JEE 2023 — Mathematics PYQ

Solution

Equation ko rearrange karte hain:

$x^2-4x+3=x[x]-[x]$

$(x-3)(x-1)=[x](x-1)$

Yahan do sthitiyan (cases) banti hain:

1. Yadi $x-1=0$: Toh $x=1$. Check karte hain: $1^2-4(1)+[1]+3=1[1]⟹1=1$. Yani $x=1$ ek solution hai.

2. Yadi $x\ne 1$: Toh hum $(x-1)$ ko cancel kar sakte hain:

   $x-3=[x]$

   Hume pata hai ki $x-[x]=\{x\}$, jahan 0 \le \{x\} < 1.

   Yahan $x-[x]=3$, jo ki sambhav nahi hai kyunki fractional part $3$ nahi ho sakta.

Atah, keval ek hi solution hai $x=1$. Chunki $x=1$ range $(-\infty ,\infty )$ mein aata hai, isliye:

Sahi Option: (D)