AP EAMCET · Maths · Limits
If \(\alpha\) and \(\beta\) are the roots of the equation \(a x^2+b x+c=0\), then \(\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}=\)
- A \(\frac{a^2(\alpha-\beta)^2}{4}\)
- B 1
- C \(\frac{a(\alpha-\beta)}{2}\)
- D \(\frac{a^2(\alpha-\beta)^2}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{a^2(\alpha-\beta)^2}{2}\)
Step-by-step Solution
Detailed explanation
\( \lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2} = \lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{\left(a x^2+b x+c\right)^2} \cdot \frac{\left(a x^2+b x+c\right)^2}{(x-\alpha)^2} \)…
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