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

Option: 1

18


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{A}=\{1,2,3, \ldots \ldots 10\}
\mathrm{B}=\{0,1,2,3,4\}
\mathrm{R}=\left\{(\mathrm{a}, \mathrm{b}) \in \mathrm{A} \times \mathrm{A}: 2(\mathrm{a}-\mathrm{b})^{2}+3(\mathrm{a}-\mathrm{b}) \in \mathrm{B}\right\}

Now \: 2(a-b)^{2}+3(a-b)=(a-b)(2(a-b)+3)

\Rightarrow \mathrm{a}=\mathrm{b}$ or $\mathrm{a}-\mathrm{b}=-2

When \mathrm{a}=\mathrm{b} \Rightarrow 10  order pairs

When \mathrm{a}-\mathrm{b}=-2 \Rightarrow 8 order pairs

Total =18

Posted by

Nehul

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