Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if 0 ≤ x^2 + 2

Q: Wait, why do we have to check $x = \pm 3$ separately? Can't we just use a general formula?

That is a great question! In continuous calculus, we love general formulas. But here, we are dealing with a finite, discrete set $A$. When the domain is small, the most 'general' formula is actually the systematic case-by-case analysis. It prevents the trap of missing boundary points or including values that don't satisfy the integer constraint. Trust the process of counting; it is the most reliable tool in your kit for discrete relations.

Q: Why do we reject the negative root when checking for reflexivity?

Actually, we aren't rejecting a root; we are testing the condition $(a, a) \in R$. For a relation to be reflexive, the pair $(a, a)$ must satisfy the inequality $0 \le a^2 + 2a \le 4$. If you plug in $a = -1$, you get $1 - 2 = -1$, which is less than $0$. Since the condition requires the value to be $\ge 0$, the pair $(-1, -1)$ fails. It is not about rejecting a root, but about verifying if the point lies within the 'safe zone' defined by the relation.

Q: Is there a faster way to find $m$ without checking every element?

You can certainly speed it up! Instead of checking every pair, define a function $f(x) = x^2 + 2x$. You are looking for all $x \in A$ such that $0 \le f(x) \le 4$. The values of $x$ that do not satisfy this are your 'missing' elements. It is essentially finding the roots of the quadratic inequalities $x^2 + 2x \ge 0$ and $x^2 + 2x \le 4$. This algebraic shortcut is much faster than manual verification!

Question

Let $A = \{-3, -2, -1, 0, 1, 2, 3\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $0 \le x^2 + 2y \le 4$. Let $l$ be the number of elements in $R$ and $m$ be the minimum number of elements required to be added in $R$ to make it a reflexive relation. Then $l + m$ is equal to

Options:

(A) 19

(B) 20

(C) 17

(D) 18

Sigma Percentile · Accepted Answer

### The Geometry of Relations: A Journey Through the Lattice

Welcome, future engineer! Today, we are not just solving a problem; we are exploring the architecture of a relation.

Often, when students see a set $A = \{-3, -2, -1, 0, 1, 2, 3\}$ and a condition like $0 \le x^2 + 2y \le 4$, they panic. They see an inequality and think, "Do I need to graph this?"

The answer is a resounding yes! But not in the way you might think. We are going to map this relation onto the discrete landscape of integers.

### Phase 1: The Corridor of Possibility

Let us look at our condition: $0 \le x^2 + 2y \le 4$. To make this manageable, we must isolate $y$.

Think of $y$ as the dependent variable, the one that must "fit" into the space created by $x$. By subtracting $x^2$ and dividing by $2$, we arrive at the following inequality:

$$-\frac{x^2}{2} \le y \le \frac{4 - x^2}{2}$$

Imagine this as a corridor. For every $x$ you choose, $y$ is trapped between two parabolic curves. Our job is to walk through every integer $x$ in our set $A$ and see which integers $y$ are trapped inside this corridor.

### Phase 2: The Systematic Hunt

Let us begin our journey at the boundaries.

When $x = \pm 3$, $x^2 = 9$. Our inequality becomes:

$$-\frac{9}{2} \le y \le \frac{4 - 9}{2} \Rightarrow -4.5 \le y \le -2.5$$

Looking at our set $A$, the only integer that fits is $y = -3$. Thus, we have two pairs: $(-3, -3)$ and $(3, -3)$.

Moving to $x = \pm 2$, $x^2 = 4$. The inequality becomes $-2 \le y \le 0$.

Here, the integers $y \in \{-2, -1, 0\}$ all satisfy the condition. With two values of $x$ and three values of $y$, we have found $2 	imes 3 = 6$ pairs.

For $x = \pm 1$, $x^2 = 1$. The inequality becomes $-0.5 \le y \le 1.5$.

The valid integers are $y \in \{0, 1\}$. That gives us $2 	imes 2 = 4$ pairs.

Finally, at the origin $x = 0$, $x^2 = 0$. The inequality is $0 \le y \le 2$.

The integers are $y \in \{0, 1, 2\}$. That is $3$ pairs. Adding them all up: $2 + 6 + 4 + 3 = 15$. We have found $l = 15$.

### Phase 3: The Mirror of Reflexivity

Now, we pivot to the concept of reflexivity. A relation is reflexive if every element "sees itself" in the relation.

Mathematically, for every $a \in A$, the pair $(a, a)$ must be in $R$. This is like checking if the line $y = x$ passes through our "corridor."

We need to check if $(a, a)$ satisfies $0 \le a^2 + 2a \le 4$ for all $a \in \{-3, -2, -1, 0, 1, 2, 3\}$.

Let us test them:

1. For $a = -3$: $(-3)^2 + 2(-3) = 9 - 6 = 3$. Since $0 \le 3 \le 4$, this is valid.
2. For $a = -2$: $(-2)^2 + 2(-2) = 4 - 4 = 0$. Since $0 \le 0 \le 4$, this is valid.
3. For $a = -1$: $(-1)^2 + 2(-1) = 1 - 2 = -1$. This is NOT in the range $[0, 4]$. We are missing $(-1, -1)$.
4. For $a = 0$: $0^2 + 2(0) = 0$. Valid.
5. For $a = 1$: $1^2 + 2(1) = 3$. Valid.
6. For $a = 2$: $2^2 + 2(2) = 8$. This is NOT in the range $[0, 4]$. We are missing $(2, 2)$.
7. For $a = 3$: $3^2 + 2(3) = 15$. This is NOT in the range $[0, 4]$. We are missing $(3, 3)$.

We found three missing pairs: $(-1, -1)$, $(2, 2)$, and $(3, 3)$. Therefore, $m = 3$.

### The Final Synthesis

We have conquered the counting of $l$ and the logic of $m$. The final step is simply the sum:

$$l + m = 15 + 3 = 18$$

Take a moment to appreciate what you have done. You didn't just calculate a number; you visualized a relation, navigated a discrete space, and applied the definition of reflexivity to find the "holes" in the structure.

This is the essence of JEE Advanced mathematics—turning a complex-looking condition into a clear, logical path. Keep this clarity, and no problem will ever be too daunting!

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Visualized Solution

Understanding Set $A$ and Relation $R$

Isolating $y$ for Systematic Counting

Case 1: $x = \pm 3$

Case 2: $x = \pm 2$

Case 3: $x = \pm 1$

Case 4: $x = 0$

Calculating Total Elements $l$

Reflexivity Check

Finding Missing Elements $m$

Final Calculation: $l + m$

The Sigma Insight: Types of Relations

The Geometry of Relations

Phase 1

Phase 2

Phase 3

The Final Synthesis

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Visualized Solution

Understanding Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y for Systematic Counting

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=±3

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=±2

Case 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=±1

Case 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Calculating Total Elements .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l