JEE PYQ: Quadratic Equation Question 29
Question 29 - 2020 (08 Jan Shift 2)
Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$. If $a = (1 + \alpha)\sum_{k=0}^{100} \alpha^{2k}$ and $b = \sum_{k=0}^{100} \alpha^{3k}$, then $a$ and $b$ are the roots of the quadratic equation:
(1) $x^2 + 101x + 100 = 0$
(2) $x^2 - 102x + 101 = 0$
(3) $x^2 - 101x + 100 = 0$
(4) $x^2 + 102x + 101 = 0$
Show Answer
Answer: (2)
Solution
Let $\alpha = \omega$, $b = 1 + \omega^3 + \omega^6 + \ldots = 101$
$a = (1 + \omega)(1 + \omega^2 + \omega^4 + \ldots + \omega^{200})$
$= (1 + \omega)\frac{(1 - (\omega^2)^{101})}{1 - \omega^2} = \frac{(\omega + 1)(\omega^{202} - 1)}{(\omega^2 - 1)}$
$\Rightarrow a = \frac{(1 + \omega)(1 - \omega)}{1 - \omega^2} = 1$
Required equation $= x^2 - (101 + 1)x + (101) \times 1 = 0$
$\Rightarrow x^2 - 102x + 101 = 0$
BETA
