Question

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. \( \frac{61}{37} \)
• B. \( \frac{121}{81} \)
• C. \( \frac{364}{243} \)
• D. \( \frac{1093}{729} \)

Answer 1

The Correct Option is C

Concept: For a quadratic equation \( ax^2 + bx + c = 0 \):

• Sum of roots \( = -\frac{b}{a} \)
• Product of roots \( = \frac{c}{a} \)

For a geometric progression (G.P.): \[ S_n = a\frac{1-r^n}{1-r}, \quad r \neq 1 \] Step 1: {Find the sum and product of the roots.} Given equation: \[ (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0 \] Here, \[ a = n^2 - 2n + 2, \quad b = -3, \quad c = (n^2 - 2n + 2)^2 \] Sum of roots: \[ S = -\frac{b}{a} = \frac{3}{n^2 - 2n + 2} \] Product of roots: \[ P = \frac{c}{a} = n^2 - 2n + 2 \] Step 2: {Find minimum value of the product.} \[ P = n^2 - 2n + 2 \] \[ = (n-1)^2 + 1 \] Minimum occurs at \( n = 1 \): \[ \alpha = 1 \] Step 3: {Find maximum value of the sum.} \[ S = \frac{3}{(n-1)^2 + 1} \] Maximum occurs when denominator is minimum: \[ \beta = 3 \] Step 4: {Form the geometric progression.} First term: \[ a = \alpha = 1 \] Common ratio: \[ r = \frac{\alpha}{\beta} = \frac{1}{3} \] Step 5: {Find the sum of first six terms.} \[ S_6 = a\frac{1-r^6}{1-r} \] \[ = 1 \cdot \frac{1-\left(\frac{1}{3}\right)^6}{1-\frac{1}{3}} \] \[ = \frac{1-\frac{1}{729}}{\frac{2}{3}} \] \[ = \frac{728}{729} \times \frac{3}{2} \] \[ = \frac{364}{243} \]