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

Let \(\alpha, \beta \in \mathrm{N}\) be roots of equation \(\mathrm{x}^2-70 \mathrm{x}+\lambda=0\), where \(\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathrm{N}\). If \(\lambda\) assumes the minimum possible value, then \(\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}\) is equal to :

  1. A \(88\)
  2. B \(80\)
  3. C \(70\)
  4. D \(60\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(60\)

Step-by-step Solution

Detailed explanation

\( x^2-70 x+\lambda=0 \) \( \alpha+\beta=70 \) \( \alpha \beta=\lambda \) \( \therefore \alpha(70-\alpha)=\lambda\) Since, \(2\) and \(3\) does not divide \(\lambda\) \(\therefore \alpha=5, \beta=65, \lambda=325\) By putting value of \(\alpha, \beta, \lambda\) we get the…
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