JEE 2023 — Mathematics PYQ

The Solution

Simplify the condition:

$adb-adc=bca-bcd$

Divide by $abcd$:

$\frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{a}⟹\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

• Reflexive: Check $(a,b)R(a,b)$.

  Does $\frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{a}$? Only if $a=b$. Not true for all $(a,b)\in \mathbb{N}\times \mathbb{N}$. Not Reflexive.

• Symmetric: If $(a,b)R(c,d)$, then $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$.

  For $(c,d)R(a,b)$, we need $\frac{1}{c}-\frac{1}{d}=\frac{1}{b}-\frac{1}{a}$.

  Multiplying the first by $-1$ gives the second. Symmetric.

• Transitive: If $(a,b)R(c,d)$ and $(c,d)R(e,f)$:

  $\frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$ and $\frac{1}{c}-\frac{1}{d}=\frac{1}{f}-\frac{1}{e}$.

  Adding them: $(\frac{1}{a}-\frac{1}{b})+(\frac{1}{c}-\frac{1}{d})=(\frac{1}{d}-\frac{1}{c})+(\frac{1}{f}-\frac{1}{e})$.

  $\frac{1}{a}-\frac{1}{b}=2(\frac{1}{d}-\frac{1}{c})+\frac{1}{f}-\frac{1}{e}$. This does not imply $(a,b)R(e,f)$. Not Transitive.

Result: Symmetric but neither reflexive nor transitive. Correct Option: (2).