Question:

Let $ A = \{-2, -1, 0, 1, 2, 3\} $. Let $ R $ be a relation on $ A $ defined by $ (x, y) \in R $ if and only if $ |x| \le |y| $. Let $ m $ be the number of reflexive elements in $ R $ 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

Show Hint

A relation is reflexive if $ (x, x) $ is in the relation for all elements $ x $ in the set. A relation is symmetric if whenever $ (x, y) $ is in the relation, $ (y, x) $ is also in the relation. To make a relation reflexive, add all missing pairs of the form $ (x, x) $. To make a relation symmetric, for every pair $ (x, y) $ in the relation, if $ (y, x) $ is not already present, add it.

Updated On: Jan 14, 2026

Hide Solution

Verified By Collegedunia

The Correct Option is A

Approach Solution - 1

The problem provides a set $A = \{-2, -1, 0, 1, 2, 3\}$ and a relation R on A defined by $xRy$ if and only if $y = \max\{x, 1\}$. We need to find the number of elements in R ($l$), the minimum number of elements to add to make R reflexive ($m$), and the minimum number of elements to add to make R symmetric ($n$). Finally, we need to compute the sum $l + m + n$.

Concept Used:

A relation R on a set A is a subset of the Cartesian product $A \times A$.

A relation R on a set A is reflexive if for every element $a \in A$, the ordered pair $(a, a)$ is in R.

A relation R on a set A is symmetric if for every ordered pair $(a, b)$ in R, the pair $(b, a)$ is also in R. That is, if $aRb$, then $bRa$.

Step-by-Step Solution:

Step 1: Determine the elements of the relation R and find its size, $l$.

The relation is defined by $y = \max\{x, 1\}$ for $x, y \in A$. We test each value of $x$ from set A:

For $x = -2$, $y = \max\{-2, 1\} = 1$. Since $1 \in A$, the pair $(-2, 1)$ is in R.
For $x = -1$, $y = \max\{-1, 1\} = 1$. Since $1 \in A$, the pair $(-1, 1)$ is in R.
For $x = 0$, $y = \max\{0, 1\} = 1$. Since $1 \in A$, the pair $(0, 1)$ is in R.
For $x = 1$, $y = \max\{1, 1\} = 1$. Since $1 \in A$, the pair $(1, 1)$ is in R.
For $x = 2$, $y = \max\{2, 1\} = 2$. Since $2 \in A$, the pair $(2, 2)$ is in R.
For $x = 3$, $y = \max\{3, 1\} = 3$. Since $3 \in A$, the pair $(3, 3)$ is in R.

So, the relation R is the set of these ordered pairs:

\[ R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\} \]

The number of elements in R is $l$. By counting the pairs, we get:

\[ l = 6 \]

Step 2: Determine the number of elements ($m$) to be added to make R reflexive.

For R to be reflexive on A, it must contain all pairs $(a, a)$ for every $a \in A$. The required set of pairs for reflexivity is:

\[ R_{\text{reflexive\_pairs}} = \{(-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)\} \]

Comparing this with the elements of R, we see that R already contains $(1, 1)$, $(2, 2)$, and $(3, 3)$.

The pairs that are missing from R are:

\[ \{(-2, -2), (-1, -1), (0, 0)\} \]

Therefore, we need to add these 3 elements to R to make it reflexive. The minimum number of elements to be added is:

\[ m = 3 \]

Step 3: Determine the number of elements ($n$) to be added to make R symmetric.

For R to be symmetric, if $(a, b) \in R$, then $(b, a)$ must also be in R. We check each pair in R:

For $(-2, 1) \in R$, we need $(1, -2)$. This is not in R.
For $(-1, 1) \in R$, we need $(1, -1)$. This is not in R.
For $(0, 1) \in R$, we need $(1, 0)$. This is not in R.
For $(1, 1) \in R$, we need $(1, 1)$, which is already in R.
For $(2, 2) \in R$, we need $(2, 2)$, which is already in R.
For $(3, 3) \in R$, we need $(3, 3)$, which is already in R.

The pairs that need to be added to make R symmetric are:

\[ \{(1, -2), (1, -1), (1, 0)\} \]

Therefore, the minimum number of elements to be added is:

\[ n = 3 \]

Step 4: Calculate the final sum $l + m + n$.

We have found the values $l=6$, $m=3$, and $n=3$. We now compute their sum.

\[ l + m + n = 6 + 3 + 3 = 12 \]

The value of $l + m + n$ is 12.

Was this answer helpful?

Hide Solution

Verified By Collegedunia

Approach Solution -2

Let the set $ A = \{-2, -1, 0, 1, 2, 3\} $
Let the relation $ R = \{(-2,1), (-1,1), (0,1), (1,1), (2,2), (3,3)\} $
\[\begin{align*} \lambda &= 6 \\m &= 3 \\n &= 3 \\\lambda + m + n &= 6 + 3 + 3 = 12 \end{align*}\]