Find the domain and range of each of the following real functions: (i) f(x) = x² (ii) f(x) = (1)/(x)

Worked Examples · Example 16

Q. Find the domain and range of each of the following real functions:

(i)

f(x) = x^2

(ii)

f(x) = \frac{1}{x}

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Concept understanding — Types of Relations

Types of Relations – From Intuition to Precision

Imagine you have a set of people in a room. A relation is simply a rule that tells you whether two people are connected in some way. "Is the brother of", "lives in the same city as", "is taller than" — each of these is a relation. The question is: what kind of connection is it?

Some relations are very special. They behave in predictable, almost perfect ways. These are the ones we study first.

The Intuition: Three Key Properties

Think about the relation "lives in the same city as" between people.

Reflexive: Does every person live in the same city as themselves? Yes — obviously. A relation is reflexive if every element is related to itself.
Symmetric: If A lives in the same city as B, does B live in the same city as A? Yes — it's mutual. A relation is symmetric if whenever A is related to B, B is also related to A.
Transitive: If A lives in the same city as B, and B lives in the same city as C, does A live in the same city as C? Yes — it chains. A relation is transitive if whenever A is related to B and B is related to C, then A is related to C.

Now think about "is taller than". It is not reflexive (no one is taller than themselves), not symmetric (if A is taller than B, B is not taller than A), but it is transitive (if A > B and B > C, then A > C). Different relations satisfy different combinations.

The Precise Definitions

Let $R$ be a relation on a set $A$ (meaning $R \subseteq A \times A$ ).

R

R

R

The Four Types of Relations You Must Know

These three properties combine to define the most important types:

Type	Reflexive?	Symmetric?	Transitive?	Example
Empty relation	No (unless $A = \emptyset$ )	Yes (vacuously)	Yes (vacuously)	$R = \emptyset$ on $A = \{1,2\}$
Universal relation	Yes	Yes	Yes	$R = A \times A$
Identity relation	Yes	Yes	Yes	$R = \{(a,a) \mid a \in A\}$
Equivalence relation	Yes	Yes	Yes	"Same city", "same remainder mod 3"

Watch out

A common mistake: thinking "symmetric" means "if (a,b) is in R, then (b,a) must also be in R" — that's correct. But it does not require that (a,b) and (b,a) are different pairs. The identity relation is symmetric because (a,a) implies (a,a).

Equivalence Relations — The Most Important Type

An equivalence relation is one that is reflexive, symmetric, and transitive all at once. It captures the idea of "sameness" or "equivalence" in some sense.

Important

An equivalence relation partitions the set into disjoint equivalence classes — groups of elements that are all related to each other. Every element belongs to exactly one class.

Example: On the set of integers, define $a \sim b$ if $a - b$ is divisible by 3. This is an equivalence relation. The classes are:

$\{\dots, -6, -3, 0, 3, 6, \dots\}$ (remainder 0) …

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Domain = the set of all real $x$ for which $f(x)$ is defined; range = the set of all values $f(x)$ actually takes.

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

(i) $f(x)=x^2$ .
- Domain: $x^2$ is defined for every real number, so $\text{Domain}=\mathbb{R}$ .
- Range: for any real $x$ , $x^2\geq0$ , so every output is $\geq 0$ . Also every non-negative real $y$ is attained, since $x=\sqrt{y}$ gives $f(x)=y$ . So $\text{Range}=[0,\infty)$ .
(ii) $f(x)=\dfrac{1}{x}$ .
- Domain: $\dfrac1x$ is undefined only at $x=0$ , so $\text{Domain}=\mathbb{R}-\{0\}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2025Set may-20251 markMCQ
$S = \{3, 4, 6\}$
›Reveal solutionSolution
(A) has the closing pair $(3,6)$ (transitive) and no $(x,x)$ pairs (not reflexive).

Step 1 — Recall the two definitions

Reflexive on $S$ : every element relates to itself, i.e. $(3,3),(4,4),(6,6)$ all present.

Transitive: whenever $(a,b)$ and $(b,c)$ are in $R$ , $(a,c)$ must also be in $R$ .

Step 2 — Test option (A)

$R=\{(3,4),(4,6),(3,6)\}$ . Chain $(3,4)\&(4,6)\Rightarrow$ need $(3,6)$ — present ✓ transitive. No self-pairs ⇒ not reflexive ✓. This satisfies the requirement.

Step 3 — Eliminate the others

(C) $\{(3,3),(4,4),(6,6)\}$ IS reflexive — excluded.

(D) $\{(3,4),(4,3)\}$ : chain $(3,4)\&(4,3)\Rightarrow$ need $(3,3)$ , absent ⇒ NOT transitive. …
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CA Foundation 2025Set sep-20251 markMCQ
$Q. For a relation R, aRb represents that a is related to b. If for all a, b, c, aRb and bRc gives that aRc, then the relation is __________. (A) Reflexive (B) Symmetric (C) Transitive (D) Asymmetric$
›Reveal solutionSolution
" $aRb$ and $bRc \Rightarrow aRc$ " is the defining property of a transitive relation.

Step 1 — Match the statement to a definition

Important

A relation $R$ on a set is transitive iff for all $a,b,c$ : if $aRb$ and $bRc$ , then $aRc$ .

The given condition is word-for-word this definition.

Step 2 — Rule out the other properties

Reflexive: every element relates to itself, $aRa$ — not what's stated.

Symmetric: $aRb \Rightarrow bRa$ — about reversing a pair, not chaining.

Asymmetric: $aRb \Rightarrow$ not $bRa$ — again about a single pair's direction.

Step 3 — Conclude …
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Q.Find the domain and range of each of the following real functions:

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