JEE Main 2023 — Mathematics Algebra

Given,

$${x}^{2}-\sqrt{2}x+2=0$$x2−2​x+2=0

$$\Rightarrow x=\frac{\sqrt{2}\pm \sqrt{6}i}{2}$$⇒x=22​±6​i​

$$\Rightarrow x=\sqrt{2}(\frac{1\pm \sqrt{3}i}{2})$$⇒x=2​(21±3​i​)

$$\Rightarrow x=\sqrt{2}(\frac{1}{2}\pm \frac{\sqrt{3}i}{2})$$⇒x=2​(21​±23​i​)

$$\Rightarrow x=\sqrt{2}(\mathrm{cos}\frac{\pi }{3}\pm i\mathrm{sin}\frac{\pi }{3})$$⇒x=2​(cos3π​±isin3π​)

So, $\alpha =\sqrt{2}(\mathrm{cos}\frac{\pi }{3}+i\mathrm{sin}\frac{\pi }{3})=\sqrt{2}{e}^{\frac{i\pi }{3}}$α=2​(cos3π​+isin3π​)=2​e3iπ​

And $\beta =\sqrt{2}(\mathrm{cos}\frac{\pi }{3}-i\mathrm{sin}\frac{\pi }{3})=\sqrt{2}{e}^{\frac{-i\pi }{3}}$β=2​(cos3π​−isin3π​)=2​e3−iπ​

Now ${\alpha }^{14}+{\beta }^{14}={(\sqrt{2}{e}^{\frac{i\pi }{3}})}^{14}+{(\sqrt{2}{e}^{\frac{-i\pi }{3}})}^{14}$α14+β14=(2​e3iπ​)14+(2​e3−iπ​)14

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}={(\sqrt{2})}^{14}({e}^{\frac{i14\pi }{3}}+{e}^{\frac{-i14\pi }{3}})$$⇒α14+β14=(2​)14(e3i14π​+e3−i14π​)

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}={(\sqrt{2})}^{14}(\mathrm{cos}\frac{14\pi }{3}+i\mathrm{sin}\frac{14\pi }{3}+\mathrm{cos}\frac{14\pi }{3}-i\mathrm{sin}\frac{14\pi }{3})$$⇒α14+β14=(2​)14(cos314π​+isin314π​+cos314π​−isin314π​)

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}={2}^{7}(2\mathrm{cos}\frac{14\pi }{3})$$⇒α14+β14=27(2cos314π​)

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}={2}^{8}(\mathrm{cos}(4\pi +\frac{2\pi }{3}))$$⇒α14+β14=28(cos(4π+32π​))

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}={2}^{8}(\mathrm{cos}\frac{2\pi }{3})$$⇒α14+β14=28(cos32π​)

$$\Rightarrow {\alpha }^{14}+{\beta }^{14}=-{2}^{8}\times \frac{1}{2}=-128$$⇒α14+β14=−28×21​=−128