Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,2,3,4,\dots ,10\}$ and $B=\{0,1,2,3,4\}$. The number of elements in the relation $R=\{(a,b)\in A\times A:2(a-b{)}^2+3(a-b)\in B\}$ is

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that at first glance looks like a simple set-theory exercise, but beneath the surface, it is a beautiful dance of algebra and logic.

We are given two sets, $A=\{1,2,\dots ,10\}$ and $B=\{0,1,2,3,4\}$, and a relation $R$ defined by the condition $2(a-b{)}^2+3(a-b)\in B$. Our mission is to count the number of elements in $R$.

The first thing that should catch your eye is the repetition of the term $(a-b)$. In mathematics, whenever you see a recurring expression, it is a massive hint to simplify.

Let us define a new variable, $x=a-b$. Since $a$ and $b$ are both integers, their difference $x$ must also be an integer. This simple substitution transforms our condition into a quadratic equation:

where $k\in \{0,1,2,3,4\}$. We are essentially looking for integer values of $x$ that satisfy this equation for any $k$ in set $B$.

Now, we must test each possible value of $k$ from set $B$. We are solving the equation $2x^2+3x-k=0$ for integer roots.

For $k=0$, we have $2x^2+3x=0$, which factors into $x(2x+3)=0$. This gives us $x=0$ or $x=-\frac{3}{2}$. Since $x$ must be an integer, we discard $-\frac{3}{2}$ and keep $x=0$.

For $k=1$, the equation is $2x^2+3x-1=0$. The discriminant is:

Since $17$ is not a perfect square, the roots are irrational; thus, no integer $x$ exists here.

For $k=2$, we have $2x^2+3x-2=0$. Factoring this gives $(2x-1)(x+2)=0$, yielding $x=\frac{1}{2}$ or $x=-2$. We keep only the integer solution, $x=-2$.

Checking $k=3$ and $k=4$ similarly reveals no integer solutions, as their discriminants ($33$ and $41$ respectively) are not perfect squares.

We have successfully identified the only two values of $x$ that satisfy our condition: $x=0$ and $x=-2$. Now, we must translate these back into pairs $(a,b)$ where $a,b\in \{1,2,\dots ,10\}$.

For $x=0$, we have $a-b=0$, or $a=b$. The valid pairs are $(1,1),(2,2),\dots ,(10,10)$. This yields $10$ pairs.

For $x=-2$, we have $a-b=-2$, or $b=a+2$. We list the possibilities: $(1,3),(2,4),(3,5),(4,6),(5,7),(6,8),(7,9),(8,10)$.

Counting these, we find $8$ pairs.

We have navigated the algebra and the constraints. The total number of elements in the relation $R$ is the sum of the pairs found in our two valid cases:

The total number of elements in the relation $R$ is $18$. It is truly elegant how a complex-looking relation collapses into a simple counting problem once you apply the right substitution.