JEE Main 2024 — 29 January, Shift 2 — Question 21

$x^{2}-\sqrt{6} x+6=0_{\searrow \beta}^{\nearrow \alpha}$

$x=\frac{\sqrt{6} \pm i \sqrt{6}}{2}=\frac{\sqrt{6}}{2}(1 \pm i)$

$\alpha=\sqrt{3}\left(\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right), \beta=\sqrt{3}\left(\mathrm{e}^{-\mathrm{i} \frac{\pi}{4}}\right)$

$\therefore \frac{\alpha^{99}}{\beta}+\alpha^{98}=\alpha^{98}\left(\frac{\alpha}{\beta}+1\right)$

$=\frac{\alpha^{98}(\alpha+\beta)}{\beta}=3^{49}\left(e^{\mathrm{i} 99 \frac{\pi}{4}}\right) \times \sqrt{2}$

$=3^{49}(-1+\mathrm{i})$

$=3^{\mathrm{n}}(\mathrm{a}+\mathrm{ib})$

$\therefore \mathrm{n}=49, \mathrm{a}=-1, \mathrm{~b}=1$

$\therefore \mathrm{n}+\mathrm{a}+\mathrm{b}=49-1+1=49$