JEE Main 2023 — Sets and Relations Question with Solution

Given,

$A=\{1,2,3,\dots ,10\} B=\{0,1,2,4\}$

$(a,b)\in A\times A$ such that

$2{\left(a-b\right)}^2+3\left(a-b\right)-k=0$

where $k\in \{0,1,2,3,4\}$

Now finding discriminant $D=9-4\times 2\left(-k\right)$

And $9-4\times 2\left(-k\right)$ a perfect square for any possible $\left(a,b\right)$ as $\left(a-b\right)$ will be a integer,

So, $9+8k$ is a perfect square

$\Rightarrow k=0$ or $k=2$

Now for $k=0$, 

$2{\left(a-b\right)}^2+3\left(a-b\right)=0$

$\Rightarrow \left(a-b\right)\left[2{\left(a-b\right)}^2+3\right]=0$

$\Rightarrow a-b=0\Rightarrow (a,b)\in \left\{\right(1,1),(2,2)\dots .(10,10\left)\right\}$

$\Rightarrow$Total $10$ elements belonging to $R.$

And, $a-b=\frac{3}{2}$ is not possible

Now for $k=2$,

$2{\left(a-b\right)}^2+3\left(a-b\right)-2=0$

$\Rightarrow a-b=-2$ or $a-b=\frac{1}{2}$ (not possible)

Now for $a-b=-2$ possible pair will be,

$\Rightarrow (a,b)\in \left\{\right(1,3),(2,4),\dots .(8,10\left)\right\}$

$\Rightarrow 8$ elements belonging to $R$

Total number of elements will be$=18$