CBSE Maths › Class 12 PYQs › Relations and Functions

Relations and Functions: 5 marks Questions (CBSE Class 12)

5 different 5 marks questions on Relations and Functions from CBSE Class 12 Maths board exams 2026, newest first.

1 mark (5)2 marks (3)3 marks (2)4 marks (1)5 marks (5)

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

Q325 marksLong AnswerRelations and Functions CBSE 2026 · 65/3/1

Show that $f : R \to R$ defined as $f (x) = \frac{x}{1 + x ^{2}}$ is one-one but not onto.

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Answer: Proved.

One-one: let $f (a) = f (b)$ , i.e. $\frac{a}{1 + a ^{2}} = \frac{b}{1 + b ^{2}}$ .
Then a and b have the same sign (or are both 0), and squaring gives $\frac{a ^{2}}{1 + a ^{2}} = \frac{b ^{2}}{1 + b ^{2}}$ , so $a^{2} + a^{2} b^{2} = b^{2} + a^{2} b^{2}$ , i.e. $a^{2} = b^{2}$ .
Same sign and $a^{2} = b^{2}$ give $a = b$ . So f is one-one.
Not onto: for every real x, $∣ x ∣ < 1 + x^{2}$ , so $- 1 < f (x) < 1$ .
Hence, e.g., $1 \in R$ has no pre-image (if $\frac{x}{1 + x ^{2}} = 1$ then $x^{2} = 1 + x^{2}$ , impossible).
So f is not onto. Proved.

Q345 marksLong AnswerRelations and Functions CBSE 2026 · 65/3/2

Show that $f : R_{+} \to [- 5, \infty)$ given by $f (x) = 4 x^{2} + 4 x - 5$ is both one-one and onto where $R_{+} = [0, \infty)$ . Also, find $p \in R_{+}$ such that $f (p) = 3$ .

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Answer: Proved;

p = 1

$f (x) = (2 x + 1)^{2} - 6$ .
One-one: if $f (a) = f (b)$ with $a, b \geq 0$ , then $(2 a + 1)^{2} = (2 b + 1)^{2}$ . Both $2 a + 1$ and $2 b + 1$ are positive, so $2 a + 1 = 2 b + 1$ , i.e. $a = b$ .
Onto: for $x \geq 0$ , $2 x + 1 \geq 1$ , so $f (x) \geq 1 - 6 = - 5$ ; f maps into $[- 5, \infty)$ .
Given $y \geq - 5$ , take $x = \frac{y + 6 - 1}{2}$ . Since $y + 6 \geq 1$ , $x \geq 0$ , and $f (x) = (y + 6)^{2} - 6 = y$ .
So every $y \in [- 5, \infty)$ has a pre-image: f is onto. Proved.
$f (p) = 3$ : $(2 p + 1)^{2} = 9$ , $2 p + 1 = 3$ (as $p \geq 0$ ), so $p = 1$ .

Q335 marksLong AnswerRelations and Functions CBSE 2026 · 65/3/3

Show that a function $f : R_{+} \to A \subset N$ , defined as $f (x) = 4 x^{2} + 12 x + 15$ is one-one. Find set A so that f is onto where $R_{+} = [0, \infty)$ . Also, find if there exists $a \in R_{+}$ such that $f (a) = 7$ . Justify.

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Answer: f is one-one; A = range of f

= [15, \infty)

(for whole-number inputs,

A = {15, 31, 55, 87, \dots}

); no such a exists, since

f (x) \geq 15

R_{+}

$f (x) = (2 x + 3)^{2} + 6$ .
One-one: if $f (a) = f (b)$ with $a, b \geq 0$ , then $(2 a + 3)^{2} = (2 b + 3)^{2}$ . Both $2 a + 3$ and $2 b + 3$ are positive, so $a = b$ .
Onto: f is onto exactly when A is its range. For $x \geq 0$ , $2 x + 3 \geq 3$ , so $f (x) \geq 15$ , and f increases without bound. So A $= {f (x) : x \in R_{+}} = [15, \infty)$ (if only whole-number inputs are meant, $A = {4 n^{2} + 12 n + 15 : n = 0, 1, 2, \dots} = {15, 31, 55, \dots}$ ).
$f (a) = 7$ means $(2 a + 3)^{2} = 1$ , so $2 a + 3 = \pm 1$ , i.e. $a = - 1$ or $a = - 2$ . Neither is in $R_{+}$ (also $7 < 15$ ). So no such $a$ exists.

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