JEE 2023 — Mathematics PYQ

Hum dono relations $S$ aur $T$ ko ek-ek karke check karenge:

1. Relation $S$ ke liye check (2 + \frac{a}{b} > 0)

• Symmetry Check:

  Maana $(a,b)=(2,-1)$.

  $2+\frac{2}{-1}=2-2=0$. Yeh $S$ mein nahi hai kyunki condition > 0 honi chahiye.

  Lekin agar hum $(a,b)=(1,-1)$ lein, toh 2 + \frac{1}{-1} = 1 > 0 (S mein hai).

  Ab check karte hain $(b,a)=(-1,1)$: 2 + \frac{-1}{1} = 1 > 0 (S mein hai).

  Lekin, agar hum $(a,b)=(4,-1)$ lein: $2+\frac{4}{-1}=-2\ngtr 0$.

  Symmetry simple counter-example: Maana $a=1,b=2$. 2 + 1/2 = 2.5 > 0. Ab $b=2,a=1$, 2 + 2/1 = 4 > 0.

  Lekin agar $a=1,b=-2/3$: 2 + (1 / -2/3) = 2 - 1.5 = 0.5 > 0.

  Ab $(b,a)=(-2/3,1)$: 2 + (-2/3 / 1) = 2 - 0.66 = 1.33 > 0.

  Asal mein $S$ symmetric nahi hai. Example: $(a,b)=(-1,2)$. 2 + (-1/2) = 1.5 > 0.

  Lekin $(b,a)=(2,-1)$: $2+(2/-1)=0\ngtr 0$.

  Isliye, $S$ symmetric nahi hai.

2. Relation $T$ ke liye check ($a^2-b^2\in \mathbb{Z}$)

• Symmetry Check:

  Agar $(a,b)\in T$, toh $a^2-b^2=k$ (jahan $k$ ek integer hai).

  $(b,a)$ ke liye: $b^2-a^2=-(a^2-b^2)=-k$.

  Kyunki $k$ ek integer hai, isliye $-k$ bhi ek integer hoga.

  Isliye, $T$ symmetric hai.

• Transitivity Check:

  Agar $(a,b)\in T$ aur $(b,c)\in T$, toh:

  $a^2-b^2=k_1$ (integer)

  $b^2-c^2=k_2$ (integer)

  Dono ko add karne par: $(a^2-b^2)+(b^2-c^2)=k_1+k_2$

  $a^2-c^2=k_1+k_2$.

  Do integers ka sum hamesha integer hota hai, isliye $(a,c)\in T$.

  Isliye, $T$ transitive bhi hai.

Conclusion

Humne dekha ki:

• $T$ symmetric hai.

• $S$ symmetric nahi hai.

Isliye sahi option (3) hai.

Sahi Answer: (3) $T$ is symmetric but $S$ is not