Q1005: Let $A$ be the set of all functions $f: Z \rightarrow Z$ and...

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

(D) Reflexive but neither symmetric nor transitive

Prev Next

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.

AI Tutor Explanation

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

AI generated content. Review strictly for academic accuracy.

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.

More from this Chapter

MCQ_SINGLE

Two newspapers A and B are published in a city. It is known that $25$% of the city populations reads A and $20$% reads B while $8$% reads both A and B. Further, $30$% of those who read A but not B look into advertisements and $40$% of those who read B but not A also look into advertisements, while $50$% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is :-

NUMERICAL

Let $A=\{1,2,3, \ldots \ldots \ldots \ldots, 100\}$. Let $R$ be a relation on $\mathrm{A}$ defined by $(x, y) \in R$ if and only if $2 x=3 y$. Let $R_1$ be a symmetric relation on $A$ such that $R \subset R_1$ and the number of elements in $R_1$ is $\mathrm{n}$. Then, the minimum value of $\mathrm{n}$ is _________.

MCQ_SINGLE

Let the relations $R_1$ and $R_2$ on the set $X = \{1, 2, 3, ..., 20\}$ be given by $R_1 = \{(x, y) : 2x - 3y = 2\}$ and $R_2 = \{(x, y) : -5x + 4y = 0\}$. If $M$ and $N$ be the minimum number of elements required to be added in $R_1$ and $R_2$, respectively, in order to make the relations symmetric, then $M + N$ equals

NUMERICAL

In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let $m$ and $n$ respectively be the least and the most number of students who studied all the three subjects. Then $\mathrm{m}+\mathrm{n}$ is equal to ___________.

MCQ_SINGLE

The relation $R = \{(a, b) : gcd(a, b) = 1, 2a \neq b, a, b \in Z\}$ is:

View All Questions