Let A = \ 1, 2, 3, 4, . . . . . . . . . . 10\ and B = \ 0, 1, 2, 3,...

Given,

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

$( a, b ) \in A \times A$ 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,</p><p>So,$9 + 8k$is a perfect square</p><p>$⇒ k = 0 or k = 2$</p><p>Now for$k = 0$, </p><p>$2{\left(a - b\right)}^{2} + 3\left(a - b\right) = 0$</p><p>$⇒ \left(a - b\right)\left[2{\left(a - b\right)}^{2} + 3\right] = 0

$⇒ a - b = 0 ⇒ ( a, b ) \in { ( 1, 1 ) , ( 2, 2 ) \ldots . ( 10, 10 ) }$

$⇒$Total $10$ elements belonging to $R .</p><p>And,$a - b = \dfrac{3}{2}$is not possible</p><p>Now for$k = 2$,</p><p>$2{\left(a - b\right)}^{2} + 3\left(a - b\right) - 2 = 0$</p><p>⇒ a - b = - 2 or$a - b = \dfrac{1}{2}$(not possible)</p><p>Now for$a - b = - 2$possible pair will be,</p><p>$⇒ ( a, b ) \in { ( 1, 3 ) , ( 2, 4 ) , \ldots . ( 8, 10 ) }$</p><p>$⇒ 8$elements belonging to$R$</p><p>Total number of elements will be$= 18