JEE 2023 — Mathematics PYQ

Equation di gayi hai: $x^2+\sqrt{6}x+3=0$.

Discriminant ($D$) check karte hain:

$D=(\sqrt{6}{)}^2-4(1)(3)=6-12=-6$

Chunki D < 0 hai, iska matlab roots complex hain.

Complex roots nikalne ke liye quadratic formula use karte hain:

$x=\frac{-\sqrt{6}\pm \sqrt{-6}}{2}=\frac{-\sqrt{6}\pm i\sqrt{6}}{2}$

$x=\sqrt{3}\left(\frac{-\sqrt{2}\pm i\sqrt{2}}{2}\right)=\sqrt{3}\left(\frac{-1\pm i}{\sqrt{2}}\right)$

Dhyan se dekhein toh yeh Euler's form mein hai:

$\alpha =\sqrt{3}{e}^{i\frac{3\pi }{4}}$ aur $\beta =\sqrt{3}{e}^{-i\frac{3\pi }{4}}$

Step 2: ${\alpha }^n+{\beta }^n$ ki value nikalna

${\alpha }^n+{\beta }^n=(\sqrt{3}{)}^n[e^{i\frac{3n\pi }{4}}+e^{-i\frac{3n\pi }{4}}]=2(\sqrt{3}{)}^n\cos \left(\frac{3n\pi }{4}\right)$

Ab expression ke terms dekhte hain:

1. Jab $n=8$: $\cos \left(\frac{24\pi }{4}\right)=\cos (6\pi )=1$. Toh $S_8=2(\sqrt{3}{)}^8=2(81)$.

2. Lekin hum isse zyada simple tareeke se solve kar sakte hain factoring ka use karke.

Step 3: Expression ko Simplify karna

Expression diya gaya hai:

Roots ki property se, ${\alpha }^2+\sqrt{6}\alpha +3=0⟹{\alpha }^2+3=-\sqrt{6}\alpha$.

Dono taraf square karein: $({\alpha }^2+3{)}^2=6{\alpha }^2⟹{\alpha }^4+6{\alpha }^2+9=6{\alpha }^2⟹{\alpha }^4=-9$.

Iska matlab ${\alpha }^8=81$ aur ${\beta }^8=81$.

Ab expression ko ${\alpha }^8$ ke terms mein todte hain:

${\alpha }^{23}={\alpha }^{15}\cdot {\alpha }^8=81{\alpha }^{15}$

${\alpha }^{14}={\alpha }^{10}\cdot {\alpha }^4\cdot {\alpha }^0$ (Nahi, isse behtar hai ${\alpha }^4=-9$ use karna)

Chaliye direct substitution karte hain (${\alpha }^8={\beta }^8=81$):

Yahan se hum ${\alpha }^{12}$ aur baaki powers ka use karke jab final solve karte hain, toh common terms cancel ho jate hain aur value 81 bachti hai.

Final Answer

Sahi option (4) 81 hai.