JEE 2023 — Mathematics PYQ

Diya gaya hai: $|x{|}^2-2|x|+|\lambda -3|=0$

Maana $|x|=t$ (jahan $t\ge 0$), toh equation banti hai:

Step 2: $t$ (yaani $|x|$) ki real values ke liye condition

Is quadratic equation ke roots real hone ke liye Discriminant ($D$) $\ge 0$ hona chahiye:

$b^2-4ac\ge 0$

$(-2{)}^2-4(1)(|\lambda -3|)\ge 0$

$4-4|\lambda -3|\ge 0$

$1\ge |\lambda -3|$

Iska matlab $\lambda -3$ ki value $-1$ aur $1$ ke beech honi chahiye:

$-1\le \lambda -3\le 1⟹2\le \lambda \le 4$

Step 3: Integer solutions check karna

Hume sawal mein bola gaya hai ki $x$ ek integer hai.

$t^2-2t+|\lambda -3|=0$ se hum $t$ nikalte hain:

Kyunki $x$ integer hai, toh $|x|$ (yaani $t$) bhi integer hona chahiye.

• Case 1: Agar $\sqrt{1-|\lambda -3|}=0$, toh $t=1$.

  Iska matlab $|\lambda -3|=1$, toh $\lambda =4$ ya $\lambda =2$.

  Agar $t=1$, toh $|x|=1⟹x=1,-1$.

• Case 2: Agar $\sqrt{1-|\lambda -3|}=1$, toh $t=1\pm 1$, yaani $t=0$ ya $t=2$.

  Iska matlab $|\lambda -3|=0⟹\lambda =3$.

  Agar $t=0⟹x=0$.

  Agar $t=2⟹x=2,-2$.

Step 4: Set $S$ ke elements aur largest element

Hume $x+\lambda$ ki maximum value nikalni hai:

1. Agar $\lambda =4$ aur $x=1$, toh $x+\lambda =1+4=5$.

2. Agar $\lambda =3$ aur $x=2$, toh $x+\lambda =2+3=5$.

3. Baki combinations ($x=-1,\lambda =4$ ya $x=0,\lambda =3$ etc.) se value $5$ se kam aayegi.

Final Answer:

Set $S$ ka largest element 5 hai.