JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
Let $\alpha,\beta$ be the roots of the quadratic equation $x^{2}+\sqrt{6}x+3=0$. Then $\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to
Quadratic Equation · Powers of Roots (αⁿ ± βⁿ Recurrence)
Solution
Concept. The roots are $\sqrt3\,e^{\pm3\pi i/4}$; then $\alpha^n+\beta^n=2\cdot3^{n/2}\cos\frac{3n\pi}{4}$. Formulas used. (1) $(re^{i\theta})^n+(re^{-i\theta})^n=2r^n\cos n\theta$Step 1 - Polar form. $\alpha\beta=3$ gives modulus $\sqrt3$; the real part is $-\frac{\sqrt6}{2}$, so $\cos\theta=-\frac{1}{\sqrt2}$, $\theta=\frac{3\pi}{4}$. Step 2 - The four sums. By (1): $$n=23:\ \cos\frac{69\pi}{4}=\cos\frac{5\pi}{4}=-\frac1{\sqrt2},\qquad n=14:\ \cos\frac{21\pi}{2}=0,$$ $$n=15:\ \cos\frac{45\pi}{4}=\cos\frac{5\pi}{4}=-\frac1{\sqrt2},\qquad n=10:\ \cos\frac{15\pi}{2}=0.$$ Step 3 - Ratio. The $n=14$ and $n=10$ terms vanish, so $$\frac{2\cdot3^{23/2}\left(-\frac1{\sqrt2}\right)}{2\cdot3^{15/2}\left(-\frac1{\sqrt2}\right)}=3^4=81.$$ Answer: (C)