Class JEE Mathematics Sets, Relations, and Functions Q #1000
COMPETENCY BASED
APPLY
4 Marks 2025 JEE Main 2025 (Online) 4th April Evening Shift MCQ SINGLE
Let $A = \{-3, -2, -1, 0, 1, 2, 3\}$ and R be a relation on A defined by $xRy$ if and only if $2x - y \in \{0, 1\}$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then $l + m + n$ is equal to:
(A) 17
(B) 18
(C) 15
(D) 16
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Correct Answer: A
Explanation
Given the relation $xRy \Leftrightarrow 2x - y \in \{0, 1\}$, this implies $y = 2x$ or $y = 2x - 1$. Given $A = \{-3, -2, -1, 0, 1, 2, 3\}$, we find $R = \{(-1, -2), (0, 0), (1, 2), (-1, -3), (0, -1), (1, 1), (2, 3)\}$. Thus, $l = |R| = 7$. For $R$ to be reflexive, we need $(a, a) \in R$ for all $a \in A$ such that $2a - a \in \{0, 1\}$. This means $a \in \{0, 1\}$. We already have $(0,0)$ and $(1,1)$ in $R$. However, we need to add pairs $(a,a)$ such that $a \in A$ and $a \notin \{0, 1\}$. These are $(-3, -3)$, $(-2, -2)$, $(-1, -1)$, $(2, 2)$, and $(3, 3)$. But $2a-a = a$ so we need $a \in \{0,1\}$ so these pairs must satisfy $a \in \{0, 1\}$. So, $a \in A$ and $2a-a = a$ which implies $a \in \{0, 1\}$, which implies $a \in \{0, 1\}$. The pairs needed are $(-3,-3)$, $(-2, -2)$, $(-1, -1)$, $(2, 2)$, $(3, 3)$. These satisfy $2a - a = a$ which gives $a \in \{0, 1\}$. This is not true. So $m = 5$. For $R$ to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. We need to add: $(-1, -2) \Rightarrow (-2, -1)$, $(1, 2) \Rightarrow (2, 1)$, $(-1, -3) \Rightarrow (-3, -1)$, $(0, -1) \Rightarrow (-1, 0)$, $(2, 3) \Rightarrow (3, 2)$. Thus, $n = 5$. Therefore, $l + m + n = 7 + 5 + 5 = 17$.

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Step-by-Step Solution

  1. Finding the elements of R:

    We have $A = \{-3, -2, -1, 0, 1, 2, 3\}$ and $2x - y \in \{0, 1\}$. We need to find all pairs $(x, y)$ such that $x, y \in A$ and $2x - y = 0$ or $2x - y = 1$.

    • If $2x - y = 0$, then $y = 2x$. The pairs are: $(-1, -2), (0, 0), (1, 2)$.
    • If $2x - y = 1$, then $y = 2x - 1$. The pairs are: $(-1, -3), (0, -1), (1, 1), (2, 3)$.

    Therefore, $R = \{(-1, -2), (0, 0), (1, 2), (-1, -3), (0, -1), (1, 1), (2, 3)\}$. So, $l = 7$.

  2. Making R reflexive:

    For R to be reflexive, it must contain all pairs $(x, x)$ for all $x \in A$. The pairs are: $(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$. R already contains $(0, 0)$ and $(1, 1)$. The pairs to be added are: $(-3, -3), (-2, -2), (-1, -1), (2, 2), (3, 3)$. So, $m = 5$.

  3. Making R symmetric:

    After making R reflexive, we have $R = \{(-1, -2), (0, 0), (1, 2), (-1, -3), (0, -1), (1, 1), (2, 3), (-3, -3), (-2, -2), (-1, -1), (2, 2), (3, 3)\}$.

    For R to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$.

    • $(-1, -2) \in R \implies (-2, -1)$ needs to be added.
    • $(1, 2) \in R \implies (2, 1)$ needs to be added.
    • $(-1, -3) \in R \implies (-3, -1)$ needs to be added.
    • $(0, -1) \in R \implies (-1, 0)$ needs to be added.
    • $(2, 3) \in R \implies (3, 2)$ needs to be added.

    So, $n = 5$.

  4. Calculating l + m + n:

    $l + m + n = 7 + 5 + 5 = 17$.

Correct Answer: 17

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AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply the definitions of relations (reflexive, symmetric) and set theory to a specific problem. They need to determine the elements of the relation R based on the given condition and then manipulate it to satisfy the reflexive and symmetric properties.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of the concepts of relations, reflexive relations, and symmetric relations, rather than just recalling facts or performing routine procedures.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It assesses the student's ability to apply the concepts of relations and functions to solve a problem, rather than just recalling definitions.

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