JEE 2023 — Mathematics PYQ

1. Given Conditions ko analyze karein:

Function ka domain $\{1,2,3,4\}$ hai aur codomain $\{a\in \mathbb{Z}:-8\le a\le 8\}$ hai.

Relation diya gaya hai: $f(n)+\frac{1}{n}f(n+1)=1$

Iska matlab:

• For $n=1$: $f(1)+f(2)=1⟹f(2)=1-f(1)$

• For $n=2$: $f(2)+\frac{1}{2}f(3)=1⟹f(3)=2(1-f(2))$

• For $n=3$: $f(3)+\frac{1}{3}f(4)=1⟹f(4)=3(1-f(3))$

2. Saare terms ko $f(1)$ ke form mein likhein:

• $f(2)=1-f(1)$

• $f(3)=2(1-(1-f(1)))=2f(1)$

• $f(4)=3(1-2f(1))=3-6f(1)$

3. Integers ki range apply karein:

Sabhi $f(n)$ integers hone chahiye aur unki value $-8$ aur $8$ ke beech honi chahiye.

$f(1)$ ek integer hai, toh baaki saare automatically integers ho jayenge. Ab range check karte hain:

1. $-8\le f(1)\le 8$

2. $-8\le 1-f(1)\le 8⟹-7\le f(1)\le 9$

3. $-8\le 2f(1)\le 8⟹-4\le f(1)\le 4$

4. $-8\le 3-6f(1)\le 8$:

   • $-11\le -6f(1)\le 5$

   • Divide by $-6$ (sign change hoga): $\frac{-5}{6}\le f(1)\le \frac{11}{6}$

   • Approximate range: $-0.83\le f(1)\le 1.83$

4. Common values dhoondhein:

Saari conditions ko satisfy karne waale integer $f(1)$ hain:

• Agar $f(1)=0$: $f(2)=1,f(3)=0,f(4)=3$. (Sahi hai)

• Agar $f(1)=1$: $f(2)=0,f(3)=2,f(4)=-3$. (Sahi hai)

Toh kul 2 aise functions possible hain.

Correct Option: (3)