JEE Mains Maths · Relations and Functions

Domain of a Function

Finding where a function is defined: the conditions set by roots, logarithms, denominators and inverse trigonometric functions, nested logarithms and greatest-integer parts, and composite functions.

All Relations and Functions notes JEE Mains Maths strategy

Why this matters

Thirty-four PYQs, the second-largest page in the chapter. Most give the domain as a union of intervals and asks for a sum of its endpoints, so a single wrong bracket or a missed exclusion costs the whole question. Three ideas cover the page.

Concept 1 of 3: The conditions that fix a domain

Write down the condition each part of the formula imposes, solve each one, and intersect the answers. An even root needs its inside

\ge0

, or

>0

if it sits in a denominator. A logarithm needs its argument

>0

and its base positive and not 1. A denominator must not be 0.

\sin^{-1}u

and

\cos^{-1}u

need

-1\le u\le1

; for

\left|\frac pq\right|\le1

with

q

of either sign, square to

p^2\le q^2

and keep

q\ne0

Definition

$\sqrt u$ : $u\ge0$ . $\frac1u$ : $u\ne0$ .
$\log_b u$ : $u>0$ , $b>0$ , $b\ne1$ .
$\sin^{-1}u,\ \cos^{-1}u$ : $-1\le u\le1$ .
The domain of a sum or product is the intersection of the parts' domains.

Inverse-sine domain

\sin^{-1}u,\ \cos^{-1}u:\ -1\le u\le1

Worked example

Find the domain of

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q71Moderate

Example 1 · Relations and Functions · Domain of a Function

If the domain of the function

f(x) =\sin^{- 1}\left( \frac{x - 1}{2x + 3} \right)

R - (\alpha,\beta)

then

12\alpha\beta

is equal to:

A logarithm in a denominator

\frac1{\log u}

needs

\log u\ne0

as well as

u>0

: the argument

u=1

is excluded. This is the usual missing point in a domain written as 'an interval minus a point'.

Drill 18 more on the conditions that fix a domain

Concept 2 of 3: Nested logarithms and greatest-integer parts

Work from the outside in.

\log_a(\log_bu)

needs

\log_bu>0

, which for

b>1

means

u>1

; a third logarithm pushes the threshold up again, to

u>b

. A base below 1 reverses each inequality. When the formula has

[x]

, put

n=[x]

, solve for the integer

n

first, and then turn each allowed

n

into the interval

[n,n+1)

Definition

$\log_a\log_b u$ (bases $>1$ ): $u>1$ .
$\log_a\log_b\log_c u$ (bases $>1$ ): $u>c$ .
$\log_{b}u\ge0$ with $0<b<1$ : $0<u\le1$ .
$[x]=n\iff n\le x<n+1$ .

Two nested logarithms

\log_a(\log_b u)\ \text{defined}\iff u>1\quad(a,b>1)

Worked example

Find the domain of

\log_2\log_3(x-1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 11 April 2023 · Q159Moderate

Example 2 · Relations and Functions · Domain of a Function

The domain of the function

f(x) =\frac{1}{\sqrt{\lbrack x\rbrack^{2}- 3\lbrack x\rbrack - 10}}

is (where [x] denotes the greatest integer less than or equal to

x

)

Turn integers back into intervals

[x]\le-3

means

x<-2

, not

x\le-3

: every

x

with integer part

-3

lies in

[-3,-2)

Drill 11 more on nested logarithms and greatest-integer parts

Concept 3 of 3: Domain of a composite function

f(g(x))

needs two things:

x

must be in the domain of

g

, and

g(x)

must be in the domain of

f

. Solve the second condition as an inequality in

x

and intersect it with the first. Simplifying

f(g(x))

first can hide a point where

g

itself is undefined.

Definition

$\operatorname{dom}(f\circ g)=\{x\in\operatorname{dom}g:\ g(x)\in\operatorname{dom}f\}$ .
Keep every exclusion of $g$ , even if it cancels later.

Domain of f∘g

\{x\in\operatorname{dom}g:\ g(x)\in\operatorname{dom}f\}

Worked example

f(x)=\sqrt x

g(x)=1-x^2

. Find the domain of

f\circ g

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q156Moderate

Example 3 · Relations and Functions · Domain of a Function

Let

f:R -\left\{ \frac{- 1}{2} \right\}\rightarrow R

and

g:R -\left\{ \frac{- 5}{2} \right\}\rightarrow R

be defined as

f(x) =\frac{2x+ 3}{2x+ 1}

and

g(x) =\frac{|x| + 1}{2x+ 5}

. Then the domain of the function fog is:

The inner function's gaps stay

g

is undefined at a point, so is

f\circ g

, whatever the simplified formula says.

Drill 2 more on domain of a composite function

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 (3)

The conditions that fix a domain
Inverse-sine domain
$\sin^{-1}u,\ \cos^{-1}u:\ -1\le u\le1$
Nested logarithms and greatest-integer parts
Two nested logarithms
$\log_a(\log_b u)\ \text{defined}\iff u>1\quad(a,b>1)$
Domain of a composite function
Domain of f∘g
$\{x\in\operatorname{dom}g:\ g(x)\in\operatorname{dom}f\}$

Watch out for (3)

A logarithm in a denominator
→ The conditions that fix a domain
Turn integers back into intervals
→ Nested logarithms and greatest-integer parts
The inner function's gaps stay
→ Domain of a composite function

Test yourself on Relations and Functions

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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