Q1002: Let $A = \{-2, -1, 0, 1, 2, 3\}$. Let R be a relation on $A$...

COMPETENCY BASED

APPLY

4 Marks 2025 JEE Main 2025 (Online) 3rd April Evening Shift MCQ SINGLE

Let $A = \{-2, -1, 0, 1, 2, 3\}$. Let R be a relation on $A$ defined by $xRy$ if and only if $y = \max\{x, 1\}$. 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) 11

(B) 12

(D) 13

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Correct Answer: B

Explanation

The relation $R$ is defined such that $y = \max(x, 1)$. Therefore, we find the following pairs:
$x = -2 \implies y = \max(-2, 1) = 1$, so $(-2, 1) \in R$.
$x = -1 \implies y = \max(-1, 1) = 1$, so $(-1, 1) \in R$.
$x = 0 \implies y = \max(0, 1) = 1$, so $(0, 1) \in R$.
$x = 1 \implies y = \max(1, 1) = 1$, so $(1, 1) \in R$.
$x = 2 \implies y = \max(2, 1) = 2$, so $(2, 2) \in R$.
$x = 3 \implies y = \max(3, 1) = 3$, so $(3, 3) \in R$.

Thus, $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and $l = 6$.

To make $R$ reflexive, we need to add $(-2, -2), (-1, -1), (0, 0)$. Thus, $m = 3$.

To make $R$ symmetric, we need to add $(1, -2), (1, -1), (1, 0)$. Thus, $n = 3$.

Therefore, $l + m + n = 6 + 3 + 3 = 12$.

AI Tutor Explanation

Step-by-Step Solution

First, we need to determine the elements of the relation R. The relation is defined by $xRy$ if and only if $y = \max\{x, 1\}$.

For $x = -2$, $y = \max\{-2, 1\} = 1$. So, $(-2, 1) \in R$.

For $x = -1$, $y = \max\{-1, 1\} = 1$. So, $(-1, 1) \in R$.

For $x = 0$, $y = \max\{0, 1\} = 1$. So, $(0, 1) \in R$.

For $x = 1$, $y = \max\{1, 1\} = 1$. So, $(1, 1) \in R$.

For $x = 2$, $y = \max\{2, 1\} = 2$. So, $(2, 2) \in R$.

For $x = 3$, $y = \max\{3, 1\} = 3$. So, $(3, 3) \in R$.

Therefore, $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$. The number of elements in R is $l = 6$.

Next, we need to find the minimum number of elements to add to R to make it reflexive. A relation is reflexive if $(x, x) \in R$ for all $x \in A$. The set $A = \{-2, -1, 0, 1, 2, 3\}$. The elements $(x, x)$ that are not in R are $(-2, -2), (-1, -1), (0, 0), (1, 1)$. Since $(1,1), (2,2), (3,3)$ are already in R, we need to add $(-2, -2), (-1, -1), (0, 0)$ to make R reflexive. Thus, $m = 3$.

Now, we need to find the minimum number of elements to add to R to make it symmetric. A relation is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$. The current relation is $R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$.

For $(-2, 1) \in R$, we need to add $(1, -2)$.

For $(-1, 1) \in R$, we need to add $(1, -1)$.

For $(0, 1) \in R$, we need to add $(1, 0)$.

For $(1, 1) \in R$, we don't need to add anything.

For $(2, 2) \in R$, we don't need to add anything.

For $(3, 3) \in R$, we don't need to add anything.

So, we need to add $(1, -2), (1, -1), (1, 0)$. Thus, $n = 3$.

Finally, we need to find $l + m + n = 6 + 3 + 3 = 12$.

Correct Answer: 12

AI Suggestion: Option B

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because it requires the student to apply the definition of relations, reflexive relations, and symmetric relations to a given set and relation.

Knowledge Dimension: CONCEPTUAL

Justification: The question tests the understanding of concepts related to relations, including the definition of a relation, reflexive relations, and symmetric relations. It requires applying these concepts to a specific example.

Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It requires the application of learned concepts to solve a problem, rather than simply recalling a definition.

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