Class JEE Mathematics Sets, Relations, and Functions Q #1016
COMPETENCY BASED
APPLY
4 Marks 2024 JEE Main 2024 (Online) 6th April Evening Shift MCQ SINGLE
Let $A = {1, 2, 3, 4, 5}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x \le 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m + n$ is equal to :
(A) $23$
(B) $26$
(C) $25$
(D) $24$
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Correct Answer: C
Explanation
Given the set $A = {1, 2, 3, 4, 5}$ and the relation $xRy$ if and only if $4x \le 5y$. We need to find the number of elements in R (denoted by m) and the minimum number of elements that need to be added to R to make it symmetric (denoted by n). First, let's find the relation R: $4x \le 5y \implies \frac{x}{y} \le \frac{5}{4} = 1.25$ Now, let's find the pairs $(x, y)$ that satisfy this condition: If $x = 1$, then $y$ can be $1, 2, 3, 4, 5$. If $x = 2$, then $y$ can be $2, 3, 4, 5$. If $x = 3$, then $y$ can be $3, 4, 5$. If $x = 4$, then $y$ can be $4, 5$. If $x = 5$, then $y$ can be $4, 5$. So, $R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)}$. The number of elements in $R$ is $m = 16$. Now, we need to find the elements to be added to R to make it symmetric. A relation is symmetric if $(x, y) \in R$ implies $(y, x) \in R$. Currently, the elements $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (5,4) \in R$. The symmetric pairs that are missing are: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$. The number of elements to be added is $n = 9$. Therefore, $m + n = 16 + 9 = 25$.

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Step-by-Step Solution

First, we need to find the elements in the relation $R$ defined by $4x \le 5y$ where $x, y \in A = \{1, 2, 3, 4, 5\}$.

We can list the pairs $(x, y)$ that satisfy the condition:

  • If $x = 1$, then $4(1) \le 5y \Rightarrow 4 \le 5y \Rightarrow y \ge \frac{4}{5}$. So, $y$ can be $1, 2, 3, 4, 5$.
  • If $x = 2$, then $4(2) \le 5y \Rightarrow 8 \le 5y \Rightarrow y \ge \frac{8}{5}$. So, $y$ can be $2, 3, 4, 5$.
  • If $x = 3$, then $4(3) \le 5y \Rightarrow 12 \le 5y \Rightarrow y \ge \frac{12}{5}$. So, $y$ can be $3, 4, 5$.
  • If $x = 4$, then $4(4) \le 5y \Rightarrow 16 \le 5y \Rightarrow y \ge \frac{16}{5}$. So, $y$ can be $4, 5$.
  • If $x = 5$, then $4(5) \le 5y \Rightarrow 20 \le 5y \Rightarrow y \ge \frac{20}{5}$. So, $y$ can be $4, 5$.

Therefore, $R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)\}$.

The number of elements in $R$ is $m = 16$.

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 $(y, x) \in R$.

We have the following pairs in $R$:

  • $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$ are already symmetric.
  • $(1, 2) \in R$, we need $(2, 1)$.
  • $(1, 3) \in R$, we need $(3, 1)$.
  • $(1, 4) \in R$, we need $(4, 1)$.
  • $(1, 5) \in R$, we need $(5, 1)$.
  • $(2, 3) \in R$, we need $(3, 2)$.
  • $(2, 4) \in R$, we need $(4, 2)$.
  • $(2, 5) \in R$, we need $(5, 2)$.
  • $(3, 4) \in R$, we need $(4, 3)$.
  • $(3, 5) \in R$, we need $(5, 3)$.
  • $(4, 5) \in R$, we need $(5, 4)$, but $(5,4)$ is already in R.

So, we need to add the following pairs: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$.

The number of elements to add is $n = 9$.

Finally, $m + n = 16 + 9 = 25$.

Correct Answer: 25

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AI Suggestion: Option C

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the definition of a relation and symmetry to a specific set and condition. The student must calculate the number of elements in the relation and then determine how many elements need to be added to make it symmetric.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concepts of relations, Cartesian products, and symmetric relations. It's not just about recalling facts but applying these concepts to a specific problem.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. The question assesses the student's ability to apply the theoretical knowledge of relations to solve a problem, rather than simply recalling definitions or theorems.

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