Consider the quadratic equation (n 2 - 2n + 2)x 2 - 3x + (n 2 - 2n + 2) 2 = 0 , n in mathbb R . Let α be the minimum value of the product of its roots and β be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is α and the common ratio is α/β , is:

Mathematics · Quadratic Equations

JEE Main 2026 — 6 April, Evening Shift — Question 22

Consider the quadratic equation $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0$ , $n \in \mathbb{R}$ . Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the G.P., whose first term is $\alpha$ and the common ratio is $\frac{\alpha}{\beta}$ , is:

Option A:
$\frac{61}{37}$
Option B:
$\frac{121}{81}$
Option C:
$\frac{364}{243}$
Correct
Option D:
$\frac{1093}{729}$

Answer: C

Step-by-step solution

$\alpha=n^{2}-2 n+2$ $\alpha=(\mathrm{n}-1)^{2}+1$ ∴ minimum value of $\alpha$ is 1 Similarly $\beta=\frac{3}{(n-1)^{2}+1}$ ∴ maximum value of $\beta$ is 3 required G.P. is $1, \frac{1}{3}, \frac{1}{3^{2}}, \ldots$ $\mathrm{S}_{6}=\frac{1\left(1-\left(\frac{1}{3}\right)^{6}\right)}{1-\frac{1}{3}}=\frac{3}{2}\left[1-\left(\frac{1}{3}\right)^{6}\right]=\frac{364}{243}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Theory of Quadratic Equations