Q27 / 75 · JEE Main 2026 (06 April Shift 2) · Mathematics
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 $\dfrac{\alpha}{\beta}$, is :
- A$\dfrac{61}{37}$
- B$\dfrac{121}{81}$
- Correct answerC$\dfrac{364}{243}$
- D$\dfrac{1093}{729}$
Solution
The given quadratic equation is $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0$.
Let $k = n^2 - 2n + 2 = (n-1)^2 + 1$. Since $(n-1)^2 \ge 0$ for all $n \in \mathbb{R}$, the minimum value of $k$ is $1$ at $n=1$.
The product of the roots is given by $P = \dfrac{(n^2 - 2n + 2)^2}{n^2 - 2n + 2} = n^2 - 2n + 2 = k$.
The minimum value of the product of the roots is $\alpha = \min(k) = 1$.
The sum of the roots is given by $S = \dfrac{3}{n^2 - 2n + 2} = \dfrac{3}{k}$.
The maximum value of the sum of the roots occurs when $k$ is minimum. Thus, $\beta = \max\left(\dfrac{3}{k}\right) = \dfrac{3}{1} = 3$.
We are given a Geometric Progression (G.P.) whose first term is $a = \alpha = 1$ and common ratio is $r = \dfrac{\alpha}{\beta} = \dfrac{1}{3}$.
The sum of the first six terms of this G.P. is:
$S_6 = a \dfrac{1 - r^6}{1 - r} = 1 \cdot \dfrac{1 - \left(\dfrac{1}{3}\right)^6}{1 - \dfrac{1}{3}}$
$S_6 = \dfrac{1 - \dfrac{1}{729}}{\dfrac{2}{3}} = \dfrac{\dfrac{728}{729}}{\dfrac{2}{3}}$
$S_6 = \dfrac{728}{729} \times \dfrac{3}{2} = \dfrac{364}{243}$
Answer: $\dfrac{364}{243}$
Let $k = n^2 - 2n + 2 = (n-1)^2 + 1$. Since $(n-1)^2 \ge 0$ for all $n \in \mathbb{R}$, the minimum value of $k$ is $1$ at $n=1$.
The product of the roots is given by $P = \dfrac{(n^2 - 2n + 2)^2}{n^2 - 2n + 2} = n^2 - 2n + 2 = k$.
The minimum value of the product of the roots is $\alpha = \min(k) = 1$.
The sum of the roots is given by $S = \dfrac{3}{n^2 - 2n + 2} = \dfrac{3}{k}$.
The maximum value of the sum of the roots occurs when $k$ is minimum. Thus, $\beta = \max\left(\dfrac{3}{k}\right) = \dfrac{3}{1} = 3$.
We are given a Geometric Progression (G.P.) whose first term is $a = \alpha = 1$ and common ratio is $r = \dfrac{\alpha}{\beta} = \dfrac{1}{3}$.
The sum of the first six terms of this G.P. is:
$S_6 = a \dfrac{1 - r^6}{1 - r} = 1 \cdot \dfrac{1 - \left(\dfrac{1}{3}\right)^6}{1 - \dfrac{1}{3}}$
$S_6 = \dfrac{1 - \dfrac{1}{729}}{\dfrac{2}{3}} = \dfrac{\dfrac{728}{729}}{\dfrac{2}{3}}$
$S_6 = \dfrac{728}{729} \times \dfrac{3}{2} = \dfrac{364}{243}$
Answer: $\dfrac{364}{243}$
Hence, the correct answer is option: 'C'.