Let A = 1, 2, 3 and consider the relation R = (1,1), (2,2), (3,3), (1,2), (2,3), (1,3) then R is:

Option c: Reflexive but not symmetric. Solution: We are given set A = 1, 2, 3 and relation R = (1,1), (2,2), (3,3), (1,2), (2,3), (1,3) . Let's test the properties of R : 1. Reflexive: For R to be reflexive, every element a A must satisfy (a, a) R . Since (1, 1), (2, 2), (3, 3) R , R is reflexive. 2. Symmetric: For R to be symmetric, (a, b) R (b, a) R . Here, (1, 2) R but (2, 1) R . Thus, R is not symmetric. 3. Transitive: For R to be transitive, (a, b) R and (b, c) R (a, c) R . Let's check: - (1, 2) R and (2, 3) R (1, 3) R . (True) Therefore, R is transitive. Thus, the relation R is reflexive and transitive, but not symmetric. Hence, Option C is the correct answer.

Let $A = \{1, 2, 3\}$ and consider the relation $R = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$ then $R$ is:

The correct answer is Option c: Reflexive but not symmetric. We are given set $A = \{1, 2, 3\}$ and relation $R = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$. Let's test the properties of $R$: 1. **Reflexive**: For $R$ to be reflexive, every element $a \in A$ must satisfy $(a, a) \in R$. Since $(1, 1), (2, 2), (3, 3) \in R$, $R$ is reflexive. 2. **Symmetric**: For $R$ to be symmetric, $(a, b) \in R \implies (b, a) \in R$. Here, $(1, 2) \in R$ but $(2, 1) \notin R$. Thus, $R$ is not symmetric. 3. **Transitive**: For $R$ to be transitive, $(a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R$. Let's check: - $(1, 2) \in R$ and $(2, 3) \in R \implies (1, 3) \in R$. (True) Therefore, $R$ is transitive. Thus, the relation $R$ is reflexive and transitive, but not symmetric. Hence, **Option C** is the correct answer.

Sets, Relations and FunctionsMCQPYQ Dec. 22Question 1945 of 217

All Questions

Let $\displaystyle A = \{1, 2, 3\}$ and consider the relation $\displaystyle R = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$ then $\displaystyle R$ is:

Options

ASymmetric and transitive

BReflexive but not transitive

CReflexive but not symmetric

DNeither symmetric, nor transitive

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Correct Answer

✅ Option c — Reflexive but not symmetric

All Options:

ASymmetric and transitive
BReflexive but not transitive
CReflexive but not symmetric
DNeither symmetric, nor transitive

Detailed Solution & Explanation

We are given set

\displaystyle A = \{1, 2, 3\}

and relation

\displaystyle R = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}

.
Let's test the properties of

\displaystyle R

:
1. **Reflexive**: For

\displaystyle R

to be reflexive, every element

\displaystyle a \in A

must satisfy

\displaystyle (a, a) \in R

.
Since

\displaystyle (1, 1), (2, 2), (3, 3) \in R

\displaystyle R

is reflexive.
2. **Symmetric**: For

\displaystyle R

to be symmetric,

\displaystyle (a, b) \in R \implies (b, a) \in R

.
Here,

\displaystyle (1, 2) \in R

but

\displaystyle (2, 1) \notin R

. Thus,

\displaystyle R

is not symmetric.
3. **Transitive**: For

\displaystyle R

to be transitive,

\displaystyle (a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R

.
Let's check:
-

\displaystyle (1, 2) \in R

and

\displaystyle (2, 3) \in R \implies (1, 3) \in R

. (True)
Therefore,

\displaystyle R

is transitive.
Thus, the relation

\displaystyle R

is reflexive and transitive, but not symmetric.
Hence, **Option C** is the correct answer.

About This Chapter: Sets, Relations and Functions

Paper

Paper 3: Quantitative Aptitude

Weightage

3-5 Marks

Key Topics

Sets, Relations, Functions

This chapter covers Sets, Relations, Functions and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 3-5 Marks weightage. Focus on understanding core concepts rather than memorizing.

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Let A={1,2,3}\displaystyle A = \{1, 2, 3\}A={1,2,3} and consider the relation R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}\displaystyle R = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}R={(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)} then R\displaystyle RR is: