Let $A=\left\{1,2,3,4,\mathrm{.....},10\right\}$ and $B=\left\{0,1,2,3,4\right\}$ . The number of  elements in the relation $R=\left\{(a,b)\in A\times A:2{(a-b)}^2+3(a-b)\in B\right\}$ is _____

A = {1, 2, 3, ..., 10} B = {0, 1, 2, 4}
$(a, b) \in A \times A such that$
2(a – b)² + 3(a – b) – k = 0
where k $\in$ {0, 1, 2, 3, 4}
We should have
9 – 4 × 2(–k) a perfect square for any possible (a, b)
i.e, 9 + 8k is perfect square
$\Rightarrow$k = 0 or k = 2
for k = 0, 2(a – b)² + 3(a – b) = 0
$\Rightarrow$ a – b = 0 $\Rightarrow$ (a, b) $\in$ {(1, 1), (2, 2) ..... (10, 10)}.
 Total 10 elements belonging to R.
$a-b=-\frac{3}{2}$is not possible
for k = 0 2(a – b) + 3(a – b) – 2 = 0
$\Rightarrow$ a – b = –2 or $a-b=\frac{1}{2}$(notpossible)
 $\Rightarrow$(a, b) $\in$ {(1, 3), (2, 4), ..... (8, 10)}
$\Rightarrow$8 element belonging to R
Total = 18

(or)
Now $2{(a-b)}^2+3(a-b)\in B$

$\Rightarrow (a-b)\left[2(a-b)+3\right]\in B$

$a=b$ or $a-b=-2$

When  $a=b\Rightarrow$10 order pairs

When  $a-b=-2\Rightarrow$ 8 order pairs

Total number of elements is R = 18.