Let A = {1, 2, 3, 4, ldots, 10} and B = {0, 1 | vedclass

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(D) Let $x = a - b$. The condition is $2x^2 + 3x \in \{0, 1, 2, 3, 4\}$.Case $1$: $2x^2 + 3x = 0 \Rightarrow x(2x + 3) = 0$. Since $a, b \in \{1, \ldo

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$18$

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(D) Let $x = a - b$. The condition is $2x^2 + 3x \in \{0, 1, 2, 3, 4\}$.Case $1$: $2x^2 + 3x = 0 \Rightarrow x(2x + 3) = 0$. Since $a, b \in \{1, \ldots, 10\}$,$x$ must be an integer. Thus,$x = 0$.If $x = 0$,then $a - b = 0 \Rightarrow a = b$. There are $10$ such pairs: $(1,1), (2,2), \ldots, (10,10)$.Case $2$: $2x^2 + 3x = 1 \Rightarrow 2x^2 + 3x - 1 = 0$. No integer solution for $x$.Case $3$: $2x^2 + 3x = 2 \Rightarrow 2x^2 + 3x - 2 = 0 \Rightarrow (2x - 1)(x + 2) = 0$. Integer solution $x = -2$.If $x = -2$,then $a - b = -2 \Rightarrow b = a + 2$. Possible pairs: $(1,3), (2,4), (3,5), (4,6), (5,7), (6,8), (7,9), (8,10)$. There are $8$ such pairs.Case $4$: $2x^2 + 3x = 3 \Rightarrow 2x^2 + 3x - 3 = 0$. No integer solution.Case $5$: $2x^2 + 3x = 4 \Rightarrow 2x^2 + 3x - 4 = 0$. No integer solution.Total number of elements $= 10 + 8 = 18$.

Let $A = \{1, 2, 3, 4, \ldots, 10\}$ 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 $.........$.

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