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
- A \(\dfrac{(\alpha-\beta)^2}{2}\)
- B \(\dfrac{a^2(\alpha-\beta)^2}{2}\)
- C 0
- D \(\dfrac{-a^2(\alpha-\beta)^2}{2}\)
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…
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