JEE 2023 — Mathematics PYQ

Step 1: Roots find karna

Quadratic formula ka use karke:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{\sqrt{2}\pm \sqrt{(\sqrt{2}{)}^2-4(1)(2)}}{2(1)}$

$x=\frac{\sqrt{2}\pm \sqrt{2-8}}{2}=\frac{\sqrt{2}\pm i\sqrt{6}}{2}$

Step 2: Roots ko Polar Form mein likhna

$\alpha ,\beta =\sqrt{2}\left(\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\right)$

$\alpha =\sqrt{2}{e}^{i\pi /3}$ aur $\beta =\sqrt{2}{e}^{-i\pi /3}$

Step 3: Power 14 calculate karna

${\alpha }^{14}=(\sqrt{2}{)}^{14}(e^{i\pi /3}{)}^{14}=2^7{e}^{i14\pi /3}$

$14\pi /3=4\pi +2\pi /3$, toh $e^{i14\pi /3}=e^{i2\pi /3}$

${\alpha }^{14}=128\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)=128\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)$

Isi tarah, ${\beta }^{14}=128\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)$

Step 4: Sum nikalna

${\alpha }^{14}+{\beta }^{14}=128\left(-\frac{1}{2}-\frac{1}{2}\right)=128(-1)=-128$

Correct Option: (3) -128