Relations and Functions: CBSE Class 12 Previous Year Questions

16 different questions from Relations and Functions (NCERT Chapter 1) asked in CBSE Class 12 Maths board exams 2026. Pick a mark group to practise, each with answers and step-by-step solutions.

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Most asked Relations and Functions questions

Q201 markAssertion–ReasonRelations and Functions CBSE 2026 · 65/4/1

Assertion (A) : A function $f : N \to N$ given by $f (x) = x^{3} + 2, \forall x \in N$ is one-one but not onto.
Reason (R) : Since $\forall y \in N$ (Codomain), there does not exist $x = (y - 2)^{1/3}$ in N (Domain) such that $f (x) = x^{3} + 2 = y$ .

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (C) Assertion (A) is true, but Reason (R) is false.

$x_{1}^{3} + 2 = x_{2}^{3} + 2 \Rightarrow x_{1} = x_{2}$ , so f is one-one.
$y = 1$ (or 2) has no pre-image in N, so f is not onto. A is true.
R says that for every $y \in N$ no such x exists, which is false: for $y = 3$ , $x = 1 \in N$ .
So A is true but R is false.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q191 markAssertion–ReasonRelations and Functions CBSE 2026 · 65/5/1

Assertion (A): A relation R on the set {1, 2, 3} defined as R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} is an equivalence relation.
Reason (R): A relation that is reflexive, symmetric and transitive is an equivalence relation.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (A) Both A and R are true and R is the correct explanation of A.

Reflexive: (1, 1), (2, 2), (3, 3) are in R.
Symmetric: (1, 2) and (2, 1) are both in R.
Transitive: (1, 2), (2, 1) give (1, 1) in R; (2, 1), (1, 2) give (2, 2) in R; all other cases are trivial.
So R is an equivalence relation; R is the definition of an equivalence relation and explains A.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q212 marksVery Short AnswerRelations and Functions CBSE 2026 · 65/2/1

Check whether $f : R - {3} \to R$ defined as $f (x) = \frac{x - 2}{x - 3}$ is onto or not.

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Answer: Not onto (1 has no pre-image).

Let $y = \frac{x - 2}{x - 3}$ . Then $x y - 3 y = x - 2 \Rightarrow x (y - 1) = 3 y - 2 \Rightarrow x = \frac{3 y - 2}{y - 1}$ .
This is not defined for $y = 1$ .
Indeed $\frac{x - 2}{x - 3} = 1$ would need $x - 2 = x - 3$ , which is impossible.
So $1 \in R$ has no pre-image; f is not onto.

Also asked in: 2026 65/2/2, 2026 65/2/3

Q21 (OR) (OR)2 marksVery Short AnswerRelations and Functions CBSE 2026 · 65/2/1

Check whether $f : Z \times Z \to Z \times Z$ (where Z is the set of integers) defined as $f (x, y) = (2 y, 3 x)$ is injective or not.

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Answer: Injective.

Let $f (x_{1}, y_{1}) = f (x_{2}, y_{2})$ .
Then $(2 y_{1}, 3 x_{1}) = (2 y_{2}, 3 x_{2})$ , so $2 y_{1} = 2 y_{2}$ and $3 x_{1} = 3 x_{2}$ .
Hence $y_{1} = y_{2}$ , $x_{1} = x_{2}$ , i.e. $(x_{1}, y_{1}) = (x_{2}, y_{2})$ .
f is injective (one-one).

Also asked in: 2026 65/2/2, 2026 65/2/3

Q23 (OR) (OR)2 marksVery Short AnswerRelations and Functions CBSE 2026 · 65/3/1

A relation R on $A = {1, 2, 3}$ is defined as $R = {(1, 1) (3, 3), (1, 2)}$ . Is R a symmetric relation ? Justify. Write the smallest relation set $R_{1}$ such that $R \cup R_{1}$ becomes an equivalence relation on the set ${1, 2, 3}$ .

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Answer: R is not symmetric;

R_{1} = {(2, 1), (2, 2)}

$(1, 2) \in R$ but $(2, 1) \in / R$ , so R is not symmetric.
For reflexivity we need $(2, 2)$ ; for symmetry we need $(2, 1)$ .
$R \cup {(2, 1), (2, 2)} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}$ is reflexive, symmetric and transitive.
So the smallest $R_{1} = {(2, 1), (2, 2)}$ .

Also asked in: 2026 65/3/2, 2026 65/3/3

Q263 marksShort AnswerRelations and Functions CBSE 2026 · 65/4/1

Let $A = R - {3}$ and $B = R - {1}$ . A function $f : A \to B$ is defined by $f (x) = (\frac{x - 2}{x - 3})$ . Find whether f is one-one and onto.

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Answer: f is one-one and onto.

One-one: $\frac{x _{1} - 2}{x _{1} - 3} = \frac{x _{2} - 2}{x _{2} - 3} \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3)$ .
$\Rightarrow - 3 x_{1} - 2 x_{2} = - 3 x_{2} - 2 x_{1} \Rightarrow x_{1} = x_{2}$ . So f is one-one.
Onto: for $y \in B$ , solve $y = \frac{x - 2}{x - 3}$ : $x = \frac{3 y - 2}{y - 1}$ , defined since $y \neq = 1$ .
$x \neq = 3$ (else $3 y - 2 = 3 y - 3$ , impossible), so $x \in A$ and $f (x) = y$ .
Hence f is onto; f is one-one and onto.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q26 (OR) (OR)3 marksShort AnswerRelations and Functions CBSE 2026 · 65/4/1

Let n be a fixed positive integer. A relation R is defined in set Z such that $R = {(x, y) : (x - y)$ is divisible by $n, x, y \in Z}$ . Determine if R is an equivalence relation.

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Answer: Yes, R is an equivalence relation.

Reflexive: $x - x = 0 = 0 \cdot n$ , so $(x, x) \in R$ for all $x \in Z$ .
Symmetric: if $x - y = k n$ then $y - x = (- k) n$ , so $(y, x) \in R$ .
Transitive: if $x - y = k n$ and $y - z = mn$ then $x - z = (k + m) n$ , so $(x, z) \in R$ .
Hence R is an equivalence relation.

Also asked in: 2026 65/4/2, 2026 65/4/3

Q364 marksCase StudyRelations and Functions CBSE 2026 · 65/5/1

A school wants the students of class XII to do a project on ‘Sustainability’ keeping the world environment in mind. They select the student participants on the basis of an essay writing competition.
7 students out of 80 are selected for the project and are categorized into two sets such that :
Girl students belong to Set A = ${G_{1}, G_{2}, G_{3}, G_{4}}$ ,
Boy students belong to Set B = ${B_{1}, B_{2}, B_{3}}$ .
Based on the above information, answer the following questions :
(i) How many relations are possible from Set A $\to$ Set B ? (1)
(ii) Let R be a relation from A $\to$ B such that R = ${(G_{1}, B_{1}), (G_{2}, B_{2}), (G_{3}, B_{2}), (G_{4}, B_{3}), (G_{1}, B_{2})}$ . Is R an injective function ? Justify your answer. (1)
(iii) (a) Let the relation R from A $\to$ A be such that R = {(x, y), x, y $\in$ A, x and y are students from the same colony in the city} Verify if R is an equivalence relation. (2)
OR (iii) (b) Verify if any function f : B $\to$ A is bijective. Give reason to support your answer. (2)

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Answer: (i)

2^{12} = 4096

(ii) No, R is not a function (

G_{1}

has two images), so it is not an injective function. (iii) (a) Yes, R is an equivalence relation. OR (iii) (b) No; since n(B) = 3 < n(A) = 4, no function from B to A can be onto, so none is bijective.

(i) $n (A \times B) = 4 \times 3 = 12$ , so the number of relations is $2^{12} = 4096$ .
(ii) $G_{1}$ is related to both $B_{1}$ and $B_{2}$ , so R is not a function and hence not an injective function.
(iii) (a) Reflexive: every student is from the same colony as himself/herself. Symmetric: if x and y are from the same colony, so are y and x. Transitive: if x, y and y, z are from the same colony, then x, z are too. So R is an equivalence relation.
(iii) (b) B has 3 elements and A has 4, so a function f : B → A has at most 3 images and cannot be onto; hence no such f is bijective.

Also asked in: 2026 65/5/2, 2026 65/5/3

Q325 marksLong AnswerRelations and Functions CBSE 2026 · 65/1/1

A relation R is defined on Z, the set of integers, as
$R = {(x, y) : ∣ x - y ∣$ is divisible by a prime number 'p', $x, y \in Z}$
check whether R is an equivalence relation or not.

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Answer: R is an equivalence relation.

Take p as a fixed prime.
Reflexive: $∣ x - x ∣ = 0$ is divisible by p, so $(x, x) \in R$ for all $x \in Z$ .
Symmetric: if $(x, y) \in R$ then p divides $∣ x - y ∣ = ∣ y - x ∣$ , so $(y, x) \in R$ .
Transitive: if $(x, y), (y, z) \in R$ then $x - y = k p$ and $y - z = m p$ for integers k, m.
Adding, $x - z = (k + m) p$ , so p divides $∣ x - z ∣$ and $(x, z) \in R$ .
R is reflexive, symmetric and transitive, so R is an equivalence relation.

Also asked in: 2026 65/1/2, 2026 65/1/3

Q32 (OR) (OR)5 marksLong AnswerRelations and Functions CBSE 2026 · 65/1/1

A function $f : R - {\frac{3}{5}} ⟶ R - {\frac{3}{5}}$ is defined as $f (x) = \frac{3 x + 2}{5 x - 3}$ . Show that f is one-one and onto.

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Answer: Proved.

One-one: let $f (a) = f (b)$ . Then $(3 a + 2) (5 b - 3) = (3 b + 2) (5 a - 3)$ .
$15 ab - 9 a + 10 b - 6 = 15 ab - 9 b + 10 a - 6 \Rightarrow - 19 a = - 19 b \Rightarrow a = b$ . So f is one-one.
Onto: let $y \in R - {\frac{3}{5}}$ and solve $y = \frac{3 x + 2}{5 x - 3}$ .
$5 x y - 3 y = 3 x + 2 \Rightarrow x (5 y - 3) = 3 y + 2 \Rightarrow x = \frac{3 y + 2}{5 y - 3}$ , defined since $y \neq = \frac{3}{5}$ .
Also $x \neq = \frac{3}{5}$ : otherwise $15 y + 10 = 15 y - 9$ , impossible.
Then $f (x) = y$ , so every y has a pre-image and f is onto.

Also asked in: 2026 65/1/2, 2026 65/1/3

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