Class JEE Mathematics Sets, Relations, and Functions Q #1005
COMPETENCY BASED
APPLY
4 Marks 2025 JEE Main 2025 (Online) 2nd April Morning Shift MCQ SINGLE
Let $A$ be the set of all functions $f: Z \rightarrow Z$ and $R$ be a relation on $A$ such that $R = {(f, g): f(0) = g(1) \text{ and } f(1) = g(0)}$. Then $R$ is :
(A) Symmetric and transitive but not reflective
(B) Symmetric but neither reflective nor transitive
(C) Transitive but neither reflexive nor symmetric
(D) Reflexive but neither symmetric nor transitive
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Correct Answer: B
Explanation
To determine if the relation $R$ is reflexive, symmetric, and transitive, we analyze each property separately.

Reflexive: For $R$ to be reflexive, $(f, f)$ must be in $R$. This means $f(0) = f(1)$ and $f(1) = f(0)$ must be true for all $f$. However, $f(0)$ is not necessarily equal to $f(1)$ for all functions $f$. Therefore, $R$ is not reflexive.

Symmetric: If $(f, g) \in R$, then $f(0) = g(1)$ and $f(1) = g(0)$. To check for symmetry, we need to verify if $(g, f) \in R$. If $f(0) = g(1)$, then $g(1) = f(0)$. And if $f(1) = g(0)$, then $g(0) = f(1)$. This shows that if $(f, g) \in R$, then $(g, f) \in R$, so $R$ is symmetric.

Transitive: If $(f, g) \in R$ and $(g, h) \in R$, then $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. For $R$ to be transitive, we need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$, so $f(0) = h(0)$. Also, $f(1) = g(0) = h(1)$, so $f(1) = h(1)$. Therefore, $f(0) = g(1)$ and $g(1) = h(0)$ implies $f(0) = h(0)$ and $f(1) = g(0)$ and $g(0) = h(1)$ implies $f(1) = h(1)$. Since $f(0)$ is not necessarily equal to $h(1)$ and $f(1)$ is not necessarily equal to $h(0)$, $R$ is not transitive. Actually $f(0) = h(0)$ and $f(1) = h(1)$ so this doesn't imply that $f(0) = h(1)$ and $f(1) = h(0)$ so the relation is not transitive

Therefore, the relation $R$ is symmetric but neither reflexive nor transitive.

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Step-by-Step Solution

Reflexivity: For $R$ to be reflexive, we need $(f, f) \in R$ for all $f \in A$. This means $f(0) = f(1)$ and $f(1) = f(0)$, which is true for all $f$. Thus, $R$ is reflexive.

Symmetry: For $R$ to be symmetric, if $(f, g) \in R$, then $(g, f) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. We need to check if $(g, f) \in R$, which means $g(0) = f(1)$ and $g(1) = f(0)$. Since $f(0) = g(1)$ and $f(1) = g(0)$, we have $g(1) = f(0)$ and $g(0) = f(1)$. Thus, $R$ is symmetric.

Transitivity: For $R$ to be transitive, if $(f, g) \in R$ and $(g, h) \in R$, then $(f, h) \in R$. Given $(f, g) \in R$, we have $f(0) = g(1)$ and $f(1) = g(0)$. Given $(g, h) \in R$, we have $g(0) = h(1)$ and $g(1) = h(0)$. We need to check if $(f, h) \in R$, which means $f(0) = h(1)$ and $f(1) = h(0)$. We have $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. Thus, $f(0) = h(0)$ and $f(1) = h(1)$. Therefore, $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation. However, the options do not include this. Let's re-examine the transitivity part.

We have $f(0) = g(1)$, $f(1) = g(0)$, $g(0) = h(1)$, and $g(1) = h(0)$. We want to show $f(0) = h(1)$ and $f(1) = h(0)$. From the given equations, $f(0) = g(1) = h(0)$ and $f(1) = g(0) = h(1)$. So, we have $f(0) = h(0)$ and $f(1) = h(1)$. This does NOT imply $f(0) = h(1)$ and $f(1) = h(0)$. Therefore, $R$ is NOT necessarily transitive.

Let's consider a counterexample for transitivity. Let $f(x) = x$, $g(x) = 1-x$, and $h(x) = x$. Then $f(0) = 0$, $f(1) = 1$, $g(0) = 1$, $g(1) = 0$, $h(0) = 0$, $h(1) = 1$. $(f, g) \in R$ since $f(0) = g(1) = 0$ and $f(1) = g(0) = 1$. $(g, h) \in R$ since $g(0) = h(1) = 1$ and $g(1) = h(0) = 0$. Now, we need to check if $(f, h) \in R$. We need $f(0) = h(1)$ and $f(1) = h(0)$. $f(0) = 0$ and $h(1) = 1$, so $f(0) \neq h(1)$. $f(1) = 1$ and $h(0) = 0$, so $f(1) \neq h(0)$. Thus, $(f, h) \notin R$, and $R$ is not transitive.

Since $R$ is reflexive and symmetric but not transitive, the correct option is (B) is incorrect and (D) is closest.

Let's re-examine reflexivity. For $R$ to be reflexive, we need $(f, f) \in R$ for all $f \in A$. This means $f(0) = f(1)$ and $f(1) = f(0)$. This is NOT true for all $f$. For example, if $f(x) = x$, then $f(0) = 0$ and $f(1) = 1$, so $f(0) \neq f(1)$. Thus, $R$ is not reflexive.

Since $R$ is symmetric but neither reflexive nor transitive, the correct option is (B).

Correct Answer: Symmetric but neither reflexive nor transitive<\/strong>

AI Suggestion: Option B

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires applying the definitions of reflexive, symmetric, and transitive relations to a specific set of functions.
Knowledge Dimension: CONCEPTUAL
Justification: The question tests the understanding of the concepts of relations, functions, and their properties like reflexivity, symmetry, and transitivity.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires applying the definitions of relations and functions to determine the properties of a given relation, going beyond simple recall of definitions.

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