JEE 2023 — Mathematics PYQ

Hame condition di gayi hai: $f(m\cdot n)=f(m)\cdot f(n)$.

1. Case for $m=1$:

   $f(1\cdot n)=f(1)\cdot f(n)⟹f(n)=f(1)\cdot f(n)$.

   Isse pata chalta hai ki $f(1)=1$ hona chahiye (kyunki $0$ set $A$ mein nahi hai).

2. Constraints check karein:

   • $f(9)=f(3\cdot 3)=f(3)\cdot f(3)=[f(3){]}^2$

     Kyunki $f(3)\in A$, toh $[f(3){]}^2$ bhi $A$ mein hona chahiye.

     $A$ mein sirf $1$ aur $9$ hi perfect squares hain.

     • Agar $f(3)=1$, toh $f(9)=1$ (1 way)

     • Agar $f(3)=3$, toh $f(9)=9$ (1 way)

       Total ways for $f(3)$ and $f(9)=2$

   • $f(8)=f(2\cdot 2\cdot 2)$ — lekin $4\notin A$, isliye hum sirf $f(2)\cdot f(2)$ nahi likh sakte. Par $f(8)$ ke liye koi restriction nahi hai siway iske ki output $A$ mein ho.

     Lekin ek cheez dhyan dein: $2\cdot 1=2\in A$ aur $1\cdot 1=1\in A$ pehle hi satisfy ho chuke hain.

     Set $A$ mein koi aur product available nahi hai (jaise $2\cdot 3=6\notin A$, $2\cdot 5=10\notin A$, etc.)

3. Independent Values:

   • $f(2),f(5),f(8)$ ke liye koi extra condition nahi hai, toh inme se har ek 6 tareekon se choose kiya ja sakta hai ($|A|=6$).

   • $f(1)$ hamesha $1$ rahega (1 way).

   • $f(3)$ aur $f(9)$ ke liye upar dekha humne 2 combinations hain.

4. Total Functions:

   $$\text{Total}=f(1)\times [f(3),f(9)]\times f(2)\times f(5)\times f(8)$$
   $$\text{Total}=1\times 2\times 6\times 6\times 6=2\times 216=432$$

Jawab: 432