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Mathematics · 2024

JEE Main · 1 February 2024, Shift 2 · Q2

Consider the relations R_1 and R_2 defined as a R_1 b ⇔ a^2+b^2=1 for all a, b ∈ R and (a, b) R_2(c, d) ⇔ a + d = b + c for all ( a, b ),( c, d ) ∈ N…

Consider the relations $\displaystyle R_1$ and $\displaystyle R_2$ defined as $\displaystyle a R_1 b \Leftrightarrow a^2+b^2=1$ for all $\displaystyle a, b \in R$ and $\displaystyle (a, b) R_2(c, d) \Leftrightarrow$ $\displaystyle \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ for all $\displaystyle (\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{N} \times \mathrm{N}$. Then
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.