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

Step 1: Understanding the Question
We are given a quadratic equation with coefficients dependent on a real number \(n\). We need to find the minimum possible value of the product of its roots (\(\alpha\)) and the maximum possible value of the sum of its roots (\(\beta\)). Using these values, we form a Geometric Progression (G.P.) and find the sum of its first six terms.
Step 2: Key Formula or Approach
For a quadratic equation \( ax^2 + bx + c = 0 \):
Sum of roots \( = -\frac{b}{a} \).
Product of roots \( = \frac{c}{a} \).
To find the minimum or maximum of a quadratic expression like \( n^2 - 2n + 2 \), we can complete the square.
Sum of first \(n\) terms of a G.P. is \( S_n = a\frac{1-r^n}{1-r} \), where \(a\) is the first term and \(r\) is the common ratio.
Step 3: Detailed Explanation
The given quadratic equation is \( (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0 \).
Let's find the sum (S) and product (P) of the roots.
\[ S = -\frac{-3}{n^2 - 2n + 2} = \frac{3}{n^2 - 2n + 2} \] \[ P = \frac{(n^2 - 2n + 2)^2}{n^2 - 2n + 2} = n^2 - 2n + 2 \] To find the min/max values of S and P, we analyze the expression \(n^2 - 2n + 2\).
By completing the square: \[ n^2 - 2n + 2 = (n^2 - 2n + 1) + 1 = (n - 1)^2 + 1 \] The minimum value of \( (n - 1)^2 \) is 0 (when \(n=1\)). So, the minimum value of \( (n - 1)^2 + 1 \) is \(0+1=1\).
Finding \(\alpha\) (minimum product):
The product of roots is \( P = (n - 1)^2 + 1 \).
The minimum value of P occurs when \( (n - 1)^2 = 0 \), so \( P_{min} = 1 \).
Therefore, \( \alpha = 1 \).
Finding \(\beta\) (maximum sum):
The sum of roots is \( S = \frac{3}{(n - 1)^2 + 1} \).
To maximize S, we must minimize its denominator, which is \( (n - 1)^2 + 1 \).
The minimum value of the denominator is 1.
So, the maximum value of S is \( S_{max} = \frac{3}{1} = 3 \).
Therefore, \( \beta = 3 \).
Calculating the sum of the G.P.:
The G.P. has:
First term \( a = \alpha = 1 \).
Common ratio \( r = \frac{\alpha}{\beta} = \frac{1}{3} \).
We need to find the sum of the first six terms, \(S_6\).
Using the formula \( S_n = a\frac{1-r^n}{1-r} \):
\[ S_6 = 1 \cdot \frac{1 - (\frac{1}{3})^6}{1 - \frac{1}{3}} \] \[ S_6 = \frac{1 - \frac{1}{729}}{\frac{2}{3}} \] \[ S_6 = \frac{\frac{729-1}{729}}{\frac{2}{3}} = \frac{\frac{728}{729}}{\frac{2}{3}} \] \[ S_6 = \frac{728}{729} \times \frac{3}{2} = \frac{364}{243} \] Step 4: Final Answer
The sum of the first six terms of the G.P. is \( \frac{364}{243} \).