A school is organising a debate competition with participants as speakers $S = \{S_1, S_2, S_3, S_4\}$ and these are judged by judges $J = \{J_1, J_2, J_3\}$. Each speaker can be assigned one judge. Let $R$ be a relation from set $S$ to $J$ defined as $R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y,\ x \in S,\ y \in J\}$. (i) How many relations can be there from $S$ to $J$? [1] (ii) A student identifies a function from $S$ to $J$ as $f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\}$. Check if it is bijective. [1] (iii)(a) How many one-one functions can be there from set $S$ to set $J$? [2] OR (iii)(b) Another student considers a relation $R_1 = \{(S_1, S_2), (S_2, S_4)\}$ in set $S$. Write minimum ordered pairs to be included in $R_1$ so that $R_1$ is reflexive but not symmetric. [2]

> [!TLDR] > (a) Relations $=2^{12}=4096$; the given $f$ is not bijective (onto but not one-one); one-one functions $S\to J$ number $0$. > (b) Add the four reflexive pairs $(S_1,S_1),(S_2,S_2),(S_3,S_3),(S_4,S_4)$ — the relation is then reflexive and already not symmetric. We have $4$ speakers $S=\{S_1,S_2,S_3,S_4\}$ and $3$ judges $J=\{J_1,J_2,J_3\}$. A relation just records (speaker, judge) pairs; a function assigns each speaker exactly one judge. ## Part (a) ### (i) How many relations from $S$ to $J$? 1. $|S|=4$ and $|J|=3$, so the Cartesian product $S\times J$ has $4\times 3 = 12$ ordered pairs. 2. A relation is **any** subset of $S\times J$. Each of the $12$ pairs is either in or out — $2$ choices each. 3. Total $=2^{12}=4096$. > [!TIP] > The number of relations from an $m$-set to an $n$-set is $2^{mn}$. ### (ii) Is $f=\{(S_1,J_1),(S_2,J_2),(S_3,J_2),(S_4,J_3)\}$ bijective? 1. Each speaker occurs once as a first coordinate, so $f$ is a valid function. 2. **One-one?** $S_2\mapsto J_2$ and $S_3\mapsto J_2$: two speakers share a judge, so $f$ is **not injective**. 3. **Onto?** $J_1,J_2,J_3$ are all hit, so $f$ is surjective. 4. A bijection must be both; since it fails injectivity, $f$ is **not bijective**. ### (iii)(a) How many one-one functions $S\to J$? 1. Injectivity requires the $4$ speakers to receive $4$ distinct judges. 2. Only $3$ judges exist; by the pigeonhole principle at least two speakers must coincide. 3. Hence no injective function exists. …

A school is organising a debate competition with participants as speakers S = \S₁, S₂, S₃, S₄\ and these are j

Question

S = \{S_1, S_2, S_3, S_4\}

(i) How many relations can be there from

S

J

? [1]

(ii) A student identifies a function from

S

J

f = \{(S_1, J_1), (S_2, J_2), (S_3, J_2), (S_4, J_3)\}

. Check if it is bijective. [1] (iii)(a) How many one-one functions can be there from set

S

to set

J

? [2]

(OR)

(iii)(b) Another student considers a relation

R_1 = \{(S_1, S_2), (S_2, S_4)\}

in set

S

. Write minimum ordered pairs to be included in

R_1

so that

R_1

is reflexive but not symmetric. [2]

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — One One Function

One-One (Injective) Function

Think of taking attendance by unique roll numbers: call a number and exactly one student responds — never two sharing a number. That is a one-one function: different inputs always land on different outputs.

The idea

A function is a machine turning inputs into outputs. It is one-one (or injective) if it never reuses an output — two different inputs can never produce the same result.

$f(x)=x+1$ is one-one: if $x_1\neq x_2$ then $x_1+1\neq x_2+1$ .
$g(x)=x^2$ on $\mathbb{R}$ is not one-one, because $g(2)=g(-2)=4$ .

Precise definition

$f:A\to B$ is one-one if for all $x_1,x_2\in A$ ,

x_1\neq x_2 \implies f(x_1)\neq f(x_2).

The contrapositive is usually easier in proofs:

f(x_1)=f(x_2)\implies x_1=x_2.

"If the outputs are equal, the inputs must have been equal."

How to check

Horizontal line test (graphs): if any horizontal line meets the graph more than once, the function is not one-one, because that line marks one output shared by several inputs.
Algebraic test: assume $f(x_1)=f(x_2)$ and try to deduce $x_1=x_2$ ; succeed and it is one-one, find a counterexample and it is not.

Tip

A strictly increasing or strictly decreasing function is automatically one-one. So a decreasing function like $f(x)=-x$ is one-one too — being one-one is about no repeated outputs, not about going up.

Why it matters …

Part (b)Concept understanding — Relation Properties

Properties of a Relation

A relation $R$ on a set $A$ pairs elements of $A$ with one another. Some relations behave in regular, predictable ways, and we name these behaviours properties. Three matter most for CBSE Class 12 — reflexive, symmetric, transitive (together they build an equivalence relation); a fourth, antisymmetric, is worth knowing for order relations.

Reflexive — everything relates to itself

$R$ is reflexive if $a\,R\,a$ for every $a\in A$ . "Has the same age as" is reflexive; "is taller than" is not. If even one element misses its self-pair, reflexivity fails: on $\{1,2,3\}$ , $\{(1,1),(2,2)\}$ is not reflexive because $(3,3)$ is absent.

Symmetric — the relation runs both ways

$R$ is symmetric if $a\,R\,b \implies b\,R\,a$ . "Is married to" is symmetric; "is taller than" is not. Symmetry does not demand that every pair be related — only that any pair which appears also appears reversed. So $\{(1,2),(2,1),(3,3)\}$ is symmetric, but $\{(1,2),(2,1),(1,3)\}$ is not, since $(3,1)$ is missing.

Transitive — relations chain

$R$ is transitive if $a\,R\,b$ and $b\,R\,c$ together force $a\,R\,c$ . "Is an ancestor of" is transitive; "is a friend of" is not. A single broken chain breaks the property: $\{(1,2),(2,3)\}$ is not transitive because $(1,3)$ is missing.

Antisymmetric — two-way ties force equality

$R$ is antisymmetric if $a\,R\,b$ and $b\,R\,a$ together force $a=b$ . The order relation $\le$ is antisymmetric: $a\le b$ and $b\le a$ give $a=b$ . It does not ban self-pairs like $(1,1)$ ; it only rules out distinct elements related both ways.

Tip

Test the properties in order of ease — reflexivity first, then symmetry, transitivity. A single counterexample is enough to disprove any of them.

Property	Condition
Reflexive	$\forall a,\ a\,R\,a$

Start your 14-day free trial to unlock the full solution →

Relations $=2^{12}=4096$ ; the given $f$ is not bijective (onto but not one-one); one-one functions $S\to J$ number $0$ .
Add the four reflexive pairs $(S_1,S_1),(S_2,S_2),(S_3,S_3),(S_4,S_4)$ — the relation is then reflexive and already not symmetric.

We have $4$ speakers $S=\{S_1,S_2,S_3,S_4\}$ and $3$ judges $J=\{J_1,J_2,J_3\}$ . A relation just records (speaker, judge) pairs; a function assigns each speaker exactly one judge.

Part (a)

(i) How many relations from $S$ to $J$ ?

$|S|=4$ and $|J|=3$ , so the Cartesian product $S\times J$ has $4\times 3 = 12$ ordered pairs.
A relation is any subset of $S\times J$ . Each of the $12$ pairs is either in or out — $2$ choices each.
Total $=2^{12}=4096$ .

Tip

The number of relations from an $m$ -set to an $n$ -set is $2^{mn}$ .

(ii) Is $f=\{(S_1,J_1),(S_2,J_2),(S_3,J_2),(S_4,J_3)\}$ bijective?

Each speaker occurs once as a first coordinate, so $f$ is a valid function.
One-one? $S_2\mapsto J_2$ and $S_3\mapsto J_2$ : two speakers share a judge, so $f$ is not injective.
Onto? $J_1,J_2,J_3$ are all hit, so $f$ is surjective.
A bijection must be both; since it fails injectivity, $f$ is not bijective.

(iii)(a) How many one-one functions $S\to J$ ?

Injectivity requires the $4$ speakers to receive $4$ distinct judges.
Only $3$ judges exist; by the pigeonhole principle at least two speakers must coincide.
Hence no injective function exists. …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 50 on this concept.

CBSE 2026Set V11 markMCQ
$R$
(a) symmetric but not transitive
(b) transitive but not symmetric
(c) symmetric and transitive
(d) neither symmetric nor transitive
›Reveal solutionSolution

$R=\{(1,1),(2,2)\}$ is both symmetric and transitive, so the answer is (c).

Symmetry: whenever $(a,b)\in R$ we need $(b,a)\in R$ . Here the only pairs are $(1,1)$ and $(2,2)$ , and each is its own reverse, so symmetry holds. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. What type of a relation is "less than" in the set of real numbers?$
(a) Only symmetric
(b) Only transitive
(c) Only reflexive
(d) Equivalence
›Reveal solutionSolution
" $<$ " on $\mathbb{R}$ is transitive only.

Consider the relation " $a < b$ " on $\mathbb{R}$ :

Reflexive? $a < a$ is false for every $a$ , so NOT reflexive.

Symmetric? $a < b$ does not imply $b < a$ , so NOT symmetric. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) Reflexive but not symmetric
(b) Reflexive but not transitive
(c) Symmetric and transitive
(d) Neither symmetric nor transitive
›Reveal solutionSolution

$R$ is reflexive (it contains every $(x,x)$ ) but not symmetric, since $(1,2)\in R$ while $(2,1)\notin R$ .

Given: $R=\{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on $A=\{1,2,3\}$ .

Reflexivity: $R$ is reflexive if $(x,x)\in R$ for every $x\in A$ . Here $(1,1),(2,2),(3,3)$ are all present, so $R$ is reflexive.

Symmetry: $R$ is symmetric if $(x,y)\in R\Rightarrow(y,x)\in R$ . Here $(1,2)\in R$ but $(2,1)\notin R$ . So $R$ is not symmetric.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 6) \in R$
›Reveal solutionSolution
Check each pair against both conditions $a=b-2$ and $b>6$ .

$R=\{(a,b): a=b-2,\ b>6\}$

$(2,4)$ : $a=b-2 \Rightarrow 2=2$ ✓, but $b>6 \Rightarrow 4>6$ ✗. Not in $R$ .

$(3,8)$ : $a=b-2 \Rightarrow 3=6$ ✗. Not in $R$ . …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $R$ is reflexive and symmetric but not transitive
(b) $R$ is reflexive and transitive but not symmetric
(c) $R$ is symmetric and transitive but not reflexive
(d) $R$ is an equivalence relation
›Reveal solutionSolution
Test $R$ against the three defining properties (reflexive, symmetric, transitive) one at a time, directly from its listed ordered pairs.

$R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}$ on $\{1,2,3,4\}$ .

Reflexive? Need $(a,a)\in R$ for every $a\in\{1,2,3,4\}$ : $(1,1)$ ✓, $(2,2)$ ✓, $(3,3)$ ✓, $(4,4)$ ✓. All present — $R$ is reflexive.

Symmetric? Need: whenever $(a,b)\in R$ , also $(b,a)\in R$ . Take $(1,2)\in R$ : is $(2,1)\in R$ ? It is not in the list. So $R$ is not symmetric.

Transitive? Need: whenever $(a,b)\in R$ and $(b,c)\in R$ , also $(a,c)\in R$ . Checking every chain:

$(1,1),(1,2)\Rightarrow(1,2)$ ✓

$(1,1),(1,3)\Rightarrow(1,3)$ ✓

$(1,2),(2,2)\Rightarrow(1,2)$ ✓

$(1,3),(3,3)\Rightarrow(1,3)$ ✓

$(1,3),(3,2)\Rightarrow(1,2)$ ✓ …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$R$
(a) $(2, 4) \in R$
(b) $(3, 8) \in R$
(c) $(6, 8) \in R$
(d) $(8, 7) \in R$
›Reveal solutionSolution

Only $(6,8)$ satisfies $a=b-2$ with $b>6$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$R$
(a) $(a, a) \in R$ for all $a \in A$
(b) $(a, a) \in R$ for atleast one $a \in A$
(c) $(a, b) \in R$ implies $(b, a) \in R$
(d) $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$
›Reveal solutionSolution

Tests the definition of a reflexive relation — correct option is (a).

A relation $R$ on a set $A$ is reflexive when every element is related to itself. Formally, $(a,a)\in R$ must hold for all $a\in A$ — not just for at least one element. Option (c) states symmetry ( $(a,b)\in R\Rightarrow(b,a)\in R$ ) and option (d) states transitivit …
Start your 14-day free trial to see the full solution →
CBSE 2025Set IX1 markMCQ
$R = \{(a, b) : a = b - 1,\ b \geq 3\}$
(a) $(2, 4) \in R$
(b) $(4, 5) \in R$
(c) $(4, 6) \in R$
(d) $(1, 3) \in R$
›Reveal solutionSolution
Only $(4,5)$ satisfies $a=b-1$ with $b\ge 3$ ; option (b).

Concept. A pair $(a,b)$ belongs to $R$ only if it satisfies both conditions: $a=b-1$ and $b\ge 3$ .

$(2,4)$ : $b-1=3\ne 2$ . ✗ …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$A=\{1,2,3,4\}$
(i) Reflexive
(ii) Symmetric
(iii) Transitive
(iv) Equivalent
›Reveal solutionSolution

Check each property directly: $R$ turns out to be symmetric only.

$A=\{1,2,3,4\}$ , $R=\{(a,b): a+b \text{ is odd}\}$ .

Reflexive? $(a,a)$ needs $a+a=2a$ to be odd — but $2a$ is always even. So $R$ is not reflexive.

Symmetric? If $a+b$ is odd, then $b+a=a+b$ is the same sum, also odd. So $(a,b)\in R \Rightarrow (b,a)\in R$ . $R$ is symmetric.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$A = \{a, b, c\}$
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation
›Reveal solutionSolution

R is symmetric (the only cross-pair (a,b)/(b,a) both appear) and not transitive ((b,a) & (a,b) would force (b,b), which is missing) — matching option (a) once we note a reflexivity caveat below.

Step 1 — Reflexive? Full reflexivity on A = {a, b, c} needs (a,a), (b,b), (c,c) all in R. Only (a,a) is present; (b,b) and (c,c) are not. So strictly, R is not fully reflexive on {a,b,c}.

Step 2 — Symmetric? The only pair with a 'partner' is (a,b), and its reverse (b,a) is also in R. There is no pair in R whose reverse is missing. So R is symmetric.

Step 3 — Transitive? Take (b,a) ∈ R and (a,b) ∈ R: transitivity would require (b,b) ∈ R. It is not. So R is not transitive.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$f(a) = f(b) \Rightarrow a = b \; \forall a, b \in A$
(a) one-one
(b) constant
(c) onto
(d) many one
›Reveal solutionSolution

The statement given is the textbook definition of a one-one (injective) function.

A function f : A → B is called one-one (or injective) if distinct elements of A always map to distinct elements of B — equivalently, if f(a) = f(b) forces a = b (no two different inputs can share an output).

This is precisely the condition stated in the question, so f is one-one.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Relation R = {(x, y) : x < y² where x, y \in R} is:$
(a) Reflexive but not symmetric.
(b) Symmetric and transitive but not Reflexive.
(c) Reflexive and Symmetric.
(d) Neither reflexive nor symmetric nor transitive.
›Reveal solutionSolution

Test each property of $R=\{(x,y): x<y^2\}$ with concrete counterexamples — all three fail.

Reflexive? Need $x<x^2$ for every $x\in\mathbb R$ . Take $x=\dfrac12$ : is $\dfrac12 < \dfrac14$ ? No. So $R$ is not reflexive.

Symmetric? Need $x<y^2 \Rightarrow y<x^2$ . Take $x=0,\,y=1$ : $0<1^2$ is true, so $(0,1)\in R$ . But is $1<0^2=0$ ? No. So $(1,0)\notin R$ , and $R$ is not symmetric.

…

Start your 14-day free trial to see the full solution →

Part (a)

(i) How many relations from SSS to JJJ?

(ii) Is f={(S1,J1),(S2,J2),(S3,J2),(S4,J3)}f=\{(S_1,J_1),(S_2,J_2),(S_3,J_2),(S_4,J_3)\}f={(S1​,J1​),(S2​,J2​),(S3​,J2​),(S4​,J3​)} bijective?

(iii)(a) How many one-one functions S→JS\to JS→J?

🎓Unlock everything free for 14 days

(i) How many relations from $S$ to $J$ ?

(ii) Is $f=\{(S_1,J_1),(S_2,J_2),(S_3,J_2),(S_4,J_3)\}$ bijective?

(iii)(a) How many one-one functions $S\to J$ ?

Unlock everything free for 14 days