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COMEDK · Maths · 16. Limits

Let \(\alpha\) and \(\beta\) be the distinct roots of \(a x^2+b x+c=0\), then \(\lim _\limits{x \rightarrow \alpha} \dfrac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}\) is equal to

  1. A \(\dfrac{(\alpha-\beta)^2}{2}\)
  2. B \(\dfrac{a^2(\alpha-\beta)^2}{2}\)
  3. C 0
  4. D \(\dfrac{-a^2(\alpha-\beta)^2}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\dfrac{a^2(\alpha-\beta)^2}{2}\)

Step-by-step Solution

Detailed explanation

The quadratic equation \(ax^2 + bx + c = 0\) has roots \(\alpha\) and \(\beta\). Thus, \(ax^2 + bx + c = a(x - \alpha)(x - \beta)\). The limit is given by \(L = \lim_{x \rightarrow \alpha} \dfrac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \alpha)^2}\). Using the identity…