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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

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 :

  1. A \(\dfrac{61}{37}\)
  2. B \(\dfrac{121}{81}\)
  3. C \(\dfrac{364}{243}\)
  4. D \(\dfrac{1093}{729}\)
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Answer & Solution

Correct Answer

(C) \(\dfrac{364}{243}\)

Step-by-step Solution

Detailed explanation

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…
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