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

**Option (a):** R is an **equivalence relation** on Z. - **Reflexive:** $|x-x| = 0$, and $0$ is divisible by every prime p ($0 = p\times 0$), so $(x,x) \in R$ for all $x \in Z$. - **Symmetric:** If p divides $|x-y|$, then since $|y-x| = |x-y|$, p also divides $|y-x|$, so $(x,y) \in R \implies (y,x) \in R$. - **Transitive:** If $p | (x-y)$ and $p | (y-z)$, then $p | [(x-y)+(y-z)] = (x-z)$, so $p | |x-z|$, giving $(x,z) \in R$. Since R is reflexive, symmetric, and transitive, **R is an equivalence relation**. **Option (b):** f is **both one-one and onto**. - **One-one:** Let $f(x_1) = f(x_2)$. Cross-multiplying $(3x_1+2)(5x_2-3) = (3x_2+2)(5x_1-3)$ and simplifying gives $19x_2 = 19x_1$, so $x_1 = x_2$. Hence f is one-one. - **Onto:** For any y in the codomain, solving $y = \frac{3x+2}{5x-3}$ for x gives $x = \frac{3y+2}{5y-3}$, which is defined (since $y\ne 3/5$) and lies in the domain $\mathbb{R}-\{3/5\}$ (substituting $x=3/5$ leads to a contradiction). Hence every y has a pre-image, so f is onto. Therefore f is a **bijection**.

Question 325 marksLong AnswerSection D

$f : \mathbb{R} - \{3/5\} \to \mathbb{R} - \{3/5\}$

Q: Why is divisibility of |x−y| by a prime always an equivalence relation?

Because it is equivalent to congruence modulo p ($x\equiv y \pmod p$), and congruence relations are always reflexive, symmetric, and transitive, since divisibility by p is preserved under negation and addition of multiples of p.

Q: What must be checked in addition to cross-multiplication when proving a rational function is onto?

After solving $y=f(x)$ for x in terms of y, you must confirm the resulting x is not equal to any value excluded from the domain — here, $x\ne 3/5$ — otherwise the function would not truly map onto the stated codomain.

Relations and Functions

Equivalence Relations and One-One Onto Functions

Official answer

|x-x| = 0

equivalence relation

reflexive symmetric transitive

congruence modulo p

one-one function

onto function

bijection

rational function

Marking scheme

1. $f(x_1)=f(x_2)$
2. $1 mark: correctly proving symmetry of R (or correct cross-multiplication and simplification for one-one).$
3. $p|(x-y)$
4. $y=f(x)$
5. $x\ne 3/5$

Hint

$x\equiv y \pmod{p}$

Quick oral answer

$x_1=x_2$

Common mistakes

1. $|x-y|$
2. $x=\frac{3y+2}{5y-3}$
3. $x_1=x_2$

Analysis and explanation

x \equiv y \pmod{p}

Interesting facts

$Congruence modulo a prime number, the basis of part (a)'s relation, is the mathematical foundation of RSA encryption, which secures much of the internet's data transmission today.$

$f(x)=\frac{ax+b}{cx+d}$

Question details

Q32 of 44 questions

Board: CBSE
Class: Class 12
Subject: Mathematics
Year: 2026
Section: Section D
Marks: 5
Type: Long Answer
Chapter: Relations and Functions
Topic: $Equivalence Relations and One-One Onto Functions$
Bloom's level: analyze
Syllabus topic: $Unit 1: Relations and Functions — Equivalence Relations and Bijective Functions$
Author: CBSE
Reviewed by: Padhantu Academic Team

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Frequently asked questions

$Why is divisibility of |x-y| by a prime always an equivalence relation?$

$x\equiv y \pmod p$

$What must be checked in addition to cross-multiplication when proving a rational function is onto?$

$y=f(x)$