph bounces off the x-axis at the origin. Sums of moduli like

|x-a|+|x-b|

read as total distance to two points — minimised anywhere between them.

Definition

$|x|=x$ for $x\ge0$ and $-x$ for $x<0$ ; it is even, with range $[0,\infty)$ .

$x+|x|=0$ for $x<0$ and $2x$ for $x\ge0$ — range $[0,\infty)$ .
$|x-a|+|x-b|$ (with $a<b$ ) has minimum value $b-a$ , attained for all $x\in[a,b]$ .

Worked example

Find the minimum value of

f(x)=|x-1|+|x-5|

Read it as the total distance from $x$ to $1$ and to $5$ .
That total is smallest when $x$ lies between them; then it equals the gap $5-1=4$ .
(For $x$ outside $[1,5]$ the total only grows.)

Answer:Minimum value

=4

, attained for all

x\in[1,5]

From the bank · past-year question

Example 6FunctionsEASY

What is the range of the function

f(x)=x+|x|

if the domain is the set of real numbers?

[Q97 · Apr · 2023]

$|x|$ is even, never odd

|-x|=|x|

, so the modulus is even. Combinations like

|x|-x^3

mix an even and an odd part and end up neither. Also:

\sqrt{x^2}=|x|

, not

x

— a common sign slip.

Drill 3 more on the modulus function and distance

Concept 7 of 7

Standard functions and their graphs

Intuition

A handful of standard shapes recur: the reciprocal

1/x

(two branches, asymptotes), the sign function

x/|x|

(a step of

\pm1

), and the exponential

a^x

(domain all reals, positive output). Knowing each one's domain, range and asymptotes answers most graph questions on sight.

Definition

Reciprocal $\dfrac{1}{x-a}$ : domain $x\neq a$ , range $y\neq0$ ; vertical asymptote $x=a$ , horizontal $y=0$ .
Sign $\dfrac{x}{|x|}$ : equals $+1$ for $x>0$ , $-1$ for $x<0$ ; undefined at 0.
Exponential $a^x\,(a>0)$ : domain $\mathbb{R}$ , range $(0,\infty)$ , continuous and differentiable everywhere.

Worked example

State the domain, range and asymptotes of

f(x)=\dfrac{1}{x+2}

Denominator zero at $x=-2$ : domain $\mathbb{R}\setminus\{-2\}$ .
It never outputs 0: range $\mathbb{R}\setminus\{0\}$ .
Vertical asymptote $x=-2$ ; horizontal asymptote $y=0$ .

Answer:Domain

x\neq-2

, range

y\neq0

; asymptotes

x=-2

and

y=0

Practice this concept4 quick reps

From the bank · past-year question

Example 7FunctionsMODERATE

Which one of the following is correct in respect of the graph of

y=\frac{1}{x-1}

[Q98 · Apr · 2020]

Drill 6 more on standard functions and their graphs

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 (2)

Even and odd functions
Even and odd tests
$f\text{ even}\iff f(-x)=f(x)\qquad f\text{ odd}\iff f(-x)=-f(x)$
Periodic functions and their period
Period after scaling the argument
$\text{period of }\sin(kx)=\dfrac{2\pi}{|k|}\qquad \text{period of }\tan(kx)=\dfrac{\pi}{|k|}\qquad \text{period of }f(ax+b)=\dfrac{T}{|a|}$

Watch out for (4)

≥ 0 under a plain root, but > 0 when the root is a denominator
→ Finding the domain (roots, denominators, logs)
Range is not the codomain
→ Finding the range
$f(0)=0$ is necessary for odd, not sufficient
→ Even and odd functions
$|x|$ is even, never odd
→ The modulus function and distance

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